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

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)*Hypergeom
etric2F1[2, (2 + m)/2, (4 + m)/2, a^2*x^2])/(4*e^2*(2 + m))

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Rubi [A]  time = 0.0302674, 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, Rules used = {82, 73, 364} \[ \frac{a (e x)^{m+2} \, _2F_1\left (2,\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{4 e^2 (m+2)}+\frac{(e x)^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{4 e (m+1)} \]

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)*Hypergeom
etric2F1[2, (2 + m)/2, (4 + m)/2, a^2*x^2])/(4*e^2*(2 + m))

Rule 82

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 73

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]

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])

Rubi steps

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

Mathematica [A]  time = 0.0282506, size = 77, normalized size = 0.9 \[ \frac{x (e x)^m \left (a (m+1) x \, _2F_1\left (2,\frac{m}{2}+1;\frac{m}{2}+2;a^2 x^2\right )+(m+2) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )\right )}{4 (m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*(a*(1 + m)*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]))/(4*(1 + m)*(2 + m))

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Maple [F]  time = 0.047, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m}}{ \left ( -2\,ax+2 \right ) ^{2} \left ( ax+1 \right ) }}\, dx \end{align*}

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. \begin{align*} \frac{1}{4} \, \int \frac{\left (e x\right )^{m}}{{\left (a x + 1\right )}{\left (a x - 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m/(-2*a*x+2)^2/(a*x+1),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. \begin{align*}{\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m/(-2*a*x+2)^2/(a*x+1),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 [C]  time = 2.4456, size = 337, normalized size = 3.92 \begin{align*} \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 (1 - m\right ) - 16 a \Gamma \left (1 - m\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 (1 - m\right ) - 16 a \Gamma \left (1 - m\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 (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} + \frac{2 a e^{m} m x x^{m} \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (1 - m\right ) - 16 a \Gamma \left (1 - m\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 (1 - m\right ) - 16 a \Gamma \left (1 - m\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 (1 - m\right ) - 16 a \Gamma \left (1 - m\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 (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} \end{align*}

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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