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

Optimal. Leaf size=192 \[ -\frac{\text{PolyLog}\left (3,1-\frac{2}{1+i a x}\right )}{2 a^4 c^2}-\frac{i \tan ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1+i a x}\right )}{a^4 c^2}-\frac{1}{4 a^4 c^2 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^2}{2 a^4 c^2 \left (a^2 x^2+1\right )}-\frac{x \tan ^{-1}(a x)}{2 a^3 c^2 \left (a^2 x^2+1\right )}-\frac{i \tan ^{-1}(a x)^3}{3 a^4 c^2}-\frac{\tan ^{-1}(a x)^2}{4 a^4 c^2}-\frac{\log \left (\frac{2}{1+i a x}\right ) \tan ^{-1}(a x)^2}{a^4 c^2} \]

[Out]

-1/(4*a^4*c^2*(1 + a^2*x^2)) - (x*ArcTan[a*x])/(2*a^3*c^2*(1 + a^2*x^2)) - ArcTan[a*x]^2/(4*a^4*c^2) + ArcTan[
a*x]^2/(2*a^4*c^2*(1 + a^2*x^2)) - ((I/3)*ArcTan[a*x]^3)/(a^4*c^2) - (ArcTan[a*x]^2*Log[2/(1 + I*a*x)])/(a^4*c
^2) - (I*ArcTan[a*x]*PolyLog[2, 1 - 2/(1 + I*a*x)])/(a^4*c^2) - PolyLog[3, 1 - 2/(1 + I*a*x)]/(2*a^4*c^2)

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Rubi [A]  time = 0.289703, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used = {4964, 4920, 4854, 4884, 4994, 6610, 4930, 4892, 261} \[ -\frac{\text{PolyLog}\left (3,1-\frac{2}{1+i a x}\right )}{2 a^4 c^2}-\frac{i \tan ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1+i a x}\right )}{a^4 c^2}-\frac{1}{4 a^4 c^2 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^2}{2 a^4 c^2 \left (a^2 x^2+1\right )}-\frac{x \tan ^{-1}(a x)}{2 a^3 c^2 \left (a^2 x^2+1\right )}-\frac{i \tan ^{-1}(a x)^3}{3 a^4 c^2}-\frac{\tan ^{-1}(a x)^2}{4 a^4 c^2}-\frac{\log \left (\frac{2}{1+i a x}\right ) \tan ^{-1}(a x)^2}{a^4 c^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcTan[a*x]^2)/(c + a^2*c*x^2)^2,x]

[Out]

-1/(4*a^4*c^2*(1 + a^2*x^2)) - (x*ArcTan[a*x])/(2*a^3*c^2*(1 + a^2*x^2)) - ArcTan[a*x]^2/(4*a^4*c^2) + ArcTan[
a*x]^2/(2*a^4*c^2*(1 + a^2*x^2)) - ((I/3)*ArcTan[a*x]^3)/(a^4*c^2) - (ArcTan[a*x]^2*Log[2/(1 + I*a*x)])/(a^4*c
^2) - (I*ArcTan[a*x]*PolyLog[2, 1 - 2/(1 + I*a*x)])/(a^4*c^2) - PolyLog[3, 1 - 2/(1 + I*a*x)]/(2*a^4*c^2)

Rule 4964

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

Rule 4920

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

Rule 4854

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

Rule 4884

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

Rule 4994

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*Arc
Tan[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p*I)/2, Int[((a + b*ArcTan[c*x])^(p - 1)*PolyLog[2, 1 - u
])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[(1 - u)^2 - (1 - (2*
I)/(I - c*x))^2, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

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 4892

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

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]

Rubi steps

\begin{align*} \int \frac{x^3 \tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx &=-\frac{\int \frac{x \tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx}{a^2}+\frac{\int \frac{x \tan ^{-1}(a x)^2}{c+a^2 c x^2} \, dx}{a^2 c}\\ &=\frac{\tan ^{-1}(a x)^2}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^3}{3 a^4 c^2}-\frac{\int \frac{\tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{a^3}-\frac{\int \frac{\tan ^{-1}(a x)^2}{i-a x} \, dx}{a^3 c^2}\\ &=-\frac{x \tan ^{-1}(a x)}{2 a^3 c^2 \left (1+a^2 x^2\right )}-\frac{\tan ^{-1}(a x)^2}{4 a^4 c^2}+\frac{\tan ^{-1}(a x)^2}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^3}{3 a^4 c^2}-\frac{\tan ^{-1}(a x)^2 \log \left (\frac{2}{1+i a x}\right )}{a^4 c^2}+\frac{\int \frac{x}{\left (c+a^2 c x^2\right )^2} \, dx}{2 a^2}+\frac{2 \int \frac{\tan ^{-1}(a x) \log \left (\frac{2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^3 c^2}\\ &=-\frac{1}{4 a^4 c^2 \left (1+a^2 x^2\right )}-\frac{x \tan ^{-1}(a x)}{2 a^3 c^2 \left (1+a^2 x^2\right )}-\frac{\tan ^{-1}(a x)^2}{4 a^4 c^2}+\frac{\tan ^{-1}(a x)^2}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^3}{3 a^4 c^2}-\frac{\tan ^{-1}(a x)^2 \log \left (\frac{2}{1+i a x}\right )}{a^4 c^2}-\frac{i \tan ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1+i a x}\right )}{a^4 c^2}+\frac{i \int \frac{\text{Li}_2\left (1-\frac{2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^3 c^2}\\ &=-\frac{1}{4 a^4 c^2 \left (1+a^2 x^2\right )}-\frac{x \tan ^{-1}(a x)}{2 a^3 c^2 \left (1+a^2 x^2\right )}-\frac{\tan ^{-1}(a x)^2}{4 a^4 c^2}+\frac{\tan ^{-1}(a x)^2}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^3}{3 a^4 c^2}-\frac{\tan ^{-1}(a x)^2 \log \left (\frac{2}{1+i a x}\right )}{a^4 c^2}-\frac{i \tan ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1+i a x}\right )}{a^4 c^2}-\frac{\text{Li}_3\left (1-\frac{2}{1+i a x}\right )}{2 a^4 c^2}\\ \end{align*}

Mathematica [A]  time = 0.178734, size = 117, normalized size = 0.61 \[ \frac{i \tan ^{-1}(a x) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(a x)}\right )-\frac{1}{2} \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(a x)}\right )+\frac{1}{3} i \tan ^{-1}(a x)^3-\tan ^{-1}(a x)^2 \log \left (1+e^{2 i \tan ^{-1}(a x)}\right )-\frac{1}{4} \tan ^{-1}(a x) \sin \left (2 \tan ^{-1}(a x)\right )+\frac{1}{8} \left (2 \tan ^{-1}(a x)^2-1\right ) \cos \left (2 \tan ^{-1}(a x)\right )}{a^4 c^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^3*ArcTan[a*x]^2)/(c + a^2*c*x^2)^2,x]

[Out]

((I/3)*ArcTan[a*x]^3 + ((-1 + 2*ArcTan[a*x]^2)*Cos[2*ArcTan[a*x]])/8 - ArcTan[a*x]^2*Log[1 + E^((2*I)*ArcTan[a
*x])] + I*ArcTan[a*x]*PolyLog[2, -E^((2*I)*ArcTan[a*x])] - PolyLog[3, -E^((2*I)*ArcTan[a*x])]/2 - (ArcTan[a*x]
*Sin[2*ArcTan[a*x]])/4)/(a^4*c^2)

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Maple [C]  time = 0.423, size = 1092, normalized size = 5.7 \begin{align*} \text{result too large to display} \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)^2,x)

[Out]

1/4*I/a^4/c^2*arctan(a*x)^2*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^
2*x^2+1)+1)^2)*csgn(I/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)+I/a^3/c^2*arctan(a*x)/(8*a*x-8*I)*x+1/4*I/a^4/c^2*arctan(
a*x)^2*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))^3+1/4*I/a^4/c^2*arctan(a*x)^2*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+
I*a*x)^2/(a^2*x^2+1)+1)^2)^3-1/4*I/a^4/c^2*arctan(a*x)^2*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^3-I/a^3/c^2*
arctan(a*x)/(8*a*x+8*I)*x+1/16/a^3/c^2/(a*x+I)*x+1/3*I/a^4/c^2*arctan(a*x)^3+1/16*I/a^4/c^2/(a*x-I)-1/16*I/a^4
/c^2/(a*x+I)-1/2/a^4/c^2*polylog(3,-(1+I*a*x)^2/(a^2*x^2+1))-1/4*arctan(a*x)^2/a^4/c^2-1/a^4/c^2*arctan(a*x)^2
*ln(2)+1/16/a^3/c^2/(a*x-I)*x+1/2/a^4/c^2*arctan(a*x)^2*ln(a^2*x^2+1)-1/a^4/c^2*arctan(a*x)/(8*a*x-8*I)+1/2*I/
a^4/c^2*arctan(a*x)^2*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1))*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^2-1/2*I/a^4
/c^2*arctan(a*x)^2*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))^2*csgn(I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+1/2*arctan(a*x)^2/
a^4/c^2/(a^2*x^2+1)-1/4*I/a^4/c^2*arctan(a*x)^2*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))*csgn(I*(1+I*a*x)^2/(a^2*x^2
+1)/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^2+1/4*I/a^4/c^2*arctan(a*x)^2*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))*csgn(I*(1+
I*a*x)/(a^2*x^2+1)^(1/2))^2-1/4*I/a^4/c^2*arctan(a*x)^2*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^2*x^
2+1)+1)^2)^2*csgn(I/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)-1/4*I/a^4/c^2*arctan(a*x)^2*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2
+1)+1))^2*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)-1/a^4/c^2*arctan(a*x)^2*ln((1+I*a*x)/(a^2*x^2+1)^(1/2))+I/a^4/
c^2*arctan(a*x)*polylog(2,-(1+I*a*x)^2/(a^2*x^2+1))-1/a^4/c^2*arctan(a*x)/(8*a*x+8*I)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{2}}\,{d x} \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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{3} \arctan \left (a x\right )^{2}}{a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{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)^2,x, algorithm="fricas")

[Out]

integral(x^3*arctan(a*x)^2/(a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{x^{3} \operatorname{atan}^{2}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \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)**2,x)

[Out]

Integral(x**3*atan(a*x)**2/(a**4*x**4 + 2*a**2*x**2 + 1), x)/c**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{2}}\,{d x} \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)^2,x, algorithm="giac")

[Out]

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