\(\int \frac {(d+e x)^m (a d e+(c d^2+a e^2) x+c d e x^2)^{-m}}{(f+g x)^{5/2}} \, dx\) [779]

Optimal result

Integrand size = 46, antiderivative size = 105 \[ \int \frac {(d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m}}{(f+g x)^{5/2}} \, dx=-\frac {2 \left (-\frac {g (a e+c d x)}{c d f-a e g}\right )^m (d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},m,-\frac {1}{2},\frac {c d (f+g x)}{c d f-a e g}\right )}{3 g (f+g x)^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {905, 72, 71} \[ \int \frac {(d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m}}{(f+g x)^{5/2}} \, dx=-\frac {2 (d+e x)^m \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{-m} \left (-\frac {g (a e+c d x)}{c d f-a e g}\right )^m \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},m,-\frac {1}{2},\frac {c d (f+g x)}{c d f-a e g}\right )}{3 g (f+g x)^{3/2}} \]

[In]

Int[(d + e*x)^m/((f + g*x)^(5/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^m),x]

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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]))

Rule 72

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 905

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/e)*x)^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*} \text {integral}& = \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)^{5/2}} \, 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 \frac {\left (-\frac {a e g}{c d f-a e g}-\frac {c d g x}{c d f-a e g}\right )^{-m}}{(f+g x)^{5/2}} \, dx \\ & = -\frac {2 \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} \, _2F_1\left (-\frac {3}{2},m;-\frac {1}{2};\frac {c d (f+g x)}{c d f-a e g}\right )}{3 g (f+g x)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int \frac {(d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m}}{(f+g x)^{5/2}} \, dx=-\frac {2 \left (\frac {g (a e+c d x)}{-c d f+a e g}\right )^m (d+e x)^m ((a e+c d x) (d+e x))^{-m} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},m,-\frac {1}{2},\frac {c d (f+g x)}{c d f-a e g}\right )}{3 g (f+g x)^{3/2}} \]

[In]

Integrate[(d + e*x)^m/((f + g*x)^(5/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^m),x]

[Out]

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

Maple [F]

\[\int \frac {\left (e x +d \right )^{m} {\left (a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}\right )}^{-m}}{\left (g x +f \right )^{\frac {5}{2}}}d x\]

[In]

int((e*x+d)^m/(g*x+f)^(5/2)/((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)^(5/2)/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m),x)

Fricas [F]

\[ \int \frac {(d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m}}{(f+g x)^{5/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{\frac {5}{2}} {\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{m}} \,d x } \]

[In]

integrate((e*x+d)^m/(g*x+f)^(5/2)/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m),x, algorithm="fricas")

[Out]

integral(sqrt(g*x + f)*(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)

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m}}{(f+g x)^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((e*x+d)**m/(g*x+f)**(5/2)/((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**m),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m}}{(f+g x)^{5/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{\frac {5}{2}} {\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{m}} \,d x } \]

[In]

integrate((e*x+d)^m/(g*x+f)^(5/2)/((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)^(5/2)*(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^m), x)

Giac [F]

\[ \int \frac {(d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m}}{(f+g x)^{5/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{\frac {5}{2}} {\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{m}} \,d x } \]

[In]

integrate((e*x+d)^m/(g*x+f)^(5/2)/((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)^(5/2)*(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^m), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m}}{(f+g x)^{5/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^m}{{\left (f+g\,x\right )}^{5/2}\,{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^m} \,d x \]

[In]

int((d + e*x)^m/((f + g*x)^(5/2)*(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)^(5/2)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^m), x)