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

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

[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]

________________________________________________________________________________________

Rubi [A]  time = 0.772699, antiderivative size = 355, normalized size of antiderivative = 1., 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 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 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^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*}

Mathematica [A]  time = 3.94454, size = 491, normalized size = 1.38 \[ \frac{a c^2 \sqrt{a^2 c x^2+c} \left (-48 \left (-i \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )+i \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )+\frac{\sqrt{a^2 x^2+1} \tan ^{-1}(a x)}{a x}+\tan ^{-1}(a x) \left (-\log \left (1-i e^{i \tan ^{-1}(a x)}\right )\right )+\tan ^{-1}(a x) \log \left (1+i e^{i \tan ^{-1}(a x)}\right )-\log \left (\sin \left (\frac{1}{2} \tan ^{-1}(a x)\right )\right )+\log \left (\cos \left (\frac{1}{2} \tan ^{-1}(a x)\right )\right )\right )+42 i \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )-42 i \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )+\frac{1}{2} \left (a^2 x^2+1\right )^{3/2}+48 \sqrt{a^2 x^2+1} \left (a x \tan ^{-1}(a x)-1\right )+\frac{3}{2} \left (a^2 x^2+1\right )^2 \cos \left (3 \tan ^{-1}(a x)\right )-\frac{3}{4} \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x) \left (-\frac{14 a x}{\sqrt{a^2 x^2+1}}+3 \log \left (1-i e^{i \tan ^{-1}(a x)}\right )-3 \log \left (1+i e^{i \tan ^{-1}(a x)}\right )+2 \sin \left (3 \tan ^{-1}(a x)\right )+4 \left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right ) \cos \left (2 \tan ^{-1}(a x)\right )+\left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right ) \cos \left (4 \tan ^{-1}(a x)\right )\right )+48 \tan ^{-1}(a x) \left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right )\right )}{48 \sqrt{a^2 x^2+1}} \]

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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Maple [A]  time = 0.353, size = 265, normalized size = 0.8 \begin{align*}{\frac{{c}^{2} \left ( 6\,\arctan \left ( ax \right ){x}^{4}{a}^{4}-2\,{a}^{3}{x}^{3}+27\,\arctan \left ( ax \right ){a}^{2}{x}^{2}-23\,ax-24\,\arctan \left ( ax \right ) \right ) }{24\,x}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}-{\frac{a{c}^{2}}{8}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) } \left ( 15\,\arctan \left ( ax \right ) \ln \left ( 1+{\frac{i \left ( 1+iax \right ) }{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) -15\,\arctan \left ( ax \right ) \ln \left ( 1-{\frac{i \left ( 1+iax \right ) }{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) -8\,\ln \left ({\frac{1+iax}{\sqrt{{a}^{2}{x}^{2}+1}}}-1 \right ) +8\,\ln \left ( 1+{\frac{1+iax}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) +15\,i{\it dilog} \left ( 1-{i \left ( 1+iax \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) -15\,i{\it dilog} \left ( 1+{i \left ( 1+iax \right ){\frac{1}{\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^2,x)

[Out]

1/24*c^2*(c*(a*x-I)*(a*x+I))^(1/2)*(6*arctan(a*x)*x^4*a^4-2*a^3*x^3+27*arctan(a*x)*a^2*x^2-23*a*x-24*arctan(a*
x))/x-1/8*(c*(a*x-I)*(a*x+I))^(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+I*a*x)/(a^2*x^2+1)^(1/2)-1)+8*ln(1+(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))-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(-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^2,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^{2}}, 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^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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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**2,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^{2}}\,{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^2,x, algorithm="giac")

[Out]

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