3.660 \(\int \frac {e^{4 \tanh ^{-1}(a x)}}{(c-\frac {c}{a^2 x^2})^3} \, dx\)

Optimal. Leaf size=111 \[ \frac {111}{16 a c^3 (1-a x)}-\frac {49}{16 a c^3 (1-a x)^2}+\frac {11}{12 a c^3 (1-a x)^3}-\frac {1}{8 a c^3 (1-a x)^4}+\frac {129 \log (1-a x)}{32 a c^3}-\frac {\log (a x+1)}{32 a c^3}+\frac {x}{c^3} \]

[Out]

x/c^3-1/8/a/c^3/(-a*x+1)^4+11/12/a/c^3/(-a*x+1)^3-49/16/a/c^3/(-a*x+1)^2+111/16/a/c^3/(-a*x+1)+129/32*ln(-a*x+
1)/a/c^3-1/32*ln(a*x+1)/a/c^3

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Rubi [A]  time = 0.17, antiderivative size = 111, 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 = {6157, 6150, 88} \[ \frac {111}{16 a c^3 (1-a x)}-\frac {49}{16 a c^3 (1-a x)^2}+\frac {11}{12 a c^3 (1-a x)^3}-\frac {1}{8 a c^3 (1-a x)^4}+\frac {129 \log (1-a x)}{32 a c^3}-\frac {\log (a x+1)}{32 a c^3}+\frac {x}{c^3} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcTanh[a*x])/(c - c/(a^2*x^2))^3,x]

[Out]

x/c^3 - 1/(8*a*c^3*(1 - a*x)^4) + 11/(12*a*c^3*(1 - a*x)^3) - 49/(16*a*c^3*(1 - a*x)^2) + 111/(16*a*c^3*(1 - a
*x)) + (129*Log[1 - a*x])/(32*a*c^3) - Log[1 + a*x]/(32*a*c^3)

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]))

Rule 6150

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

Rule 6157

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

Rubi steps

\begin {align*} \int \frac {e^{4 \tanh ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx &=-\frac {a^6 \int \frac {e^{4 \tanh ^{-1}(a x)} x^6}{\left (1-a^2 x^2\right )^3} \, dx}{c^3}\\ &=-\frac {a^6 \int \frac {x^6}{(1-a x)^5 (1+a x)} \, dx}{c^3}\\ &=-\frac {a^6 \int \left (-\frac {1}{a^6}-\frac {1}{2 a^6 (-1+a x)^5}-\frac {11}{4 a^6 (-1+a x)^4}-\frac {49}{8 a^6 (-1+a x)^3}-\frac {111}{16 a^6 (-1+a x)^2}-\frac {129}{32 a^6 (-1+a x)}+\frac {1}{32 a^6 (1+a x)}\right ) \, dx}{c^3}\\ &=\frac {x}{c^3}-\frac {1}{8 a c^3 (1-a x)^4}+\frac {11}{12 a c^3 (1-a x)^3}-\frac {49}{16 a c^3 (1-a x)^2}+\frac {111}{16 a c^3 (1-a x)}+\frac {129 \log (1-a x)}{32 a c^3}-\frac {\log (1+a x)}{32 a c^3}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 89, normalized size = 0.80 \[ \frac {2 \left (48 a^5 x^5-192 a^4 x^4-45 a^3 x^3+660 a^2 x^2-701 a x+224\right )+387 (a x-1)^4 \log (1-a x)-3 (a x-1)^4 \log (a x+1)}{96 a c^3 (a x-1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcTanh[a*x])/(c - c/(a^2*x^2))^3,x]

[Out]

(2*(224 - 701*a*x + 660*a^2*x^2 - 45*a^3*x^3 - 192*a^4*x^4 + 48*a^5*x^5) + 387*(-1 + a*x)^4*Log[1 - a*x] - 3*(
-1 + a*x)^4*Log[1 + a*x])/(96*a*c^3*(-1 + a*x)^4)

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fricas [A]  time = 0.50, size = 163, normalized size = 1.47 \[ \frac {96 \, a^{5} x^{5} - 384 \, a^{4} x^{4} - 90 \, a^{3} x^{3} + 1320 \, a^{2} x^{2} - 1402 \, a x - 3 \, {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \log \left (a x + 1\right ) + 387 \, {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \log \left (a x - 1\right ) + 448}{96 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2/(c-c/a^2/x^2)^3,x, algorithm="fricas")

[Out]

1/96*(96*a^5*x^5 - 384*a^4*x^4 - 90*a^3*x^3 + 1320*a^2*x^2 - 1402*a*x - 3*(a^4*x^4 - 4*a^3*x^3 + 6*a^2*x^2 - 4
*a*x + 1)*log(a*x + 1) + 387*(a^4*x^4 - 4*a^3*x^3 + 6*a^2*x^2 - 4*a*x + 1)*log(a*x - 1) + 448)/(a^5*c^3*x^4 -
4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*a^2*c^3*x + a*c^3)

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giac [A]  time = 0.16, size = 73, normalized size = 0.66 \[ \frac {x}{c^{3}} - \frac {\log \left ({\left | a x + 1 \right |}\right )}{32 \, a c^{3}} + \frac {129 \, \log \left ({\left | a x - 1 \right |}\right )}{32 \, a c^{3}} - \frac {333 \, a^{3} x^{3} - 852 \, a^{2} x^{2} + 749 \, a x - 224}{48 \, {\left (a x - 1\right )}^{4} a c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2/(c-c/a^2/x^2)^3,x, algorithm="giac")

[Out]

x/c^3 - 1/32*log(abs(a*x + 1))/(a*c^3) + 129/32*log(abs(a*x - 1))/(a*c^3) - 1/48*(333*a^3*x^3 - 852*a^2*x^2 +
749*a*x - 224)/((a*x - 1)^4*a*c^3)

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maple [A]  time = 0.03, size = 95, normalized size = 0.86 \[ \frac {x}{c^{3}}-\frac {1}{8 a \,c^{3} \left (a x -1\right )^{4}}-\frac {11}{12 a \,c^{3} \left (a x -1\right )^{3}}-\frac {49}{16 a \,c^{3} \left (a x -1\right )^{2}}-\frac {111}{16 a \,c^{3} \left (a x -1\right )}+\frac {129 \ln \left (a x -1\right )}{32 a \,c^{3}}-\frac {\ln \left (a x +1\right )}{32 a \,c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^4/(-a^2*x^2+1)^2/(c-c/a^2/x^2)^3,x)

[Out]

x/c^3-1/8/a/c^3/(a*x-1)^4-11/12/a/c^3/(a*x-1)^3-49/16/a/c^3/(a*x-1)^2-111/16/a/c^3/(a*x-1)+129/32/a/c^3*ln(a*x
-1)-1/32*ln(a*x+1)/a/c^3

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maxima [A]  time = 0.31, size = 107, normalized size = 0.96 \[ -\frac {333 \, a^{3} x^{3} - 852 \, a^{2} x^{2} + 749 \, a x - 224}{48 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} + \frac {x}{c^{3}} - \frac {\log \left (a x + 1\right )}{32 \, a c^{3}} + \frac {129 \, \log \left (a x - 1\right )}{32 \, a c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2/(c-c/a^2/x^2)^3,x, algorithm="maxima")

[Out]

-1/48*(333*a^3*x^3 - 852*a^2*x^2 + 749*a*x - 224)/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*a^2*c^3*x +
a*c^3) + x/c^3 - 1/32*log(a*x + 1)/(a*c^3) + 129/32*log(a*x - 1)/(a*c^3)

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mupad [B]  time = 0.91, size = 104, normalized size = 0.94 \[ \frac {x}{c^3}-\frac {\frac {749\,x}{48}-\frac {71\,a\,x^2}{4}-\frac {14}{3\,a}+\frac {111\,a^2\,x^3}{16}}{a^4\,c^3\,x^4-4\,a^3\,c^3\,x^3+6\,a^2\,c^3\,x^2-4\,a\,c^3\,x+c^3}+\frac {129\,\ln \left (a\,x-1\right )}{32\,a\,c^3}-\frac {\ln \left (a\,x+1\right )}{32\,a\,c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + 1)^4/((c - c/(a^2*x^2))^3*(a^2*x^2 - 1)^2),x)

[Out]

x/c^3 - ((749*x)/48 - (71*a*x^2)/4 - 14/(3*a) + (111*a^2*x^3)/16)/(c^3 + 6*a^2*c^3*x^2 - 4*a^3*c^3*x^3 + a^4*c
^3*x^4 - 4*a*c^3*x) + (129*log(a*x - 1))/(32*a*c^3) - log(a*x + 1)/(32*a*c^3)

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sympy [A]  time = 0.63, size = 114, normalized size = 1.03 \[ a^{6} \left (\frac {- 333 a^{3} x^{3} + 852 a^{2} x^{2} - 749 a x + 224}{48 a^{11} c^{3} x^{4} - 192 a^{10} c^{3} x^{3} + 288 a^{9} c^{3} x^{2} - 192 a^{8} c^{3} x + 48 a^{7} c^{3}} + \frac {x}{a^{6} c^{3}} + \frac {\frac {129 \log {\left (x - \frac {1}{a} \right )}}{32} - \frac {\log {\left (x + \frac {1}{a} \right )}}{32}}{a^{7} c^{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**4/(-a**2*x**2+1)**2/(c-c/a**2/x**2)**3,x)

[Out]

a**6*((-333*a**3*x**3 + 852*a**2*x**2 - 749*a*x + 224)/(48*a**11*c**3*x**4 - 192*a**10*c**3*x**3 + 288*a**9*c*
*3*x**2 - 192*a**8*c**3*x + 48*a**7*c**3) + x/(a**6*c**3) + (129*log(x - 1/a)/32 - log(x + 1/a)/32)/(a**7*c**3
))

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