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

Optimal. Leaf size=355 \[ -a c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a^2 c x^2+c}}{\sqrt {c}}\right )+\frac {15 i a c^3 \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{8 \sqrt {a^2 c x^2+c}}-\frac {15 i a c^3 \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{8 \sqrt {a^2 c x^2+c}}-\frac {15 i a c^3 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{4 \sqrt {a^2 c x^2+c}}-\frac {7}{8} a c^2 \sqrt {a^2 c x^2+c}+\frac {7}{8} a^2 c^2 x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)-\frac {c^2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{x}-\frac {1}{12} a c \left (a^2 c x^2+c\right )^{3/2}+\frac {1}{4} a^2 c x \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x) \]

[Out]

-1/12*a*c*(a^2*c*x^2+c)^(3/2)+1/4*a^2*c*x*(a^2*c*x^2+c)^(3/2)*arctan(a*x)-a*c^(5/2)*arctanh((a^2*c*x^2+c)^(1/2
)/c^(1/2))-15/4*I*a*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^2*c*x^2+c)^(1
/2)+15/8*I*a*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^2*c*x^2+c)^(1/2)-15/8*I*a*
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^2*c*x^2+c)^(1/2)-7/8*a*c^2*(a^2*c*x^2+c)
^(1/2)-c^2*arctan(a*x)*(a^2*c*x^2+c)^(1/2)/x+7/8*a^2*c^2*x*arctan(a*x)*(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.77, antiderivative size = 355, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {4950, 4944, 266, 63, 208, 4890, 4886, 4878} \[ \frac {15 i a c^3 \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 \sqrt {a^2 c x^2+c}}-\frac {15 i a c^3 \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 \sqrt {a^2 c x^2+c}}-\frac {7}{8} a c^2 \sqrt {a^2 c x^2+c}-\frac {15 i a c^3 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{4 \sqrt {a^2 c x^2+c}}+\frac {7}{8} a^2 c^2 x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)-\frac {c^2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{x}-a c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a^2 c x^2+c}}{\sqrt {c}}\right )-\frac {1}{12} a c \left (a^2 c x^2+c\right )^{3/2}+\frac {1}{4} a^2 c x \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*a*c^2*Sqrt[c + a^2*c*x^2])/8 - (a*c*(c + a^2*c*x^2)^(3/2))/12 - (c^2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/x +
(7*a^2*c^2*x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/8 + (a^2*c*x*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/4 - (((15*I)/4)*
a*c^3*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/Sqrt[c + a^2*c*x^2] - a*c^(5/2)*A
rcTanh[Sqrt[c + a^2*c*x^2]/Sqrt[c]] + (((15*I)/8)*a*c^3*Sqrt[1 + a^2*x^2]*PolyLog[2, ((-I)*Sqrt[1 + I*a*x])/Sq
rt[1 - I*a*x]])/Sqrt[c + a^2*c*x^2] - (((15*I)/8)*a*c^3*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqrt[1 + I*a*x])/Sqrt[
1 - I*a*x]])/Sqrt[c + a^2*c*x^2]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 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 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 4944

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p)/(d*f*(m + 1)), x] - Dist[(b*c*p)/(f*(m + 1)), Int[(
f*x)^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e,
c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 4950

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Dist[
d, Int[(f*x)^m*(d + e*x^2)^(q - 1)*(a + b*ArcTan[c*x])^p, x], x] + Dist[(c^2*d)/f^2, Int[(f*x)^(m + 2)*(d + e*
x^2)^(q - 1)*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[q, 0] &&
IGtQ[p, 0] && (RationalQ[m] || (EqQ[p, 1] && IntegerQ[q]))

Rubi steps

\begin {align*} \int \frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{x^2} \, dx &=c \int \frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{x^2} \, dx+\left (a^2 c\right ) \int \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x) \, dx\\ &=-\frac {1}{12} a c \left (c+a^2 c x^2\right )^{3/2}+\frac {1}{4} a^2 c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)+c^2 \int \frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{x^2} \, dx+\frac {1}{4} \left (3 a^2 c^2\right ) \int \sqrt {c+a^2 c x^2} \tan ^{-1}(a x) \, dx+\left (a^2 c^2\right ) \int \sqrt {c+a^2 c x^2} \tan ^{-1}(a x) \, dx\\ &=-\frac {7}{8} a c^2 \sqrt {c+a^2 c x^2}-\frac {1}{12} a c \left (c+a^2 c x^2\right )^{3/2}+\frac {7}{8} a^2 c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{4} a^2 c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)+c^3 \int \frac {\tan ^{-1}(a x)}{x^2 \sqrt {c+a^2 c x^2}} \, dx+\frac {1}{8} \left (3 a^2 c^3\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx+\frac {1}{2} \left (a^2 c^3\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx+\left (a^2 c^3\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx\\ &=-\frac {7}{8} a c^2 \sqrt {c+a^2 c x^2}-\frac {1}{12} a c \left (c+a^2 c x^2\right )^{3/2}-\frac {c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{x}+\frac {7}{8} a^2 c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{4} a^2 c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)+\left (a c^3\right ) \int \frac {1}{x \sqrt {c+a^2 c x^2}} \, dx+\frac {\left (3 a^2 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 {\left (a^2 c^3 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{2 \sqrt {c+a^2 c x^2}}+\frac {\left (a^2 c^3 \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 {7}{8} a c^2 \sqrt {c+a^2 c x^2}-\frac {1}{12} a c \left (c+a^2 c x^2\right )^{3/2}-\frac {c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{x}+\frac {7}{8} a^2 c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{4} a^2 c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {15 i a 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 )}{4 \sqrt {c+a^2 c x^2}}+\frac {15 i a 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 )}{8 \sqrt {c+a^2 c x^2}}-\frac {15 i a 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 )}{8 \sqrt {c+a^2 c x^2}}+\frac {1}{2} \left (a c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+a^2 c x}} \, dx,x,x^2\right )\\ &=-\frac {7}{8} a c^2 \sqrt {c+a^2 c x^2}-\frac {1}{12} a c \left (c+a^2 c x^2\right )^{3/2}-\frac {c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{x}+\frac {7}{8} a^2 c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{4} a^2 c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {15 i a 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 )}{4 \sqrt {c+a^2 c x^2}}+\frac {15 i a 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 )}{8 \sqrt {c+a^2 c x^2}}-\frac {15 i a 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 )}{8 \sqrt {c+a^2 c x^2}}+\frac {c^2 \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{a^2}+\frac {x^2}{a^2 c}} \, dx,x,\sqrt {c+a^2 c x^2}\right )}{a}\\ &=-\frac {7}{8} a c^2 \sqrt {c+a^2 c x^2}-\frac {1}{12} a c \left (c+a^2 c x^2\right )^{3/2}-\frac {c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{x}+\frac {7}{8} a^2 c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{4} a^2 c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {15 i a 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 )}{4 \sqrt {c+a^2 c x^2}}-a c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )+\frac {15 i a 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 )}{8 \sqrt {c+a^2 c x^2}}-\frac {15 i a 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 )}{8 \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 4.16, size = 491, normalized size = 1.38 \[ \frac {a c^2 \sqrt {a^2 c x^2+c} \left (-48 \left (\frac {\sqrt {a^2 x^2+1} \tan ^{-1}(a x)}{a x}-i \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )+i \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )+\tan ^{-1}(a x) \left (-\log \left (1-i e^{i \tan ^{-1}(a x)}\right )\right )+\tan ^{-1}(a x) \log \left (1+i e^{i \tan ^{-1}(a x)}\right )-\log \left (\sin \left (\frac {1}{2} \tan ^{-1}(a x)\right )\right )+\log \left (\cos \left (\frac {1}{2} \tan ^{-1}(a x)\right )\right )\right )+\frac {1}{2} \left (a^2 x^2+1\right )^{3/2}+48 \sqrt {a^2 x^2+1} \left (a x \tan ^{-1}(a x)-1\right )+\frac {3}{2} \left (a^2 x^2+1\right )^2 \cos \left (3 \tan ^{-1}(a x)\right )-\frac {3}{4} \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x) \left (-\frac {14 a x}{\sqrt {a^2 x^2+1}}+3 \log \left (1-i e^{i \tan ^{-1}(a x)}\right )-3 \log \left (1+i e^{i \tan ^{-1}(a x)}\right )+2 \sin \left (3 \tan ^{-1}(a x)\right )+4 \left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right ) \cos \left (2 \tan ^{-1}(a x)\right )+\left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right ) \cos \left (4 \tan ^{-1}(a x)\right )\right )+42 i \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )-42 i \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )+48 \tan ^{-1}(a x) \left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right )\right )}{48 \sqrt {a^2 x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*c^2*Sqrt[c + a^2*c*x^2]*((1 + a^2*x^2)^(3/2)/2 + 48*Sqrt[1 + a^2*x^2]*(-1 + a*x*ArcTan[a*x]) + (3*(1 + a^2*
x^2)^2*Cos[3*ArcTan[a*x]])/2 + 48*ArcTan[a*x]*(Log[1 - I*E^(I*ArcTan[a*x])] - Log[1 + I*E^(I*ArcTan[a*x])]) +
(42*I)*PolyLog[2, (-I)*E^(I*ArcTan[a*x])] - 48*((Sqrt[1 + a^2*x^2]*ArcTan[a*x])/(a*x) - ArcTan[a*x]*Log[1 - I*
E^(I*ArcTan[a*x])] + ArcTan[a*x]*Log[1 + I*E^(I*ArcTan[a*x])] + Log[Cos[ArcTan[a*x]/2]] - Log[Sin[ArcTan[a*x]/
2]] - I*PolyLog[2, (-I)*E^(I*ArcTan[a*x])] + I*PolyLog[2, I*E^(I*ArcTan[a*x])]) - (42*I)*PolyLog[2, I*E^(I*Arc
Tan[a*x])] - (3*(1 + a^2*x^2)^2*ArcTan[a*x]*((-14*a*x)/Sqrt[1 + a^2*x^2] + 3*Log[1 - I*E^(I*ArcTan[a*x])] + 4*
Cos[2*ArcTan[a*x]]*(Log[1 - I*E^(I*ArcTan[a*x])] - Log[1 + I*E^(I*ArcTan[a*x])]) + Cos[4*ArcTan[a*x]]*(Log[1 -
I*E^(I*ArcTan[a*x])] - Log[1 + I*E^(I*ArcTan[a*x])]) - 3*Log[1 + I*E^(I*ArcTan[a*x])] + 2*Sin[3*ArcTan[a*x]])
)/4))/(48*Sqrt[1 + a^2*x^2])

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}\right )} \sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(5/2)*arctan(a*x)/x^2,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)/x^2, 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((a^2*c*x^2+c)^(5/2)*arctan(a*x)/x^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 = 0.79, size = 265, normalized size = 0.75 \[ \frac {c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (6 \arctan \left (a x \right ) x^{4} a^{4}-2 a^{3} x^{3}+27 \arctan \left (a x \right ) x^{2} a^{2}-23 a x -24 \arctan \left (a x \right )\right )}{24 x}-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (15 \arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-15 \arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+8 \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-8 \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-1\right )+15 i \dilog \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-15 i \dilog \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right ) a \,c^{2}}{8 \sqrt {a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*c^2*(c*(a*x-I)*(I+a*x))^(1/2)*(6*arctan(a*x)*x^4*a^4-2*a^3*x^3+27*arctan(a*x)*x^2*a^2-23*a*x-24*arctan(a*
x))/x-1/8*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*x^2+1)^(1/2)*(15*arctan(a*x)*ln(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-15*a
rctan(a*x)*ln(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+8*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))-8*ln((1+I*a*x)/(a^2*x^2+1)^
(1/2)-1)+15*I*dilog(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-15*I*dilog(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2)))*a*c^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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