3.14.13 \(\int \frac {e^{n \tanh ^{-1}(a x)} x^2}{c-a^2 c x^2} \, dx\) [1313]

Optimal. Leaf size=86 \[ \frac {e^{n \tanh ^{-1}(a x)}}{a^3 c n}+\frac {2^{1+\frac {n}{2}} (1-a x)^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},-\frac {n}{2};2-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{a^3 c (2-n)} \]

[Out]

exp(n*arctanh(a*x))/a^3/c/n+2^(1+1/2*n)*(-a*x+1)^(1-1/2*n)*hypergeom([-1/2*n, 1-1/2*n],[2-1/2*n],-1/2*a*x+1/2)
/a^3/c/(2-n)

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Rubi [A]
time = 0.09, antiderivative size = 127, normalized size of antiderivative = 1.48, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6285, 92, 80, 71} \begin {gather*} \frac {2^{\frac {n}{2}+1} (1-a x)^{-n/2} \, _2F_1\left (-\frac {n}{2},-\frac {n}{2};1-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{a^3 c}+\frac {(1-n) (a x+1)^{n/2} (1-a x)^{-n/2}}{a^3 c n}-\frac {x (a x+1)^{n/2} (1-a x)^{-n/2}}{a^2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - n)*(1 + a*x)^(n/2))/(a^3*c*n*(1 - a*x)^(n/2)) - (x*(1 + a*x)^(n/2))/(a^2*c*(1 - a*x)^(n/2)) + (2^(1 + n/
2)*Hypergeometric2F1[-1/2*n, -1/2*n, 1 - n/2, (1 - a*x)/2])/(a^3*c*(1 - a*x)^(n/2))

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c
, d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

Rule 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 6285

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

Rubi steps

\begin {align*} \int \frac {e^{n \tanh ^{-1}(a x)} x^2}{c-a^2 c x^2} \, dx &=\frac {\int x^2 (1-a x)^{-1-\frac {n}{2}} (1+a x)^{-1+\frac {n}{2}} \, dx}{c}\\ &=-\frac {x (1-a x)^{-n/2} (1+a x)^{n/2}}{a^2 c}-\frac {\int (1-a x)^{-1-\frac {n}{2}} (1+a x)^{-1+\frac {n}{2}} (-1-a n x) \, dx}{a^2 c}\\ &=\frac {(1-n) (1-a x)^{-n/2} (1+a x)^{n/2}}{a^3 c n}-\frac {x (1-a x)^{-n/2} (1+a x)^{n/2}}{a^2 c}+\frac {n \int (1-a x)^{-1-\frac {n}{2}} (1+a x)^{n/2} \, dx}{a^2 c}\\ &=\frac {(1-n) (1-a x)^{-n/2} (1+a x)^{n/2}}{a^3 c n}-\frac {x (1-a x)^{-n/2} (1+a x)^{n/2}}{a^2 c}+\frac {2^{1+\frac {n}{2}} (1-a x)^{-n/2} \, _2F_1\left (-\frac {n}{2},-\frac {n}{2};1-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{a^3 c}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 63, normalized size = 0.73 \begin {gather*} \frac {e^{n \tanh ^{-1}(a x)} \left (2+n-4 e^{2 \tanh ^{-1}(a x)} n \, _2F_1\left (2,1+\frac {n}{2};2+\frac {n}{2};-e^{2 \tanh ^{-1}(a x)}\right )\right )}{a^3 c n (2+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(n*ArcTanh[a*x])*(2 + n - 4*E^(2*ArcTanh[a*x])*n*Hypergeometric2F1[2, 1 + n/2, 2 + n/2, -E^(2*ArcTanh[a*x])
]))/(a^3*c*n*(2 + n))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {{\mathrm e}^{n \arctanh \left (a x \right )} x^{2}}{-a^{2} c \,x^{2}+c}\, dx\]

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),x)

[Out]

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

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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^2/(-a^2*c*x^2+c),x, algorithm="maxima")

[Out]

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

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Fricas [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^2/(-a^2*c*x^2+c),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {x^{2} e^{n \operatorname {atanh}{\left (a x \right )}}}{a^{2} x^{2} - 1}\, dx}{c} \end {gather*}

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),x)

[Out]

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

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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^2/(-a^2*c*x^2+c),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}}{c-a^2\,c\,x^2} \,d x \end {gather*}

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),x)

[Out]

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

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