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

Optimal. Leaf size=138 \[ -\frac{3}{2} i b^2 c \text{PolyLog}\left (2,-1+\frac{2}{1-i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )+\frac{3}{4} b^3 c \text{PolyLog}\left (3,-1+\frac{2}{1-i c x^2}\right )-\frac{1}{2} i c \left (a+b \tan ^{-1}\left (c x^2\right )\right )^3-\frac{\left (a+b \tan ^{-1}\left (c x^2\right )\right )^3}{2 x^2}+\frac{3}{2} b c \log \left (2-\frac{2}{1-i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \]

[Out]

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

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Rubi [F]  time = 0.83101, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{\left (a+b \tan ^{-1}\left (c x^2\right )\right )^3}{x^3} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(a + b*ArcTan[c*x^2])^3/x^3,x]

[Out]

(3*b*c*Log[I*c*x^2]*(2*a + I*b*Log[1 - I*c*x^2])^2)/16 - ((1 - I*c*x^2)*(2*a + I*b*Log[1 - I*c*x^2])^3)/(16*x^
2) - (3*b^3*c*Log[(-I)*c*x^2]*Log[1 + I*c*x^2]^2)/16 - ((I/16)*b^3*(1 + I*c*x^2)*Log[1 + I*c*x^2]^3)/x^2 + ((3
*I)/8)*b^2*c*(2*a + I*b*Log[1 - I*c*x^2])*PolyLog[2, 1 - I*c*x^2] - (3*b^3*c*Log[1 + I*c*x^2]*PolyLog[2, 1 + I
*c*x^2])/8 + (3*b^3*c*PolyLog[3, 1 - I*c*x^2])/8 + (3*b^3*c*PolyLog[3, 1 + I*c*x^2])/8 + ((3*I)/16)*b*Defer[Su
bst][Defer[Int][(((-2*I)*a + b*Log[1 - I*c*x])^2*Log[1 + I*c*x])/x^2, x], x, x^2] - ((3*I)/16)*b^2*Defer[Subst
][Defer[Int][(((-2*I)*a + b*Log[1 - I*c*x])*Log[1 + I*c*x]^2)/x^2, x], x, x^2]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tan ^{-1}\left (c x^2\right )\right )^3}{x^3} \, dx &=\int \left (\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{8 x^3}+\frac{3 i b \left (-2 i a+b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right )}{8 x^3}-\frac{3 i b^2 \left (-2 i a+b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )}{8 x^3}+\frac{i b^3 \log ^3\left (1+i c x^2\right )}{8 x^3}\right ) \, dx\\ &=\frac{1}{8} \int \frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{x^3} \, dx+\frac{1}{8} (3 i b) \int \frac{\left (-2 i a+b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right )}{x^3} \, dx-\frac{1}{8} \left (3 i b^2\right ) \int \frac{\left (-2 i a+b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )}{x^3} \, dx+\frac{1}{8} \left (i b^3\right ) \int \frac{\log ^3\left (1+i c x^2\right )}{x^3} \, dx\\ &=\frac{1}{16} \operatorname{Subst}\left (\int \frac{(2 a+i b \log (1-i c x))^3}{x^2} \, dx,x,x^2\right )+\frac{1}{16} (3 i b) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^2} \, dx,x,x^2\right )-\frac{1}{16} \left (3 i b^2\right ) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^2} \, dx,x,x^2\right )+\frac{1}{16} \left (i b^3\right ) \operatorname{Subst}\left (\int \frac{\log ^3(1+i c x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac{\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{16 x^2}-\frac{i b^3 \left (1+i c x^2\right ) \log ^3\left (1+i c x^2\right )}{16 x^2}+\frac{1}{16} (3 i b) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^2} \, dx,x,x^2\right )-\frac{1}{16} \left (3 i b^2\right ) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^2} \, dx,x,x^2\right )+\frac{1}{16} (3 b c) \operatorname{Subst}\left (\int \frac{(2 a+i b \log (1-i c x))^2}{x} \, dx,x,x^2\right )-\frac{1}{16} \left (3 b^3 c\right ) \operatorname{Subst}\left (\int \frac{\log ^2(1+i c x)}{x} \, dx,x,x^2\right )\\ &=\frac{3}{16} b c \log \left (i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2-\frac{\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{16 x^2}-\frac{3}{16} b^3 c \log \left (-i c x^2\right ) \log ^2\left (1+i c x^2\right )-\frac{i b^3 \left (1+i c x^2\right ) \log ^3\left (1+i c x^2\right )}{16 x^2}+\frac{1}{16} (3 i b) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^2} \, dx,x,x^2\right )-\frac{1}{16} \left (3 i b^2\right ) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^2} \, dx,x,x^2\right )-\frac{1}{8} \left (3 b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log (i c x) (2 a+i b \log (1-i c x))}{1-i c x} \, dx,x,x^2\right )+\frac{1}{8} \left (3 i b^3 c^2\right ) \operatorname{Subst}\left (\int \frac{\log (-i c x) \log (1+i c x)}{1+i c x} \, dx,x,x^2\right )\\ &=\frac{3}{16} b c \log \left (i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2-\frac{\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{16 x^2}-\frac{3}{16} b^3 c \log \left (-i c x^2\right ) \log ^2\left (1+i c x^2\right )-\frac{i b^3 \left (1+i c x^2\right ) \log ^3\left (1+i c x^2\right )}{16 x^2}+\frac{1}{16} (3 i b) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^2} \, dx,x,x^2\right )-\frac{1}{16} \left (3 i b^2\right ) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^2} \, dx,x,x^2\right )-\frac{1}{8} \left (3 i b^2 c\right ) \operatorname{Subst}\left (\int \frac{(2 a+i b \log (x)) \log \left (i c \left (-\frac{i}{c}+\frac{i x}{c}\right )\right )}{x} \, dx,x,1-i c x^2\right )+\frac{1}{8} \left (3 b^3 c\right ) \operatorname{Subst}\left (\int \frac{\log (x) \log \left (-i c \left (\frac{i}{c}-\frac{i x}{c}\right )\right )}{x} \, dx,x,1+i c x^2\right )\\ &=\frac{3}{16} b c \log \left (i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2-\frac{\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{16 x^2}-\frac{3}{16} b^3 c \log \left (-i c x^2\right ) \log ^2\left (1+i c x^2\right )-\frac{i b^3 \left (1+i c x^2\right ) \log ^3\left (1+i c x^2\right )}{16 x^2}+\frac{3}{8} i b^2 c \left (2 a+i b \log \left (1-i c x^2\right )\right ) \text{Li}_2\left (1-i c x^2\right )-\frac{3}{8} b^3 c \log \left (1+i c x^2\right ) \text{Li}_2\left (1+i c x^2\right )+\frac{1}{16} (3 i b) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^2} \, dx,x,x^2\right )-\frac{1}{16} \left (3 i b^2\right ) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^2} \, dx,x,x^2\right )+\frac{1}{8} \left (3 b^3 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,1-i c x^2\right )+\frac{1}{8} \left (3 b^3 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,1+i c x^2\right )\\ &=\frac{3}{16} b c \log \left (i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2-\frac{\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{16 x^2}-\frac{3}{16} b^3 c \log \left (-i c x^2\right ) \log ^2\left (1+i c x^2\right )-\frac{i b^3 \left (1+i c x^2\right ) \log ^3\left (1+i c x^2\right )}{16 x^2}+\frac{3}{8} i b^2 c \left (2 a+i b \log \left (1-i c x^2\right )\right ) \text{Li}_2\left (1-i c x^2\right )-\frac{3}{8} b^3 c \log \left (1+i c x^2\right ) \text{Li}_2\left (1+i c x^2\right )+\frac{3}{8} b^3 c \text{Li}_3\left (1-i c x^2\right )+\frac{3}{8} b^3 c \text{Li}_3\left (1+i c x^2\right )+\frac{1}{16} (3 i b) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x))^2 \log (1+i c x)}{x^2} \, dx,x,x^2\right )-\frac{1}{16} \left (3 i b^2\right ) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x)) \log ^2(1+i c x)}{x^2} \, dx,x,x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.429957, size = 239, normalized size = 1.73 \[ \frac{1}{4} \left (6 a b^2 c \left (\tan ^{-1}\left (c x^2\right ) \left (\left (-\frac{1}{c x^2}-i\right ) \tan ^{-1}\left (c x^2\right )+2 \log \left (1-e^{2 i \tan ^{-1}\left (c x^2\right )}\right )\right )-i \text{PolyLog}\left (2,e^{2 i \tan ^{-1}\left (c x^2\right )}\right )\right )+2 b^3 c \left (3 i \tan ^{-1}\left (c x^2\right ) \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}\left (c x^2\right )}\right )+\frac{3}{2} \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}\left (c x^2\right )}\right )-\frac{\tan ^{-1}\left (c x^2\right )^3}{c x^2}+i \tan ^{-1}\left (c x^2\right )^3+3 \tan ^{-1}\left (c x^2\right )^2 \log \left (1-e^{-2 i \tan ^{-1}\left (c x^2\right )}\right )-\frac{i \pi ^3}{8}\right )-3 a^2 b c \log \left (c^2 x^4+1\right )-\frac{6 a^2 b \tan ^{-1}\left (c x^2\right )}{x^2}+12 a^2 b c \log (x)-\frac{2 a^3}{x^2}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-2*a^3)/x^2 - (6*a^2*b*ArcTan[c*x^2])/x^2 + 12*a^2*b*c*Log[x] - 3*a^2*b*c*Log[1 + c^2*x^4] + 6*a*b^2*c*(ArcT
an[c*x^2]*((-I - 1/(c*x^2))*ArcTan[c*x^2] + 2*Log[1 - E^((2*I)*ArcTan[c*x^2])]) - I*PolyLog[2, E^((2*I)*ArcTan
[c*x^2])]) + 2*b^3*c*((-I/8)*Pi^3 + I*ArcTan[c*x^2]^3 - ArcTan[c*x^2]^3/(c*x^2) + 3*ArcTan[c*x^2]^2*Log[1 - E^
((-2*I)*ArcTan[c*x^2])] + (3*I)*ArcTan[c*x^2]*PolyLog[2, E^((-2*I)*ArcTan[c*x^2])] + (3*PolyLog[3, E^((-2*I)*A
rcTan[c*x^2])])/2))/4

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Maple [F]  time = 0.332, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\arctan \left ( c{x}^{2} \right ) \right ) ^{3}}{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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+b*arctan(c*x^2))^3/x^3,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{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^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))^3/x^3,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^3, 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^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atan(c*x**2))**3/x**3, 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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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