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

Optimal. Leaf size=229 \[ -\frac{3}{4} b^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )+\frac{3}{4} b^2 \text{PolyLog}\left (3,-1+\frac{2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )-\frac{3}{4} i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2+\frac{3}{4} i b \text{PolyLog}\left (2,-1+\frac{2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2+\frac{3}{8} i b^3 \text{PolyLog}\left (4,1-\frac{2}{1+i c x^2}\right )-\frac{3}{8} i b^3 \text{PolyLog}\left (4,-1+\frac{2}{1+i c x^2}\right )+\tanh ^{-1}\left (1-\frac{2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^3 \]

[Out]

(a + b*ArcTan[c*x^2])^3*ArcTanh[1 - 2/(1 + I*c*x^2)] - ((3*I)/4)*b*(a + b*ArcTan[c*x^2])^2*PolyLog[2, 1 - 2/(1
+ I*c*x^2)] + ((3*I)/4)*b*(a + b*ArcTan[c*x^2])^2*PolyLog[2, -1 + 2/(1 + I*c*x^2)] - (3*b^2*(a + b*ArcTan[c*x
^2])*PolyLog[3, 1 - 2/(1 + I*c*x^2)])/4 + (3*b^2*(a + b*ArcTan[c*x^2])*PolyLog[3, -1 + 2/(1 + I*c*x^2)])/4 + (
(3*I)/8)*b^3*PolyLog[4, 1 - 2/(1 + I*c*x^2)] - ((3*I)/8)*b^3*PolyLog[4, -1 + 2/(1 + I*c*x^2)]

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Rubi [A]  time = 0.528502, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {5031, 4850, 4988, 4884, 4994, 4998, 6610} \[ -\frac{3}{4} b^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )+\frac{3}{4} b^2 \text{PolyLog}\left (3,-1+\frac{2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )-\frac{3}{4} i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2+\frac{3}{4} i b \text{PolyLog}\left (2,-1+\frac{2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2+\frac{3}{8} i b^3 \text{PolyLog}\left (4,1-\frac{2}{1+i c x^2}\right )-\frac{3}{8} i b^3 \text{PolyLog}\left (4,-1+\frac{2}{1+i c x^2}\right )+\tanh ^{-1}\left (1-\frac{2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^3 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x^2])^3/x,x]

[Out]

(a + b*ArcTan[c*x^2])^3*ArcTanh[1 - 2/(1 + I*c*x^2)] - ((3*I)/4)*b*(a + b*ArcTan[c*x^2])^2*PolyLog[2, 1 - 2/(1
+ I*c*x^2)] + ((3*I)/4)*b*(a + b*ArcTan[c*x^2])^2*PolyLog[2, -1 + 2/(1 + I*c*x^2)] - (3*b^2*(a + b*ArcTan[c*x
^2])*PolyLog[3, 1 - 2/(1 + I*c*x^2)])/4 + (3*b^2*(a + b*ArcTan[c*x^2])*PolyLog[3, -1 + 2/(1 + I*c*x^2)])/4 + (
(3*I)/8)*b^3*PolyLog[4, 1 - 2/(1 + I*c*x^2)] - ((3*I)/8)*b^3*PolyLog[4, -1 + 2/(1 + I*c*x^2)]

Rule 5031

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*ArcTan[c*x])^p
/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0]

Rule 4850

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

Rule 4988

Int[(ArcTanh[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[(
Log[1 + u]*(a + b*ArcTan[c*x])^p)/(d + e*x^2), x], x] - Dist[1/2, Int[(Log[1 - u]*(a + b*ArcTan[c*x])^p)/(d +
e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[u^2 - (1 - (2*I)/(I - c*x))^
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 4998

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

Rubi steps

\begin{align*} \int \frac{\left (a+b \tan ^{-1}\left (c x^2\right )\right )^3}{x} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(c x)\right )^3}{x} \, dx,x,x^2\right )\\ &=\left (a+b \tan ^{-1}\left (c x^2\right )\right )^3 \tanh ^{-1}\left (1-\frac{2}{1+i c x^2}\right )-(3 b c) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tan ^{-1}\left (c x^2\right )\right )^3 \tanh ^{-1}\left (1-\frac{2}{1+i c x^2}\right )+\frac{1}{2} (3 b c) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,x^2\right )-\frac{1}{2} (3 b c) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tan ^{-1}\left (c x^2\right )\right )^3 \tanh ^{-1}\left (1-\frac{2}{1+i c x^2}\right )-\frac{3}{4} i b \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \text{Li}_2\left (1-\frac{2}{1+i c x^2}\right )+\frac{3}{4} i b \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \text{Li}_2\left (-1+\frac{2}{1+i c x^2}\right )+\frac{1}{2} \left (3 i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,x^2\right )-\frac{1}{2} \left (3 i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tan ^{-1}\left (c x^2\right )\right )^3 \tanh ^{-1}\left (1-\frac{2}{1+i c x^2}\right )-\frac{3}{4} i b \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \text{Li}_2\left (1-\frac{2}{1+i c x^2}\right )+\frac{3}{4} i b \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \text{Li}_2\left (-1+\frac{2}{1+i c x^2}\right )-\frac{3}{4} b^2 \left (a+b \tan ^{-1}\left (c x^2\right )\right ) \text{Li}_3\left (1-\frac{2}{1+i c x^2}\right )+\frac{3}{4} b^2 \left (a+b \tan ^{-1}\left (c x^2\right )\right ) \text{Li}_3\left (-1+\frac{2}{1+i c x^2}\right )+\frac{1}{4} \left (3 b^3 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,x^2\right )-\frac{1}{4} \left (3 b^3 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (-1+\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tan ^{-1}\left (c x^2\right )\right )^3 \tanh ^{-1}\left (1-\frac{2}{1+i c x^2}\right )-\frac{3}{4} i b \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \text{Li}_2\left (1-\frac{2}{1+i c x^2}\right )+\frac{3}{4} i b \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \text{Li}_2\left (-1+\frac{2}{1+i c x^2}\right )-\frac{3}{4} b^2 \left (a+b \tan ^{-1}\left (c x^2\right )\right ) \text{Li}_3\left (1-\frac{2}{1+i c x^2}\right )+\frac{3}{4} b^2 \left (a+b \tan ^{-1}\left (c x^2\right )\right ) \text{Li}_3\left (-1+\frac{2}{1+i c x^2}\right )+\frac{3}{8} i b^3 \text{Li}_4\left (1-\frac{2}{1+i c x^2}\right )-\frac{3}{8} i b^3 \text{Li}_4\left (-1+\frac{2}{1+i c x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.195245, size = 245, normalized size = 1.07 \[ \frac{3}{8} i b \left (2 \text{PolyLog}\left (2,\frac{c x^2+i}{-c x^2+i}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2-2 \text{PolyLog}\left (2,\frac{c x^2+i}{c x^2-i}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2+b \left (-2 i \text{PolyLog}\left (3,\frac{c x^2+i}{-c x^2+i}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )+2 i \text{PolyLog}\left (3,\frac{c x^2+i}{c x^2-i}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )+b \left (\text{PolyLog}\left (4,\frac{c x^2+i}{c x^2-i}\right )-\text{PolyLog}\left (4,\frac{c x^2+i}{-c x^2+i}\right )\right )\right )\right )+\tanh ^{-1}\left (1+\frac{2 i}{c x^2-i}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*ArcTan[c*x^2])^3*ArcTanh[1 + (2*I)/(-I + c*x^2)] + ((3*I)/8)*b*(2*(a + b*ArcTan[c*x^2])^2*PolyLog[2, (I
+ c*x^2)/(I - c*x^2)] - 2*(a + b*ArcTan[c*x^2])^2*PolyLog[2, (I + c*x^2)/(-I + c*x^2)] + b*((-2*I)*(a + b*Arc
Tan[c*x^2])*PolyLog[3, (I + c*x^2)/(I - c*x^2)] + (2*I)*(a + b*ArcTan[c*x^2])*PolyLog[3, (I + c*x^2)/(-I + c*x
^2)] + b*(-PolyLog[4, (I + c*x^2)/(I - c*x^2)] + PolyLog[4, (I + c*x^2)/(-I + c*x^2)])))

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Maple [F]  time = 0.199, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\arctan \left ( c{x}^{2} \right ) \right ) ^{3}}{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \log \left (x\right ) + \frac{1}{32} \, \int \frac{28 \, b^{3} \arctan \left (c x^{2}\right )^{3} + 3 \, b^{3} \arctan \left (c x^{2}\right ) \log \left (c^{2} x^{4} + 1\right )^{2} + 96 \, a b^{2} \arctan \left (c x^{2}\right )^{2} + 96 \, a^{2} b \arctan \left (c x^{2}\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*log(x) + 1/32*integrate((28*b^3*arctan(c*x^2)^3 + 3*b^3*arctan(c*x^2)*log(c^2*x^4 + 1)^2 + 96*a*b^2*arctan
(c*x^2)^2 + 96*a^2*b*arctan(c*x^2))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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