3.358 \(\int \frac{(e x)^m}{(2-2 a x)^2 (1+a x)} \, dx\)

Optimal. Leaf size=86 \[ \frac{a (e x)^{m+2} \, _2F_1\left (2,\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{4 e^2 (m+2)}+\frac{(e x)^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{4 e (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.126994, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143 \[ \frac{a (e x)^{m+2} \, _2F_1\left (2,\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{4 e^2 (m+2)}+\frac{(e x)^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{4 e (m+1)} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 14.5979, size = 63, normalized size = 0.73 \[ \frac{a \left (e x\right )^{m + 2}{{}_{2}F_{1}\left (\begin{matrix} 2, \frac{m}{2} + 1 \\ \frac{m}{2} + 2 \end{matrix}\middle |{a^{2} x^{2}} \right )}}{4 e^{2} \left (m + 2\right )} + \frac{\left (e x\right )^{m + 1}{{}_{2}F_{1}\left (\begin{matrix} 2, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{a^{2} x^{2}} \right )}}{4 e \left (m + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

a*(e*x)**(m + 2)*hyper((2, m/2 + 1), (m/2 + 2,), a**2*x**2)/(4*e**2*(m + 2)) + (
e*x)**(m + 1)*hyper((2, m/2 + 1/2), (m/2 + 3/2,), a**2*x**2)/(4*e*(m + 1))

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Mathematica [A]  time = 0.0557756, size = 52, normalized size = 0.6 \[ \frac{x (e x)^m (\, _2F_1(1,m+1;m+2;-a x)+\, _2F_1(1,m+1;m+2;a x)+2 \, _2F_1(2,m+1;m+2;a x))}{16 (m+1)} \]

Antiderivative was successfully verified.

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

[Out]

(x*(e*x)^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]))/(16*(1 + m))

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Maple [F]  time = 0.069, size = 0, normalized size = 0. \[ \int{\frac{ \left ( ex \right ) ^{m}}{ \left ( -2\,ax+2 \right ) ^{2} \left ( ax+1 \right ) }}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \frac{1}{4} \, \int \frac{\left (e x\right )^{m}}{{\left (a x + 1\right )}{\left (a x - 1\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/4*(e*x)^m/((a*x + 1)*(a*x - 1)^2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{\left (e x\right )^{m}}{4 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/4*(e*x)^m/((a*x + 1)*(a*x - 1)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 7.67705, size = 337, normalized size = 3.92 \[ \frac{2 a e^{m} m^{2} x x^{m} \Phi \left (\frac{1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (- m + 1\right ) - 16 a \Gamma \left (- m + 1\right )} - \frac{a e^{m} m x x^{m} \Phi \left (\frac{1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (- m + 1\right ) - 16 a \Gamma \left (- m + 1\right )} + \frac{a e^{m} m x x^{m} \Phi \left (\frac{e^{i \pi }}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (- m + 1\right ) - 16 a \Gamma \left (- m + 1\right )} + \frac{2 a e^{m} m x x^{m} \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (- m + 1\right ) - 16 a \Gamma \left (- m + 1\right )} - \frac{2 e^{m} m^{2} x^{m} \Phi \left (\frac{1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (- m + 1\right ) - 16 a \Gamma \left (- m + 1\right )} + \frac{e^{m} m x^{m} \Phi \left (\frac{1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (- m + 1\right ) - 16 a \Gamma \left (- m + 1\right )} - \frac{e^{m} m x^{m} \Phi \left (\frac{e^{i \pi }}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (- m + 1\right ) - 16 a \Gamma \left (- m + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

2*a*e**m*m**2*x*x**m*lerchphi(1/(a*x), 1, m*exp_polar(I*pi))*gamma(-m)/(16*a**2*
x*gamma(-m + 1) - 16*a*gamma(-m + 1)) - a*e**m*m*x*x**m*lerchphi(1/(a*x), 1, m*e
xp_polar(I*pi))*gamma(-m)/(16*a**2*x*gamma(-m + 1) - 16*a*gamma(-m + 1)) + a*e**
m*m*x*x**m*lerchphi(exp_polar(I*pi)/(a*x), 1, m*exp_polar(I*pi))*gamma(-m)/(16*a
**2*x*gamma(-m + 1) - 16*a*gamma(-m + 1)) + 2*a*e**m*m*x*x**m*gamma(-m)/(16*a**2
*x*gamma(-m + 1) - 16*a*gamma(-m + 1)) - 2*e**m*m**2*x**m*lerchphi(1/(a*x), 1, m
*exp_polar(I*pi))*gamma(-m)/(16*a**2*x*gamma(-m + 1) - 16*a*gamma(-m + 1)) + e**
m*m*x**m*lerchphi(1/(a*x), 1, m*exp_polar(I*pi))*gamma(-m)/(16*a**2*x*gamma(-m +
1) - 16*a*gamma(-m + 1)) - e**m*m*x**m*lerchphi(exp_polar(I*pi)/(a*x), 1, m*exp
_polar(I*pi))*gamma(-m)/(16*a**2*x*gamma(-m + 1) - 16*a*gamma(-m + 1))

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\left (e x\right )^{m}}{4 \,{\left (a x + 1\right )}{\left (a x - 1\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/4*(e*x)^m/((a*x + 1)*(a*x - 1)^2),x, algorithm="giac")

[Out]

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