3.431 \(\int (c+a^2 c x^2)^{5/2} \tan ^{-1}(a x)^3 \, dx\)

Optimal. Leaf size=870 \[ -\frac {5 i \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3 c^3}{8 a \sqrt {a^2 c x^2+c}}-\frac {259 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right ) c^3}{60 a \sqrt {a^2 c x^2+c}}+\frac {15 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right ) c^3}{16 a \sqrt {a^2 c x^2+c}}-\frac {15 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right ) c^3}{16 a \sqrt {a^2 c x^2+c}}+\frac {259 i \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right ) c^3}{120 a \sqrt {a^2 c x^2+c}}-\frac {259 i \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right ) c^3}{120 a \sqrt {a^2 c x^2+c}}-\frac {15 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right ) c^3}{8 a \sqrt {a^2 c x^2+c}}+\frac {15 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right ) c^3}{8 a \sqrt {a^2 c x^2+c}}-\frac {15 i \sqrt {a^2 x^2+1} \text {Li}_4\left (-i e^{i \tan ^{-1}(a x)}\right ) c^3}{8 a \sqrt {a^2 c x^2+c}}+\frac {15 i \sqrt {a^2 x^2+1} \text {Li}_4\left (i e^{i \tan ^{-1}(a x)}\right ) c^3}{8 a \sqrt {a^2 c x^2+c}}+\frac {5}{16} x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^3 c^2-\frac {15 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2 c^2}{16 a}+\frac {17}{60} x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x) c^2-\frac {17 \sqrt {a^2 c x^2+c} c^2}{60 a}+\frac {5}{24} x \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)^3 c-\frac {5 \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)^2 c}{24 a}-\frac {\left (a^2 c x^2+c\right )^{3/2} c}{60 a}+\frac {1}{20} x \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x) c+\frac {1}{6} x \left (a^2 c x^2+c\right )^{5/2} \tan ^{-1}(a x)^3-\frac {\left (a^2 c x^2+c\right )^{5/2} \tan ^{-1}(a x)^2}{10 a} \]

[Out]

-1/60*c*(a^2*c*x^2+c)^(3/2)/a+1/20*c*x*(a^2*c*x^2+c)^(3/2)*arctan(a*x)-5/24*c*(a^2*c*x^2+c)^(3/2)*arctan(a*x)^
2/a-1/10*(a^2*c*x^2+c)^(5/2)*arctan(a*x)^2/a+5/24*c*x*(a^2*c*x^2+c)^(3/2)*arctan(a*x)^3+1/6*x*(a^2*c*x^2+c)^(5
/2)*arctan(a*x)^3-259/120*I*c^3*polylog(2,I*(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)
^(1/2)-15/8*I*c^3*polylog(4,-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)-15/16*I*c^
3*arctan(a*x)^2*polylog(2,I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)+15/16*I*c^3*a
rctan(a*x)^2*polylog(2,-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)+259/120*I*c^3*p
olylog(2,-I*(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)-5/8*I*c^3*arctan((1+I*a*x
)/(a^2*x^2+1)^(1/2))*arctan(a*x)^3*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)-15/8*c^3*arctan(a*x)*polylog(3,-I*(
1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)+15/8*c^3*arctan(a*x)*polylog(3,I*(1+I*a*x)
/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)-259/60*I*c^3*arctan(a*x)*arctan((1+I*a*x)^(1/2)/(1
-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)+15/8*I*c^3*polylog(4,I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^
2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)-17/60*c^2*(a^2*c*x^2+c)^(1/2)/a+17/60*c^2*x*arctan(a*x)*(a^2*c*x^2+c)^(1/
2)-15/16*c^2*arctan(a*x)^2*(a^2*c*x^2+c)^(1/2)/a+5/16*c^2*x*arctan(a*x)^3*(a^2*c*x^2+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.79, antiderivative size = 870, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {4880, 4890, 4888, 4181, 2531, 6609, 2282, 6589, 4886, 4878} \[ -\frac {5 i \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3 c^3}{8 a \sqrt {a^2 c x^2+c}}-\frac {259 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right ) c^3}{60 a \sqrt {a^2 c x^2+c}}+\frac {15 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right ) c^3}{16 a \sqrt {a^2 c x^2+c}}-\frac {15 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right ) c^3}{16 a \sqrt {a^2 c x^2+c}}+\frac {259 i \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right ) c^3}{120 a \sqrt {a^2 c x^2+c}}-\frac {259 i \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right ) c^3}{120 a \sqrt {a^2 c x^2+c}}-\frac {15 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right ) c^3}{8 a \sqrt {a^2 c x^2+c}}+\frac {15 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right ) c^3}{8 a \sqrt {a^2 c x^2+c}}-\frac {15 i \sqrt {a^2 x^2+1} \text {PolyLog}\left (4,-i e^{i \tan ^{-1}(a x)}\right ) c^3}{8 a \sqrt {a^2 c x^2+c}}+\frac {15 i \sqrt {a^2 x^2+1} \text {PolyLog}\left (4,i e^{i \tan ^{-1}(a x)}\right ) c^3}{8 a \sqrt {a^2 c x^2+c}}+\frac {5}{16} x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^3 c^2-\frac {15 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2 c^2}{16 a}+\frac {17}{60} x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x) c^2-\frac {17 \sqrt {a^2 c x^2+c} c^2}{60 a}+\frac {5}{24} x \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)^3 c-\frac {5 \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)^2 c}{24 a}-\frac {\left (a^2 c x^2+c\right )^{3/2} c}{60 a}+\frac {1}{20} x \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x) c+\frac {1}{6} x \left (a^2 c x^2+c\right )^{5/2} \tan ^{-1}(a x)^3-\frac {\left (a^2 c x^2+c\right )^{5/2} \tan ^{-1}(a x)^2}{10 a} \]

Antiderivative was successfully verified.

[In]

Int[(c + a^2*c*x^2)^(5/2)*ArcTan[a*x]^3,x]

[Out]

(-17*c^2*Sqrt[c + a^2*c*x^2])/(60*a) - (c*(c + a^2*c*x^2)^(3/2))/(60*a) + (17*c^2*x*Sqrt[c + a^2*c*x^2]*ArcTan
[a*x])/60 + (c*x*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/20 - (15*c^2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(16*a) - (
5*c*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x]^2)/(24*a) - ((c + a^2*c*x^2)^(5/2)*ArcTan[a*x]^2)/(10*a) + (5*c^2*x*Sqrt
[c + a^2*c*x^2]*ArcTan[a*x]^3)/16 + (5*c*x*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x]^3)/24 + (x*(c + a^2*c*x^2)^(5/2)*
ArcTan[a*x]^3)/6 - (((5*I)/8)*c^3*Sqrt[1 + a^2*x^2]*ArcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x]^3)/(a*Sqrt[c + a^2*c
*x^2]) - (((259*I)/60)*c^3*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(a*Sqrt[c +
a^2*c*x^2]) + (((15*I)/16)*c^3*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^2*PolyLog[2, (-I)*E^(I*ArcTan[a*x])])/(a*Sqrt[c +
 a^2*c*x^2]) - (((15*I)/16)*c^3*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^2*PolyLog[2, I*E^(I*ArcTan[a*x])])/(a*Sqrt[c + a
^2*c*x^2]) + (((259*I)/120)*c^3*Sqrt[1 + a^2*x^2]*PolyLog[2, ((-I)*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a*Sqrt[
c + a^2*c*x^2]) - (((259*I)/120)*c^3*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a*Sqr
t[c + a^2*c*x^2]) - (15*c^3*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[3, (-I)*E^(I*ArcTan[a*x])])/(8*a*Sqrt[c + a^
2*c*x^2]) + (15*c^3*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[3, I*E^(I*ArcTan[a*x])])/(8*a*Sqrt[c + a^2*c*x^2]) -
 (((15*I)/8)*c^3*Sqrt[1 + a^2*x^2]*PolyLog[4, (-I)*E^(I*ArcTan[a*x])])/(a*Sqrt[c + a^2*c*x^2]) + (((15*I)/8)*c
^3*Sqrt[1 + a^2*x^2]*PolyLog[4, I*E^(I*ArcTan[a*x])])/(a*Sqrt[c + a^2*c*x^2])

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
 x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4878

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> -Simp[(b*(d + e*x^2)^q)/(2*c*
q*(2*q + 1)), x] + (Dist[(2*d*q)/(2*q + 1), Int[(d + e*x^2)^(q - 1)*(a + b*ArcTan[c*x]), x], x] + Simp[(x*(d +
 e*x^2)^q*(a + b*ArcTan[c*x]))/(2*q + 1), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[q, 0]

Rule 4880

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> -Simp[(b*p*(d + e*x^2)^q
*(a + b*ArcTan[c*x])^(p - 1))/(2*c*q*(2*q + 1)), x] + (Dist[(2*d*q)/(2*q + 1), Int[(d + e*x^2)^(q - 1)*(a + b*
ArcTan[c*x])^p, x], x] + Dist[(b^2*d*p*(p - 1))/(2*q*(2*q + 1)), Int[(d + e*x^2)^(q - 1)*(a + b*ArcTan[c*x])^(
p - 2), x], x] + Simp[(x*(d + e*x^2)^q*(a + b*ArcTan[c*x])^p)/(2*q + 1), x]) /; FreeQ[{a, b, c, d, e}, x] && E
qQ[e, c^2*d] && GtQ[q, 0] && GtQ[p, 1]

Rule 4886

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(-2*I*(a + b*ArcTan[c*x])*
ArcTan[Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]])/(c*Sqrt[d]), x] + (Simp[(I*b*PolyLog[2, -((I*Sqrt[1 + I*c*x])/Sqrt[1
- I*c*x])])/(c*Sqrt[d]), x] - Simp[(I*b*PolyLog[2, (I*Sqrt[1 + I*c*x])/Sqrt[1 - I*c*x]])/(c*Sqrt[d]), x]) /; F
reeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[d, 0]

Rule 4888

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c*Sqrt[d]), Subst
[Int[(a + b*x)^p*Sec[x], x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0] &
& GtQ[d, 0]

Rule 4890

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + c^2*x^2]/Sq
rt[d + e*x^2], Int[(a + b*ArcTan[c*x])^p/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*
d] && IGtQ[p, 0] &&  !GtQ[d, 0]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rubi steps

\begin {align*} \int \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^3 \, dx &=-\frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^2}{10 a}+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^3+\frac {1}{5} c \int \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x) \, dx+\frac {1}{6} (5 c) \int \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3 \, dx\\ &=-\frac {c \left (c+a^2 c x^2\right )^{3/2}}{60 a}+\frac {1}{20} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {5 c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^2}{24 a}-\frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^2}{10 a}+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^3+\frac {1}{20} \left (3 c^2\right ) \int \sqrt {c+a^2 c x^2} \tan ^{-1}(a x) \, dx+\frac {1}{12} \left (5 c^2\right ) \int \sqrt {c+a^2 c x^2} \tan ^{-1}(a x) \, dx+\frac {1}{8} \left (5 c^2\right ) \int \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3 \, dx\\ &=-\frac {17 c^2 \sqrt {c+a^2 c x^2}}{60 a}-\frac {c \left (c+a^2 c x^2\right )^{3/2}}{60 a}+\frac {17}{60} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{20} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {15 c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{16 a}-\frac {5 c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^2}{24 a}-\frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^2}{10 a}+\frac {5}{16} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^3+\frac {1}{40} \left (3 c^3\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx+\frac {1}{24} \left (5 c^3\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx+\frac {1}{16} \left (5 c^3\right ) \int \frac {\tan ^{-1}(a x)^3}{\sqrt {c+a^2 c x^2}} \, dx+\frac {1}{8} \left (15 c^3\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx\\ &=-\frac {17 c^2 \sqrt {c+a^2 c x^2}}{60 a}-\frac {c \left (c+a^2 c x^2\right )^{3/2}}{60 a}+\frac {17}{60} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{20} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {15 c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{16 a}-\frac {5 c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^2}{24 a}-\frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^2}{10 a}+\frac {5}{16} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^3+\frac {\left (3 c^3 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{40 \sqrt {c+a^2 c x^2}}+\frac {\left (5 c^3 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{24 \sqrt {c+a^2 c x^2}}+\frac {\left (5 c^3 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx}{16 \sqrt {c+a^2 c x^2}}+\frac {\left (15 c^3 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{8 \sqrt {c+a^2 c x^2}}\\ &=-\frac {17 c^2 \sqrt {c+a^2 c x^2}}{60 a}-\frac {c \left (c+a^2 c x^2\right )^{3/2}}{60 a}+\frac {17}{60} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{20} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {15 c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{16 a}-\frac {5 c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^2}{24 a}-\frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^2}{10 a}+\frac {5}{16} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^3-\frac {259 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{60 a \sqrt {c+a^2 c x^2}}+\frac {259 i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{120 a \sqrt {c+a^2 c x^2}}-\frac {259 i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{120 a \sqrt {c+a^2 c x^2}}+\frac {\left (5 c^3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^3 \sec (x) \, dx,x,\tan ^{-1}(a x)\right )}{16 a \sqrt {c+a^2 c x^2}}\\ &=-\frac {17 c^2 \sqrt {c+a^2 c x^2}}{60 a}-\frac {c \left (c+a^2 c x^2\right )^{3/2}}{60 a}+\frac {17}{60} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{20} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {15 c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{16 a}-\frac {5 c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^2}{24 a}-\frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^2}{10 a}+\frac {5}{16} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^3-\frac {5 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3}{8 a \sqrt {c+a^2 c x^2}}-\frac {259 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{60 a \sqrt {c+a^2 c x^2}}+\frac {259 i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{120 a \sqrt {c+a^2 c x^2}}-\frac {259 i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{120 a \sqrt {c+a^2 c x^2}}-\frac {\left (15 c^3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^2 \log \left (1-i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{16 a \sqrt {c+a^2 c x^2}}+\frac {\left (15 c^3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^2 \log \left (1+i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{16 a \sqrt {c+a^2 c x^2}}\\ &=-\frac {17 c^2 \sqrt {c+a^2 c x^2}}{60 a}-\frac {c \left (c+a^2 c x^2\right )^{3/2}}{60 a}+\frac {17}{60} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{20} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {15 c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{16 a}-\frac {5 c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^2}{24 a}-\frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^2}{10 a}+\frac {5}{16} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^3-\frac {5 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3}{8 a \sqrt {c+a^2 c x^2}}-\frac {259 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{60 a \sqrt {c+a^2 c x^2}}+\frac {15 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{16 a \sqrt {c+a^2 c x^2}}-\frac {15 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{16 a \sqrt {c+a^2 c x^2}}+\frac {259 i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{120 a \sqrt {c+a^2 c x^2}}-\frac {259 i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{120 a \sqrt {c+a^2 c x^2}}-\frac {\left (15 i c^3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{8 a \sqrt {c+a^2 c x^2}}+\frac {\left (15 i c^3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{8 a \sqrt {c+a^2 c x^2}}\\ &=-\frac {17 c^2 \sqrt {c+a^2 c x^2}}{60 a}-\frac {c \left (c+a^2 c x^2\right )^{3/2}}{60 a}+\frac {17}{60} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{20} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {15 c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{16 a}-\frac {5 c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^2}{24 a}-\frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^2}{10 a}+\frac {5}{16} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^3-\frac {5 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3}{8 a \sqrt {c+a^2 c x^2}}-\frac {259 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{60 a \sqrt {c+a^2 c x^2}}+\frac {15 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{16 a \sqrt {c+a^2 c x^2}}-\frac {15 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{16 a \sqrt {c+a^2 c x^2}}+\frac {259 i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{120 a \sqrt {c+a^2 c x^2}}-\frac {259 i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{120 a \sqrt {c+a^2 c x^2}}-\frac {15 c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{8 a \sqrt {c+a^2 c x^2}}+\frac {15 c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{8 a \sqrt {c+a^2 c x^2}}+\frac {\left (15 c^3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (-i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{8 a \sqrt {c+a^2 c x^2}}-\frac {\left (15 c^3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{8 a \sqrt {c+a^2 c x^2}}\\ &=-\frac {17 c^2 \sqrt {c+a^2 c x^2}}{60 a}-\frac {c \left (c+a^2 c x^2\right )^{3/2}}{60 a}+\frac {17}{60} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{20} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {15 c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{16 a}-\frac {5 c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^2}{24 a}-\frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^2}{10 a}+\frac {5}{16} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^3-\frac {5 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3}{8 a \sqrt {c+a^2 c x^2}}-\frac {259 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{60 a \sqrt {c+a^2 c x^2}}+\frac {15 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{16 a \sqrt {c+a^2 c x^2}}-\frac {15 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{16 a \sqrt {c+a^2 c x^2}}+\frac {259 i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{120 a \sqrt {c+a^2 c x^2}}-\frac {259 i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{120 a \sqrt {c+a^2 c x^2}}-\frac {15 c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{8 a \sqrt {c+a^2 c x^2}}+\frac {15 c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{8 a \sqrt {c+a^2 c x^2}}-\frac {\left (15 i c^3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{8 a \sqrt {c+a^2 c x^2}}+\frac {\left (15 i c^3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{8 a \sqrt {c+a^2 c x^2}}\\ &=-\frac {17 c^2 \sqrt {c+a^2 c x^2}}{60 a}-\frac {c \left (c+a^2 c x^2\right )^{3/2}}{60 a}+\frac {17}{60} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{20} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {15 c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{16 a}-\frac {5 c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^2}{24 a}-\frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^2}{10 a}+\frac {5}{16} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^3-\frac {5 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3}{8 a \sqrt {c+a^2 c x^2}}-\frac {259 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{60 a \sqrt {c+a^2 c x^2}}+\frac {15 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{16 a \sqrt {c+a^2 c x^2}}-\frac {15 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{16 a \sqrt {c+a^2 c x^2}}+\frac {259 i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{120 a \sqrt {c+a^2 c x^2}}-\frac {259 i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{120 a \sqrt {c+a^2 c x^2}}-\frac {15 c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{8 a \sqrt {c+a^2 c x^2}}+\frac {15 c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{8 a \sqrt {c+a^2 c x^2}}-\frac {15 i c^3 \sqrt {1+a^2 x^2} \text {Li}_4\left (-i e^{i \tan ^{-1}(a x)}\right )}{8 a \sqrt {c+a^2 c x^2}}+\frac {15 i c^3 \sqrt {1+a^2 x^2} \text {Li}_4\left (i e^{i \tan ^{-1}(a x)}\right )}{8 a \sqrt {c+a^2 c x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 18.95, size = 4281, normalized size = 4.92 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + a^2*c*x^2)^(5/2)*ArcTan[a*x]^3,x]

[Out]

((-1/2*I)*c^2*Sqrt[c*(1 + a^2*x^2)]*(12*ArcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x] - (3*I)*Sqrt[1 + a^2*x^2]*ArcTan
[a*x]^2 + I*a*x*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^3 + 2*ArcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x]^3 - 3*(2 + ArcTan[a*
x]^2)*PolyLog[2, (-I)*E^(I*ArcTan[a*x])] + 3*(2 + ArcTan[a*x]^2)*PolyLog[2, I*E^(I*ArcTan[a*x])] - (6*I)*ArcTa
n[a*x]*PolyLog[3, (-I)*E^(I*ArcTan[a*x])] + (6*I)*ArcTan[a*x]*PolyLog[3, I*E^(I*ArcTan[a*x])] + 6*PolyLog[4, (
-I)*E^(I*ArcTan[a*x])] - 6*PolyLog[4, I*E^(I*ArcTan[a*x])]))/(a*Sqrt[1 + a^2*x^2]) + (2*c^2*((Sqrt[c*(1 + a^2*
x^2)]*(-1 + ArcTan[a*x]^2))/(4*Sqrt[1 + a^2*x^2]) + (Sqrt[c*(1 + a^2*x^2)]*(-(ArcTan[a*x]*(Log[1 - I*E^(I*ArcT
an[a*x])] - Log[1 + I*E^(I*ArcTan[a*x])])) - I*(PolyLog[2, (-I)*E^(I*ArcTan[a*x])] - PolyLog[2, I*E^(I*ArcTan[
a*x])])))/(2*Sqrt[1 + a^2*x^2]) + (Sqrt[c*(1 + a^2*x^2)]*(-1/8*(Pi^3*Log[Cot[(Pi/2 - ArcTan[a*x])/2]]) - (3*Pi
^2*((Pi/2 - ArcTan[a*x])*(Log[1 - E^(I*(Pi/2 - ArcTan[a*x]))] - Log[1 + E^(I*(Pi/2 - ArcTan[a*x]))]) + I*(Poly
Log[2, -E^(I*(Pi/2 - ArcTan[a*x]))] - PolyLog[2, E^(I*(Pi/2 - ArcTan[a*x]))])))/4 + (3*Pi*((Pi/2 - ArcTan[a*x]
)^2*(Log[1 - E^(I*(Pi/2 - ArcTan[a*x]))] - Log[1 + E^(I*(Pi/2 - ArcTan[a*x]))]) + (2*I)*(Pi/2 - ArcTan[a*x])*(
PolyLog[2, -E^(I*(Pi/2 - ArcTan[a*x]))] - PolyLog[2, E^(I*(Pi/2 - ArcTan[a*x]))]) + 2*(-PolyLog[3, -E^(I*(Pi/2
 - ArcTan[a*x]))] + PolyLog[3, E^(I*(Pi/2 - ArcTan[a*x]))])))/2 - 8*((I/64)*(Pi/2 - ArcTan[a*x])^4 + (I/4)*(Pi
/2 + (-1/2*Pi + ArcTan[a*x])/2)^4 - ((Pi/2 - ArcTan[a*x])^3*Log[1 + E^(I*(Pi/2 - ArcTan[a*x]))])/8 - (Pi^3*(I*
(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2) - Log[1 + E^((2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2))]))/8 - (Pi/2 + (-1/2
*Pi + ArcTan[a*x])/2)^3*Log[1 + E^((2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2))] + ((3*I)/8)*(Pi/2 - ArcTan[a*x])
^2*PolyLog[2, -E^(I*(Pi/2 - ArcTan[a*x]))] + (3*Pi^2*((I/2)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2)^2 - (Pi/2 + (-1
/2*Pi + ArcTan[a*x])/2)*Log[1 + E^((2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2))] + (I/2)*PolyLog[2, -E^((2*I)*(Pi
/2 + (-1/2*Pi + ArcTan[a*x])/2))]))/4 + ((3*I)/2)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2)^2*PolyLog[2, -E^((2*I)*(P
i/2 + (-1/2*Pi + ArcTan[a*x])/2))] - (3*(Pi/2 - ArcTan[a*x])*PolyLog[3, -E^(I*(Pi/2 - ArcTan[a*x]))])/4 - (3*P
i*((I/3)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2)^3 - (Pi/2 + (-1/2*Pi + ArcTan[a*x])/2)^2*Log[1 + E^((2*I)*(Pi/2 +
(-1/2*Pi + ArcTan[a*x])/2))] + I*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2)*PolyLog[2, -E^((2*I)*(Pi/2 + (-1/2*Pi + Ar
cTan[a*x])/2))] - PolyLog[3, -E^((2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2))]/2))/2 - (3*(Pi/2 + (-1/2*Pi + ArcT
an[a*x])/2)*PolyLog[3, -E^((2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2))])/2 - ((3*I)/4)*PolyLog[4, -E^(I*(Pi/2 -
ArcTan[a*x]))] - ((3*I)/4)*PolyLog[4, -E^((2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2))])))/(8*Sqrt[1 + a^2*x^2])
+ (Sqrt[c*(1 + a^2*x^2)]*ArcTan[a*x]^3)/(16*Sqrt[1 + a^2*x^2]*(Cos[ArcTan[a*x]/2] - Sin[ArcTan[a*x]/2])^4) + (
Sqrt[c*(1 + a^2*x^2)]*(2*ArcTan[a*x] - ArcTan[a*x]^2 - ArcTan[a*x]^3))/(16*Sqrt[1 + a^2*x^2]*(Cos[ArcTan[a*x]/
2] - Sin[ArcTan[a*x]/2])^2) - (Sqrt[c*(1 + a^2*x^2)]*ArcTan[a*x]^2*Sin[ArcTan[a*x]/2])/(8*Sqrt[1 + a^2*x^2]*(C
os[ArcTan[a*x]/2] - Sin[ArcTan[a*x]/2])^3) - (Sqrt[c*(1 + a^2*x^2)]*ArcTan[a*x]^3)/(16*Sqrt[1 + a^2*x^2]*(Cos[
ArcTan[a*x]/2] + Sin[ArcTan[a*x]/2])^4) + (Sqrt[c*(1 + a^2*x^2)]*ArcTan[a*x]^2*Sin[ArcTan[a*x]/2])/(8*Sqrt[1 +
 a^2*x^2]*(Cos[ArcTan[a*x]/2] + Sin[ArcTan[a*x]/2])^3) + (Sqrt[c*(1 + a^2*x^2)]*(-2*ArcTan[a*x] - ArcTan[a*x]^
2 + ArcTan[a*x]^3))/(16*Sqrt[1 + a^2*x^2]*(Cos[ArcTan[a*x]/2] + Sin[ArcTan[a*x]/2])^2) + (Sqrt[c*(1 + a^2*x^2)
]*(Sin[ArcTan[a*x]/2] - ArcTan[a*x]^2*Sin[ArcTan[a*x]/2]))/(4*Sqrt[1 + a^2*x^2]*(Cos[ArcTan[a*x]/2] + Sin[ArcT
an[a*x]/2])) + (Sqrt[c*(1 + a^2*x^2)]*(-Sin[ArcTan[a*x]/2] + ArcTan[a*x]^2*Sin[ArcTan[a*x]/2]))/(4*Sqrt[1 + a^
2*x^2]*(Cos[ArcTan[a*x]/2] - Sin[ArcTan[a*x]/2]))))/a + (c^2*((Sqrt[c*(1 + a^2*x^2)]*(50 - 19*ArcTan[a*x]^2))/
(240*Sqrt[1 + a^2*x^2]) + (19*Sqrt[c*(1 + a^2*x^2)]*(ArcTan[a*x]*(Log[1 - I*E^(I*ArcTan[a*x])] - Log[1 + I*E^(
I*ArcTan[a*x])]) + I*(PolyLog[2, (-I)*E^(I*ArcTan[a*x])] - PolyLog[2, I*E^(I*ArcTan[a*x])])))/(120*Sqrt[1 + a^
2*x^2]) + (Sqrt[c*(1 + a^2*x^2)]*((Pi^3*Log[Cot[(Pi/2 - ArcTan[a*x])/2]])/8 + (3*Pi^2*((Pi/2 - ArcTan[a*x])*(L
og[1 - E^(I*(Pi/2 - ArcTan[a*x]))] - Log[1 + E^(I*(Pi/2 - ArcTan[a*x]))]) + I*(PolyLog[2, -E^(I*(Pi/2 - ArcTan
[a*x]))] - PolyLog[2, E^(I*(Pi/2 - ArcTan[a*x]))])))/4 - (3*Pi*((Pi/2 - ArcTan[a*x])^2*(Log[1 - E^(I*(Pi/2 - A
rcTan[a*x]))] - Log[1 + E^(I*(Pi/2 - ArcTan[a*x]))]) + (2*I)*(Pi/2 - ArcTan[a*x])*(PolyLog[2, -E^(I*(Pi/2 - Ar
cTan[a*x]))] - PolyLog[2, E^(I*(Pi/2 - ArcTan[a*x]))]) + 2*(-PolyLog[3, -E^(I*(Pi/2 - ArcTan[a*x]))] + PolyLog
[3, E^(I*(Pi/2 - ArcTan[a*x]))])))/2 + 8*((I/64)*(Pi/2 - ArcTan[a*x])^4 + (I/4)*(Pi/2 + (-1/2*Pi + ArcTan[a*x]
)/2)^4 - ((Pi/2 - ArcTan[a*x])^3*Log[1 + E^(I*(Pi/2 - ArcTan[a*x]))])/8 - (Pi^3*(I*(Pi/2 + (-1/2*Pi + ArcTan[a
*x])/2) - Log[1 + E^((2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2))]))/8 - (Pi/2 + (-1/2*Pi + ArcTan[a*x])/2)^3*Log
[1 + E^((2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2))] + ((3*I)/8)*(Pi/2 - ArcTan[a*x])^2*PolyLog[2, -E^(I*(Pi/2 -
 ArcTan[a*x]))] + (3*Pi^2*((I/2)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2)^2 - (Pi/2 + (-1/2*Pi + ArcTan[a*x])/2)*Log
[1 + E^((2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2))] + (I/2)*PolyLog[2, -E^((2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x]
)/2))]))/4 + ((3*I)/2)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2)^2*PolyLog[2, -E^((2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x
])/2))] - (3*(Pi/2 - ArcTan[a*x])*PolyLog[3, -E^(I*(Pi/2 - ArcTan[a*x]))])/4 - (3*Pi*((I/3)*(Pi/2 + (-1/2*Pi +
 ArcTan[a*x])/2)^3 - (Pi/2 + (-1/2*Pi + ArcTan[a*x])/2)^2*Log[1 + E^((2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2))
] + I*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2)*PolyLog[2, -E^((2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2))] - PolyLog[3
, -E^((2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2))]/2))/2 - (3*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2)*PolyLog[3, -E^(
(2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2))])/2 - ((3*I)/4)*PolyLog[4, -E^(I*(Pi/2 - ArcTan[a*x]))] - ((3*I)/4)*
PolyLog[4, -E^((2*I)*(Pi/2 + (-1/2*Pi + ArcTan[a*x])/2))])))/(16*Sqrt[1 + a^2*x^2]) + (Sqrt[c*(1 + a^2*x^2)]*A
rcTan[a*x]^3)/(48*Sqrt[1 + a^2*x^2]*(Cos[ArcTan[a*x]/2] - Sin[ArcTan[a*x]/2])^6) + (Sqrt[c*(1 + a^2*x^2)]*(Arc
Tan[a*x] - ArcTan[a*x]^2 - 5*ArcTan[a*x]^3))/(80*Sqrt[1 + a^2*x^2]*(Cos[ArcTan[a*x]/2] - Sin[ArcTan[a*x]/2])^4
) + (Sqrt[c*(1 + a^2*x^2)]*(-2 - 52*ArcTan[a*x] + 26*ArcTan[a*x]^2 + 15*ArcTan[a*x]^3))/(480*Sqrt[1 + a^2*x^2]
*(Cos[ArcTan[a*x]/2] - Sin[ArcTan[a*x]/2])^2) - (Sqrt[c*(1 + a^2*x^2)]*ArcTan[a*x]^2*Sin[ArcTan[a*x]/2])/(40*S
qrt[1 + a^2*x^2]*(Cos[ArcTan[a*x]/2] - Sin[ArcTan[a*x]/2])^5) - (Sqrt[c*(1 + a^2*x^2)]*ArcTan[a*x]^3)/(48*Sqrt
[1 + a^2*x^2]*(Cos[ArcTan[a*x]/2] + Sin[ArcTan[a*x]/2])^6) + (Sqrt[c*(1 + a^2*x^2)]*ArcTan[a*x]^2*Sin[ArcTan[a
*x]/2])/(40*Sqrt[1 + a^2*x^2]*(Cos[ArcTan[a*x]/2] + Sin[ArcTan[a*x]/2])^5) + (Sqrt[c*(1 + a^2*x^2)]*(-ArcTan[a
*x] - ArcTan[a*x]^2 + 5*ArcTan[a*x]^3))/(80*Sqrt[1 + a^2*x^2]*(Cos[ArcTan[a*x]/2] + Sin[ArcTan[a*x]/2])^4) + (
Sqrt[c*(1 + a^2*x^2)]*(-2 + 52*ArcTan[a*x] + 26*ArcTan[a*x]^2 - 15*ArcTan[a*x]^3))/(480*Sqrt[1 + a^2*x^2]*(Cos
[ArcTan[a*x]/2] + Sin[ArcTan[a*x]/2])^2) + (Sqrt[c*(1 + a^2*x^2)]*(50*Sin[ArcTan[a*x]/2] - 19*ArcTan[a*x]^2*Si
n[ArcTan[a*x]/2]))/(240*Sqrt[1 + a^2*x^2]*(Cos[ArcTan[a*x]/2] - Sin[ArcTan[a*x]/2])) + (Sqrt[c*(1 + a^2*x^2)]*
(Sin[ArcTan[a*x]/2] - 13*ArcTan[a*x]^2*Sin[ArcTan[a*x]/2]))/(120*Sqrt[1 + a^2*x^2]*(Cos[ArcTan[a*x]/2] + Sin[A
rcTan[a*x]/2])^3) + (Sqrt[c*(1 + a^2*x^2)]*(-Sin[ArcTan[a*x]/2] + 13*ArcTan[a*x]^2*Sin[ArcTan[a*x]/2]))/(120*S
qrt[1 + a^2*x^2]*(Cos[ArcTan[a*x]/2] - Sin[ArcTan[a*x]/2])^3) + (Sqrt[c*(1 + a^2*x^2)]*(-50*Sin[ArcTan[a*x]/2]
 + 19*ArcTan[a*x]^2*Sin[ArcTan[a*x]/2]))/(240*Sqrt[1 + a^2*x^2]*(Cos[ArcTan[a*x]/2] + Sin[ArcTan[a*x]/2]))))/a

________________________________________________________________________________________

fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}\right )} \sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(5/2)*arctan(a*x)^3,x, algorithm="fricas")

[Out]

integral((a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2)*sqrt(a^2*c*x^2 + c)*arctan(a*x)^3, x)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(5/2)*arctan(a*x)^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

maple [A]  time = 0.97, size = 518, normalized size = 0.60 \[ \frac {c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (40 \arctan \left (a x \right )^{3} x^{5} a^{5}-24 \arctan \left (a x \right )^{2} x^{4} a^{4}+130 \arctan \left (a x \right )^{3} a^{3} x^{3}+12 \arctan \left (a x \right ) x^{3} a^{3}-98 \arctan \left (a x \right )^{2} x^{2} a^{2}+165 \arctan \left (a x \right )^{3} x a -4 a^{2} x^{2}+80 \arctan \left (a x \right ) x a -299 \arctan \left (a x \right )^{2}-72\right )}{240 a}+\frac {c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (75 \arctan \left (a x \right )^{3} \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-225 i \arctan \left (a x \right )^{2} \polylog \left (2, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-75 \arctan \left (a x \right )^{3} \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+225 i \arctan \left (a x \right )^{2} \polylog \left (2, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+518 \arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+450 \arctan \left (a x \right ) \polylog \left (3, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+450 i \polylog \left (4, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-518 \arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-450 \arctan \left (a x \right ) \polylog \left (3, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-450 i \polylog \left (4, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+518 i \dilog \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-518 i \dilog \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{240 a \sqrt {a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)^(5/2)*arctan(a*x)^3,x)

[Out]

1/240*c^2/a*(c*(a*x-I)*(I+a*x))^(1/2)*(40*arctan(a*x)^3*x^5*a^5-24*arctan(a*x)^2*x^4*a^4+130*arctan(a*x)^3*a^3
*x^3+12*arctan(a*x)*x^3*a^3-98*arctan(a*x)^2*x^2*a^2+165*arctan(a*x)^3*x*a-4*a^2*x^2+80*arctan(a*x)*x*a-299*ar
ctan(a*x)^2-72)+1/240*c^2*(c*(a*x-I)*(I+a*x))^(1/2)*(75*arctan(a*x)^3*ln(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-225*
I*arctan(a*x)^2*polylog(2,I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-75*arctan(a*x)^3*ln(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+
225*I*arctan(a*x)^2*polylog(2,-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+518*arctan(a*x)*ln(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/
2))+450*arctan(a*x)*polylog(3,I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+450*I*polylog(4,I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-51
8*arctan(a*x)*ln(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-450*arctan(a*x)*polylog(3,-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-45
0*I*polylog(4,-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+518*I*dilog(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-518*I*dilog(1-I*(1+
I*a*x)/(a^2*x^2+1)^(1/2)))/a/(a^2*x^2+1)^(1/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \arctan \left (a x\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(5/2)*arctan(a*x)^3,x, algorithm="maxima")

[Out]

integrate((a^2*c*x^2 + c)^(5/2)*arctan(a*x)^3, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {atan}\left (a\,x\right )}^3\,{\left (c\,a^2\,x^2+c\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a*x)^3*(c + a^2*c*x^2)^(5/2),x)

[Out]

int(atan(a*x)^3*(c + a^2*c*x^2)^(5/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \operatorname {atan}^{3}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2*c*x**2+c)**(5/2)*atan(a*x)**3,x)

[Out]

Integral((c*(a**2*x**2 + 1))**(5/2)*atan(a*x)**3, x)

________________________________________________________________________________________