3.706 \(\int \frac {x^m}{(a+b x)^3} \, dx\)

Optimal. Leaf size=29 \[ \frac {x^{m+1} \, _2F_1\left (3,m+1;m+2;-\frac {b x}{a}\right )}{a^3 (m+1)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {64} \[ \frac {x^{m+1} \, _2F_1\left (3,m+1;m+2;-\frac {b x}{a}\right )}{a^3 (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^m/(a + b*x)^3,x]

[Out]

(x^(1 + m)*Hypergeometric2F1[3, 1 + m, 2 + m, -((b*x)/a)])/(a^3*(1 + m))

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), 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*} \int \frac {x^m}{(a+b x)^3} \, dx &=\frac {x^{1+m} \, _2F_1\left (3,1+m;2+m;-\frac {b x}{a}\right )}{a^3 (1+m)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 29, normalized size = 1.00 \[ \frac {x^{m+1} \, _2F_1\left (3,m+1;m+2;-\frac {b x}{a}\right )}{a^3 (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m/(a + b*x)^3,x]

[Out]

(x^(1 + m)*Hypergeometric2F1[3, 1 + m, 2 + m, -((b*x)/a)])/(a^3*(1 + m))

________________________________________________________________________________________

fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{m}}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(b*x+a)^3,x, algorithm="fricas")

[Out]

integral(x^m/(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{{\left (b x + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(x^m/(b*x + a)^3, x)

________________________________________________________________________________________

maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{\left (b x +a \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(b*x+a)^3,x)

[Out]

int(x^m/(b*x+a)^3,x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{{\left (b x + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(b*x+a)^3,x, algorithm="maxima")

[Out]

integrate(x^m/(b*x + a)^3, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {x^m}{{\left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(a + b*x)^3,x)

[Out]

int(x^m/(a + b*x)^3, x)

________________________________________________________________________________________

sympy [C]  time = 1.45, size = 717, normalized size = 24.72 \[ \frac {a^{2} m^{3} x x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{2 a^{5} \Gamma \left (m + 2\right ) + 4 a^{4} b x \Gamma \left (m + 2\right ) + 2 a^{3} b^{2} x^{2} \Gamma \left (m + 2\right )} - \frac {a^{2} m^{2} x x^{m} \Gamma \left (m + 1\right )}{2 a^{5} \Gamma \left (m + 2\right ) + 4 a^{4} b x \Gamma \left (m + 2\right ) + 2 a^{3} b^{2} x^{2} \Gamma \left (m + 2\right )} - \frac {a^{2} m x x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{2 a^{5} \Gamma \left (m + 2\right ) + 4 a^{4} b x \Gamma \left (m + 2\right ) + 2 a^{3} b^{2} x^{2} \Gamma \left (m + 2\right )} + \frac {a^{2} m x x^{m} \Gamma \left (m + 1\right )}{2 a^{5} \Gamma \left (m + 2\right ) + 4 a^{4} b x \Gamma \left (m + 2\right ) + 2 a^{3} b^{2} x^{2} \Gamma \left (m + 2\right )} + \frac {2 a^{2} x x^{m} \Gamma \left (m + 1\right )}{2 a^{5} \Gamma \left (m + 2\right ) + 4 a^{4} b x \Gamma \left (m + 2\right ) + 2 a^{3} b^{2} x^{2} \Gamma \left (m + 2\right )} + \frac {2 a b m^{3} x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{2 a^{5} \Gamma \left (m + 2\right ) + 4 a^{4} b x \Gamma \left (m + 2\right ) + 2 a^{3} b^{2} x^{2} \Gamma \left (m + 2\right )} - \frac {a b m^{2} x^{2} x^{m} \Gamma \left (m + 1\right )}{2 a^{5} \Gamma \left (m + 2\right ) + 4 a^{4} b x \Gamma \left (m + 2\right ) + 2 a^{3} b^{2} x^{2} \Gamma \left (m + 2\right )} - \frac {2 a b m x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{2 a^{5} \Gamma \left (m + 2\right ) + 4 a^{4} b x \Gamma \left (m + 2\right ) + 2 a^{3} b^{2} x^{2} \Gamma \left (m + 2\right )} + \frac {a b x^{2} x^{m} \Gamma \left (m + 1\right )}{2 a^{5} \Gamma \left (m + 2\right ) + 4 a^{4} b x \Gamma \left (m + 2\right ) + 2 a^{3} b^{2} x^{2} \Gamma \left (m + 2\right )} + \frac {b^{2} m^{3} x^{3} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{2 a^{5} \Gamma \left (m + 2\right ) + 4 a^{4} b x \Gamma \left (m + 2\right ) + 2 a^{3} b^{2} x^{2} \Gamma \left (m + 2\right )} - \frac {b^{2} m x^{3} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{2 a^{5} \Gamma \left (m + 2\right ) + 4 a^{4} b x \Gamma \left (m + 2\right ) + 2 a^{3} b^{2} x^{2} \Gamma \left (m + 2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m/(b*x+a)**3,x)

[Out]

a**2*m**3*x*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/(2*a**5*gamma(m + 2) + 4*a**4*b*x*gamm
a(m + 2) + 2*a**3*b**2*x**2*gamma(m + 2)) - a**2*m**2*x*x**m*gamma(m + 1)/(2*a**5*gamma(m + 2) + 4*a**4*b*x*ga
mma(m + 2) + 2*a**3*b**2*x**2*gamma(m + 2)) - a**2*m*x*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m
+ 1)/(2*a**5*gamma(m + 2) + 4*a**4*b*x*gamma(m + 2) + 2*a**3*b**2*x**2*gamma(m + 2)) + a**2*m*x*x**m*gamma(m +
1)/(2*a**5*gamma(m + 2) + 4*a**4*b*x*gamma(m + 2) + 2*a**3*b**2*x**2*gamma(m + 2)) + 2*a**2*x*x**m*gamma(m +
1)/(2*a**5*gamma(m + 2) + 4*a**4*b*x*gamma(m + 2) + 2*a**3*b**2*x**2*gamma(m + 2)) + 2*a*b*m**3*x**2*x**m*lerc
hphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/(2*a**5*gamma(m + 2) + 4*a**4*b*x*gamma(m + 2) + 2*a**3*b**
2*x**2*gamma(m + 2)) - a*b*m**2*x**2*x**m*gamma(m + 1)/(2*a**5*gamma(m + 2) + 4*a**4*b*x*gamma(m + 2) + 2*a**3
*b**2*x**2*gamma(m + 2)) - 2*a*b*m*x**2*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/(2*a**5*ga
mma(m + 2) + 4*a**4*b*x*gamma(m + 2) + 2*a**3*b**2*x**2*gamma(m + 2)) + a*b*x**2*x**m*gamma(m + 1)/(2*a**5*gam
ma(m + 2) + 4*a**4*b*x*gamma(m + 2) + 2*a**3*b**2*x**2*gamma(m + 2)) + b**2*m**3*x**3*x**m*lerchphi(b*x*exp_po
lar(I*pi)/a, 1, m + 1)*gamma(m + 1)/(2*a**5*gamma(m + 2) + 4*a**4*b*x*gamma(m + 2) + 2*a**3*b**2*x**2*gamma(m
+ 2)) - b**2*m*x**3*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/(2*a**5*gamma(m + 2) + 4*a**4*
b*x*gamma(m + 2) + 2*a**3*b**2*x**2*gamma(m + 2))

________________________________________________________________________________________