3.1287 \(\int \frac {e^{\frac {1}{2} \tanh ^{-1}(a x)}}{(1-a^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=112 \[ -\frac {2 (1-10 a x) e^{\frac {1}{2} \tanh ^{-1}(a x)}}{99 a \left (1-a^2 x^2\right )^{5/2}}-\frac {256 (1-2 a x) e^{\frac {1}{2} \tanh ^{-1}(a x)}}{693 a \sqrt {1-a^2 x^2}}-\frac {32 (1-6 a x) e^{\frac {1}{2} \tanh ^{-1}(a x)}}{693 a \left (1-a^2 x^2\right )^{3/2}} \]

[Out]

-2/99*((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*(-10*a*x+1)/a/(-a^2*x^2+1)^(5/2)-32/693*((a*x+1)/(-a^2*x^2+1)^(1/2))^
(1/2)*(-6*a*x+1)/a/(-a^2*x^2+1)^(3/2)-256/693*((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*(-2*a*x+1)/a/(-a^2*x^2+1)^(1/
2)

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Rubi [A]  time = 0.12, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6136, 6135} \[ -\frac {2 (1-10 a x) e^{\frac {1}{2} \tanh ^{-1}(a x)}}{99 a \left (1-a^2 x^2\right )^{5/2}}-\frac {256 (1-2 a x) e^{\frac {1}{2} \tanh ^{-1}(a x)}}{693 a \sqrt {1-a^2 x^2}}-\frac {32 (1-6 a x) e^{\frac {1}{2} \tanh ^{-1}(a x)}}{693 a \left (1-a^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 64, normalized size = 0.57 \[ \frac {2 \left (256 a^5 x^5-128 a^4 x^4-608 a^3 x^3+272 a^2 x^2+422 a x-151\right )}{693 a (1-a x)^{11/4} (a x+1)^{9/4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(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*(1 - a*x)^(11/4)*(1 + a*x)
^(9/4))

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fricas [A]  time = 0.59, size = 104, normalized size = 0.93 \[ -\frac {2 \, {\left (256 \, a^{5} x^{5} - 128 \, a^{4} x^{4} - 608 \, a^{3} x^{3} + 272 \, a^{2} x^{2} + 422 \, a x - 151\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}}{693 \, {\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )}} \]

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*x^2+1)^(7/2),x, algorithm="fricas")

[Out]

-2/693*(256*a^5*x^5 - 128*a^4*x^4 - 608*a^3*x^3 + 272*a^2*x^2 + 422*a*x - 151)*sqrt(-a^2*x^2 + 1)*sqrt(-sqrt(-
a^2*x^2 + 1)/(a*x - 1))/(a^7*x^6 - 3*a^5*x^4 + 3*a^3*x^2 - a)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {7}{2}}}\,{d x} \]

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*x^2+1)^(7/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 86, normalized size = 0.77 \[ -\frac {2 \left (a x -1\right ) \left (a x +1\right ) \left (256 x^{5} a^{5}-128 x^{4} a^{4}-608 x^{3} a^{3}+272 a^{2} x^{2}+422 a x -151\right ) \sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}}{693 a \left (-a^{2} x^{2}+1\right )^{\frac {7}{2}}} \]

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*x^2+1)^(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*x^2+1)^(7/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {7}{2}}}\,{d x} \]

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*x^2+1)^(7/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.31, size = 165, normalized size = 1.47 \[ -\frac {\sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}}\,\left (\frac {302\,\sqrt {1-a^2\,x^2}}{693\,a^7}-\frac {844\,x\,\sqrt {1-a^2\,x^2}}{693\,a^6}-\frac {512\,x^5\,\sqrt {1-a^2\,x^2}}{693\,a^2}+\frac {256\,x^4\,\sqrt {1-a^2\,x^2}}{693\,a^3}+\frac {1216\,x^3\,\sqrt {1-a^2\,x^2}}{693\,a^4}-\frac {544\,x^2\,\sqrt {1-a^2\,x^2}}{693\,a^5}\right )}{\frac {1}{a^6}-x^6+\frac {3\,x^4}{a^2}-\frac {3\,x^2}{a^4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2)/(1 - a^2*x^2)^(7/2),x)

[Out]

-(((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2)*((302*(1 - a^2*x^2)^(1/2))/(693*a^7) - (844*x*(1 - a^2*x^2)^(1/2))/(69
3*a^6) - (512*x^5*(1 - a^2*x^2)^(1/2))/(693*a^2) + (256*x^4*(1 - a^2*x^2)^(1/2))/(693*a^3) + (1216*x^3*(1 - a^
2*x^2)^(1/2))/(693*a^4) - (544*x^2*(1 - a^2*x^2)^(1/2))/(693*a^5)))/(1/a^6 - x^6 + (3*x^4)/a^2 - (3*x^2)/a^4)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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*x**2+1)**(7/2),x)

[Out]

Timed out

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