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

Optimal. Leaf size=226 \[ -\frac{a+b \tanh ^{-1}\left (c x^2\right )}{2 e (d+e x)^2}-\frac{b c d e}{\left (c^2 d^4-e^4\right ) (d+e x)}+\frac{b c e \left (3 c^2 d^4+e^4\right ) \log (d+e x)}{\left (c^2 d^4-e^4\right )^2}+\frac{b c^{3/2} d \tan ^{-1}\left (\sqrt{c} x\right )}{\left (c d^2+e^2\right )^2}-\frac{b c^{3/2} d \tanh ^{-1}\left (\sqrt{c} x\right )}{\left (c d^2-e^2\right )^2}-\frac{b c \left (c d^2+e^2\right ) \log \left (1-c x^2\right )}{4 e \left (c d^2-e^2\right )^2}+\frac{b c \left (c d^2-e^2\right ) \log \left (c x^2+1\right )}{4 e \left (c d^2+e^2\right )^2} \]

[Out]

-((b*c*d*e)/((c^2*d^4 - e^4)*(d + e*x))) + (b*c^(3/2)*d*ArcTan[Sqrt[c]*x])/(c*d^2 + e^2)^2 - (b*c^(3/2)*d*ArcT
anh[Sqrt[c]*x])/(c*d^2 - e^2)^2 - (a + b*ArcTanh[c*x^2])/(2*e*(d + e*x)^2) + (b*c*e*(3*c^2*d^4 + e^4)*Log[d +
e*x])/(c^2*d^4 - e^4)^2 - (b*c*(c*d^2 + e^2)*Log[1 - c*x^2])/(4*e*(c*d^2 - e^2)^2) + (b*c*(c*d^2 - e^2)*Log[1
+ c*x^2])/(4*e*(c*d^2 + e^2)^2)

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Rubi [F]  time = 0.0664324, 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 = {} \[ \int \frac{a+b \tanh ^{-1}\left (c x^2\right )}{(d+e x)^3} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

-a/(2*e*(d + e*x)^2) + b*Defer[Int][ArcTanh[c*x^2]/(d + e*x)^3, x]

Rubi steps

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

Mathematica [A]  time = 0.712337, size = 379, normalized size = 1.68 \[ \frac{1}{4} \left (-\frac{2 a}{e (d+e x)^2}+\frac{b c^2 \left (c^2 d^6+3 d^2 e^4\right ) \log \left (c x^2+1\right )}{e \left (e^4-c^2 d^4\right )^2}-\frac{b c e \left (3 c^2 d^4+e^4\right ) \log \left (1-c^2 x^4\right )}{\left (e^4-c^2 d^4\right )^2}-\frac{4 b c d e}{\left (c^2 d^4-e^4\right ) (d+e x)}-\frac{b c^{3/2} d \left (-2 c^2 d^4 e+c^{5/2} d^5-4 c d^2 e^3+3 \sqrt{c} d e^4-2 e^5\right ) \log \left (1-\sqrt{c} x\right )}{e \left (e^4-c^2 d^4\right )^2}-\frac{b c^{3/2} d \left (2 c^2 d^4 e+c^{5/2} d^5+4 c d^2 e^3+3 \sqrt{c} d e^4+2 e^5\right ) \log \left (\sqrt{c} x+1\right )}{e \left (e^4-c^2 d^4\right )^2}+\frac{4 b c e \left (3 c^2 d^4+e^4\right ) \log (d+e x)}{\left (e^4-c^2 d^4\right )^2}+\frac{4 b c^{3/2} d \tan ^{-1}\left (\sqrt{c} x\right )}{\left (c d^2+e^2\right )^2}-\frac{2 b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a)/(e*(d + e*x)^2) - (4*b*c*d*e)/((c^2*d^4 - e^4)*(d + e*x)) + (4*b*c^(3/2)*d*ArcTan[Sqrt[c]*x])/(c*d^2 +
e^2)^2 - (2*b*ArcTanh[c*x^2])/(e*(d + e*x)^2) - (b*c^(3/2)*d*(c^(5/2)*d^5 - 2*c^2*d^4*e - 4*c*d^2*e^3 + 3*Sqr
t[c]*d*e^4 - 2*e^5)*Log[1 - Sqrt[c]*x])/(e*(-(c^2*d^4) + e^4)^2) - (b*c^(3/2)*d*(c^(5/2)*d^5 + 2*c^2*d^4*e + 4
*c*d^2*e^3 + 3*Sqrt[c]*d*e^4 + 2*e^5)*Log[1 + Sqrt[c]*x])/(e*(-(c^2*d^4) + e^4)^2) + (4*b*c*e*(3*c^2*d^4 + e^4
)*Log[d + e*x])/(-(c^2*d^4) + e^4)^2 + (b*c^2*(c^2*d^6 + 3*d^2*e^4)*Log[1 + c*x^2])/(e*(-(c^2*d^4) + e^4)^2) -
(b*c*e*(3*c^2*d^4 + e^4)*Log[1 - c^2*x^4])/(-(c^2*d^4) + e^4)^2)/4

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Maple [A]  time = 0.04, size = 310, normalized size = 1.4 \begin{align*} -{\frac{a}{2\, \left ( ex+d \right ) ^{2}e}}-{\frac{b{\it Artanh} \left ( c{x}^{2} \right ) }{2\, \left ( ex+d \right ) ^{2}e}}+{\frac{b{c}^{2}\ln \left ( c{x}^{2}+1 \right ){d}^{2}}{4\,e \left ( c{d}^{2}+{e}^{2} \right ) ^{2}}}-{\frac{bec\ln \left ( c{x}^{2}+1 \right ) }{4\, \left ( c{d}^{2}+{e}^{2} \right ) ^{2}}}+{\frac{bd}{ \left ( c{d}^{2}+{e}^{2} \right ) ^{2}}{c}^{{\frac{3}{2}}}\arctan \left ( x\sqrt{c} \right ) }-{\frac{b{c}^{2}\ln \left ( c{x}^{2}-1 \right ){d}^{2}}{4\,e \left ( c{d}^{2}-{e}^{2} \right ) ^{2}}}-{\frac{bec\ln \left ( c{x}^{2}-1 \right ) }{4\, \left ( c{d}^{2}-{e}^{2} \right ) ^{2}}}-{\frac{bd}{ \left ( c{d}^{2}-{e}^{2} \right ) ^{2}}{c}^{{\frac{3}{2}}}{\it Artanh} \left ( x\sqrt{c} \right ) }-{\frac{becd}{ \left ( c{d}^{2}-{e}^{2} \right ) \left ( c{d}^{2}+{e}^{2} \right ) \left ( ex+d \right ) }}+3\,{\frac{be{c}^{3}\ln \left ( ex+d \right ){d}^{4}}{ \left ( c{d}^{2}-{e}^{2} \right ) ^{2} \left ( c{d}^{2}+{e}^{2} \right ) ^{2}}}+{\frac{b{e}^{5}c\ln \left ( ex+d \right ) }{ \left ( c{d}^{2}-{e}^{2} \right ) ^{2} \left ( c{d}^{2}+{e}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a/(e*x+d)^2/e-1/2*b/(e*x+d)^2/e*arctanh(c*x^2)+1/4*b/e*c^2/(c*d^2+e^2)^2*ln(c*x^2+1)*d^2-1/4*b*e*c/(c*d^2
+e^2)^2*ln(c*x^2+1)+b*c^(3/2)*d*arctan(x*c^(1/2))/(c*d^2+e^2)^2-1/4*b/e*c^2/(c*d^2-e^2)^2*ln(c*x^2-1)*d^2-1/4*
b*e*c/(c*d^2-e^2)^2*ln(c*x^2-1)-b*c^(3/2)*d*arctanh(x*c^(1/2))/(c*d^2-e^2)^2-b*e*c*d/(c*d^2-e^2)/(c*d^2+e^2)/(
e*x+d)+3*b*e*c^3/(c*d^2-e^2)^2/(c*d^2+e^2)^2*ln(e*x+d)*d^4+b*e^5*c/(c*d^2-e^2)^2/(c*d^2+e^2)^2*ln(e*x+d)

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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))/(e*x+d)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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*arctanh(c*x^2))/(e*x+d)^3,x, algorithm="fricas")

[Out]

Timed out

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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))/(e*x+d)**3,x)

[Out]

Timed out

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Giac [B]  time = 89.4739, size = 684, normalized size = 3.03 \begin{align*} \frac{b c^{\frac{3}{2}} d \arctan \left (\sqrt{c} x\right )}{c^{2} d^{4} + 2 \, c d^{2} e^{2} + e^{4}} + \frac{b c^{2} d \arctan \left (\frac{c x}{\sqrt{-c}}\right )}{{\left (c^{2} d^{4} - 2 \, c d^{2} e^{2} + e^{4}\right )} \sqrt{-c}} - \frac{{\left (b c^{2} d^{2} + b c e^{2}\right )} \log \left (c x^{2} - 1\right )}{4 \,{\left (c^{2} d^{4} e - 2 \, c d^{2} e^{3} + e^{5}\right )}} + \frac{{\left (b c^{2} d^{2} - b c e^{2}\right )} \log \left (-c x^{2} - 1\right )}{4 \,{\left (c^{2} d^{4} e + 2 \, c d^{2} e^{3} + e^{5}\right )}} - \frac{b c^{4} d^{8} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a c^{4} d^{8} - 12 \, b c^{3} d^{4} x^{2} e^{4} \log \left (x e + d\right ) - 24 \, b c^{3} d^{5} x e^{3} \log \left (x e + d\right ) - 12 \, b c^{3} d^{6} e^{2} \log \left (x e + d\right ) + 4 \, b c^{3} d^{5} x e^{3} + 4 \, b c^{3} d^{6} e^{2} - 2 \, b c^{2} d^{4} e^{4} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) - 4 \, a c^{2} d^{4} e^{4} - 4 \, b c x^{2} e^{8} \log \left (x e + d\right ) - 8 \, b c d x e^{7} \log \left (x e + d\right ) - 4 \, b c d^{2} e^{6} \log \left (x e + d\right ) - 4 \, b c d x e^{7} - 4 \, b c d^{2} e^{6} + b e^{8} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a e^{8}}{4 \,{\left (c^{4} d^{8} x^{2} e^{3} + 2 \, c^{4} d^{9} x e^{2} + c^{4} d^{10} e - 2 \, c^{2} d^{4} x^{2} e^{7} - 4 \, c^{2} d^{5} x e^{6} - 2 \, c^{2} d^{6} e^{5} + x^{2} e^{11} + 2 \, d x e^{10} + d^{2} e^{9}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*c^(3/2)*d*arctan(sqrt(c)*x)/(c^2*d^4 + 2*c*d^2*e^2 + e^4) + b*c^2*d*arctan(c*x/sqrt(-c))/((c^2*d^4 - 2*c*d^2
*e^2 + e^4)*sqrt(-c)) - 1/4*(b*c^2*d^2 + b*c*e^2)*log(c*x^2 - 1)/(c^2*d^4*e - 2*c*d^2*e^3 + e^5) + 1/4*(b*c^2*
d^2 - b*c*e^2)*log(-c*x^2 - 1)/(c^2*d^4*e + 2*c*d^2*e^3 + e^5) - 1/4*(b*c^4*d^8*log(-(c*x^2 + 1)/(c*x^2 - 1))
+ 2*a*c^4*d^8 - 12*b*c^3*d^4*x^2*e^4*log(x*e + d) - 24*b*c^3*d^5*x*e^3*log(x*e + d) - 12*b*c^3*d^6*e^2*log(x*e
+ d) + 4*b*c^3*d^5*x*e^3 + 4*b*c^3*d^6*e^2 - 2*b*c^2*d^4*e^4*log(-(c*x^2 + 1)/(c*x^2 - 1)) - 4*a*c^2*d^4*e^4
- 4*b*c*x^2*e^8*log(x*e + d) - 8*b*c*d*x*e^7*log(x*e + d) - 4*b*c*d^2*e^6*log(x*e + d) - 4*b*c*d*x*e^7 - 4*b*c
*d^2*e^6 + b*e^8*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 2*a*e^8)/(c^4*d^8*x^2*e^3 + 2*c^4*d^9*x*e^2 + c^4*d^10*e - 2*
c^2*d^4*x^2*e^7 - 4*c^2*d^5*x*e^6 - 2*c^2*d^6*e^5 + x^2*e^11 + 2*d*x*e^10 + d^2*e^9)