3.1358 \(\int \frac {e^{n \tanh ^{-1}(a x)} x^m}{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=42 \[ \frac {x^{m+1} F_1\left (m+1;\frac {n+2}{2},1-\frac {n}{2};m+2;a x,-a x\right )}{c (m+1)} \]

[Out]

x^(1+m)*AppellF1(1+m,1+1/2*n,1-1/2*n,2+m,a*x,-a*x)/c/(1+m)

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Rubi [A]  time = 0.10, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6150, 133} \[ \frac {x^{m+1} F_1\left (m+1;\frac {n+2}{2},1-\frac {n}{2};m+2;a x,-a x\right )}{c (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*AppellF1[1 + m, (2 + n)/2, 1 - n/2, 2 + m, a*x, -(a*x)])/(c*(1 + m))

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 6150

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^m}{c-a^2 c x^2} \, dx &=\frac {\int x^m (1-a x)^{-1-\frac {n}{2}} (1+a x)^{-1+\frac {n}{2}} \, dx}{c}\\ &=\frac {x^{1+m} F_1\left (1+m;\frac {2+n}{2},1-\frac {n}{2};2+m;a x,-a x\right )}{c (1+m)}\\ \end {align*}

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Mathematica [B]  time = 0.24, size = 106, normalized size = 2.52 \[ \frac {x^m \left (e^{-2 \tanh ^{-1}(a x)}-1\right )^m \left (e^{-2 \tanh ^{-1}(a x)}+1\right )^m \left (-e^{-4 \tanh ^{-1}(a x)} \left (e^{2 \tanh ^{-1}(a x)}-1\right )^2\right )^{-m} e^{n \tanh ^{-1}(a x)} F_1\left (-\frac {n}{2};m,-m;1-\frac {n}{2};-e^{-2 \tanh ^{-1}(a x)},e^{-2 \tanh ^{-1}(a x)}\right )}{a c n} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(E^(n*ArcTanh[a*x])*(-1 + E^(-2*ArcTanh[a*x]))^m*(1 + E^(-2*ArcTanh[a*x]))^m*x^m*AppellF1[-1/2*n, m, -m, 1 - n
/2, -E^(-2*ArcTanh[a*x]), E^(-2*ArcTanh[a*x])])/(a*c*(-((-1 + E^(2*ArcTanh[a*x]))^2/E^(4*ArcTanh[a*x])))^m*n)

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {x^{m} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{2} c x^{2} - c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctanh \left (a x \right )} x^{m}}{-a^{2} c \,x^{2}+c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate(x^m*((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 = -0.02 \[ \int \frac {x^m\,{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}}{c-a^2\,c\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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