3.1349 \(\int \frac {e^{n \tanh ^{-1}(a x)} x^2}{(c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=102 \[ \frac {\left (3-n^2\right ) (n-a x) e^{n \tanh ^{-1}(a x)}}{a^3 c^2 \left (n^4-10 n^2+9\right ) \sqrt {c-a^2 c x^2}}-\frac {(n-3 a x) e^{n \tanh ^{-1}(a x)}}{a^3 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {6147, 6135} \[ \frac {\left (3-n^2\right ) (n-a x) e^{n \tanh ^{-1}(a x)}}{a^3 c^2 \left (n^4-10 n^2+9\right ) \sqrt {c-a^2 c x^2}}-\frac {(n-3 a x) e^{n \tanh ^{-1}(a x)}}{a^3 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((E^(n*ArcTanh[a*x])*(n - 3*a*x))/(a^3*c*(9 - n^2)*(c - a^2*c*x^2)^(3/2))) + (E^(n*ArcTanh[a*x])*(3 - n^2)*(n
- a*x))/(a^3*c^2*(9 - 10*n^2 + n^4)*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 6147

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^2*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((n + 2*(p + 1)*a*x)*(c
+ d*x^2)^(p + 1)*E^(n*ArcTanh[a*x]))/(a*d*(n^2 - 4*(p + 1)^2)), x] + Dist[(n^2 + 2*(p + 1))/(d*(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] && L
tQ[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)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx &=-\frac {e^{n \tanh ^{-1}(a x)} (n-3 a x)}{a^3 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}-\frac {\left (3-n^2\right ) \int \frac {e^{n \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx}{a^2 c \left (9-n^2\right )}\\ &=-\frac {e^{n \tanh ^{-1}(a x)} (n-3 a x)}{a^3 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}+\frac {e^{n \tanh ^{-1}(a x)} \left (3-n^2\right ) (n-a x)}{a^3 c^2 \left (9-10 n^2+n^4\right ) \sqrt {c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 125, normalized size = 1.23 \[ -\frac {\sqrt {1-a^2 x^2} (1-a x)^{\frac {1}{2} (-n-3)} (a x+1)^{\frac {n-3}{2}} \left (-3 a^3 x^3-a^2 n^3 x^2+a n^2 x \left (a^2 x^2+2\right )+n \left (3 a^2 x^2-2\right )\right )}{a^3 c^2 \left (n^4-10 n^2+9\right ) \sqrt {c-a^2 c x^2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^(n*ArcTanh[a*x])*x^2)/(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]*(-(a^2*n^3*x^2) - 3*a^3*x^3 + a*n^2*x*(2 +
a^2*x^2) + n*(-2 + 3*a^2*x^2)))/(a^3*c^2*(9 - 10*n^2 + n^4)*Sqrt[c - a^2*c*x^2]))

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fricas [A]  time = 0.50, size = 180, normalized size = 1.76 \[ -\frac {\sqrt {-a^{2} c x^{2} + c} {\left (2 \, a n^{2} x + {\left (a^{3} n^{2} - 3 \, a^{3}\right )} x^{3} - {\left (a^{2} n^{3} - 3 \, a^{2} n\right )} x^{2} - 2 \, n\right )} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{3} c^{3} n^{4} - 10 \, a^{3} c^{3} n^{2} + 9 \, a^{3} c^{3} + {\left (a^{7} c^{3} n^{4} - 10 \, a^{7} c^{3} n^{2} + 9 \, a^{7} c^{3}\right )} x^{4} - 2 \, {\left (a^{5} c^{3} n^{4} - 10 \, a^{5} c^{3} n^{2} + 9 \, a^{5} c^{3}\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*((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 = 96, normalized size = 0.94 \[ \frac {\left (a x -1\right ) \left (a x +1\right ) \left (a^{3} n^{2} x^{3}-a^{2} n^{3} x^{2}-3 x^{3} a^{3}+3 n \,x^{2} a^{2}+2 n^{2} x a -2 n \right ) {\mathrm e}^{n \arctanh \left (a x \right )}}{\left (n^{4}-10 n^{2}+9\right ) a^{3} \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*((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 = 1.23, size = 162, normalized size = 1.59 \[ \frac {{\mathrm {e}}^{\frac {n\,\ln \left (a\,x+1\right )}{2}-\frac {n\,\ln \left (1-a\,x\right )}{2}}\,\left (\frac {2\,n}{a^5\,c^2\,\left (n^4-10\,n^2+9\right )}-\frac {x^3\,\left (n^2-3\right )}{a^2\,c^2\,\left (n^4-10\,n^2+9\right )}-\frac {2\,n^2\,x}{a^4\,c^2\,\left (n^4-10\,n^2+9\right )}+\frac {n\,x^2\,\left (n^2-3\right )}{a^3\,c^2\,\left (n^4-10\,n^2+9\right )}\right )}{\frac {\sqrt {c-a^2\,c\,x^2}}{a^2}-x^2\,\sqrt {c-a^2\,c\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*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)*((2*n)/(a^5*c^2*(n^4 - 10*n^2 + 9)) - (x^3*(n^2 - 3))/(a^2*c^2*(
n^4 - 10*n^2 + 9)) - (2*n^2*x)/(a^4*c^2*(n^4 - 10*n^2 + 9)) + (n*x^2*(n^2 - 3))/(a^3*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 \[ \int \frac {x^{2} e^{n \operatorname {atanh}{\left (a x \right )}}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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