3.14.50 \(\int \frac {e^{n \tanh ^{-1}(a x)} x}{(c-a^2 c x^2)^{5/2}} \, dx\) [1350]

Optimal. Leaf size=133 \[ \frac {e^{n \tanh ^{-1}(a x)}}{3 a^2 c \left (c-a^2 c x^2\right )^{3/2}}+\frac {e^{n \tanh ^{-1}(a x)} n (n-3 a x)}{3 a^2 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}+\frac {2 e^{n \tanh ^{-1}(a x)} n (n-a x)}{a^2 c^2 \left (9-10 n^2+n^4\right ) \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6280, 6271, 6270} \begin {gather*} \frac {2 n (n-a x) e^{n \tanh ^{-1}(a x)}}{a^2 c^2 \left (n^4-10 n^2+9\right ) \sqrt {c-a^2 c x^2}}+\frac {n (n-3 a x) e^{n \tanh ^{-1}(a x)}}{3 a^2 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}+\frac {e^{n \tanh ^{-1}(a x)}}{3 a^2 c \left (c-a^2 c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6270

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 6271

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 6280

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

Rubi steps

\begin {align*} \int \frac {e^{n \tanh ^{-1}(a x)} x}{\left (c-a^2 c x^2\right )^{5/2}} \, dx &=\frac {e^{n \tanh ^{-1}(a x)}}{3 a^2 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {n \int \frac {e^{n \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx}{3 a}\\ &=\frac {e^{n \tanh ^{-1}(a x)}}{3 a^2 c \left (c-a^2 c x^2\right )^{3/2}}+\frac {e^{n \tanh ^{-1}(a x)} n (n-3 a x)}{3 a^2 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}-\frac {(2 n) \int \frac {e^{n \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx}{a c \left (9-n^2\right )}\\ &=\frac {e^{n \tanh ^{-1}(a x)}}{3 a^2 c \left (c-a^2 c x^2\right )^{3/2}}+\frac {e^{n \tanh ^{-1}(a x)} n (n-3 a x)}{3 a^2 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}+\frac {2 e^{n \tanh ^{-1}(a x)} n (n-a x)}{a^2 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 = 0.23, size = 114, normalized size = 0.86 \begin {gather*} \frac {(1-a x)^{\frac {1}{2} (-3-n)} (1+a x)^{\frac {1}{2} (-3+n)} \sqrt {1-a^2 x^2} \left (3+a n^3 x+a n x \left (-3+2 a^2 x^2\right )-n^2 \left (1+2 a^2 x^2\right )\right )}{a^2 c^2 \left (9-10 n^2+n^4\right ) \sqrt {c-a^2 c x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^(n*ArcTanh[a*x])*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]*(3 + a*n^3*x + a*n*x*(-3 + 2*a^2*x^2) - n^2*(
1 + 2*a^2*x^2)))/(a^2*c^2*(9 - 10*n^2 + n^4)*Sqrt[c - a^2*c*x^2])

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Maple [A]
time = 0.05, size = 86, normalized size = 0.65

method	result	size
gosper	\(-\frac {\left (a x -1\right ) \left (a x +1\right ) \left (2 a^{3} x^{3} n -2 a^{2} n^{2} x^{2}+a \,n^{3} x -3 n a x -n^{2}+3\right ) {\mathrm e}^{n \arctanh \left (a x \right )}}{a^{2} \left (n^{4}-10 n^{2}+9\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*x-1)*(a*x+1)*(2*a^3*n*x^3-2*a^2*n^2*x^2+a*n^3*x-3*a*n*x-n^2+3)*exp(n*arctanh(a*x))/a^2/(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 172, normalized size = 1.29 \begin {gather*} \frac {{\left (2 \, a^{3} n x^{3} - 2 \, a^{2} n^{2} x^{2} - n^{2} + {\left (a n^{3} - 3 \, a n\right )} x + 3\right )} \sqrt {-a^{2} c x^{2} + c} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{2} c^{3} n^{4} - 10 \, a^{2} c^{3} n^{2} + 9 \, a^{2} c^{3} + {\left (a^{6} c^{3} n^{4} - 10 \, a^{6} c^{3} n^{2} + 9 \, a^{6} c^{3}\right )} x^{4} - 2 \, {\left (a^{4} c^{3} n^{4} - 10 \, a^{4} c^{3} n^{2} + 9 \, a^{4} c^{3}\right )} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x 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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*(-(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 = 1.19, size = 163, normalized size = 1.23 \begin {gather*} -\frac {{\mathrm {e}}^{\frac {n\,\ln \left (a\,x+1\right )}{2}-\frac {n\,\ln \left (1-a\,x\right )}{2}}\,\left (\frac {n^2-3}{a^4\,c^2\,\left (n^4-10\,n^2+9\right )}+\frac {2\,n^2\,x^2}{a^2\,c^2\,\left (n^4-10\,n^2+9\right )}-\frac {2\,n\,x^3}{a\,c^2\,\left (n^4-10\,n^2+9\right )}-\frac {n\,x\,\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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