3.2150 \(\int \frac{(a+b x) (d+e x)^m}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=54 \[ -\frac{e^2 (d+e x)^{m+1} \, _2F_1\left (3,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{(m+1) (b d-a e)^3} \]

[Out]

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

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Rubi [A]  time = 0.0146187, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {27, 68} \[ -\frac{e^2 (d+e x)^{m+1} \, _2F_1\left (3,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{(m+1) (b d-a e)^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(d + e*x)^m)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{(a+b x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{(d+e x)^m}{(a+b x)^3} \, dx\\ &=-\frac{e^2 (d+e x)^{1+m} \, _2F_1\left (3,1+m;2+m;\frac{b (d+e x)}{b d-a e}\right )}{(b d-a e)^3 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.014962, size = 54, normalized size = 1. \[ \frac{e^2 (d+e x)^{m+1} \, _2F_1\left (3,m+1;m+2;-\frac{b (d+e x)}{a e-b d}\right )}{(m+1) (a e-b d)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(d + e*x)^m)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

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

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Maple [F]  time = 0.175, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) \left ( ex+d \right ) ^{m}}{ \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (e x + d\right )}^{m}}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{m}}{\left (a + b x\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x)**m/(a + b*x)**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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