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

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 = 0.0197647, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {82, 73, 364} \[ \frac{x^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{m+1}+\frac{a x^{m+2} \, _2F_1\left (2,\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{m+2} \]

Antiderivative was successfully verified.

[In]

Int[x^m/((1 - a*x)^2*(1 + a*x)),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 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{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*}

Mathematica [A]  time = 0.0060584, size = 67, normalized size = 0.96 \[ x^{m+1} \left (\frac{a x \, _2F_1\left (2,\frac{m}{2}+1;\frac{m}{2}+2;a^2 x^2\right )}{m+2}+\frac{\, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{m+1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^m/((1 - a*x)^2*(1 + a*x)),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 [F]  time = 0.046, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{m}}{ \left ( -ax+1 \right ) ^{2} \left ( ax+1 \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{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(x^m/(-a*x+1)^2/(a*x+1),x, algorithm="maxima")

[Out]

integrate(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{x^{m}}{a^{3} x^{3} - a^{2} x^{2} - a x + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(-a*x+1)^2/(a*x+1),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]  time = 2.35647, size = 313, normalized size = 4.47 \begin{align*} \frac{2 a m^{2} x x^{m} \Phi \left (\frac{1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{4 a^{2} x \Gamma \left (1 - m\right ) - 4 a \Gamma \left (1 - m\right )} - \frac{a m x x^{m} \Phi \left (\frac{1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{4 a^{2} x \Gamma \left (1 - m\right ) - 4 a \Gamma \left (1 - m\right )} + \frac{a m x x^{m} \Phi \left (\frac{e^{i \pi }}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{4 a^{2} x \Gamma \left (1 - m\right ) - 4 a \Gamma \left (1 - m\right )} + \frac{2 a m x x^{m} \Gamma \left (- m\right )}{4 a^{2} x \Gamma \left (1 - m\right ) - 4 a \Gamma \left (1 - m\right )} - \frac{2 m^{2} x^{m} \Phi \left (\frac{1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{4 a^{2} x \Gamma \left (1 - m\right ) - 4 a \Gamma \left (1 - m\right )} + \frac{m x^{m} \Phi \left (\frac{1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{4 a^{2} x \Gamma \left (1 - m\right ) - 4 a \Gamma \left (1 - m\right )} - \frac{m x^{m} \Phi \left (\frac{e^{i \pi }}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{4 a^{2} x \Gamma \left (1 - m\right ) - 4 a \Gamma \left (1 - m\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*m**2*x*x**m*lerchphi(1/(a*x), 1, m*exp_polar(I*pi))*gamma(-m)/(4*a**2*x*gamma(1 - m) - 4*a*gamma(1 - m)) -
a*m*x*x**m*lerchphi(1/(a*x), 1, m*exp_polar(I*pi))*gamma(-m)/(4*a**2*x*gamma(1 - m) - 4*a*gamma(1 - m)) + a*m
*x*x**m*lerchphi(exp_polar(I*pi)/(a*x), 1, m*exp_polar(I*pi))*gamma(-m)/(4*a**2*x*gamma(1 - m) - 4*a*gamma(1 -
m)) + 2*a*m*x*x**m*gamma(-m)/(4*a**2*x*gamma(1 - m) - 4*a*gamma(1 - m)) - 2*m**2*x**m*lerchphi(1/(a*x), 1, m*
exp_polar(I*pi))*gamma(-m)/(4*a**2*x*gamma(1 - m) - 4*a*gamma(1 - m)) + m*x**m*lerchphi(1/(a*x), 1, m*exp_pola
r(I*pi))*gamma(-m)/(4*a**2*x*gamma(1 - m) - 4*a*gamma(1 - m)) - m*x**m*lerchphi(exp_polar(I*pi)/(a*x), 1, m*ex
p_polar(I*pi))*gamma(-m)/(4*a**2*x*gamma(1 - m) - 4*a*gamma(1 - m))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{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(x^m/(-a*x+1)^2/(a*x+1),x, algorithm="giac")

[Out]

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