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

Optimal. Leaf size=385 \[ \frac{11 i c \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{60 a^4 \sqrt{a^2 c x^2+c}}-\frac{11 i c \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{60 a^4 \sqrt{a^2 c x^2+c}}+\frac{\left (a^2 c x^2+c\right )^{3/2}}{30 a^4 c}-\frac{11 \sqrt{a^2 c x^2+c}}{60 a^4}+\frac{1}{5} x^4 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)^2-\frac{x^3 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{10 a}+\frac{x^2 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)^2}{15 a^2}+\frac{x \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{12 a^3}-\frac{2 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)^2}{15 a^4}-\frac{11 i c \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{30 a^4 \sqrt{a^2 c x^2+c}} \]

[Out]

(-11*Sqrt[c + a^2*c*x^2])/(60*a^4) + (c + a^2*c*x^2)^(3/2)/(30*a^4*c) + (x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(1
2*a^3) - (x^3*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(10*a) - (2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(15*a^4) + (x^2*
Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(15*a^2) + (x^4*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/5 - (((11*I)/30)*c*Sqrt[
1 + a^2*x^2]*ArcTan[a*x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(a^4*Sqrt[c + a^2*c*x^2]) + (((11*I)/60)*c*S
qrt[1 + a^2*x^2]*PolyLog[2, ((-I)*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a^4*Sqrt[c + a^2*c*x^2]) - (((11*I)/60)*
c*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a^4*Sqrt[c + a^2*c*x^2])

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Rubi [A]  time = 1.42525, antiderivative size = 385, normalized size of antiderivative = 1., number of steps used = 26, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4950, 4952, 261, 4890, 4886, 4930, 266, 43} \[ \frac{11 i c \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{60 a^4 \sqrt{a^2 c x^2+c}}-\frac{11 i c \sqrt{a^2 x^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{60 a^4 \sqrt{a^2 c x^2+c}}+\frac{\left (a^2 c x^2+c\right )^{3/2}}{30 a^4 c}-\frac{11 \sqrt{a^2 c x^2+c}}{60 a^4}+\frac{1}{5} x^4 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)^2-\frac{x^3 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{10 a}+\frac{x^2 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)^2}{15 a^2}+\frac{x \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{12 a^3}-\frac{2 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)^2}{15 a^4}-\frac{11 i c \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}\right )}{30 a^4 \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*Sqrt[c + a^2*c*x^2])/(60*a^4) + (c + a^2*c*x^2)^(3/2)/(30*a^4*c) + (x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(1
2*a^3) - (x^3*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(10*a) - (2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(15*a^4) + (x^2*
Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(15*a^2) + (x^4*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/5 - (((11*I)/30)*c*Sqrt[
1 + a^2*x^2]*ArcTan[a*x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(a^4*Sqrt[c + a^2*c*x^2]) + (((11*I)/60)*c*S
qrt[1 + a^2*x^2]*PolyLog[2, ((-I)*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a^4*Sqrt[c + a^2*c*x^2]) - (((11*I)/60)*
c*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a^4*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 4952

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

Rule 261

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

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 4930

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^
(q + 1)*(a + b*ArcTan[c*x])^p)/(2*e*(q + 1)), x] - Dist[(b*p)/(2*c*(q + 1)), Int[(d + e*x^2)^q*(a + b*ArcTan[c
*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^3 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^2 \, dx &=c \int \frac{x^3 \tan ^{-1}(a x)^2}{\sqrt{c+a^2 c x^2}} \, dx+\left (a^2 c\right ) \int \frac{x^5 \tan ^{-1}(a x)^2}{\sqrt{c+a^2 c x^2}} \, dx\\ &=\frac{x^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 a^2}+\frac{1}{5} x^4 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^2-\frac{1}{5} (4 c) \int \frac{x^3 \tan ^{-1}(a x)^2}{\sqrt{c+a^2 c x^2}} \, dx-\frac{(2 c) \int \frac{x \tan ^{-1}(a x)^2}{\sqrt{c+a^2 c x^2}} \, dx}{3 a^2}-\frac{(2 c) \int \frac{x^2 \tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx}{3 a}-\frac{1}{5} (2 a c) \int \frac{x^4 \tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx\\ &=-\frac{x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{3 a^3}-\frac{x^3 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{10 a}-\frac{2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 a^4}+\frac{x^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^2}{15 a^2}+\frac{1}{5} x^4 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^2+\frac{1}{10} c \int \frac{x^3}{\sqrt{c+a^2 c x^2}} \, dx+\frac{c \int \frac{\tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx}{3 a^3}+\frac{(4 c) \int \frac{\tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx}{3 a^3}+\frac{c \int \frac{x}{\sqrt{c+a^2 c x^2}} \, dx}{3 a^2}+\frac{(8 c) \int \frac{x \tan ^{-1}(a x)^2}{\sqrt{c+a^2 c x^2}} \, dx}{15 a^2}+\frac{(3 c) \int \frac{x^2 \tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx}{10 a}+\frac{(8 c) \int \frac{x^2 \tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx}{15 a}\\ &=\frac{\sqrt{c+a^2 c x^2}}{3 a^4}+\frac{x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{12 a^3}-\frac{x^3 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{10 a}-\frac{2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^2}{15 a^4}+\frac{x^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^2}{15 a^2}+\frac{1}{5} x^4 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^2+\frac{1}{20} c \operatorname{Subst}\left (\int \frac{x}{\sqrt{c+a^2 c x}} \, dx,x,x^2\right )-\frac{(3 c) \int \frac{\tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx}{20 a^3}-\frac{(4 c) \int \frac{\tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx}{15 a^3}-\frac{(16 c) \int \frac{\tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx}{15 a^3}-\frac{(3 c) \int \frac{x}{\sqrt{c+a^2 c x^2}} \, dx}{20 a^2}-\frac{(4 c) \int \frac{x}{\sqrt{c+a^2 c x^2}} \, dx}{15 a^2}+\frac{\left (c \sqrt{1+a^2 x^2}\right ) \int \frac{\tan ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{3 a^3 \sqrt{c+a^2 c x^2}}+\frac{\left (4 c \sqrt{1+a^2 x^2}\right ) \int \frac{\tan ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{3 a^3 \sqrt{c+a^2 c x^2}}\\ &=-\frac{\sqrt{c+a^2 c x^2}}{12 a^4}+\frac{x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{12 a^3}-\frac{x^3 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{10 a}-\frac{2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^2}{15 a^4}+\frac{x^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^2}{15 a^2}+\frac{1}{5} x^4 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^2-\frac{10 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 )}{3 a^4 \sqrt{c+a^2 c x^2}}+\frac{5 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 )}{3 a^4 \sqrt{c+a^2 c x^2}}-\frac{5 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 )}{3 a^4 \sqrt{c+a^2 c x^2}}+\frac{1}{20} c \operatorname{Subst}\left (\int \left (-\frac{1}{a^2 \sqrt{c+a^2 c x}}+\frac{\sqrt{c+a^2 c x}}{a^2 c}\right ) \, dx,x,x^2\right )-\frac{\left (3 c \sqrt{1+a^2 x^2}\right ) \int \frac{\tan ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{20 a^3 \sqrt{c+a^2 c x^2}}-\frac{\left (4 c \sqrt{1+a^2 x^2}\right ) \int \frac{\tan ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{15 a^3 \sqrt{c+a^2 c x^2}}-\frac{\left (16 c \sqrt{1+a^2 x^2}\right ) \int \frac{\tan ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{15 a^3 \sqrt{c+a^2 c x^2}}\\ &=-\frac{11 \sqrt{c+a^2 c x^2}}{60 a^4}+\frac{\left (c+a^2 c x^2\right )^{3/2}}{30 a^4 c}+\frac{x \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{12 a^3}-\frac{x^3 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{10 a}-\frac{2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^2}{15 a^4}+\frac{x^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^2}{15 a^2}+\frac{1}{5} x^4 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)^2-\frac{11 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 )}{30 a^4 \sqrt{c+a^2 c x^2}}+\frac{11 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 )}{60 a^4 \sqrt{c+a^2 c x^2}}-\frac{11 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 )}{60 a^4 \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 1.14219, size = 360, normalized size = 0.94 \[ -\frac{\left (a^2 x^2+1\right )^2 \sqrt{c \left (a^2 x^2+1\right )} \left (-\frac{176 i \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{\left (a^2 x^2+1\right )^{5/2}}+\frac{176 i \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{\left (a^2 x^2+1\right )^{5/2}}-\frac{110 \tan ^{-1}(a x) \log \left (1-i e^{i \tan ^{-1}(a x)}\right )}{\sqrt{a^2 x^2+1}}+\frac{110 \tan ^{-1}(a x) \log \left (1+i e^{i \tan ^{-1}(a x)}\right )}{\sqrt{a^2 x^2+1}}-32 \tan ^{-1}(a x)^2+4 \tan ^{-1}(a x) \sin \left (2 \tan ^{-1}(a x)\right )-22 \tan ^{-1}(a x) \sin \left (4 \tan ^{-1}(a x)\right )+160 \tan ^{-1}(a x)^2 \cos \left (2 \tan ^{-1}(a x)\right )+72 \cos \left (2 \tan ^{-1}(a x)\right )+22 \cos \left (4 \tan ^{-1}(a x)\right )-55 \tan ^{-1}(a x) \log \left (1-i e^{i \tan ^{-1}(a x)}\right ) \cos \left (3 \tan ^{-1}(a x)\right )-11 \tan ^{-1}(a x) \log \left (1-i e^{i \tan ^{-1}(a x)}\right ) \cos \left (5 \tan ^{-1}(a x)\right )+55 \tan ^{-1}(a x) \log \left (1+i e^{i \tan ^{-1}(a x)}\right ) \cos \left (3 \tan ^{-1}(a x)\right )+11 \tan ^{-1}(a x) \log \left (1+i e^{i \tan ^{-1}(a x)}\right ) \cos \left (5 \tan ^{-1}(a x)\right )+50\right )}{960 a^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((1 + a^2*x^2)^2*Sqrt[c*(1 + a^2*x^2)]*(50 - 32*ArcTan[a*x]^2 + 72*Cos[2*ArcTan[a*x]] + 160*ArcTan[a*x]^2*Cos
[2*ArcTan[a*x]] + 22*Cos[4*ArcTan[a*x]] - (110*ArcTan[a*x]*Log[1 - I*E^(I*ArcTan[a*x])])/Sqrt[1 + a^2*x^2] - 5
5*ArcTan[a*x]*Cos[3*ArcTan[a*x]]*Log[1 - I*E^(I*ArcTan[a*x])] - 11*ArcTan[a*x]*Cos[5*ArcTan[a*x]]*Log[1 - I*E^
(I*ArcTan[a*x])] + (110*ArcTan[a*x]*Log[1 + I*E^(I*ArcTan[a*x])])/Sqrt[1 + a^2*x^2] + 55*ArcTan[a*x]*Cos[3*Arc
Tan[a*x]]*Log[1 + I*E^(I*ArcTan[a*x])] + 11*ArcTan[a*x]*Cos[5*ArcTan[a*x]]*Log[1 + I*E^(I*ArcTan[a*x])] - ((17
6*I)*PolyLog[2, (-I)*E^(I*ArcTan[a*x])])/(1 + a^2*x^2)^(5/2) + ((176*I)*PolyLog[2, I*E^(I*ArcTan[a*x])])/(1 +
a^2*x^2)^(5/2) + 4*ArcTan[a*x]*Sin[2*ArcTan[a*x]] - 22*ArcTan[a*x]*Sin[4*ArcTan[a*x]]))/(960*a^4)

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Maple [A]  time = 0.999, size = 235, normalized size = 0.6 \begin{align*}{\frac{12\, \left ( \arctan \left ( ax \right ) \right ) ^{2}{x}^{4}{a}^{4}-6\,\arctan \left ( ax \right ){x}^{3}{a}^{3}+4\, \left ( \arctan \left ( ax \right ) \right ) ^{2}{x}^{2}{a}^{2}+2\,{a}^{2}{x}^{2}+5\,\arctan \left ( ax \right ) xa-8\, \left ( \arctan \left ( ax \right ) \right ) ^{2}-9}{60\,{a}^{4}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}-{\frac{11}{60\,{a}^{4}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) } \left ( \arctan \left ( ax \right ) \ln \left ( 1+{i \left ( 1+iax \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) -\arctan \left ( ax \right ) \ln \left ( 1-{i \left ( 1+iax \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) -i{\it dilog} \left ( 1+{i \left ( 1+iax \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) +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(x^3*arctan(a*x)^2*(a^2*c*x^2+c)^(1/2),x)

[Out]

1/60/a^4*(c*(a*x-I)*(a*x+I))^(1/2)*(12*arctan(a*x)^2*x^4*a^4-6*arctan(a*x)*x^3*a^3+4*arctan(a*x)^2*x^2*a^2+2*a
^2*x^2+5*arctan(a*x)*x*a-8*arctan(a*x)^2-9)-11/60*(c*(a*x-I)*(a*x+I))^(1/2)*(arctan(a*x)*ln(1+I*(1+I*a*x)/(a^2
*x^2+1)^(1/2))-arctan(a*x)*ln(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-I*dilog(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+I*dilo
g(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2)))/a^4/(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(x^3*arctan(a*x)^2*(a^2*c*x^2+c)^(1/2),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 (\sqrt{a^{2} c x^{2} + c} x^{3} \arctan \left (a x\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \sqrt{c \left (a^{2} x^{2} + 1\right )} \operatorname{atan}^{2}{\left (a x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError