3.10.96 \(\int \frac {e^{\tanh ^{-1}(a x)} x^m}{(1-a^2 x^2)^{3/2}} \, dx\) [996]

Optimal. Leaf size=70 \[ \frac {x^{1+m} \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{1+m}+\frac {a x^{2+m} \, _2F_1\left (2,\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{2+m} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6285, 83, 74, 371} \begin {gather*} \frac {x^{m+1} \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \, _2F_1\left (2,\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{m+2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^m)/(1 - a^2*x^2)^(3/2),x]

[Out]

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

Rule 74

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

Rule 83

Int[((f_.)*(x_))^(p_.)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Dist[a, Int[(a + b*
x)^n*(c + d*x)^n*(f*x)^p, x], x] + Dist[b/f, Int[(a + b*x)^n*(c + d*x)^n*(f*x)^(p + 1), x], x] /; FreeQ[{a, b,
c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n - 1, 0] &&  !RationalQ[p] &&  !IGtQ[m, 0] && NeQ[m +
n + p + 2, 0]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 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^{\tanh ^{-1}(a x)} x^m}{\left (1-a^2 x^2\right )^{3/2}} \, dx &=\int \frac {x^m}{(1-a x)^2 (1+a x)} \, dx\\ &=a \int \frac {x^{1+m}}{(1-a x)^2 (1+a x)^2} \, dx+\int \frac {x^m}{(1-a x)^2 (1+a x)^2} \, dx\\ &=a \int \frac {x^{1+m}}{\left (1-a^2 x^2\right )^2} \, dx+\int \frac {x^m}{\left (1-a^2 x^2\right )^2} \, dx\\ &=\frac {x^{1+m} \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{1+m}+\frac {a x^{2+m} \, _2F_1\left (2,\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{2+m}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 67, normalized size = 0.96 \begin {gather*} x^{1+m} \left (\frac {a x \, _2F_1\left (2,1+\frac {m}{2};2+\frac {m}{2};a^2 x^2\right )}{2+m}+\frac {\, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{1+m}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^m)/(1 - a^2*x^2)^(3/2),x]

[Out]

x^(1 + m)*((a*x*Hypergeometric2F1[2, 1 + m/2, 2 + m/2, a^2*x^2])/(2 + m) + Hypergeometric2F1[2, (1 + m)/2, (3
+ m)/2, a^2*x^2]/(1 + m))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 5.
time = 0.06, size = 177, normalized size = 2.53

method	result	size
meijerg	\(-\frac {\left (-a^{2}\right )^{-\frac {m}{2}} \left (\frac {x^{m} \left (-a^{2}\right )^{\frac {m}{2}} \left (-2-m \right )}{\left (2+m \right ) \left (-a^{2} x^{2}+1\right )}+\frac {x^{m} \left (-a^{2}\right )^{\frac {m}{2}} m \Phi \left (a^{2} x^{2}, 1, \frac {m}{2}\right )}{2}\right )}{2 a}+\frac {\left (-a^{2}\right )^{-\frac {1}{2}-\frac {m}{2}} \left (-\frac {2 x^{1+m} \left (-a^{2}\right )^{\frac {1}{2}+\frac {m}{2}} \left (-1-m \right )}{\left (1+m \right ) \left (-2 a^{2} x^{2}+2\right )}+\frac {2 x^{1+m} \left (-a^{2}\right )^{\frac {1}{2}+\frac {m}{2}} \left (-\frac {m^{2}}{4}+\frac {1}{4}\right ) \Phi \left (a^{2} x^{2}, 1, \frac {1}{2}+\frac {m}{2}\right )}{1+m}\right )}{2}\)	\(177\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^2*x^m,x,method=_RETURNVERBOSE)

[Out]

-1/2/a*(-a^2)^(-1/2*m)*(1/(2+m)*x^m*(-a^2)^(1/2*m)*(-2-m)/(-a^2*x^2+1)+1/2*x^m*(-a^2)^(1/2*m)*m*LerchPhi(a^2*x
^2,1,1/2*m))+1/2*(-a^2)^(-1/2-1/2*m)*(-2/(1+m)*x^(1+m)*(-a^2)^(1/2+1/2*m)*(-1-m)/(-2*a^2*x^2+2)+2/(1+m)*x^(1+m
)*(-a^2)^(1/2+1/2*m)*(-1/4*m^2+1/4)*LerchPhi(a^2*x^2,1,1/2+1/2*m))

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

[Out]

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

[Out]

integral(x^m/(a^3*x^3 - a^2*x^2 - a*x + 1), x)

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Sympy [C] Result contains complex when optimal does not.
time = 3.04, size = 673, normalized size = 9.61 \begin {gather*} - \frac {a^{2} m^{2} x^{3} x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) - 8 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {a^{2} x^{3} x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) - 8 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + a \left (- \frac {a^{2} m^{2} x^{4} x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + 1\right ) \Gamma \left (\frac {m}{2} + 1\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + 2\right ) - 8 \Gamma \left (\frac {m}{2} + 2\right )} - \frac {2 a^{2} m x^{4} x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + 1\right ) \Gamma \left (\frac {m}{2} + 1\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + 2\right ) - 8 \Gamma \left (\frac {m}{2} + 2\right )} + \frac {m^{2} x^{2} x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + 1\right ) \Gamma \left (\frac {m}{2} + 1\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + 2\right ) - 8 \Gamma \left (\frac {m}{2} + 2\right )} + \frac {2 m x^{2} x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + 1\right ) \Gamma \left (\frac {m}{2} + 1\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + 2\right ) - 8 \Gamma \left (\frac {m}{2} + 2\right )} - \frac {2 m x^{2} x^{m} \Gamma \left (\frac {m}{2} + 1\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + 2\right ) - 8 \Gamma \left (\frac {m}{2} + 2\right )} - \frac {4 x^{2} x^{m} \Gamma \left (\frac {m}{2} + 1\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + 2\right ) - 8 \Gamma \left (\frac {m}{2} + 2\right )}\right ) + \frac {m^{2} x x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) - 8 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} - \frac {2 m x x^{m} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) - 8 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} - \frac {x x^{m} \Phi \left (a^{2} x^{2} e^{2 i \pi }, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) - 8 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} - \frac {2 x x^{m} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) - 8 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*m**2*x**3*x**m*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m
/2 + 3/2) - 8*gamma(m/2 + 3/2)) + a**2*x**3*x**m*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2
+ 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/2)) + a*(-a**2*m**2*x**4*x**m*lerchphi(a**2*x**2*exp_p
olar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2)) - 2*a**2*m*x**4*x**m*
lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2
)) + m**2*x**2*x**m*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 +
2) - 8*gamma(m/2 + 2)) + 2*m*x**2*x**m*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**
2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2)) - 2*m*x**2*x**m*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamma
(m/2 + 2)) - 4*x**2*x**m*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2))) + m**2*x*x**m*lerchph
i(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/
2)) - 2*m*x*x**m*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/2)) - x*x**m*lerchphi(a**2*x
**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/2)) - 2*
x*x**m*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/2))

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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