3.721 \(\int \frac {e^{-\tanh ^{-1}(a x)}}{(c-\frac {c}{a^2 x^2})^{5/2}} \, dx\)

Optimal. Leaf size=265 \[ \frac {\left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (1-a x) \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {\left (1-a^2 x^2\right )^{5/2}}{a^6 x^5 (a x+1) \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {\left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (a x+1)^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {7 \left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {23 \left (1-a^2 x^2\right )^{5/2} \log (a x+1)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {\left (1-a^2 x^2\right )^{5/2}}{a^5 x^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \]

[Out]

(-a^2*x^2+1)^(5/2)/a^5/(c-c/a^2/x^2)^(5/2)/x^4+1/8*(-a^2*x^2+1)^(5/2)/a^6/(c-c/a^2/x^2)^(5/2)/x^5/(-a*x+1)+1/8
*(-a^2*x^2+1)^(5/2)/a^6/(c-c/a^2/x^2)^(5/2)/x^5/(a*x+1)^2-(-a^2*x^2+1)^(5/2)/a^6/(c-c/a^2/x^2)^(5/2)/x^5/(a*x+
1)+7/16*(-a^2*x^2+1)^(5/2)*ln(-a*x+1)/a^6/(c-c/a^2/x^2)^(5/2)/x^5-23/16*(-a^2*x^2+1)^(5/2)*ln(a*x+1)/a^6/(c-c/
a^2/x^2)^(5/2)/x^5

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Rubi [A]  time = 0.21, antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6160, 6150, 88} \[ \frac {\left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (1-a x) \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {\left (1-a^2 x^2\right )^{5/2}}{a^6 x^5 (a x+1) \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {\left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (a x+1)^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {\left (1-a^2 x^2\right )^{5/2}}{a^5 x^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {7 \left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {23 \left (1-a^2 x^2\right )^{5/2} \log (a x+1)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a*x]*(c - c/(a^2*x^2))^(5/2)),x]

[Out]

(1 - a^2*x^2)^(5/2)/(a^5*(c - c/(a^2*x^2))^(5/2)*x^4) + (1 - a^2*x^2)^(5/2)/(8*a^6*(c - c/(a^2*x^2))^(5/2)*x^5
*(1 - a*x)) + (1 - a^2*x^2)^(5/2)/(8*a^6*(c - c/(a^2*x^2))^(5/2)*x^5*(1 + a*x)^2) - (1 - a^2*x^2)^(5/2)/(a^6*(
c - c/(a^2*x^2))^(5/2)*x^5*(1 + a*x)) + (7*(1 - a^2*x^2)^(5/2)*Log[1 - a*x])/(16*a^6*(c - c/(a^2*x^2))^(5/2)*x
^5) - (23*(1 - a^2*x^2)^(5/2)*Log[1 + a*x])/(16*a^6*(c - c/(a^2*x^2))^(5/2)*x^5)

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 6160

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

Rubi steps

\begin {align*} \int \frac {e^{-\tanh ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx &=\frac {\left (1-a^2 x^2\right )^{5/2} \int \frac {e^{-\tanh ^{-1}(a x)} x^5}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {\left (1-a^2 x^2\right )^{5/2} \int \frac {x^5}{(1-a x)^2 (1+a x)^3} \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {\left (1-a^2 x^2\right )^{5/2} \int \left (\frac {1}{a^5}+\frac {1}{8 a^5 (-1+a x)^2}+\frac {7}{16 a^5 (-1+a x)}-\frac {1}{4 a^5 (1+a x)^3}+\frac {1}{a^5 (1+a x)^2}-\frac {23}{16 a^5 (1+a x)}\right ) \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {\left (1-a^2 x^2\right )^{5/2}}{a^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^4}+\frac {\left (1-a^2 x^2\right )^{5/2}}{8 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1-a x)}+\frac {\left (1-a^2 x^2\right )^{5/2}}{8 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1+a x)^2}-\frac {\left (1-a^2 x^2\right )^{5/2}}{a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1+a x)}+\frac {7 \left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}-\frac {23 \left (1-a^2 x^2\right )^{5/2} \log (1+a x)}{16 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 88, normalized size = 0.33 \[ \frac {\left (1-a^2 x^2\right )^{5/2} \left (2 \left (8 a x+\frac {1}{1-a x}-\frac {8}{a x+1}+\frac {1}{(a x+1)^2}\right )+7 \log (1-a x)-23 \log (a x+1)\right )}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^ArcTanh[a*x]*(c - c/(a^2*x^2))^(5/2)),x]

[Out]

((1 - a^2*x^2)^(5/2)*(2*(8*a*x + (1 - a*x)^(-1) + (1 + a*x)^(-2) - 8/(1 + a*x)) + 7*Log[1 - a*x] - 23*Log[1 +
a*x]))/(16*a^6*(c - c/(a^2*x^2))^(5/2)*x^5)

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fricas [F]  time = 1.23, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} a^{6} x^{6} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{7} c^{3} x^{7} + a^{6} c^{3} x^{6} - 3 \, a^{5} c^{3} x^{5} - 3 \, a^{4} c^{3} x^{4} + 3 \, a^{3} c^{3} x^{3} + 3 \, a^{2} c^{3} x^{2} - a c^{3} x - c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*x^2 + 1)*a^6*x^6*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^7*c^3*x^7 + a^6*c^3*x^6 - 3*a^5*c^3*x^5
- 3*a^4*c^3*x^4 + 3*a^3*c^3*x^3 + 3*a^2*c^3*x^2 - a*c^3*x - c^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a x + 1\right )} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(-a^2*x^2 + 1)/((a*x + 1)*(c - c/(a^2*x^2))^(5/2)), x)

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maple [A]  time = 0.06, size = 167, normalized size = 0.63 \[ \frac {\sqrt {-a^{2} x^{2}+1}\, \left (a x -1\right ) \left (16 x^{4} a^{4}+7 \ln \left (a x -1\right ) x^{3} a^{3}-23 a^{3} x^{3} \ln \left (a x +1\right )+16 x^{3} a^{3}+7 \ln \left (a x -1\right ) x^{2} a^{2}-23 \ln \left (a x +1\right ) x^{2} a^{2}-34 a^{2} x^{2}-7 \ln \left (a x -1\right ) x a +23 a x \ln \left (a x +1\right )-18 a x -7 \ln \left (a x -1\right )+23 \ln \left (a x +1\right )+12\right )}{16 a^{6} x^{5} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(5/2),x)

[Out]

1/16*(-a^2*x^2+1)^(1/2)*(a*x-1)*(16*x^4*a^4+7*ln(a*x-1)*x^3*a^3-23*a^3*x^3*ln(a*x+1)+16*x^3*a^3+7*ln(a*x-1)*x^
2*a^2-23*ln(a*x+1)*x^2*a^2-34*a^2*x^2-7*ln(a*x-1)*x*a+23*a*x*ln(a*x+1)-18*a*x-7*ln(a*x-1)+23*ln(a*x+1)+12)/a^6
/x^5/(c*(a^2*x^2-1)/a^2/x^2)^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a x + 1\right )} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-a^2*x^2 + 1)/((a*x + 1)*(c - c/(a^2*x^2))^(5/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {1-a^2\,x^2}}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}\,\left (a\,x+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - a^2*x^2)^(1/2)/((c - c/(a^2*x^2))^(5/2)*(a*x + 1)),x)

[Out]

int((1 - a^2*x^2)^(1/2)/((c - c/(a^2*x^2))^(5/2)*(a*x + 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{\left (- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )\right )^{\frac {5}{2}} \left (a x + 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(c-c/a**2/x**2)**(5/2),x)

[Out]

Integral(sqrt(-(a*x - 1)*(a*x + 1))/((-c*(-1 + 1/(a*x))*(1 + 1/(a*x)))**(5/2)*(a*x + 1)), x)

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