3.740 \(\int \frac{e^{-3 \tanh ^{-1}(a x)}}{(c-\frac{c}{a^2 x^2})^{7/2}} \, dx\)

Optimal. Leaf size=359 \[ \frac{\left (1-a^2 x^2\right )^{7/2}}{32 a^8 x^7 (1-a x) \left (c-\frac{c}{a^2 x^2}\right )^{7/2}}-\frac{75 \left (1-a^2 x^2\right )^{7/2}}{16 a^8 x^7 (a x+1) \left (c-\frac{c}{a^2 x^2}\right )^{7/2}}+\frac{59 \left (1-a^2 x^2\right )^{7/2}}{32 a^8 x^7 (a x+1)^2 \left (c-\frac{c}{a^2 x^2}\right )^{7/2}}-\frac{\left (1-a^2 x^2\right )^{7/2}}{2 a^8 x^7 (a x+1)^3 \left (c-\frac{c}{a^2 x^2}\right )^{7/2}}+\frac{\left (1-a^2 x^2\right )^{7/2}}{16 a^8 x^7 (a x+1)^4 \left (c-\frac{c}{a^2 x^2}\right )^{7/2}}+\frac{\left (1-a^2 x^2\right )^{7/2}}{a^7 x^6 \left (c-\frac{c}{a^2 x^2}\right )^{7/2}}+\frac{9 \left (1-a^2 x^2\right )^{7/2} \log (1-a x)}{64 a^8 x^7 \left (c-\frac{c}{a^2 x^2}\right )^{7/2}}-\frac{201 \left (1-a^2 x^2\right )^{7/2} \log (a x+1)}{64 a^8 x^7 \left (c-\frac{c}{a^2 x^2}\right )^{7/2}} \]

[Out]

(1 - a^2*x^2)^(7/2)/(a^7*(c - c/(a^2*x^2))^(7/2)*x^6) + (1 - a^2*x^2)^(7/2)/(32*a^8*(c - c/(a^2*x^2))^(7/2)*x^
7*(1 - a*x)) + (1 - a^2*x^2)^(7/2)/(16*a^8*(c - c/(a^2*x^2))^(7/2)*x^7*(1 + a*x)^4) - (1 - a^2*x^2)^(7/2)/(2*a
^8*(c - c/(a^2*x^2))^(7/2)*x^7*(1 + a*x)^3) + (59*(1 - a^2*x^2)^(7/2))/(32*a^8*(c - c/(a^2*x^2))^(7/2)*x^7*(1
+ a*x)^2) - (75*(1 - a^2*x^2)^(7/2))/(16*a^8*(c - c/(a^2*x^2))^(7/2)*x^7*(1 + a*x)) + (9*(1 - a^2*x^2)^(7/2)*L
og[1 - a*x])/(64*a^8*(c - c/(a^2*x^2))^(7/2)*x^7) - (201*(1 - a^2*x^2)^(7/2)*Log[1 + a*x])/(64*a^8*(c - c/(a^2
*x^2))^(7/2)*x^7)

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Rubi [A]  time = 0.251201, antiderivative size = 359, normalized size of antiderivative = 1., 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} \[ \frac{\left (1-a^2 x^2\right )^{7/2}}{32 a^8 x^7 (1-a x) \left (c-\frac{c}{a^2 x^2}\right )^{7/2}}-\frac{75 \left (1-a^2 x^2\right )^{7/2}}{16 a^8 x^7 (a x+1) \left (c-\frac{c}{a^2 x^2}\right )^{7/2}}+\frac{59 \left (1-a^2 x^2\right )^{7/2}}{32 a^8 x^7 (a x+1)^2 \left (c-\frac{c}{a^2 x^2}\right )^{7/2}}-\frac{\left (1-a^2 x^2\right )^{7/2}}{2 a^8 x^7 (a x+1)^3 \left (c-\frac{c}{a^2 x^2}\right )^{7/2}}+\frac{\left (1-a^2 x^2\right )^{7/2}}{16 a^8 x^7 (a x+1)^4 \left (c-\frac{c}{a^2 x^2}\right )^{7/2}}+\frac{\left (1-a^2 x^2\right )^{7/2}}{a^7 x^6 \left (c-\frac{c}{a^2 x^2}\right )^{7/2}}+\frac{9 \left (1-a^2 x^2\right )^{7/2} \log (1-a x)}{64 a^8 x^7 \left (c-\frac{c}{a^2 x^2}\right )^{7/2}}-\frac{201 \left (1-a^2 x^2\right )^{7/2} \log (a x+1)}{64 a^8 x^7 \left (c-\frac{c}{a^2 x^2}\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*(c - c/(a^2*x^2))^(7/2)),x]

[Out]

(1 - a^2*x^2)^(7/2)/(a^7*(c - c/(a^2*x^2))^(7/2)*x^6) + (1 - a^2*x^2)^(7/2)/(32*a^8*(c - c/(a^2*x^2))^(7/2)*x^
7*(1 - a*x)) + (1 - a^2*x^2)^(7/2)/(16*a^8*(c - c/(a^2*x^2))^(7/2)*x^7*(1 + a*x)^4) - (1 - a^2*x^2)^(7/2)/(2*a
^8*(c - c/(a^2*x^2))^(7/2)*x^7*(1 + a*x)^3) + (59*(1 - a^2*x^2)^(7/2))/(32*a^8*(c - c/(a^2*x^2))^(7/2)*x^7*(1
+ a*x)^2) - (75*(1 - a^2*x^2)^(7/2))/(16*a^8*(c - c/(a^2*x^2))^(7/2)*x^7*(1 + a*x)) + (9*(1 - a^2*x^2)^(7/2)*L
og[1 - a*x])/(64*a^8*(c - c/(a^2*x^2))^(7/2)*x^7) - (201*(1 - a^2*x^2)^(7/2)*Log[1 + a*x])/(64*a^8*(c - c/(a^2
*x^2))^(7/2)*x^7)

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]

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 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]))

Rubi steps

\begin{align*} \int \frac{e^{-3 \tanh ^{-1}(a x)}}{\left (c-\frac{c}{a^2 x^2}\right )^{7/2}} \, dx &=\frac{\left (1-a^2 x^2\right )^{7/2} \int \frac{e^{-3 \tanh ^{-1}(a x)} x^7}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{\left (c-\frac{c}{a^2 x^2}\right )^{7/2} x^7}\\ &=\frac{\left (1-a^2 x^2\right )^{7/2} \int \frac{x^7}{(1-a x)^2 (1+a x)^5} \, dx}{\left (c-\frac{c}{a^2 x^2}\right )^{7/2} x^7}\\ &=\frac{\left (1-a^2 x^2\right )^{7/2} \int \left (\frac{1}{a^7}+\frac{1}{32 a^7 (-1+a x)^2}+\frac{9}{64 a^7 (-1+a x)}-\frac{1}{4 a^7 (1+a x)^5}+\frac{3}{2 a^7 (1+a x)^4}-\frac{59}{16 a^7 (1+a x)^3}+\frac{75}{16 a^7 (1+a x)^2}-\frac{201}{64 a^7 (1+a x)}\right ) \, dx}{\left (c-\frac{c}{a^2 x^2}\right )^{7/2} x^7}\\ &=\frac{\left (1-a^2 x^2\right )^{7/2}}{a^7 \left (c-\frac{c}{a^2 x^2}\right )^{7/2} x^6}+\frac{\left (1-a^2 x^2\right )^{7/2}}{32 a^8 \left (c-\frac{c}{a^2 x^2}\right )^{7/2} x^7 (1-a x)}+\frac{\left (1-a^2 x^2\right )^{7/2}}{16 a^8 \left (c-\frac{c}{a^2 x^2}\right )^{7/2} x^7 (1+a x)^4}-\frac{\left (1-a^2 x^2\right )^{7/2}}{2 a^8 \left (c-\frac{c}{a^2 x^2}\right )^{7/2} x^7 (1+a x)^3}+\frac{59 \left (1-a^2 x^2\right )^{7/2}}{32 a^8 \left (c-\frac{c}{a^2 x^2}\right )^{7/2} x^7 (1+a x)^2}-\frac{75 \left (1-a^2 x^2\right )^{7/2}}{16 a^8 \left (c-\frac{c}{a^2 x^2}\right )^{7/2} x^7 (1+a x)}+\frac{9 \left (1-a^2 x^2\right )^{7/2} \log (1-a x)}{64 a^8 \left (c-\frac{c}{a^2 x^2}\right )^{7/2} x^7}-\frac{201 \left (1-a^2 x^2\right )^{7/2} \log (1+a x)}{64 a^8 \left (c-\frac{c}{a^2 x^2}\right )^{7/2} x^7}\\ \end{align*}

Mathematica [A]  time = 0.121583, size = 146, normalized size = 0.41 \[ \frac{\sqrt{1-a^2 x^2} \left (-2 \left (32 a^6 x^6+96 a^5 x^5-87 a^4 x^4-309 a^3 x^3-59 a^2 x^2+207 a x+104\right )-9 (a x-1) (a x+1)^4 \log (1-a x)+201 (a x-1) (a x+1)^4 \log (a x+1)\right )}{64 a^2 c^3 x (a x-1) (a x+1)^4 \sqrt{c-\frac{c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*(c - c/(a^2*x^2))^(7/2)),x]

[Out]

(Sqrt[1 - a^2*x^2]*(-2*(104 + 207*a*x - 59*a^2*x^2 - 309*a^3*x^3 - 87*a^4*x^4 + 96*a^5*x^5 + 32*a^6*x^6) - 9*(
-1 + a*x)*(1 + a*x)^4*Log[1 - a*x] + 201*(-1 + a*x)*(1 + a*x)^4*Log[1 + a*x]))/(64*a^2*c^3*Sqrt[c - c/(a^2*x^2
)]*x*(-1 + a*x)*(1 + a*x)^4)

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Maple [A]  time = 0.173, size = 248, normalized size = 0.7 \begin{align*}{\frac{ \left ( ax-1 \right ) ^{2} \left ( -64\,{x}^{6}{a}^{6}+201\,\ln \left ( ax+1 \right ){x}^{5}{a}^{5}-9\,\ln \left ( ax-1 \right ){x}^{5}{a}^{5}-192\,{x}^{5}{a}^{5}+603\,\ln \left ( ax+1 \right ){a}^{4}{x}^{4}-27\,\ln \left ( ax-1 \right ){a}^{4}{x}^{4}+174\,{x}^{4}{a}^{4}+402\,{a}^{3}{x}^{3}\ln \left ( ax+1 \right ) -18\,\ln \left ( ax-1 \right ){x}^{3}{a}^{3}+618\,{x}^{3}{a}^{3}-402\,\ln \left ( ax+1 \right ){a}^{2}{x}^{2}+18\,\ln \left ( ax-1 \right ){a}^{2}{x}^{2}+118\,{a}^{2}{x}^{2}-603\,ax\ln \left ( ax+1 \right ) +27\,\ln \left ( ax-1 \right ) xa-414\,ax-201\,\ln \left ( ax+1 \right ) +9\,\ln \left ( ax-1 \right ) -208 \right ) }{ \left ( 64\,ax+64 \right ){a}^{8}{x}^{7}}\sqrt{-{a}^{2}{x}^{2}+1} \left ({\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}} \right ) ^{-{\frac{7}{2}}}} \end{align*}

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)^(7/2),x)

[Out]

1/64*(-a^2*x^2+1)^(1/2)*(a*x-1)^2*(-64*x^6*a^6+201*ln(a*x+1)*x^5*a^5-9*ln(a*x-1)*x^5*a^5-192*x^5*a^5+603*ln(a*
x+1)*a^4*x^4-27*ln(a*x-1)*a^4*x^4+174*x^4*a^4+402*a^3*x^3*ln(a*x+1)-18*ln(a*x-1)*x^3*a^3+618*x^3*a^3-402*ln(a*
x+1)*a^2*x^2+18*ln(a*x-1)*a^2*x^2+118*a^2*x^2-603*a*x*ln(a*x+1)+27*ln(a*x-1)*x*a-414*a*x-201*ln(a*x+1)+9*ln(a*
x-1)-208)/(a*x+1)/a^8/x^7/(c*(a^2*x^2-1)/a^2/x^2)^(7/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{7}{2}}}\,{d x} \end{align*}

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)^(7/2),x, algorithm="maxima")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)/((a*x + 1)^3*(c - c/(a^2*x^2))^(7/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-a^{2} x^{2} + 1} a^{8} x^{8} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{9} c^{4} x^{9} + 3 \, a^{8} c^{4} x^{8} - 8 \, a^{6} c^{4} x^{6} - 6 \, a^{5} c^{4} x^{5} + 6 \, a^{4} c^{4} x^{4} + 8 \, a^{3} c^{4} x^{3} - 3 \, a c^{4} x - c^{4}}, x\right ) \end{align*}

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)^(7/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*x^2 + 1)*a^8*x^8*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^9*c^4*x^9 + 3*a^8*c^4*x^8 - 8*a^6*c^4*
x^6 - 6*a^5*c^4*x^5 + 6*a^4*c^4*x^4 + 8*a^3*c^4*x^3 - 3*a*c^4*x - c^4), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{7}{2}}}\,{d x} \end{align*}

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)^(7/2),x, algorithm="giac")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)/((a*x + 1)^3*(c - c/(a^2*x^2))^(7/2)), x)