3.8.39 \(\int \frac {e^{-3 \tanh ^{-1}(a x)}}{(c-\frac {c}{a^2 x^2})^{5/2}} \, dx\) [739]

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

[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 = 0.15, 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 = {6295, 6285, 90} \begin {gather*} \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}} \end {gather*}

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 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 6285

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 6295

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/x^(2*p))*(1 + c*(x^2/d))^p*E^(n*ArcTanh[a*x]), 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 = 0.11, size = 110, normalized size = 0.41 \begin {gather*} \frac {\sqrt {1-a^2 x^2} \left (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)\right )}{48 a^2 c^2 \sqrt {c-\frac {c}{a^2 x^2}} x (1+a x)^3} \end {gather*}

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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Maple [A]
time = 0.05, size = 176, normalized size = 0.66

method	result	size
default	\(\frac {\left (-48 a^{4} x^{4}+147 \ln \left (a x +1\right ) a^{3} x^{3}-3 \ln \left (a x -1\right ) a^{3} x^{3}-144 a^{3} x^{3}+441 \ln \left (a x +1\right ) a^{2} x^{2}-9 \ln \left (a x -1\right ) a^{2} x^{2}+42 a^{2} x^{2}+441 \ln \left (a x +1\right ) a x -9 \ln \left (a x -1\right ) a x +270 a x +147 \ln \left (a x +1\right )-3 \ln \left (a x -1\right )+140\right ) \left (a x -1\right )^{2} \sqrt {-a^{2} x^{2}+1}}{48 \left (a x +1\right ) a^{6} x^{5} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )^{\frac {5}{2}}}\)	\(176\)

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,method=_RETURNVERBOSE)

[Out]

1/48*(-48*a^4*x^4+147*ln(a*x+1)*a^3*x^3-3*ln(a*x-1)*a^3*x^3-144*a^3*x^3+441*ln(a*x+1)*a^2*x^2-9*ln(a*x-1)*a^2*
x^2+42*a^2*x^2+441*ln(a*x+1)*a*x-9*ln(a*x-1)*a*x+270*a*x+147*ln(a*x+1)-3*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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

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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