3.415 \(\int \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3 \, dx\)

Optimal. Leaf size=626 \[ \frac {3 i c \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {3 i c \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a \sqrt {a^2 c x^2+c}}+\frac {3 i c \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{2 a \sqrt {a^2 c x^2+c}}-\frac {3 i c \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{2 a \sqrt {a^2 c x^2+c}}-\frac {3 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}+\frac {3 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {3 i c \sqrt {a^2 x^2+1} \text {Li}_4\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}+\frac {3 i c \sqrt {a^2 x^2+1} \text {Li}_4\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {i c \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3}{a \sqrt {a^2 c x^2+c}}+\frac {1}{2} x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^3-\frac {3 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{2 a}-\frac {6 i c \sqrt {a^2 x^2+1} \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right ) \tan ^{-1}(a x)}{a \sqrt {a^2 c x^2+c}} \]

[Out]

-I*c*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)-6*I*c*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)+3/2*I*c*arctan(a*x)^2*polyl
og(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)-3/2*I*c*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)+3*I*c*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)-3*I*c*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)-3*c*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)+3*c*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)-3*I*c*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)+3*I*c*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)-3/2*arctan(a*x)^2*(a^2*c*x^2+c)^(1/2)
/a+1/2*x*arctan(a*x)^3*(a^2*c*x^2+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.36, antiderivative size = 626, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4880, 4890, 4888, 4181, 2531, 6609, 2282, 6589, 4886} \[ \frac {3 i c \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {3 i c \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a \sqrt {a^2 c x^2+c}}+\frac {3 i c \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{2 a \sqrt {a^2 c x^2+c}}-\frac {3 i c \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \text {PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{2 a \sqrt {a^2 c x^2+c}}-\frac {3 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}+\frac {3 c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {3 i c \sqrt {a^2 x^2+1} \text {PolyLog}\left (4,-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}+\frac {3 i c \sqrt {a^2 x^2+1} \text {PolyLog}\left (4,i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {i c \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3}{a \sqrt {a^2 c x^2+c}}+\frac {1}{2} x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^3-\frac {3 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{2 a}-\frac {6 i c \sqrt {a^2 x^2+1} \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right ) \tan ^{-1}(a x)}{a \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^3,x]

[Out]

(-3*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(2*a) + (x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^3)/2 - (I*c*Sqrt[1 + a^2*x^2
]*ArcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x]^3)/(a*Sqrt[c + a^2*c*x^2]) - ((6*I)*c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*Ar
cTan[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(a*Sqrt[c + a^2*c*x^2]) + (((3*I)/2)*c*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]) - (((3*I)/2)*c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^2*Pol
yLog[2, I*E^(I*ArcTan[a*x])])/(a*Sqrt[c + a^2*c*x^2]) + ((3*I)*c*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]) - ((3*I)*c*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqrt[1 + I*a*x])/S
qrt[1 - I*a*x]])/(a*Sqrt[c + a^2*c*x^2]) - (3*c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[3, (-I)*E^(I*ArcTan[a*x]
)])/(a*Sqrt[c + a^2*c*x^2]) + (3*c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[3, I*E^(I*ArcTan[a*x])])/(a*Sqrt[c +
a^2*c*x^2]) - ((3*I)*c*Sqrt[1 + a^2*x^2]*PolyLog[4, (-I)*E^(I*ArcTan[a*x])])/(a*Sqrt[c + a^2*c*x^2]) + ((3*I)*
c*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 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 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3 \, dx &=-\frac {3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3+\frac {1}{2} c \int \frac {\tan ^{-1}(a x)^3}{\sqrt {c+a^2 c x^2}} \, dx+(3 c) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx\\ &=-\frac {3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3+\frac {\left (c \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx}{2 \sqrt {c+a^2 c x^2}}+\frac {\left (3 c \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3-\frac {6 i c \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 )}{a \sqrt {c+a^2 c x^2}}+\frac {3 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {3 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a \sqrt {c+a^2 c x^2}}+\frac {\left (c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^3 \sec (x) \, dx,x,\tan ^{-1}(a x)\right )}{2 a \sqrt {c+a^2 c x^2}}\\ &=-\frac {3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3-\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3}{a \sqrt {c+a^2 c x^2}}-\frac {6 i c \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 )}{a \sqrt {c+a^2 c x^2}}+\frac {3 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {3 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {\left (3 c \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 )}{2 a \sqrt {c+a^2 c x^2}}+\frac {\left (3 c \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 )}{2 a \sqrt {c+a^2 c x^2}}\\ &=-\frac {3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3-\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3}{a \sqrt {c+a^2 c x^2}}-\frac {6 i c \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 )}{a \sqrt {c+a^2 c x^2}}+\frac {3 i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{2 a \sqrt {c+a^2 c x^2}}-\frac {3 i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{2 a \sqrt {c+a^2 c x^2}}+\frac {3 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {3 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {\left (3 i c \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 )}{a \sqrt {c+a^2 c x^2}}+\frac {\left (3 i c \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 )}{a \sqrt {c+a^2 c x^2}}\\ &=-\frac {3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3-\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3}{a \sqrt {c+a^2 c x^2}}-\frac {6 i c \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 )}{a \sqrt {c+a^2 c x^2}}+\frac {3 i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{2 a \sqrt {c+a^2 c x^2}}-\frac {3 i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{2 a \sqrt {c+a^2 c x^2}}+\frac {3 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {3 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {3 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}+\frac {3 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}+\frac {\left (3 c \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 )}{a \sqrt {c+a^2 c x^2}}-\frac {\left (3 c \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 )}{a \sqrt {c+a^2 c x^2}}\\ &=-\frac {3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3-\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3}{a \sqrt {c+a^2 c x^2}}-\frac {6 i c \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 )}{a \sqrt {c+a^2 c x^2}}+\frac {3 i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{2 a \sqrt {c+a^2 c x^2}}-\frac {3 i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{2 a \sqrt {c+a^2 c x^2}}+\frac {3 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {3 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {3 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}+\frac {3 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {\left (3 i c \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 )}{a \sqrt {c+a^2 c x^2}}+\frac {\left (3 i c \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 )}{a \sqrt {c+a^2 c x^2}}\\ &=-\frac {3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3-\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3}{a \sqrt {c+a^2 c x^2}}-\frac {6 i c \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 )}{a \sqrt {c+a^2 c x^2}}+\frac {3 i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{2 a \sqrt {c+a^2 c x^2}}-\frac {3 i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{2 a \sqrt {c+a^2 c x^2}}+\frac {3 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {3 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {3 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}+\frac {3 c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {3 i c \sqrt {1+a^2 x^2} \text {Li}_4\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}+\frac {3 i c \sqrt {1+a^2 x^2} \text {Li}_4\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.84, size = 258, normalized size = 0.41 \[ -\frac {i \sqrt {c \left (a^2 x^2+1\right )} \left (i a x \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^3-3 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2-6 i \tan ^{-1}(a x) \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )+6 i \tan ^{-1}(a x) \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )-3 \left (\tan ^{-1}(a x)^2+2\right ) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )+3 \left (\tan ^{-1}(a x)^2+2\right ) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )+6 \text {Li}_4\left (-i e^{i \tan ^{-1}(a x)}\right )-6 \text {Li}_4\left (i e^{i \tan ^{-1}(a x)}\right )+2 \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^3+12 \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)\right )}{2 a \sqrt {a^2 x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^3,x]

[Out]

((-1/2*I)*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)*ArcTan[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])

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^3*(a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

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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(arctan(a*x)^3*(a^2*c*x^2+c)^(1/2),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

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maple [A]  time = 0.89, size = 422, normalized size = 0.67 \[ \frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \arctan \left (a x \right )^{2} \left (\arctan \left (a x \right ) x a -3\right )}{2 a}+\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (\arctan \left (a x \right )^{3} \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-3 i \arctan \left (a x \right )^{2} \polylog \left (2, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\arctan \left (a x \right )^{3} \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+3 i \arctan \left (a x \right )^{2} \polylog \left (2, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+6 \arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+6 \arctan \left (a x \right ) \polylog \left (3, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+6 i \polylog \left (4, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-6 \arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-6 \arctan \left (a x \right ) \polylog \left (3, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-6 i \polylog \left (4, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+6 i \dilog \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-6 i \dilog \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{2 a \sqrt {a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(a*x)^3*(a^2*c*x^2+c)^(1/2),x)

[Out]

1/2/a*(c*(a*x-I)*(I+a*x))^(1/2)*arctan(a*x)^2*(arctan(a*x)*x*a-3)+1/2*(c*(a*x-I)*(I+a*x))^(1/2)*(arctan(a*x)^3
*ln(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-3*I*arctan(a*x)^2*polylog(2,I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-arctan(a*x)^3*
ln(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+3*I*arctan(a*x)^2*polylog(2,-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+6*arctan(a*x)*
ln(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+6*arctan(a*x)*polylog(3,I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+6*I*polylog(4,I*(1+
I*a*x)/(a^2*x^2+1)^(1/2))-6*arctan(a*x)*ln(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-6*arctan(a*x)*polylog(3,-I*(1+I*a*
x)/(a^2*x^2+1)^(1/2))-6*I*polylog(4,-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+6*I*dilog(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))
-6*I*dilog(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2)))/a/(a^2*x^2+1)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^3*(a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a*x)^3*(c + a^2*c*x^2)^(1/2),x)

[Out]

int(atan(a*x)^3*(c + a^2*c*x^2)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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