3.31 \(\int \frac{(a+b \tanh ^{-1}(c x^2))^2}{(d+e x)^2} \, dx\)

Optimal. Leaf size=22 \[ \text{Unintegrable}\left (\frac{\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{(d+e x)^2},x\right ) \]

[Out]

Unintegrable[(a + b*ArcTanh[c*x^2])^2/(d + e*x)^2, x]

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Rubi [A]  time = 0.368842, 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 = {} \[ \int \frac{\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{(d+e x)^2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(a + b*ArcTanh[c*x^2])^2/(d + e*x)^2,x]

[Out]

-(a^2/(e*(d + e*x))) + (2*a*b*Sqrt[c]*ArcTan[Sqrt[c]*x])/(c*d^2 + e^2) - (2*a*b*Sqrt[c]*ArcTanh[Sqrt[c]*x])/(c
*d^2 - e^2) - (2*a*b*ArcTanh[c*x^2])/(e*(d + e*x)) + (4*a*b*c*d*e*Log[d + e*x])/(c^2*d^4 - e^4) - (a*b*c*d*Log
[1 - c*x^2])/(e*(c*d^2 - e^2)) + (a*b*c*d*Log[1 + c*x^2])/(e*(c*d^2 + e^2)) + b^2*Defer[Int][ArcTanh[c*x^2]^2/
(d + e*x)^2, x]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{(d+e x)^2} \, dx &=\int \left (\frac{a^2}{(d+e x)^2}+\frac{2 a b \tanh ^{-1}\left (c x^2\right )}{(d+e x)^2}+\frac{b^2 \tanh ^{-1}\left (c x^2\right )^2}{(d+e x)^2}\right ) \, dx\\ &=-\frac{a^2}{e (d+e x)}+(2 a b) \int \frac{\tanh ^{-1}\left (c x^2\right )}{(d+e x)^2} \, dx+b^2 \int \frac{\tanh ^{-1}\left (c x^2\right )^2}{(d+e x)^2} \, dx\\ &=-\frac{a^2}{e (d+e x)}-\frac{2 a b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}+b^2 \int \frac{\tanh ^{-1}\left (c x^2\right )^2}{(d+e x)^2} \, dx+\frac{(2 a b) \int \frac{2 c x}{(d+e x) \left (1-c^2 x^4\right )} \, dx}{e}\\ &=-\frac{a^2}{e (d+e x)}-\frac{2 a b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}+b^2 \int \frac{\tanh ^{-1}\left (c x^2\right )^2}{(d+e x)^2} \, dx+\frac{(4 a b c) \int \frac{x}{(d+e x) \left (1-c^2 x^4\right )} \, dx}{e}\\ &=-\frac{a^2}{e (d+e x)}-\frac{2 a b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}+b^2 \int \frac{\tanh ^{-1}\left (c x^2\right )^2}{(d+e x)^2} \, dx+\frac{(4 a b c) \int \left (-\frac{d e^3}{\left (-c^2 d^4+e^4\right ) (d+e x)}+\frac{e-c d x}{2 \left (c d^2-e^2\right ) \left (-1+c x^2\right )}+\frac{e+c d x}{2 \left (c d^2+e^2\right ) \left (1+c x^2\right )}\right ) \, dx}{e}\\ &=-\frac{a^2}{e (d+e x)}-\frac{2 a b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}+\frac{4 a b c d e \log (d+e x)}{c^2 d^4-e^4}+b^2 \int \frac{\tanh ^{-1}\left (c x^2\right )^2}{(d+e x)^2} \, dx+\frac{(2 a b c) \int \frac{e-c d x}{-1+c x^2} \, dx}{e \left (c d^2-e^2\right )}+\frac{(2 a b c) \int \frac{e+c d x}{1+c x^2} \, dx}{e \left (c d^2+e^2\right )}\\ &=-\frac{a^2}{e (d+e x)}-\frac{2 a b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}+\frac{4 a b c d e \log (d+e x)}{c^2 d^4-e^4}+b^2 \int \frac{\tanh ^{-1}\left (c x^2\right )^2}{(d+e x)^2} \, dx+\frac{(2 a b c) \int \frac{1}{-1+c x^2} \, dx}{c d^2-e^2}-\frac{\left (2 a b c^2 d\right ) \int \frac{x}{-1+c x^2} \, dx}{e \left (c d^2-e^2\right )}+\frac{(2 a b c) \int \frac{1}{1+c x^2} \, dx}{c d^2+e^2}+\frac{\left (2 a b c^2 d\right ) \int \frac{x}{1+c x^2} \, dx}{e \left (c d^2+e^2\right )}\\ &=-\frac{a^2}{e (d+e x)}+\frac{2 a b \sqrt{c} \tan ^{-1}\left (\sqrt{c} x\right )}{c d^2+e^2}-\frac{2 a b \sqrt{c} \tanh ^{-1}\left (\sqrt{c} x\right )}{c d^2-e^2}-\frac{2 a b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}+\frac{4 a b c d e \log (d+e x)}{c^2 d^4-e^4}-\frac{a b c d \log \left (1-c x^2\right )}{e \left (c d^2-e^2\right )}+\frac{a b c d \log \left (1+c x^2\right )}{e \left (c d^2+e^2\right )}+b^2 \int \frac{\tanh ^{-1}\left (c x^2\right )^2}{(d+e x)^2} \, dx\\ \end{align*}

Mathematica [A]  time = 42.4755, size = 0, normalized size = 0. \[ \int \frac{\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{(d+e x)^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 0.457, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b{\it Artanh} \left ( c{x}^{2} \right ) \right ) ^{2}}{ \left ( ex+d \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a+b*arctanh(c*x^2))^2/(e*x+d)^2,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*arctanh(c*x^2))^2/(e*x+d)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \operatorname{artanh}\left (c x^{2}\right )^{2} + 2 \, a b \operatorname{artanh}\left (c x^{2}\right ) + a^{2}}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))^2/(e*x+d)^2,x, algorithm="fricas")

[Out]

integral((b^2*arctanh(c*x^2)^2 + 2*a*b*arctanh(c*x^2) + a^2)/(e^2*x^2 + 2*d*e*x + d^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))^2/(e*x+d)^2,x, algorithm="giac")

[Out]

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