Integrand size = 27, antiderivative size = 461 \[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx=-\frac {c (3 c e f+c d g-2 b e g) (d+e x)^{1+m}}{e^2 g^4 (1+m)}+\frac {c^2 (d+e x)^{2+m}}{e^2 g^3 (2+m)}+\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}+\frac {\left (c f^2-b f g+a g^2\right ) (c f (8 d g-e f (7+m))+g (a e g (1-m)-b (4 d g-e f (3+m)))) (d+e x)^{1+m}}{2 g^4 (e f-d g)^2 (f+g x)}+\frac {\left (c^2 f^2 \left (12 d^2 g^2-8 d e f g (3+m)+e^2 f^2 \left (12+7 m+m^2\right )\right )-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (1+m))-b^2 \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-b f \left (6 d^2 g^2-6 d e f g (2+m)+e^2 f^2 \left (6+5 m+m^2\right )\right )\right )\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {g (d+e x)}{e f-d g}\right )}{2 g^4 (e f-d g)^3 (1+m)} \]
-c*(-2*b*e*g+c*d*g+3*c*e*f)*(e*x+d)^(1+m)/e^2/g^4/(1+m)+c^2*(e*x+d)^(2+m)/ e^2/g^3/(2+m)+1/2*(a*g^2-b*f*g+c*f^2)^2*(e*x+d)^(1+m)/g^4/(-d*g+e*f)/(g*x+ f)^2+1/2*(a*g^2-b*f*g+c*f^2)*(c*f*(8*d*g-e*f*(7+m))+g*(a*e*g*(1-m)-b*(4*d* g-e*f*(3+m))))*(e*x+d)^(1+m)/g^4/(-d*g+e*f)^2/(g*x+f)+1/2*(c^2*f^2*(12*d^2 *g^2-8*d*e*f*g*(3+m)+e^2*f^2*(m^2+7*m+12))-g^2*(a^2*e^2*g^2*(1-m)*m-2*a*b* e*g*m*(2*d*g-e*f*(1+m))-b^2*(2*d^2*g^2-4*d*e*f*g*(1+m)+e^2*f^2*(m^2+3*m+2) ))+2*c*g*(a*g*(2*d^2*g^2-4*d*e*f*g*(1+m)+e^2*f^2*(m^2+3*m+2))-b*f*(6*d^2*g ^2-6*d*e*f*g*(2+m)+e^2*f^2*(m^2+5*m+6))))*(e*x+d)^(1+m)*hypergeom([1, 1+m] ,[2+m],-g*(e*x+d)/(-d*g+e*f))/g^4/(-d*g+e*f)^3/(1+m)
Time = 1.18 (sec) , antiderivative size = 257, normalized size of antiderivative = 0.56 \[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx=\frac {(d+e x)^{1+m} \left (-\frac {c (3 c e f+c d g-2 b e g)}{e^2 (1+m)}+\frac {c^2 g (d+e x)}{e^2 (2+m)}+\frac {\left (6 c^2 f^2+b^2 g^2+2 c g (-3 b f+a g)\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {g (d+e x)}{-e f+d g}\right )}{(e f-d g) (1+m)}-\frac {2 e (2 c f-b g) \left (c f^2+g (-b f+a g)\right ) \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {g (d+e x)}{-e f+d g}\right )}{(e f-d g)^2 (1+m)}+\frac {e^2 \left (c f^2+g (-b f+a g)\right )^2 \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,\frac {g (d+e x)}{-e f+d g}\right )}{(e f-d g)^3 (1+m)}\right )}{g^4} \]
((d + e*x)^(1 + m)*(-((c*(3*c*e*f + c*d*g - 2*b*e*g))/(e^2*(1 + m))) + (c^ 2*g*(d + e*x))/(e^2*(2 + m)) + ((6*c^2*f^2 + b^2*g^2 + 2*c*g*(-3*b*f + a*g ))*Hypergeometric2F1[1, 1 + m, 2 + m, (g*(d + e*x))/(-(e*f) + d*g)])/((e*f - d*g)*(1 + m)) - (2*e*(2*c*f - b*g)*(c*f^2 + g*(-(b*f) + a*g))*Hypergeom etric2F1[2, 1 + m, 2 + m, (g*(d + e*x))/(-(e*f) + d*g)])/((e*f - d*g)^2*(1 + m)) + (e^2*(c*f^2 + g*(-(b*f) + a*g))^2*Hypergeometric2F1[3, 1 + m, 2 + m, (g*(d + e*x))/(-(e*f) + d*g)])/((e*f - d*g)^3*(1 + m))))/g^4
Time = 1.88 (sec) , antiderivative size = 504, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1193, 2124, 1195, 2009}
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.
| \(\displaystyle \int \frac {\left (a+b x+c x^2\right )^2 (d+e x)^m}{(f+g x)^3} \, dx\) |
| \(\Big \downarrow \) 1193 |
| \(\displaystyle \frac {\int \frac {(d+e x)^m \left (-2 c^2 \left (d-\frac {e f}{g}\right ) x^3-\frac {2 c (c f-2 b g) (e f-d g) x^2}{g^2}+\frac {2 (e f-d g) \left (c^2 f^2+b^2 g^2-2 c g (b f-a g)\right ) x}{g^3}+\frac {c^2 (2 d g-e f (m+1)) f^3-2 c g (b f-a g) (2 d g-e f (m+1)) f+g^2 \left (f (2 d g-e f (m+1)) b^2-2 a g (2 d g-e f (m+1)) b+a^2 e g^2 (1-m)\right )}{g^4}\right )}{(f+g x)^2}dx}{2 (e f-d g)}+\frac {(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2}{2 g^4 (f+g x)^2 (e f-d g)}\) |
| \(\Big \downarrow \) 2124 |
| \(\displaystyle \frac {\frac {\int \frac {(d+e x)^m \left (\frac {2 c^2 x^2 (e f-d g)^2}{g^2}-\frac {4 c (c f-b g) x (e f-d g)^2}{g^3}+\frac {c^2 \left (e^2 \left (m^2+7 m+6\right ) f^2-4 d e g (2 m+3) f+6 d^2 g^2\right ) f^2-g^2 \left (-\left (\left (e^2 \left (m^2+3 m+2\right ) f^2-4 d e g (m+1) f+2 d^2 g^2\right ) b^2\right )-2 a e g m (2 d g-e f (m+1)) b+a^2 e^2 g^2 (1-m) m\right )+2 c g \left (a g \left (e^2 \left (m^2+3 m+2\right ) f^2-4 d e g (m+1) f+2 d^2 g^2\right )-b f \left (e^2 \left (m^2+5 m+4\right ) f^2-2 d e g (3 m+4) f+4 d^2 g^2\right )\right )}{g^4}\right )}{f+g x}dx}{e f-d g}-\frac {(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right ) (g (-a e g (1-m)+4 b d g-b e f (m+3))-c f (8 d g-e f (m+7)))}{g^4 (f+g x) (e f-d g)}}{2 (e f-d g)}+\frac {(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2}{2 g^4 (f+g x)^2 (e f-d g)}\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle \frac {\frac {\int \left (-\frac {2 c (e f-d g)^2 (3 c e f+c d g-2 b e g) (d+e x)^m}{e g^4}+\frac {\left (c^2 \left (e^2 \left (m^2+7 m+12\right ) f^2-8 d e g (m+3) f+12 d^2 g^2\right ) f^2-g^2 \left (-\left (\left (e^2 \left (m^2+3 m+2\right ) f^2-4 d e g (m+1) f+2 d^2 g^2\right ) b^2\right )-2 a e g m (2 d g-e f (m+1)) b+a^2 e^2 g^2 (1-m) m\right )+2 c g \left (a g \left (e^2 \left (m^2+3 m+2\right ) f^2-4 d e g (m+1) f+2 d^2 g^2\right )-b f \left (e^2 \left (m^2+5 m+6\right ) f^2-6 d e g (m+2) f+6 d^2 g^2\right )\right )\right ) (d+e x)^m}{g^4 (f+g x)}+\frac {2 c^2 (e f-d g)^2 (d+e x)^{m+1}}{e g^3}\right )dx}{e f-d g}-\frac {(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right ) (g (-a e g (1-m)+4 b d g-b e f (m+3))-c f (8 d g-e f (m+7)))}{g^4 (f+g x) (e f-d g)}}{2 (e f-d g)}+\frac {(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2}{2 g^4 (f+g x)^2 (e f-d g)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\frac {(d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {g (d+e x)}{e f-d g}\right ) \left (-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (m+1))-\left (b^2 \left (2 d^2 g^2-4 d e f g (m+1)+e^2 f^2 \left (m^2+3 m+2\right )\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (m+1)+e^2 f^2 \left (m^2+3 m+2\right )\right )-b f \left (6 d^2 g^2-6 d e f g (m+2)+e^2 f^2 \left (m^2+5 m+6\right )\right )\right )+c^2 f^2 \left (12 d^2 g^2-8 d e f g (m+3)+e^2 f^2 \left (m^2+7 m+12\right )\right )\right )}{g^4 (m+1) (e f-d g)}-\frac {2 c (e f-d g)^2 (d+e x)^{m+1} (-2 b e g+c d g+3 c e f)}{e^2 g^4 (m+1)}+\frac {2 c^2 (e f-d g)^2 (d+e x)^{m+2}}{e^2 g^3 (m+2)}}{e f-d g}-\frac {(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right ) (g (-a e g (1-m)+4 b d g-b e f (m+3))-c f (8 d g-e f (m+7)))}{g^4 (f+g x) (e f-d g)}}{2 (e f-d g)}+\frac {(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2}{2 g^4 (f+g x)^2 (e f-d g)}\) |
((c*f^2 - b*f*g + a*g^2)^2*(d + e*x)^(1 + m))/(2*g^4*(e*f - d*g)*(f + g*x) ^2) + (-(((c*f^2 - b*f*g + a*g^2)*(g*(4*b*d*g - a*e*g*(1 - m) - b*e*f*(3 + m)) - c*f*(8*d*g - e*f*(7 + m)))*(d + e*x)^(1 + m))/(g^4*(e*f - d*g)*(f + g*x))) + ((-2*c*(e*f - d*g)^2*(3*c*e*f + c*d*g - 2*b*e*g)*(d + e*x)^(1 + m))/(e^2*g^4*(1 + m)) + (2*c^2*(e*f - d*g)^2*(d + e*x)^(2 + m))/(e^2*g^3*( 2 + m)) + ((c^2*f^2*(12*d^2*g^2 - 8*d*e*f*g*(3 + m) + e^2*f^2*(12 + 7*m + m^2)) - g^2*(a^2*e^2*g^2*(1 - m)*m - 2*a*b*e*g*m*(2*d*g - e*f*(1 + m)) - b ^2*(2*d^2*g^2 - 4*d*e*f*g*(1 + m) + e^2*f^2*(2 + 3*m + m^2))) + 2*c*g*(a*g *(2*d^2*g^2 - 4*d*e*f*g*(1 + m) + e^2*f^2*(2 + 3*m + m^2)) - b*f*(6*d^2*g^ 2 - 6*d*e*f*g*(2 + m) + e^2*f^2*(6 + 5*m + m^2))))*(d + e*x)^(1 + m)*Hyper geometric2F1[1, 1 + m, 2 + m, -((g*(d + e*x))/(e*f - d*g))])/(g^4*(e*f - d *g)*(1 + m)))/(e*f - d*g))/(2*(e*f - d*g))
3.10.29.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p, d + e*x, x]}, Simp[R*(d + e*x)^(m + 1)*((f + g*x)^(n + 1)/((m + 1)*(e*f - d*g)) ), x] + Simp[1/((m + 1)*(e*f - d*g)) Int[(d + e*x)^(m + 1)*(f + g*x)^n*Ex pandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /; FreeQ[{a , b, c, d, e, f, g, n}, x] && IGtQ[p, 0] && ILtQ[2*m, -2] && !IntegerQ[n] && !(EqQ[m, -2] && EqQ[p, 1] && EqQ[2*c*d - b*e, 0])
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : > With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px , a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - a*d))), x] + Simp[1/((m + 1)*(b*c - a*d)) Int[(a + b*x)^(m + 1)*(c + d*x )^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] || ! ILtQ[n, -1])
\[\int \frac {\left (e x +d \right )^{m} \left (c \,x^{2}+b x +a \right )^{2}}{\left (g x +f \right )^{3}}d x\]
\[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{2} {\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{3}} \,d x } \]
integral((c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*(e*x + d)^m/(g^3*x^3 + 3*f*g^2*x^2 + 3*f^2*g*x + f^3), x)
\[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx=\int \frac {\left (d + e x\right )^{m} \left (a + b x + c x^{2}\right )^{2}}{\left (f + g x\right )^{3}}\, dx \]
\[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{2} {\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{3}} \,d x } \]
\[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{2} {\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx=\int \frac {{\left (d+e\,x\right )}^m\,{\left (c\,x^2+b\,x+a\right )}^2}{{\left (f+g\,x\right )}^3} \,d x \]