3.1325 \(\int \frac {e^{n \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=127 \[ -\frac {(n-4 a x) e^{n \tanh ^{-1}(a x)}}{a c^3 \left (16-n^2\right ) \left (1-a^2 x^2\right )^2}-\frac {12 (n-2 a x) e^{n \tanh ^{-1}(a x)}}{a c^3 \left (4-n^2\right ) \left (16-n^2\right ) \left (1-a^2 x^2\right )}+\frac {24 e^{n \tanh ^{-1}(a x)}}{a c^3 n \left (n^4-20 n^2+64\right )} \]

[Out]

24*exp(n*arctanh(a*x))/a/c^3/n/(n^4-20*n^2+64)-exp(n*arctanh(a*x))*(-4*a*x+n)/a/c^3/(-n^2+16)/(-a^2*x^2+1)^2-1
2*exp(n*arctanh(a*x))*(-2*a*x+n)/a/c^3/(n^4-20*n^2+64)/(-a^2*x^2+1)

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Rubi [A]  time = 0.13, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6136, 6137} \[ -\frac {(n-4 a x) e^{n \tanh ^{-1}(a x)}}{a c^3 \left (16-n^2\right ) \left (1-a^2 x^2\right )^2}-\frac {12 (n-2 a x) e^{n \tanh ^{-1}(a x)}}{a c^3 \left (4-n^2\right ) \left (16-n^2\right ) \left (1-a^2 x^2\right )}+\frac {24 e^{n \tanh ^{-1}(a x)}}{a c^3 n \left (n^4-20 n^2+64\right )} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTanh[a*x])/(c - a^2*c*x^2)^3,x]

[Out]

(24*E^(n*ArcTanh[a*x]))/(a*c^3*n*(64 - 20*n^2 + n^4)) - (E^(n*ArcTanh[a*x])*(n - 4*a*x))/(a*c^3*(16 - n^2)*(1
- a^2*x^2)^2) - (12*E^(n*ArcTanh[a*x])*(n - 2*a*x))/(a*c^3*(4 - n^2)*(16 - n^2)*(1 - a^2*x^2))

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]

Rule 6137

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

Rubi steps

\begin {align*} \int \frac {e^{n \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx &=-\frac {e^{n \tanh ^{-1}(a x)} (n-4 a x)}{a c^3 \left (16-n^2\right ) \left (1-a^2 x^2\right )^2}+\frac {12 \int \frac {e^{n \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx}{c \left (16-n^2\right )}\\ &=-\frac {e^{n \tanh ^{-1}(a x)} (n-4 a x)}{a c^3 \left (16-n^2\right ) \left (1-a^2 x^2\right )^2}-\frac {12 e^{n \tanh ^{-1}(a x)} (n-2 a x)}{a c^3 \left (4-n^2\right ) \left (16-n^2\right ) \left (1-a^2 x^2\right )}+\frac {24 \int \frac {e^{n \tanh ^{-1}(a x)}}{c-a^2 c x^2} \, dx}{c^2 \left (64-20 n^2+n^4\right )}\\ &=\frac {24 e^{n \tanh ^{-1}(a x)}}{a c^3 n \left (64-20 n^2+n^4\right )}-\frac {e^{n \tanh ^{-1}(a x)} (n-4 a x)}{a c^3 \left (16-n^2\right ) \left (1-a^2 x^2\right )^2}-\frac {12 e^{n \tanh ^{-1}(a x)} (n-2 a x)}{a c^3 \left (4-n^2\right ) \left (16-n^2\right ) \left (1-a^2 x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 111, normalized size = 0.87 \[ \frac {(1-a x)^{-\frac {n}{2}-2} (a x+1)^{\frac {n}{2}-2} \left (4 n^2 \left (3 a^2 x^2-4\right )-8 a n x \left (3 a^2 x^2-5\right )+24 \left (a^2 x^2-1\right )^2-4 a n^3 x+n^4\right )}{a c^3 (n-4) (n-2) n (n+2) (n+4)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(n*ArcTanh[a*x])/(c - a^2*c*x^2)^3,x]

[Out]

((1 - a*x)^(-2 - n/2)*(1 + a*x)^(-2 + n/2)*(n^4 - 4*a*n^3*x + 24*(-1 + a^2*x^2)^2 - 8*a*n*x*(-5 + 3*a^2*x^2) +
4*n^2*(-4 + 3*a^2*x^2)))/(a*c^3*(-4 + n)*(-2 + n)*n*(2 + n)*(4 + n))

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fricas [A]  time = 0.55, size = 174, normalized size = 1.37 \[ \frac {{\left (24 \, a^{4} x^{4} - 24 \, a^{3} n x^{3} + n^{4} + 12 \, {\left (a^{2} n^{2} - 4 \, a^{2}\right )} x^{2} - 16 \, n^{2} - 4 \, {\left (a n^{3} - 10 \, a n\right )} x + 24\right )} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a c^{3} n^{5} - 20 \, a c^{3} n^{3} + 64 \, a c^{3} n + {\left (a^{5} c^{3} n^{5} - 20 \, a^{5} c^{3} n^{3} + 64 \, a^{5} c^{3} n\right )} x^{4} - 2 \, {\left (a^{3} c^{3} n^{5} - 20 \, a^{3} c^{3} n^{3} + 64 \, a^{3} c^{3} n\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/(-a^2*c*x^2+c)^3,x, algorithm="fricas")

[Out]

(24*a^4*x^4 - 24*a^3*n*x^3 + n^4 + 12*(a^2*n^2 - 4*a^2)*x^2 - 16*n^2 - 4*(a*n^3 - 10*a*n)*x + 24)*((a*x + 1)/(
a*x - 1))^(1/2*n)/(a*c^3*n^5 - 20*a*c^3*n^3 + 64*a*c^3*n + (a^5*c^3*n^5 - 20*a^5*c^3*n^3 + 64*a^5*c^3*n)*x^4 -
2*(a^3*c^3*n^5 - 20*a^3*c^3*n^3 + 64*a^3*c^3*n)*x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (a^{2} c x^{2} - c\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/(-a^2*c*x^2+c)^3,x, algorithm="giac")

[Out]

integrate(-((a*x + 1)/(a*x - 1))^(1/2*n)/(a^2*c*x^2 - c)^3, x)

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maple [A]  time = 0.03, size = 101, normalized size = 0.80 \[ \frac {\left (24 x^{4} a^{4}-24 x^{3} a^{3} n +12 a^{2} n^{2} x^{2}-4 a \,n^{3} x -48 a^{2} x^{2}+n^{4}+40 n a x -16 n^{2}+24\right ) {\mathrm e}^{n \arctanh \left (a x \right )}}{\left (a^{2} x^{2}-1\right )^{2} c^{3} a \left (n^{2}-16\right ) \left (n^{2}-4\right ) n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(a*x))/(-a^2*c*x^2+c)^3,x)

[Out]

(24*a^4*x^4-24*a^3*n*x^3+12*a^2*n^2*x^2-4*a*n^3*x-48*a^2*x^2+n^4+40*a*n*x-16*n^2+24)*exp(n*arctanh(a*x))/(a^2*
x^2-1)^2/c^3/a/(n^2-16)/(n^2-4)/n

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (a^{2} c x^{2} - c\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/(-a^2*c*x^2+c)^3,x, algorithm="maxima")

[Out]

-integrate(((a*x + 1)/(a*x - 1))^(1/2*n)/(a^2*c*x^2 - c)^3, x)

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mupad [B]  time = 1.39, size = 179, normalized size = 1.41 \[ \frac {{\left (a\,x+1\right )}^{n/2}\,\left (\frac {24\,x^4}{a\,c^3\,n\,\left (n^4-20\,n^2+64\right )}-\frac {4\,x\,\left (n^2-10\right )}{a^4\,c^3\,\left (n^4-20\,n^2+64\right )}-\frac {24\,x^3}{a^2\,c^3\,\left (n^4-20\,n^2+64\right )}+\frac {n^4-16\,n^2+24}{a^5\,c^3\,n\,\left (n^4-20\,n^2+64\right )}+\frac {x^2\,\left (12\,n^2-48\right )}{a^3\,c^3\,n\,\left (n^4-20\,n^2+64\right )}\right )}{{\left (1-a\,x\right )}^{n/2}\,\left (\frac {1}{a^4}+x^4-\frac {2\,x^2}{a^2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*atanh(a*x))/(c - a^2*c*x^2)^3,x)

[Out]

((a*x + 1)^(n/2)*((24*x^4)/(a*c^3*n*(n^4 - 20*n^2 + 64)) - (4*x*(n^2 - 10))/(a^4*c^3*(n^4 - 20*n^2 + 64)) - (2
4*x^3)/(a^2*c^3*(n^4 - 20*n^2 + 64)) + (n^4 - 16*n^2 + 24)/(a^5*c^3*n*(n^4 - 20*n^2 + 64)) + (x^2*(12*n^2 - 48
))/(a^3*c^3*n*(n^4 - 20*n^2 + 64))))/((1 - a*x)^(n/2)*(1/a^4 + x^4 - (2*x^2)/a^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(exp(n*atanh(a*x))/(-a**2*c*x**2+c)**3,x)

[Out]

Timed out

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