\(\int \frac {x^m}{(1-a x)^2 (1+a x)} \, dx\) [499]

Optimal result

Integrand size = 19, antiderivative size = 70 \[ \int \frac {x^m}{(1-a x)^2 (1+a x)} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{1+m}+\frac {a x^{2+m} \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{2+m} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {83, 74, 371} \[ \int \frac {x^m}{(1-a x)^2 (1+a x)} \, dx=\frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (2,\frac {m+1}{2},\frac {m+3}{2},a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \operatorname {Hypergeometric2F1}\left (2,\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{m+2} \]

[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 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*} \text {integral}& = 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] (verified)

Time = 0.00 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.96 \[ \int \frac {x^m}{(1-a x)^2 (1+a x)} \, dx=x^{1+m} \left (\frac {a x \operatorname {Hypergeometric2F1}\left (2,1+\frac {m}{2},2+\frac {m}{2},a^2 x^2\right )}{2+m}+\frac {\operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{1+m}\right ) \]

[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))

Maple [F]

\[\int \frac {x^{m}}{\left (-a x +1\right )^{2} \left (a x +1\right )}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.35 (sec) , antiderivative size = 313, normalized size of antiderivative = 4.47 \[ \int \frac {x^m}{(1-a x)^2 (1+a x)} \, dx=\frac {2 a m^{2} x x^{m} \Phi \left (\frac {1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{4 a^{2} x \Gamma \left (1 - m\right ) - 4 a \Gamma \left (1 - m\right )} - \frac {a m x x^{m} \Phi \left (\frac {1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{4 a^{2} x \Gamma \left (1 - m\right ) - 4 a \Gamma \left (1 - m\right )} + \frac {a m x x^{m} \Phi \left (\frac {e^{i \pi }}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{4 a^{2} x \Gamma \left (1 - m\right ) - 4 a \Gamma \left (1 - m\right )} + \frac {2 a m x x^{m} \Gamma \left (- m\right )}{4 a^{2} x \Gamma \left (1 - m\right ) - 4 a \Gamma \left (1 - m\right )} - \frac {2 m^{2} x^{m} \Phi \left (\frac {1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{4 a^{2} x \Gamma \left (1 - m\right ) - 4 a \Gamma \left (1 - m\right )} + \frac {m x^{m} \Phi \left (\frac {1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{4 a^{2} x \Gamma \left (1 - m\right ) - 4 a \Gamma \left (1 - m\right )} - \frac {m x^{m} \Phi \left (\frac {e^{i \pi }}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{4 a^{2} x \Gamma \left (1 - m\right ) - 4 a \Gamma \left (1 - m\right )} \]

[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))

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^m}{(1-a x)^2 (1+a x)} \, dx=\int \frac {x^m}{{\left (a\,x-1\right )}^2\,\left (a\,x+1\right )} \,d x \]

[In]

int(x^m/((a*x - 1)^2*(a*x + 1)),x)

[Out]

int(x^m/((a*x - 1)^2*(a*x + 1)), x)