\(\int \frac {x^m}{(a+b x)^2} \, dx\) [705]

Optimal result

Integrand size = 11, antiderivative size = 29 \[ \int \frac {x^m}{(a+b x)^2} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {b x}{a}\right )}{a^2 (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {66} \[ \int \frac {x^m}{(a+b x)^2} \, dx=\frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,-\frac {b x}{a}\right )}{a^2 (m+1)} \]

[In]

Int[x^m/(a + b*x)^2,x]

[Out]

(x^(1 + m)*Hypergeometric2F1[2, 1 + m, 2 + m, -((b*x)/a)])/(a^2*(1 + m))

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))

Rubi steps \begin{align*} \text {integral}& = \frac {x^{1+m} \, _2F_1\left (2,1+m;2+m;-\frac {b x}{a}\right )}{a^2 (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {x^m}{(a+b x)^2} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {b x}{a}\right )}{a^2 (1+m)} \]

[In]

Integrate[x^m/(a + b*x)^2,x]

[Out]

(x^(1 + m)*Hypergeometric2F1[2, 1 + m, 2 + m, -((b*x)/a)])/(a^2*(1 + m))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

integrate(x^m/(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(x^m/(b^2*x^2 + 2*a*b*x + a^2), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.58 (sec) , antiderivative size = 262, normalized size of antiderivative = 9.03 \[ \int \frac {x^m}{(a+b x)^2} \, dx=- \frac {a m^{2} x^{m + 1} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} - \frac {a m x^{m + 1} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} + \frac {a m x^{m + 1} \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} + \frac {a x^{m + 1} \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} - \frac {b m^{2} x x^{m + 1} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} - \frac {b m x x^{m + 1} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} \]

[In]

integrate(x**m/(b*x+a)**2,x)

[Out]

-a*m**2*x**(m + 1)*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/(a**3*gamma(m + 2) + a**2*b*x*gamma(
m + 2)) - a*m*x**(m + 1)*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/(a**3*gamma(m + 2) + a**2*b*x*
gamma(m + 2)) + a*m*x**(m + 1)*gamma(m + 1)/(a**3*gamma(m + 2) + a**2*b*x*gamma(m + 2)) + a*x**(m + 1)*gamma(m
+ 1)/(a**3*gamma(m + 2) + a**2*b*x*gamma(m + 2)) - b*m**2*x*x**(m + 1)*lerchphi(b*x*exp_polar(I*pi)/a, 1, m +
1)*gamma(m + 1)/(a**3*gamma(m + 2) + a**2*b*x*gamma(m + 2)) - b*m*x*x**(m + 1)*lerchphi(b*x*exp_polar(I*pi)/a
, 1, m + 1)*gamma(m + 1)/(a**3*gamma(m + 2) + a**2*b*x*gamma(m + 2))

Maxima [F]

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

[In]

integrate(x^m/(b*x+a)^2,x, algorithm="maxima")

[Out]

integrate(x^m/(b*x + a)^2, x)

Giac [F]

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

[In]

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

[Out]

integrate(x^m/(b*x + a)^2, x)

Mupad [F(-1)]

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

[In]

int(x^m/(a + b*x)^2,x)

[Out]

int(x^m/(a + b*x)^2, x)