3.1304 \(\int \frac {e^{\frac {1}{2} \tanh ^{-1}(a x)} x^2}{(c-a^2 c x^2)^{9/8}} \, dx\)

Optimal. Leaf size=41 \[ \frac {4 (2-a x) e^{\frac {1}{2} \tanh ^{-1}(a x)}}{3 a^3 c \sqrt [8]{c-a^2 c x^2}} \]

[Out]

4/3*((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*(-a*x+2)/a^3/c/(-a^2*c*x^2+c)^(1/8)

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Rubi [A]  time = 0.12, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6146} \[ \frac {4 (2-a x) e^{\frac {1}{2} \tanh ^{-1}(a x)}}{3 a^3 c \sqrt [8]{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^(ArcTanh[a*x]/2)*x^2)/(c - a^2*c*x^2)^(9/8),x]

[Out]

(4*E^(ArcTanh[a*x]/2)*(2 - a*x))/(3*a^3*c*(c - a^2*c*x^2)^(1/8))

Rule 6146

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

Rubi steps

\begin {align*} \int \frac {e^{\frac {1}{2} \tanh ^{-1}(a x)} x^2}{\left (c-a^2 c x^2\right )^{9/8}} \, dx &=\frac {4 e^{\frac {1}{2} \tanh ^{-1}(a x)} (2-a x)}{3 a^3 c \sqrt [8]{c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 63, normalized size = 1.54 \[ -\frac {4 (a x-2) \sqrt [8]{a x+1} \sqrt [8]{1-a^2 x^2}}{3 a^3 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^(ArcTanh[a*x]/2)*x^2)/(c - a^2*c*x^2)^(9/8),x]

[Out]

(-4*(-2 + a*x)*(1 + a*x)^(1/8)*(1 - a^2*x^2)^(1/8))/(3*a^3*c*(1 - a*x)^(3/8)*(c - a^2*c*x^2)^(1/8))

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fricas [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)*x^2/(-a^2*c*x^2+c)^(9/8),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {9}{8}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^2/(-a^2*c*x^2+c)^(9/8),x, algorithm="giac")

[Out]

integrate(x^2*sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1))/(-a^2*c*x^2 + c)^(9/8), x)

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maple [A]  time = 0.03, size = 54, normalized size = 1.32 \[ \frac {4 \left (a x -1\right ) \left (a x +1\right ) \left (a x -2\right ) \sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}}{3 a^{3} \left (-a^{2} c \,x^{2}+c \right )^{\frac {9}{8}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^2/(-a^2*c*x^2+c)^(9/8),x)

[Out]

4/3*(a*x-1)*(a*x+1)*(a*x-2)*((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)/a^3/(-a^2*c*x^2+c)^(9/8)

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maxima [A]  time = 1.06, size = 28, normalized size = 0.68 \[ -\frac {4 \, {\left (a x + 1\right )}^{\frac {1}{8}} {\left (a x - 2\right )}}{3 \, {\left (-a x + 1\right )}^{\frac {3}{8}} a^{3} c^{\frac {9}{8}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^2/(-a^2*c*x^2+c)^(9/8),x, algorithm="maxima")

[Out]

-4/3*(a*x + 1)^(1/8)*(a*x - 2)/((-a*x + 1)^(3/8)*a^3*c^(9/8))

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mupad [B]  time = 1.26, size = 52, normalized size = 1.27 \[ \frac {\left (\frac {8}{3\,a^3\,c}-\frac {4\,x}{3\,a^2\,c}\right )\,\sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}}}{{\left (c-a^2\,c\,x^2\right )}^{1/8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2))/(c - a^2*c*x^2)^(9/8),x)

[Out]

((8/(3*a^3*c) - (4*x)/(3*a^2*c))*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2))/(c - a^2*c*x^2)^(1/8)

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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)*x**2/(-a**2*c*x**2+c)**(9/8),x)

[Out]

Timed out

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