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

Optimal. Leaf size=103 \[ -\frac{(d+e x)^m (f+g x)^{n+1} (a e+c d x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{-m} \, _2F_1\left (1,-m+n+2;n+2;\frac{c d (f+g x)}{c d f-a e g}\right )}{(n+1) (c d f-a e g)} \]

[Out]

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

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Rubi [A]  time = 0.0854224, antiderivative size = 107, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, integrand size = 44, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.068, Rules used = {891, 70, 69} \[ \frac{(d+e x)^m (f+g x)^{n+1} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{-m} \left (-\frac{g (a e+c d x)}{c d f-a e g}\right )^m \, _2F_1\left (m,n+1;n+2;\frac{c d (f+g x)}{c d f-a e g}\right )}{g (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-((g*(a*e + c*d*x))/(c*d*f - a*e*g)))^m*(d + e*x)^m*(f + g*x)^(1 + n)*Hypergeometric2F1[m, 1 + n, 2 + n, (c*
d*(f + g*x))/(c*d*f - a*e*g)])/(g*(1 + n)*(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 70

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

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

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

Mathematica [A]  time = 0.0545406, size = 95, normalized size = 0.92 \[ \frac{(d+e x)^m (f+g x)^{n+1} ((d+e x) (a e+c d x))^{-m} \left (\frac{g (a e+c d x)}{a e g-c d f}\right )^m \, _2F_1\left (m,n+1;n+2;\frac{c d (f+g x)}{c d f-a e g}\right )}{g (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.717, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( gx+f \right ) ^{n} \left ( ex+d \right ) ^{m}}{ \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)^n/((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)^n/((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 )}^{n}}{{\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)^n/((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)^n/(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 x + f\right )}^{n}}{{\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)^n/((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*x + f)^n/(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)**n/((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 )}^{n}}{{\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)^n/((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)^n/(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^m, x)