Integrand size = 34, antiderivative size = 200 \[ \int \left (a+\frac {c e x}{f}+c x^2\right )^p \left (\frac {a f}{c}+e x+f x^2\right )^q \, dx=-\frac {2^{1+p+q} \sqrt {c} \left (-\frac {\sqrt {c} \left (e-\frac {\sqrt {c e^2-4 a f^2}}{\sqrt {c}}+2 f x\right )}{\sqrt {c e^2-4 a f^2}}\right )^{-1-p-q} \left (a+\frac {c e x}{f}+c x^2\right )^p \left (\frac {a f}{c}+e x+f x^2\right )^{1+q} \operatorname {Hypergeometric2F1}\left (-p-q,1+p+q,2+p+q,\frac {\sqrt {c} \left (e+\frac {\sqrt {c e^2-4 a f^2}}{\sqrt {c}}+2 f x\right )}{2 \sqrt {c e^2-4 a f^2}}\right )}{\sqrt {c e^2-4 a f^2} (1+p+q)} \]
-2^(1+p+q)*(a+c*e*x/f+c*x^2)^p*(a*f/c+e*x+f*x^2)^(1+q)*hypergeom([-p-q, 1+ p+q],[2+p+q],1/2*c^(1/2)*(e+2*f*x+(-4*a*f^2+c*e^2)^(1/2)/c^(1/2))/(-4*a*f^ 2+c*e^2)^(1/2))*c^(1/2)*(-c^(1/2)*(e+2*f*x-(-4*a*f^2+c*e^2)^(1/2)/c^(1/2)) /(-4*a*f^2+c*e^2)^(1/2))^(-1-p-q)/(1+p+q)/(-4*a*f^2+c*e^2)^(1/2)
Time = 0.40 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.86 \[ \int \left (a+\frac {c e x}{f}+c x^2\right )^p \left (\frac {a f}{c}+e x+f x^2\right )^q \, dx=\frac {2^{-1+p+q} \left (\frac {a f}{c}+x (e+f x)\right )^q \left (a+\frac {c x (e+f x)}{f}\right )^p \left (-\sqrt {c e^2-4 a f^2}+\sqrt {c} (e+2 f x)\right ) \left (1+\frac {\sqrt {c} (e+2 f x)}{\sqrt {c e^2-4 a f^2}}\right )^{-p-q} \operatorname {Hypergeometric2F1}\left (-p-q,1+p+q,2+p+q,\frac {1}{2}-\frac {\sqrt {c} (e+2 f x)}{2 \sqrt {c e^2-4 a f^2}}\right )}{\sqrt {c} f (1+p+q)} \]
(2^(-1 + p + q)*((a*f)/c + x*(e + f*x))^q*(a + (c*x*(e + f*x))/f)^p*(-Sqrt [c*e^2 - 4*a*f^2] + Sqrt[c]*(e + 2*f*x))*(1 + (Sqrt[c]*(e + 2*f*x))/Sqrt[c *e^2 - 4*a*f^2])^(-p - q)*Hypergeometric2F1[-p - q, 1 + p + q, 2 + p + q, 1/2 - (Sqrt[c]*(e + 2*f*x))/(2*Sqrt[c*e^2 - 4*a*f^2])])/(Sqrt[c]*f*(1 + p + q))
Time = 0.31 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {1296, 1096}
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.
\(\displaystyle \int \left (a+\frac {c e x}{f}+c x^2\right )^p \left (\frac {a f}{c}+e x+f x^2\right )^q \, dx\) |
\(\Big \downarrow \) 1296 |
\(\displaystyle \left (a+\frac {c e x}{f}+c x^2\right )^p \left (\frac {a f}{c}+e x+f x^2\right )^{-p} \int \left (f x^2+e x+\frac {a f}{c}\right )^{p+q}dx\) |
\(\Big \downarrow \) 1096 |
\(\displaystyle -\frac {\sqrt {c} 2^{p+q+1} \left (a+\frac {c e x}{f}+c x^2\right )^p \left (\frac {a f}{c}+e x+f x^2\right )^{q+1} \left (-\frac {\sqrt {c} \left (-\frac {\sqrt {c e^2-4 a f^2}}{\sqrt {c}}+e+2 f x\right )}{\sqrt {c e^2-4 a f^2}}\right )^{-p-q-1} \operatorname {Hypergeometric2F1}\left (-p-q,p+q+1,p+q+2,\frac {\sqrt {c} \left (e+2 f x+\frac {\sqrt {c e^2-4 a f^2}}{\sqrt {c}}\right )}{2 \sqrt {c e^2-4 a f^2}}\right )}{(p+q+1) \sqrt {c e^2-4 a f^2}}\) |
-((2^(1 + p + q)*Sqrt[c]*(-((Sqrt[c]*(e - Sqrt[c*e^2 - 4*a*f^2]/Sqrt[c] + 2*f*x))/Sqrt[c*e^2 - 4*a*f^2]))^(-1 - p - q)*(a + (c*e*x)/f + c*x^2)^p*((a *f)/c + e*x + f*x^2)^(1 + q)*Hypergeometric2F1[-p - q, 1 + p + q, 2 + p + q, (Sqrt[c]*(e + Sqrt[c*e^2 - 4*a*f^2]/Sqrt[c] + 2*f*x))/(2*Sqrt[c*e^2 - 4 *a*f^2])])/(Sqrt[c*e^2 - 4*a*f^2]*(1 + p + q)))
3.1.10.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(-(a + b*x + c*x^2)^(p + 1)/(q*(p + 1)*((q - b - 2*c*x) /(2*q))^(p + 1)))*Hypergeometric2F1[-p, p + 1, p + 2, (b + q + 2*c*x)/(2*q) ], x]] /; FreeQ[{a, b, c, p}, x] && !IntegerQ[4*p] && !IntegerQ[3*p]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_) ^2)^(q_.), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x + c*x^2)^FracPart[p]/(d ^IntPart[p]*(d + e*x + f*x^2)^FracPart[p])) Int[(d + e*x + f*x^2)^(p + q) , x], x] /; FreeQ[{a, b, c, d, e, f, p, q}, x] && EqQ[c*d - a*f, 0] && EqQ[ b*d - a*e, 0] && !IntegerQ[p] && !IntegerQ[q] && !GtQ[c/f, 0]
\[\int \left (a +\frac {c e x}{f}+c \,x^{2}\right )^{p} \left (\frac {a f}{c}+e x +f \,x^{2}\right )^{q}d x\]
\[ \int \left (a+\frac {c e x}{f}+c x^2\right )^p \left (\frac {a f}{c}+e x+f x^2\right )^q \, dx=\int { {\left (c x^{2} + \frac {c e x}{f} + a\right )}^{p} {\left (f x^{2} + e x + \frac {a f}{c}\right )}^{q} \,d x } \]
\[ \int \left (a+\frac {c e x}{f}+c x^2\right )^p \left (\frac {a f}{c}+e x+f x^2\right )^q \, dx=\int \left (a + \frac {c e x}{f} + c x^{2}\right )^{p} \left (\frac {a f}{c} + e x + f x^{2}\right )^{q}\, dx \]
\[ \int \left (a+\frac {c e x}{f}+c x^2\right )^p \left (\frac {a f}{c}+e x+f x^2\right )^q \, dx=\int { {\left (c x^{2} + \frac {c e x}{f} + a\right )}^{p} {\left (f x^{2} + e x + \frac {a f}{c}\right )}^{q} \,d x } \]
\[ \int \left (a+\frac {c e x}{f}+c x^2\right )^p \left (\frac {a f}{c}+e x+f x^2\right )^q \, dx=\int { {\left (c x^{2} + \frac {c e x}{f} + a\right )}^{p} {\left (f x^{2} + e x + \frac {a f}{c}\right )}^{q} \,d x } \]
Timed out. \[ \int \left (a+\frac {c e x}{f}+c x^2\right )^p \left (\frac {a f}{c}+e x+f x^2\right )^q \, dx=\int {\left (e\,x+f\,x^2+\frac {a\,f}{c}\right )}^q\,{\left (a+c\,x^2+\frac {c\,e\,x}{f}\right )}^p \,d x \]