Integrand size = 26, antiderivative size = 123 \[ \int \frac {\sqrt {a+b x} (e+f x)^n}{\sqrt {c+d x}} \, dx=\frac {2 (a+b x)^{3/2} \sqrt {\frac {b (c+d x)}{b c-a d}} (e+f x)^n \left (\frac {b (e+f x)}{b e-a f}\right )^{-n} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2},-n,\frac {5}{2},-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{3 b \sqrt {c+d x}} \]
2/3*(b*x+a)^(3/2)*(f*x+e)^n*AppellF1(3/2,1/2,-n,5/2,-d*(b*x+a)/(-a*d+b*c), -f*(b*x+a)/(-a*f+b*e))*(b*(d*x+c)/(-a*d+b*c))^(1/2)/b/((b*(f*x+e)/(-a*f+b* e))^n)/(d*x+c)^(1/2)
Time = 1.80 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {a+b x} (e+f x)^n}{\sqrt {c+d x}} \, dx=\frac {2 (a+b x)^{3/2} \sqrt {\frac {b (c+d x)}{b c-a d}} (e+f x)^n \left (\frac {b (e+f x)}{b e-a f}\right )^{-n} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2},-n,\frac {5}{2},\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )}{3 b \sqrt {c+d x}} \]
(2*(a + b*x)^(3/2)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*(e + f*x)^n*AppellF1[3/ 2, 1/2, -n, 5/2, (d*(a + b*x))/(-(b*c) + a*d), (f*(a + b*x))/(-(b*e) + a*f )])/(3*b*Sqrt[c + d*x]*((b*(e + f*x))/(b*e - a*f))^n)
Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {157, 156, 155}
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.
\(\displaystyle \int \frac {\sqrt {a+b x} (e+f x)^n}{\sqrt {c+d x}} \, dx\) |
\(\Big \downarrow \) 157 |
\(\displaystyle \frac {\sqrt {\frac {b (c+d x)}{b c-a d}} \int \frac {\sqrt {a+b x} (e+f x)^n}{\sqrt {\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}}}dx}{\sqrt {c+d x}}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {(e+f x)^n \sqrt {\frac {b (c+d x)}{b c-a d}} \left (\frac {b (e+f x)}{b e-a f}\right )^{-n} \int \frac {\sqrt {a+b x} \left (\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}\right )^n}{\sqrt {\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}}}dx}{\sqrt {c+d x}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {2 (a+b x)^{3/2} (e+f x)^n \sqrt {\frac {b (c+d x)}{b c-a d}} \left (\frac {b (e+f x)}{b e-a f}\right )^{-n} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2},-n,\frac {5}{2},-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{3 b \sqrt {c+d x}}\) |
(2*(a + b*x)^(3/2)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*(e + f*x)^n*AppellF1[3/ 2, 1/2, -n, 5/2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f) )])/(3*b*Sqrt[c + d*x]*((b*(e + f*x))/(b*e - a*f))^n)
3.32.72.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ (b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] && GtQ[Sim plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] && !(GtQ[Simpl ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d *x, a + b*x]) && !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c - e*d)], 0] && SimplerQ[e + f*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p ]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]) Int[(a + b*x)^m*(c + d*x)^n*Si mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & GtQ[Simplify[b/(b*c - a*d)], 0] && !GtQ[Simplify[b/(b*e - a*f)], 0]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[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*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & !GtQ[Simplify[b/(b*c - a*d)], 0] && !SimplerQ[c + d*x, a + b*x] && !Si mplerQ[e + f*x, a + b*x]
\[\int \frac {\left (f x +e \right )^{n} \sqrt {b x +a}}{\sqrt {d x +c}}d x\]
\[ \int \frac {\sqrt {a+b x} (e+f x)^n}{\sqrt {c+d x}} \, dx=\int { \frac {\sqrt {b x + a} {\left (f x + e\right )}^{n}}{\sqrt {d x + c}} \,d x } \]
\[ \int \frac {\sqrt {a+b x} (e+f x)^n}{\sqrt {c+d x}} \, dx=\int \frac {\sqrt {a + b x} \left (e + f x\right )^{n}}{\sqrt {c + d x}}\, dx \]
\[ \int \frac {\sqrt {a+b x} (e+f x)^n}{\sqrt {c+d x}} \, dx=\int { \frac {\sqrt {b x + a} {\left (f x + e\right )}^{n}}{\sqrt {d x + c}} \,d x } \]
\[ \int \frac {\sqrt {a+b x} (e+f x)^n}{\sqrt {c+d x}} \, dx=\int { \frac {\sqrt {b x + a} {\left (f x + e\right )}^{n}}{\sqrt {d x + c}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a+b x} (e+f x)^n}{\sqrt {c+d x}} \, dx=\int \frac {{\left (e+f\,x\right )}^n\,\sqrt {a+b\,x}}{\sqrt {c+d\,x}} \,d x \]