Integrand size = 46, antiderivative size = 104 \[ \int \frac {(d+e x)^{5/2} (f+g x)^n}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=-\frac {(a e+c d x) (d+e x)^{5/2} (f+g x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,-\frac {1}{2}+n,2+n,\frac {c d (f+g x)}{c d f-a e g}\right )}{(c d f-a e g) (1+n) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \]
-(c*d*x+a*e)*(e*x+d)^(5/2)*(g*x+f)^(1+n)*hypergeom([1, -1/2+n],[2+n],c*d*( g*x+f)/(-a*e*g+c*d*f))/(-a*e*g+c*d*f)/(1+n)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x ^2)^(5/2)
Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.96 \[ \int \frac {(d+e x)^{5/2} (f+g x)^n}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=-\frac {2 (d+e x)^{3/2} (f+g x)^n \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-n,-\frac {1}{2},\frac {g (a e+c d x)}{-c d f+a e g}\right )}{3 c d ((a e+c d x) (d+e x))^{3/2}} \]
(-2*(d + e*x)^(3/2)*(f + g*x)^n*Hypergeometric2F1[-3/2, -n, -1/2, (g*(a*e + c*d*x))/(-(c*d*f) + a*e*g)])/(3*c*d*((a*e + c*d*x)*(d + e*x))^(3/2)*((c* d*(f + g*x))/(c*d*f - a*e*g))^n)
Time = 0.28 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.17, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1268, 80, 79}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(d+e x)^{5/2} (f+g x)^n}{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1268 |
\(\displaystyle \frac {\sqrt {d+e x} \sqrt {a e+c d x} \int \frac {(f+g x)^n}{(a e+c d x)^{5/2}}dx}{\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {\sqrt {d+e x} (f+g x)^n \sqrt {a e+c d x} \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \int \frac {\left (\frac {c d f}{c d f-a e g}+\frac {c d g x}{c d f-a e g}\right )^n}{(a e+c d x)^{5/2}}dx}{\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {2 \sqrt {d+e x} (f+g x)^n \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-n,-\frac {1}{2},-\frac {g (a e+c d x)}{c d f-a e g}\right )}{3 c d (a e+c d x) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\) |
(-2*Sqrt[d + e*x]*(f + g*x)^n*Hypergeometric2F1[-3/2, -n, -1/2, -((g*(a*e + c*d*x))/(c*d*f - a*e*g))])/(3*c*d*(a*e + c*d*x)*((c*d*(f + g*x))/(c*d*f - a*e*g))^n*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])
3.8.61.3.1 Defintions of rubi rules used
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] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(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] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[((d_) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(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 ] && EqQ[c*d^2 - b*d*e + a*e^2, 0]
\[\int \frac {\left (e x +d \right )^{\frac {5}{2}} \left (g x +f \right )^{n}}{{\left (a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}\right )}^{\frac {5}{2}}}d x\]
\[ \int \frac {(d+e x)^{5/2} (f+g x)^n}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {5}{2}} {\left (g x + f\right )}^{n}}{{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{2}}} \,d x } \]
integrate((e*x+d)^(5/2)*(g*x+f)^n/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2), x, algorithm="fricas")
integral(sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)*(g*x + f)^n/(c^3*d^3*e*x^4 + a^3*d*e^3 + (c^3*d^4 + 3*a*c^2*d^2*e^2)*x^3 + 3*(a*c ^2*d^3*e + a^2*c*d*e^3)*x^2 + (3*a^2*c*d^2*e^2 + a^3*e^4)*x), x)
Timed out. \[ \int \frac {(d+e x)^{5/2} (f+g x)^n}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(d+e x)^{5/2} (f+g x)^n}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {5}{2}} {\left (g x + f\right )}^{n}}{{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{2}}} \,d x } \]
integrate((e*x+d)^(5/2)*(g*x+f)^n/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2), x, algorithm="maxima")
\[ \int \frac {(d+e x)^{5/2} (f+g x)^n}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {5}{2}} {\left (g x + f\right )}^{n}}{{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{2}}} \,d x } \]
integrate((e*x+d)^(5/2)*(g*x+f)^n/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2), x, algorithm="giac")
Timed out. \[ \int \frac {(d+e x)^{5/2} (f+g x)^n}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\int \frac {{\left (f+g\,x\right )}^n\,{\left (d+e\,x\right )}^{5/2}}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/2}} \,d x \]