3.739 \(\int \frac {e^{-3 \tanh ^{-1}(a x)}}{(c-\frac {c}{a^2 x^2})^{5/2}} \, dx\)

Optimal. Leaf size=267 \[ \frac {31 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (a x+1) \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {9 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (a x+1)^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {\left (1-a^2 x^2\right )^{5/2}}{6 a^6 x^5 (a x+1)^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {\left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {49 \left (1-a^2 x^2\right )^{5/2} \log (a x+1)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {\left (1-a^2 x^2\right )^{5/2}}{a^5 x^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \]

[Out]

-(-a^2*x^2+1)^(5/2)/a^5/(c-c/a^2/x^2)^(5/2)/x^4+1/6*(-a^2*x^2+1)^(5/2)/a^6/(c-c/a^2/x^2)^(5/2)/x^5/(a*x+1)^3-9
/8*(-a^2*x^2+1)^(5/2)/a^6/(c-c/a^2/x^2)^(5/2)/x^5/(a*x+1)^2+31/8*(-a^2*x^2+1)^(5/2)/a^6/(c-c/a^2/x^2)^(5/2)/x^
5/(a*x+1)-1/16*(-a^2*x^2+1)^(5/2)*ln(-a*x+1)/a^6/(c-c/a^2/x^2)^(5/2)/x^5+49/16*(-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 = 0.22, antiderivative size = 267, 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} \[ \frac {31 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (a x+1) \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {9 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (a x+1)^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {\left (1-a^2 x^2\right )^{5/2}}{6 a^6 x^5 (a x+1)^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {\left (1-a^2 x^2\right )^{5/2}}{a^5 x^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {\left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {49 \left (1-a^2 x^2\right )^{5/2} \log (a x+1)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*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)/(6*a^6*(c - c/(a^2*x^2))^(5/2)*
x^5*(1 + a*x)^3) - (9*(1 - a^2*x^2)^(5/2))/(8*a^6*(c - c/(a^2*x^2))^(5/2)*x^5*(1 + a*x)^2) + (31*(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)*Log[1 - a*x])/(16*a^6*(c - c/(a^2
*x^2))^(5/2)*x^5) + (49*(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^{-3 \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^{-3 \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) (1+a x)^4} \, 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}{16 a^5 (-1+a x)}-\frac {1}{2 a^5 (1+a x)^4}+\frac {9}{4 a^5 (1+a x)^3}-\frac {31}{8 a^5 (1+a x)^2}+\frac {49}{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}}{6 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1+a x)^3}-\frac {9 \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 {31 \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} \log (1-a x)}{16 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}+\frac {49 \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 = 0.09, size = 110, normalized size = 0.41 \[ \frac {\sqrt {1-a^2 x^2} \left (-48 a^4 x^4-144 a^3 x^3+42 a^2 x^2+270 a x-3 (a x+1)^3 \log (1-a x)+147 (a x+1)^3 \log (a x+1)+140\right )}{48 a^2 c^2 x (a x+1)^3 \sqrt {c-\frac {c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*(c - c/(a^2*x^2))^(5/2)),x]

[Out]

(Sqrt[1 - a^2*x^2]*(140 + 270*a*x + 42*a^2*x^2 - 144*a^3*x^3 - 48*a^4*x^4 - 3*(1 + a*x)^3*Log[1 - a*x] + 147*(
1 + a*x)^3*Log[1 + a*x]))/(48*a^2*c^2*Sqrt[c - c/(a^2*x^2)]*x*(1 + a*x)^3)

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} a^{6} x^{6} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{7} c^{3} x^{7} + 3 \, a^{6} c^{3} x^{6} + a^{5} c^{3} x^{5} - 5 \, a^{4} c^{3} x^{4} - 5 \, a^{3} c^{3} x^{3} + a^{2} c^{3} x^{2} + 3 \, a c^{3} x + c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="giac")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)/((a*x + 1)^3*(c - c/(a^2*x^2))^(5/2)), x)

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maple [A]  time = 0.06, size = 176, normalized size = 0.66 \[ -\frac {\left (48 x^{4} a^{4}+3 \ln \left (a x -1\right ) x^{3} a^{3}-147 a^{3} x^{3} \ln \left (a x +1\right )+144 x^{3} a^{3}+9 \ln \left (a x -1\right ) x^{2} a^{2}-441 \ln \left (a x +1\right ) x^{2} a^{2}-42 a^{2} x^{2}+9 \ln \left (a x -1\right ) x a -441 a x \ln \left (a x +1\right )-270 a x +3 \ln \left (a x -1\right )-147 \ln \left (a x +1\right )-140\right ) \left (a x -1\right )^{2} \sqrt {-a^{2} x^{2}+1}}{48 \left (a x +1\right ) a^{6} x^{5} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^(5/2),x)

[Out]

-1/48*(48*x^4*a^4+3*ln(a*x-1)*x^3*a^3-147*a^3*x^3*ln(a*x+1)+144*x^3*a^3+9*ln(a*x-1)*x^2*a^2-441*ln(a*x+1)*x^2*
a^2-42*a^2*x^2+9*ln(a*x-1)*x*a-441*a*x*ln(a*x+1)-270*a*x+3*ln(a*x-1)-147*ln(a*x+1)-140)*(a*x-1)^2*(-a^2*x^2+1)
^(1/2)/(a*x+1)/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 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="maxima")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)/((a*x + 1)^3*(c - c/(a^2*x^2))^(5/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}\,{\left (a\,x+1\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - a^2*x^2)^(3/2)/((c - c/(a^2*x^2))^(5/2)*(a*x + 1)^3),x)

[Out]

int((1 - a^2*x^2)^(3/2)/((c - c/(a^2*x^2))^(5/2)*(a*x + 1)^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\left (- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )\right )^{\frac {5}{2}} \left (a x + 1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/(c-c/a**2/x**2)**(5/2),x)

[Out]

Integral((-(a*x - 1)*(a*x + 1))**(3/2)/((-c*(-1 + 1/(a*x))*(1 + 1/(a*x)))**(5/2)*(a*x + 1)**3), x)

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