3.5.100 \(\int \frac {e^{-2 \tanh ^{-1}(a x)}}{(c-\frac {c}{a x})^3} \, dx\) [500]

Optimal. Leaf size=58 \[ -\frac {x}{c^3}-\frac {1}{2 a c^3 (1-a x)}-\frac {5 \log (1-a x)}{4 a c^3}+\frac {\log (1+a x)}{4 a c^3} \]

[Out]

-x/c^3-1/2/a/c^3/(-a*x+1)-5/4*ln(-a*x+1)/a/c^3+1/4*ln(a*x+1)/a/c^3

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Rubi [A]
time = 0.08, antiderivative size = 58, normalized size of antiderivative = 1.00, 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 = {6266, 6264, 90} \begin {gather*} -\frac {1}{2 a c^3 (1-a x)}-\frac {5 \log (1-a x)}{4 a c^3}+\frac {\log (a x+1)}{4 a c^3}-\frac {x}{c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a*x])*(c - c/(a*x))^3),x]

[Out]

-(x/c^3) - 1/(2*a*c^3*(1 - a*x)) - (5*Log[1 - a*x])/(4*a*c^3) + Log[1 + a*x]/(4*a*c^3)

Rule 90

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]))

Rule 6264

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 6266

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]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 57, normalized size = 0.98 \begin {gather*} -\frac {x}{c^3}+\frac {1}{2 a c^3 (-1+a x)}-\frac {5 \log (1-a x)}{4 a c^3}+\frac {\log (1+a x)}{4 a c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*(c - c/(a*x))^3),x]

[Out]

-(x/c^3) + 1/(2*a*c^3*(-1 + a*x)) - (5*Log[1 - a*x])/(4*a*c^3) + Log[1 + a*x]/(4*a*c^3)

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Maple [A]
time = 0.38, size = 49, normalized size = 0.84

method	result	size
default	\(\frac {a^{3} \left (-\frac {x}{a^{3}}+\frac {\ln \left (a x +1\right )}{4 a^{4}}-\frac {5 \ln \left (a x -1\right )}{4 a^{4}}+\frac {1}{2 a^{4} \left (a x -1\right )}\right )}{c^{3}}\)	\(49\)
risch	\(-\frac {x}{c^{3}}+\frac {1}{2 a \,c^{3} \left (a x -1\right )}+\frac {\ln \left (-a x -1\right )}{4 a \,c^{3}}-\frac {5 \ln \left (a x -1\right )}{4 a \,c^{3}}\)	\(52\)
norman	\(\frac {-\frac {a^{3} x^{4}}{c}-\frac {3 x}{2 c}+\frac {a \,x^{2}}{c}+\frac {3 a^{2} x^{3}}{2 c}}{\left (a x +1\right ) c^{2} \left (a x -1\right )^{2}}-\frac {5 \ln \left (a x -1\right )}{4 a \,c^{3}}+\frac {\ln \left (a x +1\right )}{4 a \,c^{3}}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3/c^3*(-x/a^3+1/4/a^4*ln(a*x+1)-5/4/a^4*ln(a*x-1)+1/2/a^4/(a*x-1))

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Maxima [A]
time = 0.26, size = 54, normalized size = 0.93 \begin {gather*} \frac {1}{2 \, {\left (a^{2} c^{3} x - a c^{3}\right )}} - \frac {x}{c^{3}} + \frac {\log \left (a x + 1\right )}{4 \, a c^{3}} - \frac {5 \, \log \left (a x - 1\right )}{4 \, a c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(a^2*c^3*x - a*c^3) - x/c^3 + 1/4*log(a*x + 1)/(a*c^3) - 5/4*log(a*x - 1)/(a*c^3)

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Fricas [A]
time = 0.35, size = 59, normalized size = 1.02 \begin {gather*} -\frac {4 \, a^{2} x^{2} - 4 \, a x - {\left (a x - 1\right )} \log \left (a x + 1\right ) + 5 \, {\left (a x - 1\right )} \log \left (a x - 1\right ) - 2}{4 \, {\left (a^{2} c^{3} x - a c^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*a^2*x^2 - 4*a*x - (a*x - 1)*log(a*x + 1) + 5*(a*x - 1)*log(a*x - 1) - 2)/(a^2*c^3*x - a*c^3)

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Sympy [A]
time = 0.28, size = 58, normalized size = 1.00 \begin {gather*} - a^{3} \left (- \frac {1}{2 a^{5} c^{3} x - 2 a^{4} c^{3}} + \frac {x}{a^{3} c^{3}} + \frac {\frac {5 \log {\left (x - \frac {1}{a} \right )}}{4} - \frac {\log {\left (x + \frac {1}{a} \right )}}{4}}{a^{4} c^{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**3*(-1/(2*a**5*c**3*x - 2*a**4*c**3) + x/(a**3*c**3) + (5*log(x - 1/a)/4 - log(x + 1/a)/4)/(a**4*c**3))

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Giac [A]
time = 0.41, size = 85, normalized size = 1.47 \begin {gather*} -\frac {{\left (a x + 1\right )} {\left (\frac {9}{a x + 1} - 4\right )}}{4 \, a c^{3} {\left (\frac {2}{a x + 1} - 1\right )}} + \frac {\log \left (\frac {{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2} {\left | a \right |}}\right )}{a c^{3}} - \frac {5 \, \log \left ({\left | -\frac {2}{a x + 1} + 1 \right |}\right )}{4 \, a c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a*x + 1)*(9/(a*x + 1) - 4)/(a*c^3*(2/(a*x + 1) - 1)) + log(abs(a*x + 1)/((a*x + 1)^2*abs(a)))/(a*c^3) -
5/4*log(abs(-2/(a*x + 1) + 1))/(a*c^3)

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Mupad [B]
time = 0.10, size = 53, normalized size = 0.91 \begin {gather*} \frac {\ln \left (a\,x+1\right )}{4\,a\,c^3}-\frac {1}{2\,a\,\left (c^3-a\,c^3\,x\right )}-\frac {5\,\ln \left (a\,x-1\right )}{4\,a\,c^3}-\frac {x}{c^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a*x + 1)/(4*a*c^3) - 1/(2*a*(c^3 - a*c^3*x)) - (5*log(a*x - 1))/(4*a*c^3) - x/c^3

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