3.495 \(\int e^{-2 \tanh ^{-1}(a x)} (c-\frac{c}{a x})^3 \, dx\)

Optimal. Leaf size=55 \[ \frac{c^3}{2 a^3 x^2}-\frac{5 c^3}{a^2 x}-\frac{11 c^3 \log (x)}{a}+\frac{16 c^3 \log (a x+1)}{a}+c^3 (-x) \]

[Out]

c^3/(2*a^3*x^2) - (5*c^3)/(a^2*x) - c^3*x - (11*c^3*Log[x])/a + (16*c^3*Log[1 + a*x])/a

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Rubi [A]  time = 0.116294, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6131, 6129, 88} \[ \frac{c^3}{2 a^3 x^2}-\frac{5 c^3}{a^2 x}-\frac{11 c^3 \log (x)}{a}+\frac{16 c^3 \log (a x+1)}{a}+c^3 (-x) \]

Antiderivative was successfully verified.

[In]

Int[(c - c/(a*x))^3/E^(2*ArcTanh[a*x]),x]

[Out]

c^3/(2*a^3*x^2) - (5*c^3)/(a^2*x) - c^3*x - (11*c^3*Log[x])/a + (16*c^3*Log[1 + a*x])/a

Rule 6131

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 + (c*x)/d)
^p*E^(n*ArcTanh[a*x]))/x^p, x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]

Rule 6129

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[c^p, Int[(u*(1 + (d*x)/c)
^p*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && (IntegerQ
[p] || GtQ[c, 0])

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int e^{-2 \tanh ^{-1}(a x)} \left (c-\frac{c}{a x}\right )^3 \, dx &=-\frac{c^3 \int \frac{e^{-2 \tanh ^{-1}(a x)} (1-a x)^3}{x^3} \, dx}{a^3}\\ &=-\frac{c^3 \int \frac{(1-a x)^4}{x^3 (1+a x)} \, dx}{a^3}\\ &=-\frac{c^3 \int \left (a^3+\frac{1}{x^3}-\frac{5 a}{x^2}+\frac{11 a^2}{x}-\frac{16 a^3}{1+a x}\right ) \, dx}{a^3}\\ &=\frac{c^3}{2 a^3 x^2}-\frac{5 c^3}{a^2 x}-c^3 x-\frac{11 c^3 \log (x)}{a}+\frac{16 c^3 \log (1+a x)}{a}\\ \end{align*}

Mathematica [A]  time = 0.124412, size = 57, normalized size = 1.04 \[ \frac{c^3}{2 a^3 x^2}-\frac{5 c^3}{a^2 x}-\frac{11 c^3 \log (a x)}{a}+\frac{16 c^3 \log (a x+1)}{a}+c^3 (-x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c - c/(a*x))^3/E^(2*ArcTanh[a*x]),x]

[Out]

c^3/(2*a^3*x^2) - (5*c^3)/(a^2*x) - c^3*x - (11*c^3*Log[a*x])/a + (16*c^3*Log[1 + a*x])/a

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Maple [A]  time = 0.037, size = 54, normalized size = 1. \begin{align*}{\frac{{c}^{3}}{2\,{x}^{2}{a}^{3}}}-5\,{\frac{{c}^{3}}{{a}^{2}x}}-{c}^{3}x-11\,{\frac{{c}^{3}\ln \left ( x \right ) }{a}}+16\,{\frac{{c}^{3}\ln \left ( ax+1 \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c/a/x)^3/(a*x+1)^2*(-a^2*x^2+1),x)

[Out]

1/2*c^3/x^2/a^3-5*c^3/a^2/x-c^3*x-11*c^3*ln(x)/a+16*c^3*ln(a*x+1)/a

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Maxima [A]  time = 0.966151, size = 70, normalized size = 1.27 \begin{align*} -c^{3} x + \frac{16 \, c^{3} \log \left (a x + 1\right )}{a} - \frac{11 \, c^{3} \log \left (x\right )}{a} - \frac{10 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^3/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="maxima")

[Out]

-c^3*x + 16*c^3*log(a*x + 1)/a - 11*c^3*log(x)/a - 1/2*(10*a*c^3*x - c^3)/(a^3*x^2)

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Fricas [A]  time = 2.22177, size = 142, normalized size = 2.58 \begin{align*} -\frac{2 \, a^{3} c^{3} x^{3} - 32 \, a^{2} c^{3} x^{2} \log \left (a x + 1\right ) + 22 \, a^{2} c^{3} x^{2} \log \left (x\right ) + 10 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^3/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="fricas")

[Out]

-1/2*(2*a^3*c^3*x^3 - 32*a^2*c^3*x^2*log(a*x + 1) + 22*a^2*c^3*x^2*log(x) + 10*a*c^3*x - c^3)/(a^3*x^2)

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Sympy [A]  time = 0.629576, size = 44, normalized size = 0.8 \begin{align*} - c^{3} x - \frac{c^{3} \left (11 \log{\left (x \right )} - 16 \log{\left (x + \frac{1}{a} \right )}\right )}{a} - \frac{10 a c^{3} x - c^{3}}{2 a^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)**3/(a*x+1)**2*(-a**2*x**2+1),x)

[Out]

-c**3*x - c**3*(11*log(x) - 16*log(x + 1/a))/a - (10*a*c**3*x - c**3)/(2*a**3*x**2)

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Giac [A]  time = 1.16233, size = 135, normalized size = 2.45 \begin{align*} -\frac{5 \, c^{3} \log \left (\frac{{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2}{\left | a \right |}}\right )}{a} - \frac{11 \, c^{3} \log \left ({\left | -\frac{1}{a x + 1} + 1 \right |}\right )}{a} - \frac{{\left (2 \, c^{3} + \frac{7 \, c^{3}}{a x + 1} - \frac{10 \, c^{3}}{{\left (a x + 1\right )}^{2}}\right )}{\left (a x + 1\right )}}{2 \, a{\left (\frac{1}{a x + 1} - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^3/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="giac")

[Out]

-5*c^3*log(abs(a*x + 1)/((a*x + 1)^2*abs(a)))/a - 11*c^3*log(abs(-1/(a*x + 1) + 1))/a - 1/2*(2*c^3 + 7*c^3/(a*
x + 1) - 10*c^3/(a*x + 1)^2)*(a*x + 1)/(a*(1/(a*x + 1) - 1)^2)