3.1128 \(\int \frac {e^{2 \tanh ^{-1}(a x)} x^m}{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=25 \[ \frac {x^{m+1} \, _2F_1(2,m+1;m+2;a x)}{c (m+1)} \]

[Out]

x^(1+m)*hypergeom([2, 1+m],[2+m],a*x)/c/(1+m)

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Rubi [A]  time = 0.08, antiderivative size = 25, 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, 64} \[ \frac {x^{m+1} \, _2F_1(2,m+1;m+2;a x)}{c (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*Hypergeometric2F1[2, 1 + m, 2 + m, a*x])/(c*(1 + m))

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 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^{2 \tanh ^{-1}(a x)} x^m}{c-a^2 c x^2} \, dx &=\frac {\int \frac {x^m}{(1-a x)^2} \, dx}{c}\\ &=\frac {x^{1+m} \, _2F_1(2,1+m;2+m;a x)}{c (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 25, normalized size = 1.00 \[ \frac {x^{m+1} \, _2F_1(2,m+1;m+2;a x)}{c (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*Hypergeometric2F1[2, 1 + m, 2 + m, a*x])/(c*(1 + m))

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{m}}{a^{2} c x^{2} - 2 \, a c x + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*x^m/(-a^2*c*x^2+c),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*x^m/(-a^2*c*x^2+c),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.42, size = 0, normalized size = 0.00 \[ \int \frac {\left (a x +1\right )^{2} x^{m}}{\left (-a^{2} x^{2}+1\right ) \left (-a^{2} c \,x^{2}+c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*x^m/(-a^2*c*x^2+c),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ -\int \frac {x^m\,{\left (a\,x+1\right )}^2}{\left (c-a^2\,c\,x^2\right )\,\left (a^2\,x^2-1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x^{m}}{a^{2} x^{2} - 2 a x + 1}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**m/(a**2*x**2 - 2*a*x + 1), x)/c

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