3.1294 \(\int \frac{e^{\frac{1}{2} \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=83 \[ -\frac{16 (1-2 a x) e^{\frac{1}{2} \tanh ^{-1}(a x)}}{35 a c^2 \sqrt{c-a^2 c x^2}}-\frac{2 (1-6 a x) e^{\frac{1}{2} \tanh ^{-1}(a x)}}{35 a c \left (c-a^2 c x^2\right )^{3/2}} \]

[Out]

(-2*E^(ArcTanh[a*x]/2)*(1 - 6*a*x))/(35*a*c*(c - a^2*c*x^2)^(3/2)) - (16*E^(ArcTanh[a*x]/2)*(1 - 2*a*x))/(35*a
*c^2*Sqrt[c - a^2*c*x^2])

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Rubi [A]  time = 0.093848, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {6136, 6135} \[ -\frac{16 (1-2 a x) e^{\frac{1}{2} \tanh ^{-1}(a x)}}{35 a c^2 \sqrt{c-a^2 c x^2}}-\frac{2 (1-6 a x) e^{\frac{1}{2} \tanh ^{-1}(a x)}}{35 a c \left (c-a^2 c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*E^(ArcTanh[a*x]/2)*(1 - 6*a*x))/(35*a*c*(c - a^2*c*x^2)^(3/2)) - (16*E^(ArcTanh[a*x]/2)*(1 - 2*a*x))/(35*a
*c^2*Sqrt[c - a^2*c*x^2])

Rule 6136

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((n + 2*a*(p + 1)*x)*(c + d*x^2
)^(p + 1)*E^(n*ArcTanh[a*x]))/(a*c*(n^2 - 4*(p + 1)^2)), x] - Dist[(2*(p + 1)*(2*p + 3))/(c*(n^2 - 4*(p + 1)^2
)), Int[(c + d*x^2)^(p + 1)*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] && LtQ[p
, -1] &&  !IntegerQ[n] && NeQ[n^2 - 4*(p + 1)^2, 0] && IntegerQ[2*p]

Rule 6135

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[((n - a*x)*E^(n*ArcTanh[a*x]))
/(a*c*(n^2 - 1)*Sqrt[c + d*x^2]), x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] &&  !IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{e^{\frac{1}{2} \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx &=-\frac{2 e^{\frac{1}{2} \tanh ^{-1}(a x)} (1-6 a x)}{35 a c \left (c-a^2 c x^2\right )^{3/2}}+\frac{24 \int \frac{e^{\frac{1}{2} \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx}{35 c}\\ &=-\frac{2 e^{\frac{1}{2} \tanh ^{-1}(a x)} (1-6 a x)}{35 a c \left (c-a^2 c x^2\right )^{3/2}}-\frac{16 e^{\frac{1}{2} \tanh ^{-1}(a x)} (1-2 a x)}{35 a c^2 \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0366685, size = 80, normalized size = 0.96 \[ -\frac{2 \sqrt{1-a^2 x^2} \left (16 a^3 x^3-8 a^2 x^2-22 a x+9\right )}{35 a c^2 (1-a x)^{7/4} (a x+1)^{5/4} \sqrt{c-a^2 c x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*Sqrt[1 - a^2*x^2]*(9 - 22*a*x - 8*a^2*x^2 + 16*a^3*x^3))/(35*a*c^2*(1 - a*x)^(7/4)*(1 + a*x)^(5/4)*Sqrt[c
- a^2*c*x^2])

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Maple [A]  time = 0.03, size = 71, normalized size = 0.9 \begin{align*}{\frac{ \left ( 2\,ax-2 \right ) \left ( ax+1 \right ) \left ( 16\,{x}^{3}{a}^{3}-8\,{a}^{2}{x}^{2}-22\,ax+9 \right ) }{35\,a}\sqrt{{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(a*x-1)*(a*x+1)*(16*a^3*x^3-8*a^2*x^2-22*a*x+9)*((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)/a/(-a^2*c*x^2+c)^(5/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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

[Out]

Timed out

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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