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]

-2/35*((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*(-6*a*x+1)/a/c/(-a^2*c*x^2+c)^(3/2)-16/35*((a*x+1)/(-a^2*x^2+1)^(1/2)
)^(1/2)*(-2*a*x+1)/a/c^2/(-a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 83, normalized size of antiderivative = 1.00, 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 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 (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*}

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Mathematica [A]  time = 0.04, 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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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)/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

Timed out

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

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*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)

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maple [A]  time = 0.03, size = 71, normalized size = 0.86 \[ \frac {2 \left (a x -1\right ) \left (a x +1\right ) \left (16 x^{3} a^{3}-8 a^{2} x^{2}-22 a x +9\right ) \sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}}{35 a \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}} \]

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*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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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 >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [B]  time = 1.33, size = 97, normalized size = 1.17 \[ -\frac {\sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}}\,\left (\frac {18}{35\,a^3\,c^2}+\frac {32\,x^3}{35\,c^2}-\frac {44\,x}{35\,a^2\,c^2}-\frac {16\,x^2}{35\,a\,c^2}\right )}{\frac {\sqrt {c-a^2\,c\,x^2}}{a^2}-x^2\,\sqrt {c-a^2\,c\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*c*x**2+c)**(5/2),x)

[Out]

Timed out

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