3.498 \(\int \frac{x^m (1+a x)}{\left (1-a^2 x^2\right )^2} \, dx\)

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

[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)

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Rubi [A]  time = 0.117214, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19 \[ \frac{x^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{m+1}+\frac{a x^{m+2} \, _2F_1\left (2,\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{m+2} \]

Antiderivative was successfully verified.

[In]  Int[(x^m*(1 + a*x))/(1 - a^2*x^2)^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)

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Rubi in Sympy [A]  time = 13.8242, size = 51, normalized size = 0.73 \[ \frac{a x^{m + 2}{{}_{2}F_{1}\left (\begin{matrix} 2, \frac{m}{2} + 1 \\ \frac{m}{2} + 2 \end{matrix}\middle |{a^{2} x^{2}} \right )}}{m + 2} + \frac{x^{m + 1}{{}_{2}F_{1}\left (\begin{matrix} 2, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{a^{2} x^{2}} \right )}}{m + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**m*(a*x+1)/(-a**2*x**2+1)**2,x)

[Out]

a*x**(m + 2)*hyper((2, m/2 + 1), (m/2 + 2,), a**2*x**2)/(m + 2) + x**(m + 1)*hyp
er((2, m/2 + 1/2), (m/2 + 3/2,), a**2*x**2)/(m + 1)

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Mathematica [A]  time = 0.0622908, size = 51, normalized size = 0.73 \[ \frac{x^{m+1} (\, _2F_1(1,m+1;m+2;-a x)+\, _2F_1(1,m+1;m+2;a x)+2 \, _2F_1(2,m+1;m+2;a x))}{4 (m+1)} \]

Antiderivative was successfully verified.

[In]  Integrate[(x^m*(1 + a*x))/(1 - a^2*x^2)^2,x]

[Out]

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

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Maple [C]  time = 0.17, size = 177, normalized size = 2.5 \[ -{\frac{1}{2\,a} \left ( -{a}^{2} \right ) ^{-{\frac{m}{2}}} \left ({\frac{{x}^{m} \left ( -2-m \right ) }{ \left ( 2+m \right ) \left ( -{a}^{2}{x}^{2}+1 \right ) } \left ( -{a}^{2} \right ) ^{{\frac{m}{2}}}}+{\frac{{x}^{m}m}{2} \left ( -{a}^{2} \right ) ^{{\frac{m}{2}}}{\it LerchPhi} \left ({a}^{2}{x}^{2},1,{\frac{m}{2}} \right ) } \right ) }+{\frac{1}{2} \left ( -{a}^{2} \right ) ^{-{\frac{1}{2}}-{\frac{m}{2}}} \left ( -2\,{\frac{{x}^{1+m} \left ( -{a}^{2} \right ) ^{1/2+m/2} \left ( -1-m \right ) }{ \left ( 1+m \right ) \left ( -2\,{a}^{2}{x}^{2}+2 \right ) }}+2\,{\frac{{x}^{1+m} \left ( -{a}^{2} \right ) ^{1/2+m/2} \left ( -1/4\,{m}^{2}+1/4 \right ){\it LerchPhi} \left ({a}^{2}{x}^{2},1,1/2+m/2 \right ) }{1+m}} \right ) } \]

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]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((a*x + 1)*x^m/(a^2*x^2 - 1)^2,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., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{x^{m}}{a^{3} x^{3} - a^{2} x^{2} - a x + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 12.9497, size = 673, normalized size = 9.61 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x**m*(a*x+1)/(-a**2*x**2+1)**2,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*l
erchphi(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_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*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*ex
p_polar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamm
a(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*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*m*x*x**m*gam
ma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/2)) - x*x**m*lerch
phi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**2*x**2*gam
ma(m/2 + 3/2) - 8*gamma(m/2 + 3/2)) - 2*x*x**m*gamma(m/2 + 1/2)/(8*a**2*x**2*gam
ma(m/2 + 3/2) - 8*gamma(m/2 + 3/2))

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (a x + 1\right )} x^{m}}{{\left (a^{2} x^{2} - 1\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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