Integrand size = 28, antiderivative size = 123 \[ \int \frac {(c+d x)^{2/5} (e+f x)^{3/5}}{\sqrt {a+b x}} \, dx=\frac {2 \sqrt {a+b x} (c+d x)^{2/5} (e+f x)^{3/5} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {2}{5},-\frac {3}{5},\frac {3}{2},-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b \left (\frac {b (c+d x)}{b c-a d}\right )^{2/5} \left (\frac {b (e+f x)}{b e-a f}\right )^{3/5}} \]
2*(d*x+c)^(2/5)*(f*x+e)^(3/5)*AppellF1(1/2,-2/5,-3/5,3/2,-d*(b*x+a)/(-a*d+ b*c),-f*(b*x+a)/(-a*f+b*e))*(b*x+a)^(1/2)/b/(b*(d*x+c)/(-a*d+b*c))^(2/5)/( b*(f*x+e)/(-a*f+b*e))^(3/5)
Leaf count is larger than twice the leaf count of optimal. \(536\) vs. \(2(123)=246\).
Time = 22.62 (sec) , antiderivative size = 536, normalized size of antiderivative = 4.36 \[ \int \frac {(c+d x)^{2/5} (e+f x)^{3/5}}{\sqrt {a+b x}} \, dx=\frac {2 \sqrt {a+b x} \left (15 b^2 (c+d x) (e+f x)-2 (a+b x) \left (-\frac {3 b^2 (3 b d e+2 b c f-5 a d f) (c+d x) (e+f x)}{d f (a+b x)^2}+\frac {(b c-a d) (b e-a f) (3 b d e+2 b c f-5 a d f) \left (\frac {b (c+d x)}{d (a+b x)}\right )^{3/5} \left (\frac {b (e+f x)}{f (a+b x)}\right )^{2/5} \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{5},\frac {2}{5},\frac {5}{2},\frac {-b c+a d}{d (a+b x)},\frac {-b e+a f}{f (a+b x)}\right )}{d f (a+b x)^2}+\frac {9 \left (25 a^2 d^2 f^2-10 a b d f (3 d e+2 c f)+b^2 \left (3 d^2 e^2+24 c d e f-2 c^2 f^2\right )\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{5},\frac {2}{5},\frac {3}{2},\frac {-b c+a d}{d (a+b x)},\frac {-b e+a f}{f (a+b x)}\right )}{15 d f (a+b x) \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{5},\frac {2}{5},\frac {3}{2},\frac {-b c+a d}{d (a+b x)},\frac {-b e+a f}{f (a+b x)}\right )+(-4 b d e+4 a d f) \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{5},\frac {7}{5},\frac {5}{2},\frac {-b c+a d}{d (a+b x)},\frac {-b e+a f}{f (a+b x)}\right )+6 (-b c+a d) f \operatorname {AppellF1}\left (\frac {3}{2},\frac {8}{5},\frac {2}{5},\frac {5}{2},\frac {-b c+a d}{d (a+b x)},\frac {-b e+a f}{f (a+b x)}\right )}\right )\right )}{45 b^3 (c+d x)^{3/5} (e+f x)^{2/5}} \]
(2*Sqrt[a + b*x]*(15*b^2*(c + d*x)*(e + f*x) - 2*(a + b*x)*((-3*b^2*(3*b*d *e + 2*b*c*f - 5*a*d*f)*(c + d*x)*(e + f*x))/(d*f*(a + b*x)^2) + ((b*c - a *d)*(b*e - a*f)*(3*b*d*e + 2*b*c*f - 5*a*d*f)*((b*(c + d*x))/(d*(a + b*x)) )^(3/5)*((b*(e + f*x))/(f*(a + b*x)))^(2/5)*AppellF1[3/2, 3/5, 2/5, 5/2, ( -(b*c) + a*d)/(d*(a + b*x)), (-(b*e) + a*f)/(f*(a + b*x))])/(d*f*(a + b*x) ^2) + (9*(25*a^2*d^2*f^2 - 10*a*b*d*f*(3*d*e + 2*c*f) + b^2*(3*d^2*e^2 + 2 4*c*d*e*f - 2*c^2*f^2))*AppellF1[1/2, 3/5, 2/5, 3/2, (-(b*c) + a*d)/(d*(a + b*x)), (-(b*e) + a*f)/(f*(a + b*x))])/(15*d*f*(a + b*x)*AppellF1[1/2, 3/ 5, 2/5, 3/2, (-(b*c) + a*d)/(d*(a + b*x)), (-(b*e) + a*f)/(f*(a + b*x))] + (-4*b*d*e + 4*a*d*f)*AppellF1[3/2, 3/5, 7/5, 5/2, (-(b*c) + a*d)/(d*(a + b*x)), (-(b*e) + a*f)/(f*(a + b*x))] + 6*(-(b*c) + a*d)*f*AppellF1[3/2, 8/ 5, 2/5, 5/2, (-(b*c) + a*d)/(d*(a + b*x)), (-(b*e) + a*f)/(f*(a + b*x))])) ))/(45*b^3*(c + d*x)^(3/5)*(e + f*x)^(2/5))
Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, 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.
\(\displaystyle \int \frac {(c+d x)^{2/5} (e+f x)^{3/5}}{\sqrt {a+b x}} \, dx\) |
\(\Big \downarrow \) 157 |
\(\displaystyle \frac {(c+d x)^{2/5} \int \frac {\left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{2/5} (e+f x)^{3/5}}{\sqrt {a+b x}}dx}{\left (\frac {b (c+d x)}{b c-a d}\right )^{2/5}}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {(c+d x)^{2/5} (e+f x)^{3/5} \int \frac {\left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{2/5} \left (\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}\right )^{3/5}}{\sqrt {a+b x}}dx}{\left (\frac {b (c+d x)}{b c-a d}\right )^{2/5} \left (\frac {b (e+f x)}{b e-a f}\right )^{3/5}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {2 \sqrt {a+b x} (c+d x)^{2/5} (e+f x)^{3/5} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {2}{5},-\frac {3}{5},\frac {3}{2},-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b \left (\frac {b (c+d x)}{b c-a d}\right )^{2/5} \left (\frac {b (e+f x)}{b e-a f}\right )^{3/5}}\) |
(2*Sqrt[a + b*x]*(c + d*x)^(2/5)*(e + f*x)^(3/5)*AppellF1[1/2, -2/5, -3/5, 3/2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*((b* (c + d*x))/(b*c - a*d))^(2/5)*((b*(e + f*x))/(b*e - a*f))^(3/5))
3.32.71.3.1 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]
\[\int \frac {\left (d x +c \right )^{\frac {2}{5}} \left (f x +e \right )^{\frac {3}{5}}}{\sqrt {b x +a}}d x\]
\[ \int \frac {(c+d x)^{2/5} (e+f x)^{3/5}}{\sqrt {a+b x}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {2}{5}} {\left (f x + e\right )}^{\frac {3}{5}}}{\sqrt {b x + a}} \,d x } \]
\[ \int \frac {(c+d x)^{2/5} (e+f x)^{3/5}}{\sqrt {a+b x}} \, dx=\int \frac {\left (c + d x\right )^{\frac {2}{5}} \left (e + f x\right )^{\frac {3}{5}}}{\sqrt {a + b x}}\, dx \]
\[ \int \frac {(c+d x)^{2/5} (e+f x)^{3/5}}{\sqrt {a+b x}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {2}{5}} {\left (f x + e\right )}^{\frac {3}{5}}}{\sqrt {b x + a}} \,d x } \]
\[ \int \frac {(c+d x)^{2/5} (e+f x)^{3/5}}{\sqrt {a+b x}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {2}{5}} {\left (f x + e\right )}^{\frac {3}{5}}}{\sqrt {b x + a}} \,d x } \]
Timed out. \[ \int \frac {(c+d x)^{2/5} (e+f x)^{3/5}}{\sqrt {a+b x}} \, dx=\int \frac {{\left (e+f\,x\right )}^{3/5}\,{\left (c+d\,x\right )}^{2/5}}{\sqrt {a+b\,x}} \,d x \]