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

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

[Out]

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

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Rubi [A]  time = 0.0052433, antiderivative size = 29, normalized size of antiderivative = 1., 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} \[ \frac{x^{m+1} \, _2F_1\left (2,m+1;m+2;-\frac{b x}{a}\right )}{a^2 (m+1)} \]

Antiderivative was successfully verified.

[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 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)^2} \, dx &=\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]  time = 0.0050386, size = 29, normalized size = 1. \[ \frac{x^{m+1} \, _2F_1\left (2,m+1;m+2;-\frac{b x}{a}\right )}{a^2 (m+1)} \]

Antiderivative was successfully verified.

[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))

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Maple [F]  time = 0.032, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{m}}{ \left ( bx+a \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{{\left (b x + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m/(b*x + a)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{m}}{b^{2} x^{2} + 2 \, a b x + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [C]  time = 0.841628, size = 262, normalized size = 9.03 \begin{align*} - \frac{a m^{2} x x^{m} \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 x^{m} \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 x^{m} \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} + \frac{a x x^{m} \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^{2} x^{m} \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^{2} x^{m} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*m**2*x*x**m*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*x**m*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*x**m*gamma(m + 1)/(a**3*gamma(m + 2) + a**2*b*x*gamma(m + 2)) + a*x*x**m*gamma(m + 1)/(a**3*gamm
a(m + 2) + a**2*b*x*gamma(m + 2)) - b*m**2*x**2*x**m*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**2*x**m*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))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{{\left (b x + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m/(b*x + a)^2, x)