3.774 \(\int \frac{(d+e x)^m (a d e+(c d^2+a e^2) x+c d e x^2)^{-m}}{(f+g x)^3} \, dx\)

Optimal. Leaf size=105 \[ \frac{c^2 d^2 (d+e x)^m (a e+c d x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{-m} \, _2F_1\left (3,1-m;2-m;-\frac{g (a e+c d x)}{c d f-a e g}\right )}{(1-m) (c d f-a e g)^3} \]

[Out]

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

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Rubi [A]  time = 0.0747857, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 44, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used = {891, 68} \[ \frac{c^2 d^2 (d+e x)^m (a e+c d x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{-m} \, _2F_1\left (3,1-m;2-m;-\frac{g (a e+c d x)}{c d f-a e g}\right )}{(1-m) (c d f-a e g)^3} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^m/((f + g*x)^3*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^m),x]

[Out]

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

Rule 891

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Dist[(a + b*x + c*x^2)^FracPart[p]/((d + e*x)^FracPart[p]*(a/d + (c*x)/e)^FracPart[p]), Int[(d + e*x)^(m + p)*
(f + g*x)^n*(a/d + (c*x)/e)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2
- 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] &&  !IGtQ[m, 0] &&  !IGtQ[n, 0]

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

Mathematica [A]  time = 0.0374521, size = 88, normalized size = 0.84 \[ -\frac{c^2 d^2 (d+e x)^{m-1} ((d+e x) (a e+c d x))^{1-m} \, _2F_1\left (3,1-m;2-m;\frac{g (a e+c d x)}{a e g-c d f}\right )}{(m-1) (c d f-a e g)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^m/((f + g*x)^3*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^m),x]

[Out]

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

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Maple [F]  time = 1.732, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{m}}{ \left ( gx+f \right ) ^{3} \left ( ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2} \right ) ^{m}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^m/(g*x+f)^3/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m),x)

[Out]

int((e*x+d)^m/(g*x+f)^3/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{3}{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{m}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(g*x+f)^3/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m),x, algorithm="maxima")

[Out]

integrate((e*x + d)^m/((g*x + f)^3*(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^m), 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}}{{\left (g^{3} x^{3} + 3 \, f g^{2} x^{2} + 3 \, f^{2} g x + f^{3}\right )}{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{m}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(g*x+f)^3/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m),x, algorithm="fricas")

[Out]

integral((e*x + d)^m/((g^3*x^3 + 3*f*g^2*x^2 + 3*f^2*g*x + f^3)*(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^m), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m/(g*x+f)**3/((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**m),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{3}{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{m}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(g*x+f)^3/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m),x, algorithm="giac")

[Out]

integrate((e*x + d)^m/((g*x + f)^3*(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^m), x)