3.1295 \(\int \frac{e^{\frac{1}{2} \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^{7/2}} \, dx\)

Optimal. Leaf size=124 \[ -\frac{256 (1-2 a x) e^{\frac{1}{2} \tanh ^{-1}(a x)}}{693 a c^3 \sqrt{c-a^2 c x^2}}-\frac{32 (1-6 a x) e^{\frac{1}{2} \tanh ^{-1}(a x)}}{693 a c^2 \left (c-a^2 c x^2\right )^{3/2}}-\frac{2 (1-10 a x) e^{\frac{1}{2} \tanh ^{-1}(a x)}}{99 a c \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

(-2*E^(ArcTanh[a*x]/2)*(1 - 10*a*x))/(99*a*c*(c - a^2*c*x^2)^(5/2)) - (32*E^(ArcTanh[a*x]/2)*(1 - 6*a*x))/(693
*a*c^2*(c - a^2*c*x^2)^(3/2)) - (256*E^(ArcTanh[a*x]/2)*(1 - 2*a*x))/(693*a*c^3*Sqrt[c - a^2*c*x^2])

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Rubi [A]  time = 0.143283, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {6136, 6135} \[ -\frac{256 (1-2 a x) e^{\frac{1}{2} \tanh ^{-1}(a x)}}{693 a c^3 \sqrt{c-a^2 c x^2}}-\frac{32 (1-6 a x) e^{\frac{1}{2} \tanh ^{-1}(a x)}}{693 a c^2 \left (c-a^2 c x^2\right )^{3/2}}-\frac{2 (1-10 a x) e^{\frac{1}{2} \tanh ^{-1}(a x)}}{99 a c \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*E^(ArcTanh[a*x]/2)*(1 - 10*a*x))/(99*a*c*(c - a^2*c*x^2)^(5/2)) - (32*E^(ArcTanh[a*x]/2)*(1 - 6*a*x))/(693
*a*c^2*(c - a^2*c*x^2)^(3/2)) - (256*E^(ArcTanh[a*x]/2)*(1 - 2*a*x))/(693*a*c^3*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 )^{7/2}} \, dx &=-\frac{2 e^{\frac{1}{2} \tanh ^{-1}(a x)} (1-10 a x)}{99 a c \left (c-a^2 c x^2\right )^{5/2}}+\frac{80 \int \frac{e^{\frac{1}{2} \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx}{99 c}\\ &=-\frac{2 e^{\frac{1}{2} \tanh ^{-1}(a x)} (1-10 a x)}{99 a c \left (c-a^2 c x^2\right )^{5/2}}-\frac{32 e^{\frac{1}{2} \tanh ^{-1}(a x)} (1-6 a x)}{693 a c^2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{128 \int \frac{e^{\frac{1}{2} \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx}{231 c^2}\\ &=-\frac{2 e^{\frac{1}{2} \tanh ^{-1}(a x)} (1-10 a x)}{99 a c \left (c-a^2 c x^2\right )^{5/2}}-\frac{32 e^{\frac{1}{2} \tanh ^{-1}(a x)} (1-6 a x)}{693 a c^2 \left (c-a^2 c x^2\right )^{3/2}}-\frac{256 e^{\frac{1}{2} \tanh ^{-1}(a x)} (1-2 a x)}{693 a c^3 \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0473499, size = 96, normalized size = 0.77 \[ \frac{2 \sqrt{1-a^2 x^2} \left (256 a^5 x^5-128 a^4 x^4-608 a^3 x^3+272 a^2 x^2+422 a x-151\right )}{693 a c^3 (1-a x)^{11/4} (a x+1)^{9/4} \sqrt{c-a^2 c x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Sqrt[1 - a^2*x^2]*(-151 + 422*a*x + 272*a^2*x^2 - 608*a^3*x^3 - 128*a^4*x^4 + 256*a^5*x^5))/(693*a*c^3*(1 -
a*x)^(11/4)*(1 + a*x)^(9/4)*Sqrt[c - a^2*c*x^2])

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Maple [A]  time = 0.03, size = 87, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,ax-2 \right ) \left ( ax+1 \right ) \left ( 256\,{x}^{5}{a}^{5}-128\,{x}^{4}{a}^{4}-608\,{x}^{3}{a}^{3}+272\,{a}^{2}{x}^{2}+422\,ax-151 \right ) }{693\,a}\sqrt{{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{-{\frac{7}{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)^(7/2),x)

[Out]

-2/693*(a*x-1)*(a*x+1)*(256*a^5*x^5-128*a^4*x^4-608*a^3*x^3+272*a^2*x^2+422*a*x-151)*((a*x+1)/(-a^2*x^2+1)^(1/
2))^(1/2)/a/(-a^2*c*x^2+c)^(7/2)

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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)^(7/2),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)^(7/2),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)**(7/2),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{7}{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)^(7/2),x, algorithm="giac")

[Out]

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