3.1351 \(\int \frac {e^{n \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=102 \[ -\frac {6 (n-a x) e^{n \tanh ^{-1}(a x)}}{a c^2 \left (1-n^2\right ) \left (9-n^2\right ) \sqrt {c-a^2 c x^2}}-\frac {(n-3 a x) e^{n \tanh ^{-1}(a x)}}{a c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}} \]

[Out]

-exp(n*arctanh(a*x))*(-3*a*x+n)/a/c/(-n^2+9)/(-a^2*c*x^2+c)^(3/2)-6*exp(n*arctanh(a*x))*(-a*x+n)/a/c^2/(n^4-10
*n^2+9)/(-a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6136, 6135} \[ -\frac {6 (n-a x) e^{n \tanh ^{-1}(a x)}}{a c^2 \left (1-n^2\right ) \left (9-n^2\right ) \sqrt {c-a^2 c x^2}}-\frac {(n-3 a x) e^{n \tanh ^{-1}(a x)}}{a c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((E^(n*ArcTanh[a*x])*(n - 3*a*x))/(a*c*(9 - n^2)*(c - a^2*c*x^2)^(3/2))) - (6*E^(n*ArcTanh[a*x])*(n - a*x))/(
a*c^2*(1 - n^2)*(9 - n^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^{n \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx &=-\frac {e^{n \tanh ^{-1}(a x)} (n-3 a x)}{a c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}+\frac {6 \int \frac {e^{n \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx}{c \left (9-n^2\right )}\\ &=-\frac {e^{n \tanh ^{-1}(a x)} (n-3 a x)}{a c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}-\frac {6 e^{n \tanh ^{-1}(a x)} (n-a x)}{a c^2 \left (1-n^2\right ) \left (9-n^2\right ) \sqrt {c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 123, normalized size = 1.21 \[ -\frac {\sqrt {1-a^2 x^2} (1-a x)^{\frac {1}{2} (-n-3)} (a x+1)^{\frac {n-3}{2}} \left (6 a^3 x^3+n \left (7-6 a^2 x^2\right )+3 a n^2 x-9 a x-n^3\right )}{a c^2 (n-3) (n-1) (n+1) (n+3) \sqrt {c-a^2 c x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(((1 - a*x)^((-3 - n)/2)*(1 + a*x)^((-3 + n)/2)*Sqrt[1 - a^2*x^2]*(-n^3 - 9*a*x + 3*a*n^2*x + 6*a^3*x^3 + n*(
7 - 6*a^2*x^2)))/(a*c^2*(-3 + n)*(-1 + n)*(1 + n)*(3 + n)*Sqrt[c - a^2*c*x^2]))

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fricas [A]  time = 0.52, size = 165, normalized size = 1.62 \[ -\frac {{\left (6 \, a^{3} x^{3} - 6 \, a^{2} n x^{2} - n^{3} + 3 \, {\left (a n^{2} - 3 \, a\right )} x + 7 \, n\right )} \sqrt {-a^{2} c x^{2} + c} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a c^{3} n^{4} - 10 \, a c^{3} n^{2} + {\left (a^{5} c^{3} n^{4} - 10 \, a^{5} c^{3} n^{2} + 9 \, a^{5} c^{3}\right )} x^{4} + 9 \, a c^{3} - 2 \, {\left (a^{3} c^{3} n^{4} - 10 \, a^{3} c^{3} n^{2} + 9 \, a^{3} c^{3}\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

-(6*a^3*x^3 - 6*a^2*n*x^2 - n^3 + 3*(a*n^2 - 3*a)*x + 7*n)*sqrt(-a^2*c*x^2 + c)*((a*x + 1)/(a*x - 1))^(1/2*n)/
(a*c^3*n^4 - 10*a*c^3*n^2 + (a^5*c^3*n^4 - 10*a^5*c^3*n^2 + 9*a^5*c^3)*x^4 + 9*a*c^3 - 2*(a^3*c^3*n^4 - 10*a^3
*c^3*n^2 + 9*a^3*c^3)*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 84, normalized size = 0.82 \[ \frac {\left (a x -1\right ) \left (a x +1\right ) \left (6 x^{3} a^{3}-6 n \,x^{2} a^{2}+3 n^{2} x a -n^{3}-9 a x +7 n \right ) {\mathrm e}^{n \arctanh \left (a x \right )}}{a \left (n^{4}-10 n^{2}+9\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(a*x))/(-a^2*c*x^2+c)^(5/2),x)

[Out]

(a*x-1)*(a*x+1)*(6*a^3*x^3-6*a^2*n*x^2+3*a*n^2*x-n^3-9*a*x+7*n)*exp(n*arctanh(a*x))/a/(n^4-10*n^2+9)/(-a^2*c*x
^2+c)^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.21, size = 160, normalized size = 1.57 \[ -\frac {{\mathrm {e}}^{\frac {n\,\ln \left (a\,x+1\right )}{2}-\frac {n\,\ln \left (1-a\,x\right )}{2}}\,\left (\frac {6\,x^3}{c^2\,\left (n^4-10\,n^2+9\right )}+\frac {7\,n-n^3}{a^3\,c^2\,\left (n^4-10\,n^2+9\right )}+\frac {3\,x\,\left (n^2-3\right )}{a^2\,c^2\,\left (n^4-10\,n^2+9\right )}-\frac {6\,n\,x^2}{a\,c^2\,\left (n^4-10\,n^2+9\right )}\right )}{\frac {\sqrt {c-a^2\,c\,x^2}}{a^2}-x^2\,\sqrt {c-a^2\,c\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*atanh(a*x))/(c - a^2*c*x^2)^(5/2),x)

[Out]

-(exp((n*log(a*x + 1))/2 - (n*log(1 - a*x))/2)*((6*x^3)/(c^2*(n^4 - 10*n^2 + 9)) + (7*n - n^3)/(a^3*c^2*(n^4 -
10*n^2 + 9)) + (3*x*(n^2 - 3))/(a^2*c^2*(n^4 - 10*n^2 + 9)) - (6*n*x^2)/(a*c^2*(n^4 - 10*n^2 + 9))))/((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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{n \operatorname {atanh}{\left (a x \right )}}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atanh(a*x))/(-a**2*c*x**2+c)**(5/2),x)

[Out]

Integral(exp(n*atanh(a*x))/(-c*(a*x - 1)*(a*x + 1))**(5/2), x)

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