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

Optimal. Leaf size=125 \[ -\frac{3}{2} b^2 c \text{PolyLog}\left (2,\frac{2}{c x^2+1}-1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac{3}{4} b^3 c \text{PolyLog}\left (3,\frac{2}{c x^2+1}-1\right )+\frac{1}{2} c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3-\frac{\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{2 x^2}+\frac{3}{2} b c \log \left (2-\frac{2}{c x^2+1}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \]

[Out]

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

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Rubi [F]  time = 0.780071, 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 = {} \[ \int \frac{\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{x^3} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(a + b*ArcTanh[c*x^2])^3/x^3,x]

[Out]

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

Rubi steps

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

Mathematica [C]  time = 0.414398, size = 222, normalized size = 1.78 \[ \frac{1}{4} \left (6 a b^2 c \left (\tanh ^{-1}\left (c x^2\right ) \left (\left (1-\frac{1}{c x^2}\right ) \tanh ^{-1}\left (c x^2\right )+2 \log \left (1-e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )\right )-\text{PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )\right )+2 b^3 c \left (3 \tanh ^{-1}\left (c x^2\right ) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}\left (c x^2\right )}\right )-\frac{3}{2} \text{PolyLog}\left (3,e^{2 \tanh ^{-1}\left (c x^2\right )}\right )-\frac{\tanh ^{-1}\left (c x^2\right )^3}{c x^2}-\tanh ^{-1}\left (c x^2\right )^3+3 \tanh ^{-1}\left (c x^2\right )^2 \log \left (1-e^{2 \tanh ^{-1}\left (c x^2\right )}\right )+\frac{i \pi ^3}{8}\right )-3 a^2 b c \log \left (1-c^2 x^4\right )-\frac{6 a^2 b \tanh ^{-1}\left (c x^2\right )}{x^2}+12 a^2 b c \log (x)-\frac{2 a^3}{x^2}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.194, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b{\it Artanh} \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*arctanh(c*x^2))^3/x^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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