Integrand size = 25, antiderivative size = 265 \[ \int (d+e x)^m (f+g x)^n \left (a+b x+c x^2\right ) \, dx=\frac {(b e g (3+m+n)-c (e f (2+m)+d g (4+m+2 n))) (d+e x)^{1+m} (f+g x)^{1+n}}{e^2 g^2 (2+m+n) (3+m+n)}+\frac {c (d+e x)^{2+m} (f+g x)^{1+n}}{e^2 g (3+m+n)}+\frac {\left (g (2+m+n) \left (a e^2 g (3+m+n)-c d (e f (2+m)+d g (1+n))\right )-(e f (1+m)+d g (1+n)) (b e g (3+m+n)-c (e f (2+m)+d g (4+m+2 n)))\right ) (d+e x)^{1+m} (f+g x)^n \left (\frac {e (f+g x)}{e f-d g}\right )^{-n} \operatorname {Hypergeometric2F1}\left (1+m,-n,2+m,-\frac {g (d+e x)}{e f-d g}\right )}{e^3 g^2 (1+m) (2+m+n) (3+m+n)} \]
(b*e*g*(3+m+n)-c*(e*f*(2+m)+d*g*(4+m+2*n)))*(e*x+d)^(1+m)*(g*x+f)^(1+n)/e^ 2/g^2/(2+m+n)/(3+m+n)+c*(e*x+d)^(2+m)*(g*x+f)^(1+n)/e^2/g/(3+m+n)+(g*(2+m+ n)*(a*e^2*g*(3+m+n)-c*d*(e*f*(2+m)+d*g*(1+n)))-(e*f*(1+m)+d*g*(1+n))*(b*e* g*(3+m+n)-c*(e*f*(2+m)+d*g*(4+m+2*n))))*(e*x+d)^(1+m)*(g*x+f)^n*hypergeom( [-n, 1+m],[2+m],-g*(e*x+d)/(-d*g+e*f))/e^3/g^2/(1+m)/(2+m+n)/(3+m+n)/((e*( g*x+f)/(-d*g+e*f))^n)
Time = 0.22 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.71 \[ \int (d+e x)^m (f+g x)^n \left (a+b x+c x^2\right ) \, dx=\frac {(d+e x)^{1+m} (f+g x)^n \left (\frac {e (f+g x)}{e f-d g}\right )^{-n} \left (c (e f-d g)^2 \operatorname {Hypergeometric2F1}\left (1+m,-2-n,2+m,\frac {g (d+e x)}{-e f+d g}\right )+e \left (-\left ((2 c f-b g) (e f-d g) \operatorname {Hypergeometric2F1}\left (1+m,-1-n,2+m,\frac {g (d+e x)}{-e f+d g}\right )\right )+e \left (c f^2+g (-b f+a g)\right ) \operatorname {Hypergeometric2F1}\left (1+m,-n,2+m,\frac {g (d+e x)}{-e f+d g}\right )\right )\right )}{e^3 g^2 (1+m)} \]
((d + e*x)^(1 + m)*(f + g*x)^n*(c*(e*f - d*g)^2*Hypergeometric2F1[1 + m, - 2 - n, 2 + m, (g*(d + e*x))/(-(e*f) + d*g)] + e*(-((2*c*f - b*g)*(e*f - d* g)*Hypergeometric2F1[1 + m, -1 - n, 2 + m, (g*(d + e*x))/(-(e*f) + d*g)]) + e*(c*f^2 + g*(-(b*f) + a*g))*Hypergeometric2F1[1 + m, -n, 2 + m, (g*(d + e*x))/(-(e*f) + d*g)])))/(e^3*g^2*(1 + m)*((e*(f + g*x))/(e*f - d*g))^n)
Time = 0.41 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1194, 90, 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.
| \(\displaystyle \int \left (a+b x+c x^2\right ) (d+e x)^m (f+g x)^n \, dx\) |
| \(\Big \downarrow \) 1194 |
| \(\displaystyle \frac {\int (d+e x)^m (f+g x)^n \left (a g (m+n+3) e^2-(c e f (m+2)-b e g (m+n+3)+c d g (m+2 n+4)) x e-c d (e f (m+2)+d g (n+1))\right )dx}{e^2 g (m+n+3)}+\frac {c (d+e x)^{m+2} (f+g x)^{n+1}}{e^2 g (m+n+3)}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {\left (a e^2 g (m+n+3)+\frac {(d g (n+1)+e f (m+1)) (-b e g (m+n+3)+c d g (m+2 n+4)+c e f (m+2))}{g (m+n+2)}-c d (d g (n+1)+e f (m+2))\right ) \int (d+e x)^m (f+g x)^ndx-\frac {(d+e x)^{m+1} (f+g x)^{n+1} (-b e g (m+n+3)+c d g (m+2 n+4)+c e f (m+2))}{g (m+n+2)}}{e^2 g (m+n+3)}+\frac {c (d+e x)^{m+2} (f+g x)^{n+1}}{e^2 g (m+n+3)}\) |
| \(\Big \downarrow \) 80 |
| \(\displaystyle \frac {(f+g x)^n \left (\frac {e (f+g x)}{e f-d g}\right )^{-n} \left (a e^2 g (m+n+3)+\frac {(d g (n+1)+e f (m+1)) (-b e g (m+n+3)+c d g (m+2 n+4)+c e f (m+2))}{g (m+n+2)}-c d (d g (n+1)+e f (m+2))\right ) \int (d+e x)^m \left (\frac {e f}{e f-d g}+\frac {e g x}{e f-d g}\right )^ndx-\frac {(d+e x)^{m+1} (f+g x)^{n+1} (-b e g (m+n+3)+c d g (m+2 n+4)+c e f (m+2))}{g (m+n+2)}}{e^2 g (m+n+3)}+\frac {c (d+e x)^{m+2} (f+g x)^{n+1}}{e^2 g (m+n+3)}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle \frac {\frac {(d+e x)^{m+1} (f+g x)^n \left (\frac {e (f+g x)}{e f-d g}\right )^{-n} \operatorname {Hypergeometric2F1}\left (m+1,-n,m+2,-\frac {g (d+e x)}{e f-d g}\right ) \left (a e^2 g (m+n+3)+\frac {(d g (n+1)+e f (m+1)) (-b e g (m+n+3)+c d g (m+2 n+4)+c e f (m+2))}{g (m+n+2)}-c d (d g (n+1)+e f (m+2))\right )}{e (m+1)}-\frac {(d+e x)^{m+1} (f+g x)^{n+1} (-b e g (m+n+3)+c d g (m+2 n+4)+c e f (m+2))}{g (m+n+2)}}{e^2 g (m+n+3)}+\frac {c (d+e x)^{m+2} (f+g x)^{n+1}}{e^2 g (m+n+3)}\) |
(c*(d + e*x)^(2 + m)*(f + g*x)^(1 + n))/(e^2*g*(3 + m + n)) + (-(((c*e*f*( 2 + m) - b*e*g*(3 + m + n) + c*d*g*(4 + m + 2*n))*(d + e*x)^(1 + m)*(f + g *x)^(1 + n))/(g*(2 + m + n))) + ((a*e^2*g*(3 + m + n) - c*d*(e*f*(2 + m) + d*g*(1 + n)) + ((e*f*(1 + m) + d*g*(1 + n))*(c*e*f*(2 + m) - b*e*g*(3 + m + n) + c*d*g*(4 + m + 2*n)))/(g*(2 + m + n)))*(d + e*x)^(1 + m)*(f + g*x) ^n*Hypergeometric2F1[1 + m, -n, 2 + m, -((g*(d + e*x))/(e*f - d*g))])/(e*( 1 + m)*((e*(f + g*x))/(e*f - d*g))^n))/(e^2*g*(3 + m + n))
3.10.53.3.1 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[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[c^p*(d + e*x)^(m + 2*p)*((f + g*x )^(n + 1)/(g*e^(2*p)*(m + n + 2*p + 1))), x] + Simp[1/(g*e^(2*p)*(m + n + 2 *p + 1)) Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^( 2*p)*(a + b*x + c*x^2)^p - c^p*(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p) *(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ [p, 0] && !IntegerQ[m] && !IntegerQ[n] && NeQ[m + n + 2*p + 1, 0]
\[\int \left (e x +d \right )^{m} \left (g x +f \right )^{n} \left (c \,x^{2}+b x +a \right )d x\]
\[ \int (d+e x)^m (f+g x)^n \left (a+b x+c x^2\right ) \, dx=\int { {\left (c x^{2} + b x + a\right )} {\left (e x + d\right )}^{m} {\left (g x + f\right )}^{n} \,d x } \]
Exception generated. \[ \int (d+e x)^m (f+g x)^n \left (a+b x+c x^2\right ) \, dx=\text {Exception raised: HeuristicGCDFailed} \]
\[ \int (d+e x)^m (f+g x)^n \left (a+b x+c x^2\right ) \, dx=\int { {\left (c x^{2} + b x + a\right )} {\left (e x + d\right )}^{m} {\left (g x + f\right )}^{n} \,d x } \]
\[ \int (d+e x)^m (f+g x)^n \left (a+b x+c x^2\right ) \, dx=\int { {\left (c x^{2} + b x + a\right )} {\left (e x + d\right )}^{m} {\left (g x + f\right )}^{n} \,d x } \]
Timed out. \[ \int (d+e x)^m (f+g x)^n \left (a+b x+c x^2\right ) \, dx=\int {\left (f+g\,x\right )}^n\,{\left (d+e\,x\right )}^m\,\left (c\,x^2+b\,x+a\right ) \,d x \]