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

Optimal. Leaf size=130 \[ -\frac{a+b \tanh ^{-1}(c x)}{2 e (d+e x)^2}+\frac{b c}{2 \left (c^2 d^2-e^2\right ) (d+e x)}-\frac{b c^3 d \log (d+e x)}{\left (c^2 d^2-e^2\right )^2}-\frac{b c^2 \log (1-c x)}{4 e (c d+e)^2}+\frac{b c^2 \log (c x+1)}{4 e (c d-e)^2} \]

[Out]

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

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Rubi [A]  time = 0.12943, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {5926, 710, 801} \[ -\frac{a+b \tanh ^{-1}(c x)}{2 e (d+e x)^2}+\frac{b c}{2 \left (c^2 d^2-e^2\right ) (d+e x)}-\frac{b c^3 d \log (d+e x)}{\left (c^2 d^2-e^2\right )^2}-\frac{b c^2 \log (1-c x)}{4 e (c d+e)^2}+\frac{b c^2 \log (c x+1)}{4 e (c d-e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5926

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b
*ArcTanh[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 - c^2*x^2), x], x] /; FreeQ
[{a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 710

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 +
a*e^2)), x] + Dist[c/(c*d^2 + a*e^2), Int[((d + e*x)^(m + 1)*(d - e*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d,
e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

Mathematica [A]  time = 0.149741, size = 133, normalized size = 1.02 \[ \frac{1}{4} \left (-\frac{2 a}{e (d+e x)^2}+\frac{2 b c}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac{4 b c^3 d \log (d+e x)}{\left (e^2-c^2 d^2\right )^2}-\frac{b c^2 \log (1-c x)}{e (c d+e)^2}+\frac{b c^2 \log (c x+1)}{e (e-c d)^2}-\frac{2 b \tanh ^{-1}(c x)}{e (d+e x)^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a)/(e*(d + e*x)^2) + (2*b*c)/((c^2*d^2 - e^2)*(d + e*x)) - (2*b*ArcTanh[c*x])/(e*(d + e*x)^2) - (b*c^2*Lo
g[1 - c*x])/(e*(c*d + e)^2) + (b*c^2*Log[1 + c*x])/(e*(-(c*d) + e)^2) - (4*b*c^3*d*Log[d + e*x])/(-(c^2*d^2) +
e^2)^2)/4

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Maple [A]  time = 0.039, size = 154, normalized size = 1.2 \begin{align*} -{\frac{a{c}^{2}}{2\, \left ( cxe+cd \right ) ^{2}e}}-{\frac{{c}^{2}b{\it Artanh} \left ( cx \right ) }{2\, \left ( cxe+cd \right ) ^{2}e}}+{\frac{{c}^{2}b}{ \left ( 2\,cd+2\,e \right ) \left ( cd-e \right ) \left ( cxe+cd \right ) }}-{\frac{b{c}^{3}d\ln \left ( cxe+cd \right ) }{ \left ( cd+e \right ) ^{2} \left ( cd-e \right ) ^{2}}}-{\frac{{c}^{2}b\ln \left ( cx-1 \right ) }{4\,e \left ( cd+e \right ) ^{2}}}+{\frac{{c}^{2}b\ln \left ( cx+1 \right ) }{4\, \left ( cd-e \right ) ^{2}e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.970935, size = 257, normalized size = 1.98 \begin{align*} -\frac{1}{4} \,{\left ({\left (\frac{4 \, c^{2} d \log \left (e x + d\right )}{c^{4} d^{4} - 2 \, c^{2} d^{2} e^{2} + e^{4}} - \frac{c \log \left (c x + 1\right )}{c^{2} d^{2} e - 2 \, c d e^{2} + e^{3}} + \frac{c \log \left (c x - 1\right )}{c^{2} d^{2} e + 2 \, c d e^{2} + e^{3}} - \frac{2}{c^{2} d^{3} - d e^{2} +{\left (c^{2} d^{2} e - e^{3}\right )} x}\right )} c + \frac{2 \, \operatorname{artanh}\left (c x\right )}{e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e}\right )} b - \frac{a}{2 \,{\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.3707, size = 933, normalized size = 7.18 \begin{align*} -\frac{2 \, a c^{4} d^{4} - 2 \, b c^{3} d^{3} e - 4 \, a c^{2} d^{2} e^{2} + 2 \, b c d e^{3} + 2 \, a e^{4} - 2 \,{\left (b c^{3} d^{2} e^{2} - b c e^{4}\right )} x -{\left (b c^{4} d^{4} + 2 \, b c^{3} d^{3} e + b c^{2} d^{2} e^{2} +{\left (b c^{4} d^{2} e^{2} + 2 \, b c^{3} d e^{3} + b c^{2} e^{4}\right )} x^{2} + 2 \,{\left (b c^{4} d^{3} e + 2 \, b c^{3} d^{2} e^{2} + b c^{2} d e^{3}\right )} x\right )} \log \left (c x + 1\right ) +{\left (b c^{4} d^{4} - 2 \, b c^{3} d^{3} e + b c^{2} d^{2} e^{2} +{\left (b c^{4} d^{2} e^{2} - 2 \, b c^{3} d e^{3} + b c^{2} e^{4}\right )} x^{2} + 2 \,{\left (b c^{4} d^{3} e - 2 \, b c^{3} d^{2} e^{2} + b c^{2} d e^{3}\right )} x\right )} \log \left (c x - 1\right ) + 4 \,{\left (b c^{3} d e^{3} x^{2} + 2 \, b c^{3} d^{2} e^{2} x + b c^{3} d^{3} e\right )} \log \left (e x + d\right ) +{\left (b c^{4} d^{4} - 2 \, b c^{2} d^{2} e^{2} + b e^{4}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{4 \,{\left (c^{4} d^{6} e - 2 \, c^{2} d^{4} e^{3} + d^{2} e^{5} +{\left (c^{4} d^{4} e^{3} - 2 \, c^{2} d^{2} e^{5} + e^{7}\right )} x^{2} + 2 \,{\left (c^{4} d^{5} e^{2} - 2 \, c^{2} d^{3} e^{4} + d e^{6}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*a*c^4*d^4 - 2*b*c^3*d^3*e - 4*a*c^2*d^2*e^2 + 2*b*c*d*e^3 + 2*a*e^4 - 2*(b*c^3*d^2*e^2 - b*c*e^4)*x -
(b*c^4*d^4 + 2*b*c^3*d^3*e + b*c^2*d^2*e^2 + (b*c^4*d^2*e^2 + 2*b*c^3*d*e^3 + b*c^2*e^4)*x^2 + 2*(b*c^4*d^3*e
+ 2*b*c^3*d^2*e^2 + b*c^2*d*e^3)*x)*log(c*x + 1) + (b*c^4*d^4 - 2*b*c^3*d^3*e + b*c^2*d^2*e^2 + (b*c^4*d^2*e^2
- 2*b*c^3*d*e^3 + b*c^2*e^4)*x^2 + 2*(b*c^4*d^3*e - 2*b*c^3*d^2*e^2 + b*c^2*d*e^3)*x)*log(c*x - 1) + 4*(b*c^3
*d*e^3*x^2 + 2*b*c^3*d^2*e^2*x + b*c^3*d^3*e)*log(e*x + d) + (b*c^4*d^4 - 2*b*c^2*d^2*e^2 + b*e^4)*log(-(c*x +
1)/(c*x - 1)))/(c^4*d^6*e - 2*c^2*d^4*e^3 + d^2*e^5 + (c^4*d^4*e^3 - 2*c^2*d^2*e^5 + e^7)*x^2 + 2*(c^4*d^5*e^
2 - 2*c^2*d^3*e^4 + d*e^6)*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))/(e*x+d)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.7422, size = 791, normalized size = 6.08 \begin{align*} \frac{b c^{4} d^{2} x^{2} e^{2} \log \left (c x + 1\right ) + 2 \, b c^{4} d^{3} x e \log \left (c x + 1\right ) - b c^{4} d^{2} x^{2} e^{2} \log \left (c x - 1\right ) - 2 \, b c^{4} d^{3} x e \log \left (c x - 1\right ) + b c^{4} d^{4} \log \left (c x + 1\right ) - b c^{4} d^{4} \log \left (c x - 1\right ) - b c^{4} d^{4} \log \left (-\frac{c x + 1}{c x - 1}\right ) - 2 \, a c^{4} d^{4} + 2 \, b c^{3} d x^{2} e^{3} \log \left (c x + 1\right ) + 4 \, b c^{3} d^{2} x e^{2} \log \left (c x + 1\right ) + 2 \, b c^{3} d^{3} e \log \left (c x + 1\right ) + 2 \, b c^{3} d x^{2} e^{3} \log \left (c x - 1\right ) + 4 \, b c^{3} d^{2} x e^{2} \log \left (c x - 1\right ) + 2 \, b c^{3} d^{3} e \log \left (c x - 1\right ) - 4 \, b c^{3} d x^{2} e^{3} \log \left (x e + d\right ) - 8 \, b c^{3} d^{2} x e^{2} \log \left (x e + d\right ) - 4 \, b c^{3} d^{3} e \log \left (x e + d\right ) + 2 \, b c^{3} d^{2} x e^{2} + 2 \, b c^{3} d^{3} e + b c^{2} x^{2} e^{4} \log \left (c x + 1\right ) + 2 \, b c^{2} d x e^{3} \log \left (c x + 1\right ) + b c^{2} d^{2} e^{2} \log \left (c x + 1\right ) - b c^{2} x^{2} e^{4} \log \left (c x - 1\right ) - 2 \, b c^{2} d x e^{3} \log \left (c x - 1\right ) - b c^{2} d^{2} e^{2} \log \left (c x - 1\right ) + 2 \, b c^{2} d^{2} e^{2} \log \left (-\frac{c x + 1}{c x - 1}\right ) + 4 \, a c^{2} d^{2} e^{2} - 2 \, b c x e^{4} - 2 \, b c d e^{3} - b e^{4} \log \left (-\frac{c x + 1}{c x - 1}\right ) - 2 \, a e^{4}}{4 \,{\left (c^{4} d^{4} x^{2} e^{3} + 2 \, c^{4} d^{5} x e^{2} + c^{4} d^{6} e - 2 \, c^{2} d^{2} x^{2} e^{5} - 4 \, c^{2} d^{3} x e^{4} - 2 \, c^{2} d^{4} e^{3} + x^{2} e^{7} + 2 \, d x e^{6} + d^{2} e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*c^4*d^2*x^2*e^2*log(c*x + 1) + 2*b*c^4*d^3*x*e*log(c*x + 1) - b*c^4*d^2*x^2*e^2*log(c*x - 1) - 2*b*c^4*
d^3*x*e*log(c*x - 1) + b*c^4*d^4*log(c*x + 1) - b*c^4*d^4*log(c*x - 1) - b*c^4*d^4*log(-(c*x + 1)/(c*x - 1)) -
2*a*c^4*d^4 + 2*b*c^3*d*x^2*e^3*log(c*x + 1) + 4*b*c^3*d^2*x*e^2*log(c*x + 1) + 2*b*c^3*d^3*e*log(c*x + 1) +
2*b*c^3*d*x^2*e^3*log(c*x - 1) + 4*b*c^3*d^2*x*e^2*log(c*x - 1) + 2*b*c^3*d^3*e*log(c*x - 1) - 4*b*c^3*d*x^2*e
^3*log(x*e + d) - 8*b*c^3*d^2*x*e^2*log(x*e + d) - 4*b*c^3*d^3*e*log(x*e + d) + 2*b*c^3*d^2*x*e^2 + 2*b*c^3*d^
3*e + b*c^2*x^2*e^4*log(c*x + 1) + 2*b*c^2*d*x*e^3*log(c*x + 1) + b*c^2*d^2*e^2*log(c*x + 1) - b*c^2*x^2*e^4*l
og(c*x - 1) - 2*b*c^2*d*x*e^3*log(c*x - 1) - b*c^2*d^2*e^2*log(c*x - 1) + 2*b*c^2*d^2*e^2*log(-(c*x + 1)/(c*x
- 1)) + 4*a*c^2*d^2*e^2 - 2*b*c*x*e^4 - 2*b*c*d*e^3 - b*e^4*log(-(c*x + 1)/(c*x - 1)) - 2*a*e^4)/(c^4*d^4*x^2*
e^3 + 2*c^4*d^5*x*e^2 + c^4*d^6*e - 2*c^2*d^2*x^2*e^5 - 4*c^2*d^3*x*e^4 - 2*c^2*d^4*e^3 + x^2*e^7 + 2*d*x*e^6
+ d^2*e^5)