\(\int \frac {(d+e x)^m}{(a d e+(c d^2+a e^2) x+c d e x^2)^4} \, dx\) [2090]

Optimal result

Integrand size = 35, antiderivative size = 65 \[ \int \frac {(d+e x)^m}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^4} \, dx=-\frac {e^3 (d+e x)^{-3+m} \operatorname {Hypergeometric2F1}\left (4,-3+m,-2+m,\frac {c d (d+e x)}{c d^2-a e^2}\right )}{\left (c d^2-a e^2\right )^4 (3-m)} \]

[Out]

-e^3*(e*x+d)^(-3+m)*hypergeom([4, -3+m],[-2+m],c*d*(e*x+d)/(-a*e^2+c*d^2))/(-a*e^2+c*d^2)^4/(3-m)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {640, 70} \[ \int \frac {(d+e x)^m}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^4} \, dx=-\frac {e^3 (d+e x)^{m-3} \operatorname {Hypergeometric2F1}\left (4,m-3,m-2,\frac {c d (d+e x)}{c d^2-a e^2}\right )}{(3-m) \left (c d^2-a e^2\right )^4} \]

[In]

Int[(d + e*x)^m/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^4,x]

[Out]

-((e^3*(d + e*x)^(-3 + m)*Hypergeometric2F1[4, -3 + m, -2 + m, (c*d*(d + e*x))/(c*d^2 - a*e^2)])/((c*d^2 - a*e
^2)^4*(3 - m)))

Rule 70

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

Rule 640

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {(d+e x)^{-4+m}}{(a e+c d x)^4} \, dx \\ & = -\frac {e^3 (d+e x)^{-3+m} \, _2F_1\left (4,-3+m;-2+m;\frac {c d (d+e x)}{c d^2-a e^2}\right )}{\left (c d^2-a e^2\right )^4 (3-m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97 \[ \int \frac {(d+e x)^m}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^4} \, dx=\frac {e^3 (d+e x)^{-3+m} \operatorname {Hypergeometric2F1}\left (4,-3+m,-2+m,-\frac {c d (d+e x)}{-c d^2+a e^2}\right )}{\left (-c d^2+a e^2\right )^4 (-3+m)} \]

[In]

Integrate[(d + e*x)^m/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^4,x]

[Out]

(e^3*(d + e*x)^(-3 + m)*Hypergeometric2F1[4, -3 + m, -2 + m, -((c*d*(d + e*x))/(-(c*d^2) + a*e^2))])/((-(c*d^2
) + a*e^2)^4*(-3 + m))

Maple [F]

\[\int \frac {\left (e x +d \right )^{m}}{{\left (a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}\right )}^{4}}d x\]

[In]

int((e*x+d)^m/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^4,x)

[Out]

int((e*x+d)^m/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^4,x)

Fricas [F]

\[ \int \frac {(d+e x)^m}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^4} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{4}} \,d x } \]

[In]

integrate((e*x+d)^m/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^4,x, algorithm="fricas")

[Out]

integral((e*x + d)^m/(c^4*d^4*e^4*x^8 + a^4*d^4*e^4 + 4*(c^4*d^5*e^3 + a*c^3*d^3*e^5)*x^7 + 2*(3*c^4*d^6*e^2 +
8*a*c^3*d^4*e^4 + 3*a^2*c^2*d^2*e^6)*x^6 + 4*(c^4*d^7*e + 6*a*c^3*d^5*e^3 + 6*a^2*c^2*d^3*e^5 + a^3*c*d*e^7)*
x^5 + (c^4*d^8 + 16*a*c^3*d^6*e^2 + 36*a^2*c^2*d^4*e^4 + 16*a^3*c*d^2*e^6 + a^4*e^8)*x^4 + 4*(a*c^3*d^7*e + 6*
a^2*c^2*d^5*e^3 + 6*a^3*c*d^3*e^5 + a^4*d*e^7)*x^3 + 2*(3*a^2*c^2*d^6*e^2 + 8*a^3*c*d^4*e^4 + 3*a^4*d^2*e^6)*x
^2 + 4*(a^3*c*d^5*e^3 + a^4*d^3*e^5)*x), x)

Sympy [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^m}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^4} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

integrate((e*x+d)**m/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**4,x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

\[ \int \frac {(d+e x)^m}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^4} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{4}} \,d x } \]

[In]

integrate((e*x+d)^m/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^4,x, algorithm="maxima")

[Out]

integrate((e*x + d)^m/(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^4, x)

Giac [F]

\[ \int \frac {(d+e x)^m}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^4} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{4}} \,d x } \]

[In]

integrate((e*x+d)^m/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^4,x, algorithm="giac")

[Out]

integrate((e*x + d)^m/(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^m}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^4} \, dx=\int \frac {{\left (d+e\,x\right )}^m}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^4} \,d x \]

[In]

int((d + e*x)^m/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^4,x)

[Out]

int((d + e*x)^m/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^4, x)