3.223 \(\int \frac{(c+a^2 c x^2)^{5/2} \tan ^{-1}(a x)}{x^4} \, dx\)

Optimal. Leaf size=372 \[ \frac{5 i a^3 c^3 \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{2 \sqrt{a^2 c x^2+c}}-\frac{5 i a^3 c^3 \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{2 \sqrt{a^2 c x^2+c}}-\frac{1}{2} a^3 c^2 \sqrt{a^2 c x^2+c}-\frac{a c^2 \sqrt{a^2 c x^2+c}}{6 x^2}+\frac{1}{2} a^4 c^2 x \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)-\frac{5 i a^3 c^3 \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{\sqrt{a^2 c x^2+c}}-\frac{2 a^2 c^2 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{x}-\frac{13}{6} a^3 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a^2 c x^2+c}}{\sqrt{c}}\right )-\frac{c \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)}{3 x^3} \]

[Out]

-(a^3*c^2*Sqrt[c + a^2*c*x^2])/2 - (a*c^2*Sqrt[c + a^2*c*x^2])/(6*x^2) - (2*a^2*c^2*Sqrt[c + a^2*c*x^2]*ArcTan
[a*x])/x + (a^4*c^2*x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/2 - (c*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/(3*x^3) - ((5
*I)*a^3*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] - (13*a
^3*c^(5/2)*ArcTanh[Sqrt[c + a^2*c*x^2]/Sqrt[c]])/6 + (((5*I)/2)*a^3*c^3*Sqrt[1 + a^2*x^2]*PolyLog[2, ((-I)*Sqr
t[1 + I*a*x])/Sqrt[1 - I*a*x]])/Sqrt[c + a^2*c*x^2] - (((5*I)/2)*a^3*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]

________________________________________________________________________________________

Rubi [A]  time = 0.975109, antiderivative size = 372, normalized size of antiderivative = 1., number of steps used = 25, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used = {4950, 4944, 266, 47, 63, 208, 4890, 4886, 4878} \[ \frac{5 i a^3 c^3 \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{2 \sqrt{a^2 c x^2+c}}-\frac{5 i a^3 c^3 \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{2 \sqrt{a^2 c x^2+c}}-\frac{1}{2} a^3 c^2 \sqrt{a^2 c x^2+c}-\frac{a c^2 \sqrt{a^2 c x^2+c}}{6 x^2}+\frac{1}{2} a^4 c^2 x \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)-\frac{5 i a^3 c^3 \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{\sqrt{a^2 c x^2+c}}-\frac{2 a^2 c^2 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{x}-\frac{13}{6} a^3 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a^2 c x^2+c}}{\sqrt{c}}\right )-\frac{c \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((c + a^2*c*x^2)^(5/2)*ArcTan[a*x])/x^4,x]

[Out]

-(a^3*c^2*Sqrt[c + a^2*c*x^2])/2 - (a*c^2*Sqrt[c + a^2*c*x^2])/(6*x^2) - (2*a^2*c^2*Sqrt[c + a^2*c*x^2]*ArcTan
[a*x])/x + (a^4*c^2*x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/2 - (c*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/(3*x^3) - ((5
*I)*a^3*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] - (13*a
^3*c^(5/2)*ArcTanh[Sqrt[c + a^2*c*x^2]/Sqrt[c]])/6 + (((5*I)/2)*a^3*c^3*Sqrt[1 + a^2*x^2]*PolyLog[2, ((-I)*Sqr
t[1 + I*a*x])/Sqrt[1 - I*a*x]])/Sqrt[c + a^2*c*x^2] - (((5*I)/2)*a^3*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 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]))

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 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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{x^4} \, dx &=c \int \frac{\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{x^4} \, dx+\left (a^2 c\right ) \int \frac{\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{x^2} \, dx\\ &=c^2 \int \frac{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{x^4} \, dx+2 \left (\left (a^2 c^2\right ) \int \frac{\sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{x^2} \, dx\right )+\left (a^4 c^2\right ) \int \sqrt{c+a^2 c x^2} \tan ^{-1}(a x) \, dx\\ &=-\frac{1}{2} a^3 c^2 \sqrt{c+a^2 c x^2}+\frac{1}{2} a^4 c^2 x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)-\frac{c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 x^3}+\frac{1}{3} \left (a c^2\right ) \int \frac{\sqrt{c+a^2 c x^2}}{x^3} \, dx+\frac{1}{2} \left (a^4 c^3\right ) \int \frac{\tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx+2 \left (\left (a^2 c^3\right ) \int \frac{\tan ^{-1}(a x)}{x^2 \sqrt{c+a^2 c x^2}} \, dx+\left (a^4 c^3\right ) \int \frac{\tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx\right )\\ &=-\frac{1}{2} a^3 c^2 \sqrt{c+a^2 c x^2}+\frac{1}{2} a^4 c^2 x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)-\frac{c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 x^3}+\frac{1}{6} \left (a c^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+a^2 c x}}{x^2} \, dx,x,x^2\right )+\frac{\left (a^4 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}}+2 \left (-\frac{a^2 c^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{x}+\left (a^3 c^3\right ) \int \frac{1}{x \sqrt{c+a^2 c x^2}} \, dx+\frac{\left (a^4 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}}\right )\\ &=-\frac{1}{2} a^3 c^2 \sqrt{c+a^2 c x^2}-\frac{a c^2 \sqrt{c+a^2 c x^2}}{6 x^2}+\frac{1}{2} a^4 c^2 x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)-\frac{c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 x^3}-\frac{i a^3 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 )}{\sqrt{c+a^2 c x^2}}+\frac{i a^3 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 )}{2 \sqrt{c+a^2 c x^2}}-\frac{i a^3 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 )}{2 \sqrt{c+a^2 c x^2}}+\frac{1}{12} \left (a^3 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+a^2 c x}} \, dx,x,x^2\right )+2 \left (-\frac{a^2 c^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{x}-\frac{2 i a^3 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 )}{\sqrt{c+a^2 c x^2}}+\frac{i a^3 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 )}{\sqrt{c+a^2 c x^2}}-\frac{i a^3 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 )}{\sqrt{c+a^2 c x^2}}+\frac{1}{2} \left (a^3 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+a^2 c x}} \, dx,x,x^2\right )\right )\\ &=-\frac{1}{2} a^3 c^2 \sqrt{c+a^2 c x^2}-\frac{a c^2 \sqrt{c+a^2 c x^2}}{6 x^2}+\frac{1}{2} a^4 c^2 x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)-\frac{c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 x^3}-\frac{i a^3 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 )}{\sqrt{c+a^2 c x^2}}+\frac{i a^3 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 )}{2 \sqrt{c+a^2 c x^2}}-\frac{i a^3 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 )}{2 \sqrt{c+a^2 c x^2}}+\frac{1}{6} \left (a c^2\right ) \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 )+2 \left (-\frac{a^2 c^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{x}-\frac{2 i a^3 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 )}{\sqrt{c+a^2 c x^2}}+\frac{i a^3 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 )}{\sqrt{c+a^2 c x^2}}-\frac{i a^3 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 )}{\sqrt{c+a^2 c x^2}}+\left (a c^2\right ) \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 )\right )\\ &=-\frac{1}{2} a^3 c^2 \sqrt{c+a^2 c x^2}-\frac{a c^2 \sqrt{c+a^2 c x^2}}{6 x^2}+\frac{1}{2} a^4 c^2 x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)-\frac{c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 x^3}-\frac{i a^3 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 )}{\sqrt{c+a^2 c x^2}}-\frac{1}{6} a^3 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+a^2 c x^2}}{\sqrt{c}}\right )+\frac{i a^3 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 )}{2 \sqrt{c+a^2 c x^2}}-\frac{i a^3 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 )}{2 \sqrt{c+a^2 c x^2}}+2 \left (-\frac{a^2 c^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{x}-\frac{2 i a^3 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 )}{\sqrt{c+a^2 c x^2}}-a^3 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+a^2 c x^2}}{\sqrt{c}}\right )+\frac{i a^3 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 )}{\sqrt{c+a^2 c x^2}}-\frac{i a^3 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 )}{\sqrt{c+a^2 c x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.980372, size = 313, normalized size = 0.84 \[ \frac{c^2 \sqrt{a^2 c x^2+c} \left (15 i a^3 x^3 \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )-15 i a^3 x^3 \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )-3 a^3 x^3 \sqrt{a^2 x^2+1}-a x \sqrt{a^2 x^2+1}+3 a^4 x^4 \sqrt{a^2 x^2+1} \tan ^{-1}(a x)-14 a^2 x^2 \sqrt{a^2 x^2+1} \tan ^{-1}(a x)-2 \sqrt{a^2 x^2+1} \tan ^{-1}(a x)-a^3 x^3 \tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right )+15 a^3 x^3 \tan ^{-1}(a x) \log \left (1-i e^{i \tan ^{-1}(a x)}\right )-15 a^3 x^3 \tan ^{-1}(a x) \log \left (1+i e^{i \tan ^{-1}(a x)}\right )+12 a^3 x^3 \log \left (\sin \left (\frac{1}{2} \tan ^{-1}(a x)\right )\right )-12 a^3 x^3 \log \left (\cos \left (\frac{1}{2} \tan ^{-1}(a x)\right )\right )\right )}{6 x^3 \sqrt{a^2 x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((c + a^2*c*x^2)^(5/2)*ArcTan[a*x])/x^4,x]

[Out]

(c^2*Sqrt[c + a^2*c*x^2]*(-(a*x*Sqrt[1 + a^2*x^2]) - 3*a^3*x^3*Sqrt[1 + a^2*x^2] - 2*Sqrt[1 + a^2*x^2]*ArcTan[
a*x] - 14*a^2*x^2*Sqrt[1 + a^2*x^2]*ArcTan[a*x] + 3*a^4*x^4*Sqrt[1 + a^2*x^2]*ArcTan[a*x] - a^3*x^3*ArcTanh[Sq
rt[1 + a^2*x^2]] + 15*a^3*x^3*ArcTan[a*x]*Log[1 - I*E^(I*ArcTan[a*x])] - 15*a^3*x^3*ArcTan[a*x]*Log[1 + I*E^(I
*ArcTan[a*x])] - 12*a^3*x^3*Log[Cos[ArcTan[a*x]/2]] + 12*a^3*x^3*Log[Sin[ArcTan[a*x]/2]] + (15*I)*a^3*x^3*Poly
Log[2, (-I)*E^(I*ArcTan[a*x])] - (15*I)*a^3*x^3*PolyLog[2, I*E^(I*ArcTan[a*x])]))/(6*x^3*Sqrt[1 + a^2*x^2])

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Maple [A]  time = 0.487, size = 270, normalized size = 0.7 \begin{align*}{\frac{{c}^{2} \left ( 3\,\arctan \left ( ax \right ){x}^{4}{a}^{4}-3\,{a}^{3}{x}^{3}-14\,\arctan \left ( ax \right ){a}^{2}{x}^{2}-ax-2\,\arctan \left ( ax \right ) \right ) }{6\,{x}^{3}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}+{{\frac{i}{6}}{a}^{3}{c}^{2}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) } \left ( 15\,i\arctan \left ( ax \right ) \ln \left ( 1+{i \left ( 1+iax \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) -15\,i\arctan \left ( ax \right ) \ln \left ( 1-{i \left ( 1+iax \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) -13\,i\ln \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-1 \right ) +13\,i\ln \left ( 1+{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) -15\,{\it dilog} \left ( 1-{\frac{i \left ( 1+iax \right ) }{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) +15\,{\it dilog} \left ( 1+{\frac{i \left ( 1+iax \right ) }{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)^(5/2)*arctan(a*x)/x^4,x)

[Out]

1/6*c^2*(c*(a*x-I)*(a*x+I))^(1/2)*(3*arctan(a*x)*x^4*a^4-3*a^3*x^3-14*arctan(a*x)*a^2*x^2-a*x-2*arctan(a*x))/x
^3+1/6*I*a^3*c^2*(c*(a*x-I)*(a*x+I))^(1/2)*(15*I*arctan(a*x)*ln(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-15*I*arctan(a
*x)*ln(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-13*I*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)-1)+13*I*ln(1+(1+I*a*x)/(a^2*x^2+1)
^(1/2))-15*dilog(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+15*dilog(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2)))/(a^2*x^2+1)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(5/2)*arctan(a*x)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(5/2)*arctan(a*x)/x^4,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^4, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2*c*x**2+c)**(5/2)*atan(a*x)/x**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{2}} \arctan \left (a x\right )}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(5/2)*arctan(a*x)/x^4,x, algorithm="giac")

[Out]

integrate((a^2*c*x^2 + c)^(5/2)*arctan(a*x)/x^4, x)