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

Optimal. Leaf size=80 \[ \frac {x^{m+1} \, _2F_1\left (\frac {5}{2},\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{c^2 (m+1)}+\frac {a x^{m+2} \, _2F_1\left (\frac {5}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{c^2 (m+2)} \]

[Out]

x^(1+m)*hypergeom([5/2, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/c^2/(1+m)+a*x^(2+m)*hypergeom([5/2, 1+1/2*m],[2+1/2*m]
,a^2*x^2)/c^2/(2+m)

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Rubi [A]  time = 0.11, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6148, 808, 364} \[ \frac {x^{m+1} \, _2F_1\left (\frac {5}{2},\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{c^2 (m+1)}+\frac {a x^{m+2} \, _2F_1\left (\frac {5}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{c^2 (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*Hypergeometric2F1[5/2, (1 + m)/2, (3 + m)/2, a^2*x^2])/(c^2*(1 + m)) + (a*x^(2 + m)*Hypergeometric2
F1[5/2, (2 + m)/2, (4 + m)/2, a^2*x^2])/(c^2*(2 + m))

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 6148

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

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)} x^m}{\left (c-a^2 c x^2\right )^2} \, dx &=\frac {\int \frac {x^m (1+a x)}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2}\\ &=\frac {\int \frac {x^m}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2}+\frac {a \int \frac {x^{1+m}}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2}\\ &=\frac {x^{1+m} \, _2F_1\left (\frac {5}{2},\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{c^2 (1+m)}+\frac {a x^{2+m} \, _2F_1\left (\frac {5}{2},\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{c^2 (2+m)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 82, normalized size = 1.02 \[ \frac {\frac {x^{m+1} \, _2F_1\left (\frac {5}{2},\frac {m+1}{2};\frac {m+1}{2}+1;a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \, _2F_1\left (\frac {5}{2},\frac {m+2}{2};\frac {m+2}{2}+1;a^2 x^2\right )}{m+2}}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((x^(1 + m)*Hypergeometric2F1[5/2, (1 + m)/2, 1 + (1 + m)/2, a^2*x^2])/(1 + m) + (a*x^(2 + m)*Hypergeometric2F
1[5/2, (2 + m)/2, 1 + (2 + m)/2, a^2*x^2])/(2 + m))/c^2

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} x^{m}}{a^{5} c^{2} x^{5} - a^{4} c^{2} x^{4} - 2 \, a^{3} c^{2} x^{3} + 2 \, a^{2} c^{2} x^{2} + a c^{2} x - c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-a^2*x^2 + 1)*x^m/(a^5*c^2*x^5 - a^4*c^2*x^4 - 2*a^3*c^2*x^3 + 2*a^2*c^2*x^2 + a*c^2*x - c^2),
x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.32, size = 0, normalized size = 0.00 \[ \int \frac {\left (a x +1\right ) x^{m}}{\sqrt {-a^{2} x^{2}+1}\, \left (-a^{2} c \,x^{2}+c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^m\,\left (a\,x+1\right )}{{\left (c-a^2\,c\,x^2\right )}^2\,\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x^{m}}{a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 2 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x x^{m}}{a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 2 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(x**m/(a**4*x**4*sqrt(-a**2*x**2 + 1) - 2*a**2*x**2*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x)
+ Integral(a*x*x**m/(a**4*x**4*sqrt(-a**2*x**2 + 1) - 2*a**2*x**2*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1))
, x))/c**2

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