3.1286 \(\int \frac {e^{\frac {1}{2} \tanh ^{-1}(a x)}}{(1-a^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=75 \[ -\frac {2 (1-6 a x) e^{\frac {1}{2} \tanh ^{-1}(a x)}}{35 a \left (1-a^2 x^2\right )^{3/2}}-\frac {16 (1-2 a x) e^{\frac {1}{2} \tanh ^{-1}(a x)}}{35 a \sqrt {1-a^2 x^2}} \]

[Out]

-2/35*((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*(-6*a*x+1)/a/(-a^2*x^2+1)^(3/2)-16/35*((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 = 0.08, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6136, 6135} \[ -\frac {2 (1-6 a x) e^{\frac {1}{2} \tanh ^{-1}(a x)}}{35 a \left (1-a^2 x^2\right )^{3/2}}-\frac {16 (1-2 a x) e^{\frac {1}{2} \tanh ^{-1}(a x)}}{35 a \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*E^(ArcTanh[a*x]/2)*(1 - 6*a*x))/(35*a*(1 - a^2*x^2)^(3/2)) - (16*E^(ArcTanh[a*x]/2)*(1 - 2*a*x))/(35*a*Sqr
t[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 )^{5/2}} \, dx &=-\frac {2 e^{\frac {1}{2} \tanh ^{-1}(a x)} (1-6 a x)}{35 a \left (1-a^2 x^2\right )^{3/2}}+\frac {24}{35} \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)} (1-6 a x)}{35 a \left (1-a^2 x^2\right )^{3/2}}-\frac {16 e^{\frac {1}{2} \tanh ^{-1}(a x)} (1-2 a x)}{35 a \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 48, normalized size = 0.64 \[ -\frac {2 \left (16 a^3 x^3-8 a^2 x^2-22 a x+9\right )}{35 a (1-a x)^{7/4} (a x+1)^{5/4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.69, size = 78, normalized size = 1.04 \[ -\frac {2 \, {\left (16 \, a^{3} x^{3} - 8 \, a^{2} x^{2} - 22 \, a x + 9\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}}{35 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )}} \]

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)^(5/2),x, algorithm="fricas")

[Out]

-2/35*(16*a^3*x^3 - 8*a^2*x^2 - 22*a*x + 9)*sqrt(-a^2*x^2 + 1)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))/(a^5*x^4 -
2*a^3*x^2 + a)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}}}\,{d x} \]

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)^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 70, normalized size = 0.93 \[ \frac {2 \left (a x -1\right ) \left (a x +1\right ) \left (16 x^{3} a^{3}-8 a^{2} x^{2}-22 a x +9\right ) \sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}}{35 a \left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}} \]

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)^(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*x^2+1)^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}}}\,{d x} \]

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)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.15, size = 115, normalized size = 1.53 \[ -\frac {\sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}}\,\left (\frac {18\,\sqrt {1-a^2\,x^2}}{35\,a^5}-\frac {44\,x\,\sqrt {1-a^2\,x^2}}{35\,a^4}+\frac {32\,x^3\,\sqrt {1-a^2\,x^2}}{35\,a^2}-\frac {16\,x^2\,\sqrt {1-a^2\,x^2}}{35\,a^3}\right )}{\frac {1}{a^4}+x^4-\frac {2\,x^2}{a^2}} \]

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)^(5/2),x)

[Out]

-(((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2)*((18*(1 - a^2*x^2)^(1/2))/(35*a^5) - (44*x*(1 - a^2*x^2)^(1/2))/(35*a^
4) + (32*x^3*(1 - a^2*x^2)^(1/2))/(35*a^2) - (16*x^2*(1 - a^2*x^2)^(1/2))/(35*a^3)))/(1/a^4 + x^4 - (2*x^2)/a^
2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {a x + 1}{\sqrt {- a^{2} x^{2} + 1}}}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

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)**(5/2),x)

[Out]

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

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