3.90 \(\int \frac{(a+b \tan ^{-1}(c x^2))^3}{x^5} \, dx\)

Optimal. Leaf size=149 \[ -\frac{3}{4} i b^3 c^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i c x^2}\right )+\frac{3}{2} b^2 c^2 \log \left (2-\frac{2}{1-i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )-\frac{3}{4} i b c^2 \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2-\frac{1}{4} c^2 \left (a+b \tan ^{-1}\left (c x^2\right )\right )^3-\frac{3 b c \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2}{4 x^2}-\frac{\left (a+b \tan ^{-1}\left (c x^2\right )\right )^3}{4 x^4} \]

[Out]

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

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Rubi [F]  time = 1.65887, 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 = {} \[ \int \frac{\left (a+b \tan ^{-1}\left (c x^2\right )\right )^3}{x^5} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(a + b*ArcTan[c*x^2])^3/x^5,x]

[Out]

(3*a*b^2*c^2*Log[x])/4 - (3*b*c*(1 - I*c*x^2)*(2*a + I*b*Log[1 - I*c*x^2])^2)/(32*x^2) + ((3*I)/32)*b*c^2*Log[
I*c*x^2]*(2*a + I*b*Log[1 - I*c*x^2])^2 - (c^2*(2*a + I*b*Log[1 - I*c*x^2])^3)/32 - (2*a + I*b*Log[1 - I*c*x^2
])^3/(32*x^4) + (3*b^3*c*(1 + I*c*x^2)*Log[1 + I*c*x^2]^2)/(32*x^2) + ((3*I)/32)*b^3*c^2*Log[(-I)*c*x^2]*Log[1
+ I*c*x^2]^2 - (I/32)*b^3*c^2*Log[1 + I*c*x^2]^3 - ((I/32)*b^3*Log[1 + I*c*x^2]^3)/x^4 + ((3*I)/16)*b^3*c^2*P
olyLog[2, (-I)*c*x^2] - ((3*I)/16)*b^3*c^2*PolyLog[2, I*c*x^2] - (3*b^2*c^2*(2*a + I*b*Log[1 - I*c*x^2])*PolyL
og[2, 1 - I*c*x^2])/16 + ((3*I)/16)*b^3*c^2*Log[1 + I*c*x^2]*PolyLog[2, 1 + I*c*x^2] + ((3*I)/16)*b^3*c^2*Poly
Log[3, 1 - I*c*x^2] - ((3*I)/16)*b^3*c^2*PolyLog[3, 1 + I*c*x^2] + ((3*I)/16)*b*Defer[Subst][Defer[Int][(((-2*
I)*a + b*Log[1 - I*c*x])^2*Log[1 + I*c*x])/x^3, 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^3, x], x, x^2]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tan ^{-1}\left (c x^2\right )\right )^3}{x^5} \, dx &=\int \left (\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{8 x^5}+\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^5}-\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^5}+\frac{i b^3 \log ^3\left (1+i c x^2\right )}{8 x^5}\right ) \, dx\\ &=\frac{1}{8} \int \frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{x^5} \, 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^5} \, 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^5} \, dx+\frac{1}{8} \left (i b^3\right ) \int \frac{\log ^3\left (1+i c x^2\right )}{x^5} \, dx\\ &=\frac{1}{16} \operatorname{Subst}\left (\int \frac{(2 a+i b \log (1-i c x))^3}{x^3} \, 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^3} \, 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^3} \, 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^3} \, dx,x,x^2\right )\\ &=-\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{32 x^4}-\frac{i b^3 \log ^3\left (1+i c x^2\right )}{32 x^4}+\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^3} \, 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^3} \, dx,x,x^2\right )+\frac{1}{32} (3 b c) \operatorname{Subst}\left (\int \frac{(2 a+i b \log (1-i c x))^2}{x^2 (1-i c x)} \, dx,x,x^2\right )-\frac{1}{32} \left (3 b^3 c\right ) \operatorname{Subst}\left (\int \frac{\log ^2(1+i c x)}{x^2 (1+i c x)} \, dx,x,x^2\right )\\ &=-\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{32 x^4}-\frac{i b^3 \log ^3\left (1+i c x^2\right )}{32 x^4}+\frac{1}{32} (3 i b) \operatorname{Subst}\left (\int \frac{(2 a+i b \log (x))^2}{x \left (-\frac{i}{c}+\frac{i x}{c}\right )^2} \, dx,x,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^3} \, 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^3} \, dx,x,x^2\right )+\frac{1}{32} \left (3 i b^3\right ) \operatorname{Subst}\left (\int \frac{\log ^2(x)}{x \left (\frac{i}{c}-\frac{i x}{c}\right )^2} \, dx,x,1+i c x^2\right )\\ &=-\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{32 x^4}-\frac{i b^3 \log ^3\left (1+i c x^2\right )}{32 x^4}+\frac{1}{32} (3 i b) \operatorname{Subst}\left (\int \frac{(2 a+i b \log (x))^2}{\left (-\frac{i}{c}+\frac{i x}{c}\right )^2} \, dx,x,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^3} \, 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^3} \, dx,x,x^2\right )+\frac{1}{32} \left (3 i b^3\right ) \operatorname{Subst}\left (\int \frac{\log ^2(x)}{\left (\frac{i}{c}-\frac{i x}{c}\right )^2} \, dx,x,1+i c x^2\right )-\frac{1}{32} (3 b c) \operatorname{Subst}\left (\int \frac{(2 a+i b \log (x))^2}{x \left (-\frac{i}{c}+\frac{i x}{c}\right )} \, dx,x,1-i c x^2\right )+\frac{1}{32} \left (3 b^3 c\right ) \operatorname{Subst}\left (\int \frac{\log ^2(x)}{x \left (\frac{i}{c}-\frac{i x}{c}\right )} \, dx,x,1+i c x^2\right )\\ &=-\frac{3 b c \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{32 x^2}-\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{32 x^4}+\frac{3 b^3 c \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{32 x^2}-\frac{i b^3 \log ^3\left (1+i c x^2\right )}{32 x^4}+\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^3} \, 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^3} \, dx,x,x^2\right )-\frac{1}{32} (3 b c) \operatorname{Subst}\left (\int \frac{(2 a+i b \log (x))^2}{-\frac{i}{c}+\frac{i x}{c}} \, dx,x,1-i c x^2\right )+\frac{1}{16} \left (3 i b^2 c\right ) \operatorname{Subst}\left (\int \frac{2 a+i b \log (x)}{-\frac{i}{c}+\frac{i x}{c}} \, dx,x,1-i c x^2\right )+\frac{1}{32} \left (3 b^3 c\right ) \operatorname{Subst}\left (\int \frac{\log ^2(x)}{\frac{i}{c}-\frac{i x}{c}} \, dx,x,1+i c x^2\right )-\frac{1}{16} \left (3 b^3 c\right ) \operatorname{Subst}\left (\int \frac{\log (x)}{\frac{i}{c}-\frac{i x}{c}} \, dx,x,1+i c x^2\right )-\frac{1}{32} \left (3 i b c^2\right ) \operatorname{Subst}\left (\int \frac{(2 a+i b \log (x))^2}{x} \, dx,x,1-i c x^2\right )-\frac{1}{32} \left (3 i b^3 c^2\right ) \operatorname{Subst}\left (\int \frac{\log ^2(x)}{x} \, dx,x,1+i c x^2\right )\\ &=\frac{3}{4} a b^2 c^2 \log (x)-\frac{3 b c \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{32 x^2}+\frac{3}{32} i b c^2 \log \left (i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2-\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{32 x^4}+\frac{3 b^3 c \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{32 x^2}+\frac{3}{32} i b^3 c^2 \log \left (-i c x^2\right ) \log ^2\left (1+i c x^2\right )-\frac{i b^3 \log ^3\left (1+i c x^2\right )}{32 x^4}+\frac{3}{16} i b^3 c^2 \text{Li}_2\left (-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^3} \, 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^3} \, dx,x,x^2\right )-\frac{1}{16} \left (3 b^3 c\right ) \operatorname{Subst}\left (\int \frac{\log (x)}{-\frac{i}{c}+\frac{i x}{c}} \, dx,x,1-i c x^2\right )-\frac{1}{32} \left (3 c^2\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,2 a+i b \log \left (1-i c x^2\right )\right )+\frac{1}{16} \left (3 b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-x) (2 a+i b \log (x))}{x} \, dx,x,1-i c x^2\right )-\frac{1}{32} \left (3 i b^3 c^2\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,\log \left (1+i c x^2\right )\right )-\frac{1}{16} \left (3 i b^3 c^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-x) \log (x)}{x} \, dx,x,1+i c x^2\right )\\ &=\frac{3}{4} a b^2 c^2 \log (x)-\frac{3 b c \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{32 x^2}+\frac{3}{32} i b c^2 \log \left (i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2-\frac{1}{32} c^2 \left (2 a+i b \log \left (1-i c x^2\right )\right )^3-\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{32 x^4}+\frac{3 b^3 c \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{32 x^2}+\frac{3}{32} i b^3 c^2 \log \left (-i c x^2\right ) \log ^2\left (1+i c x^2\right )-\frac{1}{32} i b^3 c^2 \log ^3\left (1+i c x^2\right )-\frac{i b^3 \log ^3\left (1+i c x^2\right )}{32 x^4}+\frac{3}{16} i b^3 c^2 \text{Li}_2\left (-i c x^2\right )-\frac{3}{16} i b^3 c^2 \text{Li}_2\left (i c x^2\right )-\frac{3}{16} b^2 c^2 \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}{16} i b^3 c^2 \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^3} \, 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^3} \, dx,x,x^2\right )+\frac{1}{16} \left (3 i b^3 c^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,1-i c x^2\right )-\frac{1}{16} \left (3 i b^3 c^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,1+i c x^2\right )\\ &=\frac{3}{4} a b^2 c^2 \log (x)-\frac{3 b c \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{32 x^2}+\frac{3}{32} i b c^2 \log \left (i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2-\frac{1}{32} c^2 \left (2 a+i b \log \left (1-i c x^2\right )\right )^3-\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{32 x^4}+\frac{3 b^3 c \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{32 x^2}+\frac{3}{32} i b^3 c^2 \log \left (-i c x^2\right ) \log ^2\left (1+i c x^2\right )-\frac{1}{32} i b^3 c^2 \log ^3\left (1+i c x^2\right )-\frac{i b^3 \log ^3\left (1+i c x^2\right )}{32 x^4}+\frac{3}{16} i b^3 c^2 \text{Li}_2\left (-i c x^2\right )-\frac{3}{16} i b^3 c^2 \text{Li}_2\left (i c x^2\right )-\frac{3}{16} b^2 c^2 \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}{16} i b^3 c^2 \log \left (1+i c x^2\right ) \text{Li}_2\left (1+i c x^2\right )+\frac{3}{16} i b^3 c^2 \text{Li}_3\left (1-i c x^2\right )-\frac{3}{16} i b^3 c^2 \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^3} \, 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^3} \, dx,x,x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.334389, size = 196, normalized size = 1.32 \[ -\frac{3 i b^3 c^2 x^4 \text{PolyLog}\left (2,e^{2 i \tan ^{-1}\left (c x^2\right )}\right )+a \left (a \left (a+3 b c x^2\right )-6 b^2 c^2 x^4 \log \left (\frac{c x^2}{\sqrt{c^2 x^4+1}}\right )\right )+3 b^2 \tan ^{-1}\left (c x^2\right )^2 \left (a c^2 x^4+a+b c x^2 \left (1+i c x^2\right )\right )+3 b \tan ^{-1}\left (c x^2\right ) \left (a \left (a c^2 x^4+a+2 b c x^2\right )-2 b^2 c^2 x^4 \log \left (1-e^{2 i \tan ^{-1}\left (c x^2\right )}\right )\right )+b^3 \left (c^2 x^4+1\right ) \tan ^{-1}\left (c x^2\right )^3}{4 x^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTan[c*x^2])^3/x^5,x]

[Out]

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

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Maple [F]  time = 0.661, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\arctan \left ( c{x}^{2} \right ) \right ) ^{3}}{{x}^{5}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x^2))^3/x^5,x)

[Out]

int((a+b*arctan(c*x^2))^3/x^5,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^5,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^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))^3/x^5,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^5, 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^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x**2))**3/x**5,x)

[Out]

Integral((a + b*atan(c*x**2))**3/x**5, 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^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))^3/x^5,x, algorithm="giac")

[Out]

integrate((b*arctan(c*x^2) + a)^3/x^5, x)