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

Optimal. Leaf size=101 \[ \frac {c d (d+e x)^m (a e+c d x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{-m} \, _2F_1\left (2,1-m;2-m;-\frac {g (a e+c d x)}{c d f-a e g}\right )}{(1-m) (c d f-a e g)^2} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 101, normalized size of antiderivative = 1.00, 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} \[ \frac {c d (d+e x)^m (a e+c d x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{-m} \, _2F_1\left (2,1-m;2-m;-\frac {g (a e+c d x)}{c d f-a e g}\right )}{(1-m) (c d f-a e g)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]

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

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Mathematica [A]  time = 0.03, size = 84, normalized size = 0.83 \[ -\frac {c d (d+e x)^{m-1} ((d+e x) (a e+c d x))^{1-m} \, _2F_1\left (2,1-m;2-m;\frac {g (a e+c d x)}{a e g-c d f}\right )}{(m-1) (c d f-a e g)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x + d\right )}^{m}}{{\left (g^{2} x^{2} + 2 \, f g x + f^{2}\right )} {\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{m}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{2} {\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{m}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x \right )^{-m} \left (e x +d \right )^{m}}{\left (g x +f \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{2} {\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{m}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^m}{{\left (f+g\,x\right )}^2\,{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^m} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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