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

Optimal. Leaf size=139 \[ -\frac{3}{4} b^3 c^2 \text{PolyLog}\left (2,\frac{2}{c x^2+1}-1\right )+\frac{3}{2} b^2 c^2 \log \left (2-\frac{2}{c x^2+1}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\frac{3}{4} b c^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\frac{1}{4} c^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3-\frac{3 b c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{4 x^2}-\frac{\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{4 x^4} \]

[Out]

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

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Rubi [F]  time = 1.55794, 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 = {} \[ \int \frac{\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{x^5} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(a + b*ArcTanh[c*x^2])^3/x^5,x]

[Out]

(3*a*b^2*c^2*Log[x])/4 - (3*b*c*(1 - c*x^2)*(2*a - b*Log[1 - c*x^2])^2)/(32*x^2) + (3*b*c^2*Log[c*x^2]*(2*a -
b*Log[1 - c*x^2])^2)/32 + (c^2*(2*a - b*Log[1 - c*x^2])^3)/32 - (2*a - b*Log[1 - c*x^2])^3/(32*x^4) - (3*b^3*c
*(1 + c*x^2)*Log[1 + c*x^2]^2)/(32*x^2) - (3*b^3*c^2*Log[-(c*x^2)]*Log[1 + c*x^2]^2)/32 + (b^3*c^2*Log[1 + c*x
^2]^3)/32 - (b^3*Log[1 + c*x^2]^3)/(32*x^4) - (3*b^3*c^2*PolyLog[2, -(c*x^2)])/16 + (3*b^3*c^2*PolyLog[2, c*x^
2])/16 - (3*b^2*c^2*(2*a - b*Log[1 - c*x^2])*PolyLog[2, 1 - c*x^2])/16 - (3*b^3*c^2*Log[1 + c*x^2]*PolyLog[2,
1 + c*x^2])/16 - (3*b^3*c^2*PolyLog[3, 1 - c*x^2])/16 + (3*b^3*c^2*PolyLog[3, 1 + c*x^2])/16 + (3*b*Defer[Subs
t][Defer[Int][((-2*a + b*Log[1 - c*x])^2*Log[1 + c*x])/x^3, x], x, x^2])/16 - (3*b^2*Defer[Subst][Defer[Int][(
(-2*a + b*Log[1 - c*x])*Log[1 + c*x]^2)/x^3, x], x, x^2])/16

Rubi steps

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

Mathematica [A]  time = 0.290196, size = 218, normalized size = 1.57 \[ \frac{-6 b^3 c^2 x^4 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )+a \left (-2 a^2-3 a b c^2 x^4 \log \left (1-c x^2\right )+3 a b c^2 x^4 \log \left (c x^2+1\right )-6 a b c x^2+12 b^2 c^2 x^4 \log \left (\frac{c x^2}{\sqrt{1-c^2 x^4}}\right )\right )-6 b \tanh ^{-1}\left (c x^2\right ) \left (a^2+2 a b c x^2-2 b^2 c^2 x^4 \log \left (1-e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )\right )+6 b^2 \left (c x^2-1\right ) \tanh ^{-1}\left (c x^2\right )^2 \left (a c x^2+a+b c x^2\right )+2 b^3 \left (c^2 x^4-1\right ) \tanh ^{-1}\left (c x^2\right )^3}{8 x^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(6*b^2*(-1 + c*x^2)*(a + a*c*x^2 + b*c*x^2)*ArcTanh[c*x^2]^2 + 2*b^3*(-1 + c^2*x^4)*ArcTanh[c*x^2]^3 - 6*b*Arc
Tanh[c*x^2]*(a^2 + 2*a*b*c*x^2 - 2*b^2*c^2*x^4*Log[1 - E^(-2*ArcTanh[c*x^2])]) + a*(-2*a^2 - 6*a*b*c*x^2 - 3*a
*b*c^2*x^4*Log[1 - c*x^2] + 3*a*b*c^2*x^4*Log[1 + c*x^2] + 12*b^2*c^2*x^4*Log[(c*x^2)/Sqrt[1 - c^2*x^4]]) - 6*
b^3*c^2*x^4*PolyLog[2, E^(-2*ArcTanh[c*x^2])])/(8*x^4)

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Maple [F]  time = 0.231, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b{\it Artanh} \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*arctanh(c*x^2))^3/x^5,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{3}{8} \,{\left ({\left (c \log \left (c x^{2} + 1\right ) - c \log \left (c x^{2} - 1\right ) - \frac{2}{x^{2}}\right )} c - \frac{2 \, \operatorname{artanh}\left (c x^{2}\right )}{x^{4}}\right )} a^{2} b + \frac{3}{16} \,{\left ({\left (2 \,{\left (\log \left (c x^{2} - 1\right ) - 2\right )} \log \left (c x^{2} + 1\right ) - \log \left (c x^{2} + 1\right )^{2} - \log \left (c x^{2} - 1\right )^{2} - 4 \, \log \left (c x^{2} - 1\right ) + 16 \, \log \left (x\right )\right )} c^{2} + 4 \,{\left (c \log \left (c x^{2} + 1\right ) - c \log \left (c x^{2} - 1\right ) - \frac{2}{x^{2}}\right )} c \operatorname{artanh}\left (c x^{2}\right )\right )} a b^{2} - \frac{1}{32} \, b^{3}{\left (\frac{{\left (c^{2} x^{4} - 1\right )} \log \left (-c x^{2} + 1\right )^{3} + 3 \,{\left (2 \, c x^{2} -{\left (c^{2} x^{4} - 1\right )} \log \left (c x^{2} + 1\right )\right )} \log \left (-c x^{2} + 1\right )^{2}}{x^{4}} + 4 \, \int -\frac{{\left (c x^{2} - 1\right )} \log \left (c x^{2} + 1\right )^{3} + 3 \,{\left (2 \, c^{2} x^{4} -{\left (c x^{2} - 1\right )} \log \left (c x^{2} + 1\right )^{2} -{\left (c^{3} x^{6} - c x^{2}\right )} \log \left (c x^{2} + 1\right )\right )} \log \left (-c x^{2} + 1\right )}{c x^{7} - x^{5}}\,{d x}\right )} - \frac{3 \, a b^{2} \operatorname{artanh}\left (c x^{2}\right )^{2}}{4 \, x^{4}} - \frac{a^{3}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*((c*log(c*x^2 + 1) - c*log(c*x^2 - 1) - 2/x^2)*c - 2*arctanh(c*x^2)/x^4)*a^2*b + 3/16*((2*(log(c*x^2 - 1)
- 2)*log(c*x^2 + 1) - log(c*x^2 + 1)^2 - log(c*x^2 - 1)^2 - 4*log(c*x^2 - 1) + 16*log(x))*c^2 + 4*(c*log(c*x^2
+ 1) - c*log(c*x^2 - 1) - 2/x^2)*c*arctanh(c*x^2))*a*b^2 - 1/32*b^3*(((c^2*x^4 - 1)*log(-c*x^2 + 1)^3 + 3*(2*
c*x^2 - (c^2*x^4 - 1)*log(c*x^2 + 1))*log(-c*x^2 + 1)^2)/x^4 + 4*integrate(-((c*x^2 - 1)*log(c*x^2 + 1)^3 + 3*
(2*c^2*x^4 - (c*x^2 - 1)*log(c*x^2 + 1)^2 - (c^3*x^6 - c*x^2)*log(c*x^2 + 1))*log(-c*x^2 + 1))/(c*x^7 - x^5),
x)) - 3/4*a*b^2*arctanh(c*x^2)^2/x^4 - 1/4*a^3/x^4

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{3} \operatorname{artanh}\left (c x^{2}\right )^{3} + 3 \, a b^{2} \operatorname{artanh}\left (c x^{2}\right )^{2} + 3 \, a^{2} b \operatorname{artanh}\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*arctanh(c*x^2))^3/x^5,x, algorithm="fricas")

[Out]

integral((b^3*arctanh(c*x^2)^3 + 3*a*b^2*arctanh(c*x^2)^2 + 3*a^2*b*arctanh(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{atanh}{\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*atanh(c*x**2))**3/x**5,x)

[Out]

Integral((a + b*atanh(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 \operatorname{artanh}\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*arctanh(c*x^2))^3/x^5,x, algorithm="giac")

[Out]

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