3.6.1 \(\int \frac {e^{-2 \tanh ^{-1}(a x)}}{(c-\frac {c}{a x})^4} \, dx\) [501]

Optimal. Leaf size=76 \[ -\frac {x}{c^4}+\frac {1}{4 a c^4 (1-a x)^2}-\frac {7}{4 a c^4 (1-a x)}-\frac {17 \log (1-a x)}{8 a c^4}+\frac {\log (1+a x)}{8 a c^4} \]

[Out]

-x/c^4+1/4/a/c^4/(-a*x+1)^2-7/4/a/c^4/(-a*x+1)-17/8*ln(-a*x+1)/a/c^4+1/8*ln(a*x+1)/a/c^4

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 76, 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 {7}{4 a c^4 (1-a x)}+\frac {1}{4 a c^4 (1-a x)^2}-\frac {17 \log (1-a x)}{8 a c^4}+\frac {\log (a x+1)}{8 a c^4}-\frac {x}{c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a*x])*(c - c/(a*x))^4),x]

[Out]

-(x/c^4) + 1/(4*a*c^4*(1 - a*x)^2) - 7/(4*a*c^4*(1 - a*x)) - (17*Log[1 - a*x])/(8*a*c^4) + Log[1 + a*x]/(8*a*c
^4)

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 )^4} \, dx &=\frac {a^4 \int \frac {e^{-2 \tanh ^{-1}(a x)} x^4}{(1-a x)^4} \, dx}{c^4}\\ &=\frac {a^4 \int \frac {x^4}{(1-a x)^3 (1+a x)} \, dx}{c^4}\\ &=\frac {a^4 \int \left (-\frac {1}{a^4}-\frac {1}{2 a^4 (-1+a x)^3}-\frac {7}{4 a^4 (-1+a x)^2}-\frac {17}{8 a^4 (-1+a x)}+\frac {1}{8 a^4 (1+a x)}\right ) \, dx}{c^4}\\ &=-\frac {x}{c^4}+\frac {1}{4 a c^4 (1-a x)^2}-\frac {7}{4 a c^4 (1-a x)}-\frac {17 \log (1-a x)}{8 a c^4}+\frac {\log (1+a x)}{8 a c^4}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.05, size = 69, normalized size = 0.91 \begin {gather*} \frac {-12+6 a x+16 a^2 x^2-8 a^3 x^3-17 (-1+a x)^2 \log (1-a x)+(-1+a x)^2 \log (1+a x)}{8 a c^4 (-1+a x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*(c - c/(a*x))^4),x]

[Out]

(-12 + 6*a*x + 16*a^2*x^2 - 8*a^3*x^3 - 17*(-1 + a*x)^2*Log[1 - a*x] + (-1 + a*x)^2*Log[1 + a*x])/(8*a*c^4*(-1
+ a*x)^2)

________________________________________________________________________________________

Maple [A]
time = 0.43, size = 61, normalized size = 0.80

method	result	size
default	\(\frac {a^{4} \left (-\frac {x}{a^{4}}+\frac {\ln \left (a x +1\right )}{8 a^{5}}-\frac {17 \ln \left (a x -1\right )}{8 a^{5}}+\frac {1}{4 \left (a x -1\right )^{2} a^{5}}+\frac {7}{4 \left (a x -1\right ) a^{5}}\right )}{c^{4}}\)	\(61\)
risch	\(-\frac {x}{c^{4}}+\frac {\frac {7 c^{4} x}{4}-\frac {3 c^{4}}{2 a}}{c^{8} \left (a x -1\right )^{2}}+\frac {\ln \left (-a x -1\right )}{8 a \,c^{4}}-\frac {17 \ln \left (a x -1\right )}{8 a \,c^{4}}\)	\(63\)
norman	\(\frac {-\frac {7 a \,x^{2}}{2 c}-\frac {a^{4} x^{5}}{c}+\frac {9 x}{4 c}-\frac {5 a^{2} x^{3}}{4 c}+\frac {7 a^{3} x^{4}}{2 c}}{\left (a x +1\right ) c^{3} \left (a x -1\right )^{3}}-\frac {17 \ln \left (a x -1\right )}{8 a \,c^{4}}+\frac {\ln \left (a x +1\right )}{8 c^{4} a}\)	\(97\)

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)^4,x,method=_RETURNVERBOSE)

[Out]

a^4/c^4*(-x/a^4+1/8/a^5*ln(a*x+1)-17/8/a^5*ln(a*x-1)+1/4/(a*x-1)^2/a^5+7/4/(a*x-1)/a^5)

________________________________________________________________________________________

Maxima [A]
time = 0.26, size = 70, normalized size = 0.92 \begin {gather*} \frac {7 \, a x - 6}{4 \, {\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} - \frac {x}{c^{4}} + \frac {\log \left (a x + 1\right )}{8 \, a c^{4}} - \frac {17 \, \log \left (a x - 1\right )}{8 \, a c^{4}} \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)^4,x, algorithm="maxima")

[Out]

1/4*(7*a*x - 6)/(a^3*c^4*x^2 - 2*a^2*c^4*x + a*c^4) - x/c^4 + 1/8*log(a*x + 1)/(a*c^4) - 17/8*log(a*x - 1)/(a*
c^4)

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 93, normalized size = 1.22 \begin {gather*} -\frac {8 \, a^{3} x^{3} - 16 \, a^{2} x^{2} - 6 \, a x - {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) + 17 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 12}{8 \, {\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\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)^4,x, algorithm="fricas")

[Out]

-1/8*(8*a^3*x^3 - 16*a^2*x^2 - 6*a*x - (a^2*x^2 - 2*a*x + 1)*log(a*x + 1) + 17*(a^2*x^2 - 2*a*x + 1)*log(a*x -
1) + 12)/(a^3*c^4*x^2 - 2*a^2*c^4*x + a*c^4)

________________________________________________________________________________________

Sympy [A]
time = 0.30, size = 75, normalized size = 0.99 \begin {gather*} - a^{4} \left (\frac {- 7 a x + 6}{4 a^{7} c^{4} x^{2} - 8 a^{6} c^{4} x + 4 a^{5} c^{4}} + \frac {x}{a^{4} c^{4}} + \frac {\frac {17 \log {\left (x - \frac {1}{a} \right )}}{8} - \frac {\log {\left (x + \frac {1}{a} \right )}}{8}}{a^{5} c^{4}}\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)**4,x)

[Out]

-a**4*((-7*a*x + 6)/(4*a**7*c**4*x**2 - 8*a**6*c**4*x + 4*a**5*c**4) + x/(a**4*c**4) + (17*log(x - 1/a)/8 - lo
g(x + 1/a)/8)/(a**5*c**4))

________________________________________________________________________________________

Giac [A]
time = 0.40, size = 95, normalized size = 1.25 \begin {gather*} \frac {2 \, \log \left (\frac {{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2} {\left | a \right |}}\right )}{a c^{4}} - \frac {17 \, \log \left ({\left | -\frac {2}{a x + 1} + 1 \right |}\right )}{8 \, a c^{4}} + \frac {{\left (a x + 1\right )} {\left (\frac {77}{a x + 1} - \frac {88}{{\left (a x + 1\right )}^{2}} - 16\right )}}{16 \, a c^{4} {\left (\frac {2}{a x + 1} - 1\right )}^{2}} \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)^4,x, algorithm="giac")

[Out]

2*log(abs(a*x + 1)/((a*x + 1)^2*abs(a)))/(a*c^4) - 17/8*log(abs(-2/(a*x + 1) + 1))/(a*c^4) + 1/16*(a*x + 1)*(7
7/(a*x + 1) - 88/(a*x + 1)^2 - 16)/(a*c^4*(2/(a*x + 1) - 1)^2)

________________________________________________________________________________________

Mupad [B]
time = 0.10, size = 68, normalized size = 0.89 \begin {gather*} \frac {\frac {7\,x}{4}-\frac {3}{2\,a}}{a^2\,c^4\,x^2-2\,a\,c^4\,x+c^4}-\frac {x}{c^4}-\frac {17\,\ln \left (a\,x-1\right )}{8\,a\,c^4}+\frac {\ln \left (a\,x+1\right )}{8\,a\,c^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a^2*x^2 - 1)/((c - c/(a*x))^4*(a*x + 1)^2),x)

[Out]

((7*x)/4 - 3/(2*a))/(c^4 + a^2*c^4*x^2 - 2*a*c^4*x) - x/c^4 - (17*log(a*x - 1))/(8*a*c^4) + log(a*x + 1)/(8*a*
c^4)

________________________________________________________________________________________