3.1277 \(\int \frac{e^{-3 \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=182 \[ -\frac{\sqrt{1-a^2 x^2}}{8 a c^2 (a x+1) \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2}}{8 a c^2 (a x+1)^2 \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2}}{6 a c^2 (a x+1)^3 \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{8 a c^2 \sqrt{c-a^2 c x^2}} \]

[Out]

-Sqrt[1 - a^2*x^2]/(6*a*c^2*(1 + a*x)^3*Sqrt[c - a^2*c*x^2]) - Sqrt[1 - a^2*x^2]/(8*a*c^2*(1 + a*x)^2*Sqrt[c -
a^2*c*x^2]) - Sqrt[1 - a^2*x^2]/(8*a*c^2*(1 + a*x)*Sqrt[c - a^2*c*x^2]) + (Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(8
*a*c^2*Sqrt[c - a^2*c*x^2])

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Rubi [A]  time = 0.110792, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6143, 6140, 44, 207} \[ -\frac{\sqrt{1-a^2 x^2}}{8 a c^2 (a x+1) \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2}}{8 a c^2 (a x+1)^2 \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2}}{6 a c^2 (a x+1)^3 \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{8 a c^2 \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 - a^2*x^2]/(6*a*c^2*(1 + a*x)^3*Sqrt[c - a^2*c*x^2]) - Sqrt[1 - a^2*x^2]/(8*a*c^2*(1 + a*x)^2*Sqrt[c -
a^2*c*x^2]) - Sqrt[1 - a^2*x^2]/(8*a*c^2*(1 + a*x)*Sqrt[c - a^2*c*x^2]) + (Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(8
*a*c^2*Sqrt[c - a^2*c*x^2])

Rule 6143

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

Rule 6140

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

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0610418, size = 73, normalized size = 0.4 \[ \frac{\sqrt{1-a^2 x^2} \left (-3 a^2 x^2-9 a x+3 (a x+1)^3 \tanh ^{-1}(a x)-10\right )}{24 a c^2 (a x+1)^3 \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(-10 - 9*a*x - 3*a^2*x^2 + 3*(1 + a*x)^3*ArcTanh[a*x]))/(24*a*c^2*(1 + a*x)^3*Sqrt[c - a^2*
c*x^2])

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Maple [A]  time = 0.092, size = 159, normalized size = 0.9 \begin{align*} -{\frac{3\,{a}^{3}{x}^{3}\ln \left ( ax+1 \right ) -3\,\ln \left ( ax-1 \right ){x}^{3}{a}^{3}+9\,\ln \left ( ax+1 \right ){a}^{2}{x}^{2}-9\,\ln \left ( ax-1 \right ){a}^{2}{x}^{2}-6\,{a}^{2}{x}^{2}+9\,ax\ln \left ( ax+1 \right ) -9\,\ln \left ( ax-1 \right ) xa-18\,ax+3\,\ln \left ( ax+1 \right ) -3\,\ln \left ( ax-1 \right ) -20}{ \left ( 48\,{a}^{2}{x}^{2}-48 \right ){c}^{3}a \left ( ax+1 \right ) ^{3}}\sqrt{-{a}^{2}{x}^{2}+1}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }} \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)/(-a^2*c*x^2+c)^(5/2),x)

[Out]

-1/48*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(3*a^3*x^3*ln(a*x+1)-3*ln(a*x-1)*x^3*a^3+9*ln(a*x+1)*a^2*x^2-9
*ln(a*x-1)*a^2*x^2-6*a^2*x^2+9*a*x*ln(a*x+1)-9*ln(a*x-1)*x*a-18*a*x+3*ln(a*x+1)-3*ln(a*x-1)-20)/(a^2*x^2-1)/c^
3/a/(a*x+1)^3

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Maxima [A]  time = 1.00669, size = 126, normalized size = 0.69 \begin{align*} -\frac{3 \, a^{2} \sqrt{c} x^{2} + 9 \, a \sqrt{c} x + 10 \, \sqrt{c}}{24 \,{\left (a^{4} c^{3} x^{3} + 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x + a c^{3}\right )}} + \frac{\log \left (a x + 1\right )}{16 \, a c^{\frac{5}{2}}} - \frac{\log \left (a x - 1\right )}{16 \, a c^{\frac{5}{2}}} \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)/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

-1/24*(3*a^2*sqrt(c)*x^2 + 9*a*sqrt(c)*x + 10*sqrt(c))/(a^4*c^3*x^3 + 3*a^3*c^3*x^2 + 3*a^2*c^3*x + a*c^3) + 1
/16*log(a*x + 1)/(a*c^(5/2)) - 1/16*log(a*x - 1)/(a*c^(5/2))

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Fricas [A]  time = 2.92127, size = 969, normalized size = 5.32 \begin{align*} \left [\frac{3 \,{\left (a^{5} x^{5} + 3 \, a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a^{2} x^{2} - 3 \, a x - 1\right )} \sqrt{c} \log \left (-\frac{a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} - 4 \,{\left (a^{3} x^{3} + a x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} \sqrt{c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right ) - 4 \,{\left (10 \, a^{3} x^{3} + 27 \, a^{2} x^{2} + 21 \, a x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{96 \,{\left (a^{6} c^{3} x^{5} + 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x - a c^{3}\right )}}, \frac{3 \,{\left (a^{5} x^{5} + 3 \, a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a^{2} x^{2} - 3 \, a x - 1\right )} \sqrt{-c} \arctan \left (\frac{2 \, \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} a \sqrt{-c} x}{a^{4} c x^{4} - c}\right ) - 2 \,{\left (10 \, a^{3} x^{3} + 27 \, a^{2} x^{2} + 21 \, a x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{48 \,{\left (a^{6} c^{3} x^{5} + 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x - a c^{3}\right )}}\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)/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[1/96*(3*(a^5*x^5 + 3*a^4*x^4 + 2*a^3*x^3 - 2*a^2*x^2 - 3*a*x - 1)*sqrt(c)*log(-(a^6*c*x^6 + 5*a^4*c*x^4 - 5*a
^2*c*x^2 - 4*(a^3*x^3 + a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - c)/(a^6*x^6 - 3*a^4*x^4 + 3*a^2
*x^2 - 1)) - 4*(10*a^3*x^3 + 27*a^2*x^2 + 21*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^6*c^3*x^5 + 3*a^
5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^3*c^3*x^2 - 3*a^2*c^3*x - a*c^3), 1/48*(3*(a^5*x^5 + 3*a^4*x^4 + 2*a^3*x^3 - 2
*a^2*x^2 - 3*a*x - 1)*sqrt(-c)*arctan(2*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*a*sqrt(-c)*x/(a^4*c*x^4 - c))
- 2*(10*a^3*x^3 + 27*a^2*x^2 + 21*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^6*c^3*x^5 + 3*a^5*c^3*x^4 +
2*a^4*c^3*x^3 - 2*a^3*c^3*x^2 - 3*a^2*c^3*x - a*c^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{5}{2}} \left (a x + 1\right )^{3}}\, dx \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)/(-a**2*c*x**2+c)**(5/2),x)

[Out]

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

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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^{2} c x^{2} + c\right )}^{\frac{5}{2}}{\left (a x + 1\right )}^{3}}\,{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)/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

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