3.1.59 \(\int \frac {(e x)^m}{(2-2 a x)^2 (1+a x)} \, dx\) [59]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 86, normalized size of antiderivative = 1.00, 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 = {83, 74, 371} \begin {gather*} \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)} \end {gather*}

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 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])

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*}

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Mathematica [A]
time = 0.08, size = 77, normalized size = 0.90 \begin {gather*} \frac {x (e x)^m \left (a (1+m) x \, _2F_1\left (2,1+\frac {m}{2};2+\frac {m}{2};a^2 x^2\right )+(2+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )\right )}{4 (1+m) (2+m)} \end {gather*}

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 = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m}}{\left (-2 a x +2\right )^{2} \left (a x +1\right )}\, dx\]

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 = 0.00, 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((e*x)^m/(-2*a*x+2)^2/(a*x+1),x, algorithm="maxima")

[Out]

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

[Out]

integral(1/4*(x*e)^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 = 1.28, size = 337, normalized size = 3.92 \begin {gather*} \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 {gather*}

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 = 0.00, 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((e*x)^m/(-2*a*x+2)^2/(a*x+1),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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