Integrand size = 25, antiderivative size = 344 \[ \int (d+e x)^2 (f+g x) \left (a+b x+c x^2\right )^p \, dx=\frac {g (d+e x)^2 \left (a+b x+c x^2\right )^{1+p}}{2 c (2+p)}+\frac {(b e (2+p) (b e g (3+p)-2 c (d g+e f (2+p)))-2 c (3+2 p) (e g (a e+b d (2+p))-c d (d g+2 e f (2+p)))-2 c e (1+p) (b e g (3+p)-2 c (d g+e f (2+p))) x) \left (a+b x+c x^2\right )^{1+p}}{4 c^3 (1+p) (2+p) (3+2 p)}-\frac {2^{-3-2 p} \left (b^3 e^2 g (3+p)-4 c^3 d^2 f (3+2 p)-2 b c e (3 a e g+b (e f+2 d g) (2+p))+2 c^2 (2 a e (e f+2 d g)+b d (2 e f+d g) (3+2 p))\right ) (b+2 c x) \left (a+b x+c x^2\right )^p \left (-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{c^4 (3+2 p)} \] Output:
1/2*g*(e*x+d)^2*(c*x^2+b*x+a)^(p+1)/c/(2+p)+1/4*(b*e*(2+p)*(b*e*g*(3+p)-2* c*(d*g+e*f*(2+p)))-2*c*(3+2*p)*(e*g*(a*e+b*d*(2+p))-c*d*(d*g+2*e*f*(2+p))) -2*c*e*(p+1)*(b*e*g*(3+p)-2*c*(d*g+e*f*(2+p)))*x)*(c*x^2+b*x+a)^(p+1)/c^3/ (p+1)/(2+p)/(3+2*p)-2^(-3-2*p)*(b^3*e^2*g*(3+p)-4*c^3*d^2*f*(3+2*p)-2*b*c* e*(3*a*e*g+b*(2*d*g+e*f)*(2+p))+2*c^2*(2*a*e*(2*d*g+e*f)+b*d*(d*g+2*e*f)*( 3+2*p)))*(2*c*x+b)*(c*x^2+b*x+a)^p*hypergeom([1/2, -p],[3/2],(2*c*x+b)^2/( -4*a*c+b^2))/c^4/(3+2*p)/((-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^p)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 2.83 (sec) , antiderivative size = 659, normalized size of antiderivative = 1.92 \[ \int (d+e x)^2 (f+g x) \left (a+b x+c x^2\right )^p \, dx=\frac {2^{-2-p} \left (\frac {b-\sqrt {b^2-4 a c}}{2 c}+x\right )^{-p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{c}\right )^p \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} (a+x (b+c x))^p \left (6 c d (2 e f+d g) (1+p) x^2 \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^p \operatorname {AppellF1}\left (2,-p,-p,3,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+4 c e (e f+2 d g) (1+p) x^3 \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^p \operatorname {AppellF1}\left (3,-p,-p,4,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+3 c e^2 g (1+p) x^4 \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^p \operatorname {AppellF1}\left (4,-p,-p,5,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )-3\ 2^{1+p} d^2 f \left (-b+\sqrt {b^2-4 a c}-2 c x\right ) \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^p \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^p \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {-b+\sqrt {b^2-4 a c}-2 c x}{2 \sqrt {b^2-4 a c}}\right )\right )}{3 c (1+p)} \] Input:
Integrate[(d + e*x)^2*(f + g*x)*(a + b*x + c*x^2)^p,x]
Output:
(2^(-2 - p)*((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/c)^p*(a + x*(b + c*x))^p*(6*c *d*(2*e*f + d*g)*(1 + p)*x^2*((b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4 *a*c])^p*AppellF1[2, -p, -p, 3, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x)/ (-b + Sqrt[b^2 - 4*a*c])] + 4*c*e*(e*f + 2*d*g)*(1 + p)*x^3*((b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c])^p*AppellF1[3, -p, -p, 4, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x)/(-b + Sqrt[b^2 - 4*a*c])] + 3*c*e^2*g*(1 + p)*x^4*((b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c])^p*AppellF1[4, - p, -p, 5, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x)/(-b + Sqrt[b^2 - 4*a*c ])] - 3*2^(1 + p)*d^2*f*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x)*((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x )/(b + Sqrt[b^2 - 4*a*c]))^p*Hypergeometric2F1[-p, 1 + p, 2 + p, (-b + Sqr t[b^2 - 4*a*c] - 2*c*x)/(2*Sqrt[b^2 - 4*a*c])]))/(3*c*(1 + p)*((b - Sqrt[b ^2 - 4*a*c])/(2*c) + x)^p*((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c])^p*((b + Sq rt[b^2 - 4*a*c] + 2*c*x)/(b + Sqrt[b^2 - 4*a*c]))^p)
Time = 1.00 (sec) , antiderivative size = 391, normalized size of antiderivative = 1.14, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1236, 1225, 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.
| \(\displaystyle \int (d+e x)^2 (f+g x) \left (a+b x+c x^2\right )^p \, dx\) |
| \(\Big \downarrow \) 1236 |
| \(\displaystyle \frac {\int (d+e x) (2 c d f (p+2)-g (2 a e+b d (p+1))-(b e g (p+3)-2 c (d g+e f (p+2))) x) \left (c x^2+b x+a\right )^pdx}{2 c (p+2)}+\frac {g (d+e x)^2 \left (a+b x+c x^2\right )^{p+1}}{2 c (p+2)}\) |
| \(\Big \downarrow \) 1225 |
| \(\displaystyle \frac {\frac {\left (a+b x+c x^2\right )^{p+1} (-2 c (2 p+3) (e g (a e+b d (p+2))-c d (d g+2 e f (p+2)))-2 c e (p+1) x (b e g (p+3)-2 c (d g+e f (p+2)))+b e (p+2) (b e g (p+3)-2 c (d g+e f (p+2))))}{2 c^2 (p+1) (2 p+3)}-\frac {(p+2) \left (2 c^2 (2 a e (2 d g+e f)+b d (2 p+3) (d g+2 e f))-2 b c e (3 a e g+b (p+2) (2 d g+e f))+b^3 e^2 g (p+3)-4 c^3 d^2 f (2 p+3)\right ) \int \left (c x^2+b x+a\right )^pdx}{2 c^2 (2 p+3)}}{2 c (p+2)}+\frac {g (d+e x)^2 \left (a+b x+c x^2\right )^{p+1}}{2 c (p+2)}\) |
| \(\Big \downarrow \) 1096 |
| \(\displaystyle \frac {\frac {2^p (p+2) \left (a+b x+c x^2\right )^{p+1} \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right ) \left (2 c^2 (2 a e (2 d g+e f)+b d (2 p+3) (d g+2 e f))-2 b c e (3 a e g+b (p+2) (2 d g+e f))+b^3 e^2 g (p+3)-4 c^3 d^2 f (2 p+3)\right )}{c^2 (p+1) (2 p+3) \sqrt {b^2-4 a c}}+\frac {\left (a+b x+c x^2\right )^{p+1} (-2 c (2 p+3) (e g (a e+b d (p+2))-c d (d g+2 e f (p+2)))-2 c e (p+1) x (b e g (p+3)-2 c (d g+e f (p+2)))+b e (p+2) (b e g (p+3)-2 c (d g+e f (p+2))))}{2 c^2 (p+1) (2 p+3)}}{2 c (p+2)}+\frac {g (d+e x)^2 \left (a+b x+c x^2\right )^{p+1}}{2 c (p+2)}\) |
Input:
Int[(d + e*x)^2*(f + g*x)*(a + b*x + c*x^2)^p,x]
Output:
(g*(d + e*x)^2*(a + b*x + c*x^2)^(1 + p))/(2*c*(2 + p)) + (((b*e*(2 + p)*( b*e*g*(3 + p) - 2*c*(d*g + e*f*(2 + p))) - 2*c*(3 + 2*p)*(e*g*(a*e + b*d*( 2 + p)) - c*d*(d*g + 2*e*f*(2 + p))) - 2*c*e*(1 + p)*(b*e*g*(3 + p) - 2*c* (d*g + e*f*(2 + p)))*x)*(a + b*x + c*x^2)^(1 + p))/(2*c^2*(1 + p)*(3 + 2*p )) + (2^p*(2 + p)*(b^3*e^2*g*(3 + p) - 4*c^3*d^2*f*(3 + 2*p) - 2*b*c*e*(3* a*e*g + b*(e*f + 2*d*g)*(2 + p)) + 2*c^2*(2*a*e*(e*f + 2*d*g) + b*d*(2*e*f + d*g)*(3 + 2*p)))*(-((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c])) ^(-1 - p)*(a + b*x + c*x^2)^(1 + p)*Hypergeometric2F1[-p, 1 + p, 2 + p, (b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(2*Sqrt[b^2 - 4*a*c])])/(c^2*Sqrt[b^2 - 4*a* c]*(1 + p)*(3 + 2*p)))/(2*c*(2 + p))
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[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c , d, e, f, g, p}, x] && !LeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Simp[1/(c*(m + 2*p + 2)) Int[(d + e*x)^(m - 1 )*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m *(c*e*f + c*d*g - b*e*g) + e*(p + 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[ {a, b, c, d, e, f, g, p}, x] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (Intege rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) && !(IGtQ[m, 0] && EqQ[f, 0])
\[\int \left (e x +d \right )^{2} \left (g x +f \right ) \left (c \,x^{2}+b x +a \right )^{p}d x\]
Input:
int((e*x+d)^2*(g*x+f)*(c*x^2+b*x+a)^p,x)
Output:
int((e*x+d)^2*(g*x+f)*(c*x^2+b*x+a)^p,x)
\[ \int (d+e x)^2 (f+g x) \left (a+b x+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (g x + f\right )} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \] Input:
integrate((e*x+d)^2*(g*x+f)*(c*x^2+b*x+a)^p,x, algorithm="fricas")
Output:
integral((e^2*g*x^3 + d^2*f + (e^2*f + 2*d*e*g)*x^2 + (2*d*e*f + d^2*g)*x) *(c*x^2 + b*x + a)^p, x)
Timed out. \[ \int (d+e x)^2 (f+g x) \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \] Input:
integrate((e*x+d)**2*(g*x+f)*(c*x**2+b*x+a)**p,x)
Output:
Timed out
\[ \int (d+e x)^2 (f+g x) \left (a+b x+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (g x + f\right )} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \] Input:
integrate((e*x+d)^2*(g*x+f)*(c*x^2+b*x+a)^p,x, algorithm="maxima")
Output:
integrate((e*x + d)^2*(g*x + f)*(c*x^2 + b*x + a)^p, x)
\[ \int (d+e x)^2 (f+g x) \left (a+b x+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (g x + f\right )} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \] Input:
integrate((e*x+d)^2*(g*x+f)*(c*x^2+b*x+a)^p,x, algorithm="giac")
Output:
integrate((e*x + d)^2*(g*x + f)*(c*x^2 + b*x + a)^p, x)
Timed out. \[ \int (d+e x)^2 (f+g x) \left (a+b x+c x^2\right )^p \, dx=\int \left (f+g\,x\right )\,{\left (d+e\,x\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \] Input:
int((f + g*x)*(d + e*x)^2*(a + b*x + c*x^2)^p,x)
Output:
int((f + g*x)*(d + e*x)^2*(a + b*x + c*x^2)^p, x)
\[ \int (d+e x)^2 (f+g x) \left (a+b x+c x^2\right )^p \, dx=\text {too large to display} \] Input:
int((e*x+d)^2*(g*x+f)*(c*x^2+b*x+a)^p,x)
Output:
(4*(a + b*x + c*x**2)**p*a**2*b*c*e**2*g*p**2 + 20*(a + b*x + c*x**2)**p*a **2*b*c*e**2*g*p + 18*(a + b*x + c*x**2)**p*a**2*b*c*e**2*g - 16*(a + b*x + c*x**2)**p*a**2*c**2*d*e*g*p**2 - 48*(a + b*x + c*x**2)**p*a**2*c**2*d*e *g*p - 32*(a + b*x + c*x**2)**p*a**2*c**2*d*e*g - 8*(a + b*x + c*x**2)**p* a**2*c**2*e**2*f*p**2 - 24*(a + b*x + c*x**2)**p*a**2*c**2*e**2*f*p - 16*( a + b*x + c*x**2)**p*a**2*c**2*e**2*f - (a + b*x + c*x**2)**p*a*b**3*e**2* g*p**2 - 5*(a + b*x + c*x**2)**p*a*b**3*e**2*g*p - 6*(a + b*x + c*x**2)**p *a*b**3*e**2*g + 4*(a + b*x + c*x**2)**p*a*b**2*c*d*e*g*p**2 + 16*(a + b*x + c*x**2)**p*a*b**2*c*d*e*g*p + 16*(a + b*x + c*x**2)**p*a*b**2*c*d*e*g + 2*(a + b*x + c*x**2)**p*a*b**2*c*e**2*f*p**2 + 8*(a + b*x + c*x**2)**p*a* b**2*c*e**2*f*p + 8*(a + b*x + c*x**2)**p*a*b**2*c*e**2*f - 4*(a + b*x + c *x**2)**p*a*b**2*c*e**2*g*p**3*x - 20*(a + b*x + c*x**2)**p*a*b**2*c*e**2* g*p**2*x - 18*(a + b*x + c*x**2)**p*a*b**2*c*e**2*g*p*x - 4*(a + b*x + c*x **2)**p*a*b*c**2*d**2*g*p**2 - 14*(a + b*x + c*x**2)**p*a*b*c**2*d**2*g*p - 12*(a + b*x + c*x**2)**p*a*b*c**2*d**2*g - 8*(a + b*x + c*x**2)**p*a*b*c **2*d*e*f*p**2 - 28*(a + b*x + c*x**2)**p*a*b*c**2*d*e*f*p - 24*(a + b*x + c*x**2)**p*a*b*c**2*d*e*f + 16*(a + b*x + c*x**2)**p*a*b*c**2*d*e*g*p**3* x + 48*(a + b*x + c*x**2)**p*a*b*c**2*d*e*g*p**2*x + 32*(a + b*x + c*x**2) **p*a*b*c**2*d*e*g*p*x + 8*(a + b*x + c*x**2)**p*a*b*c**2*e**2*f*p**3*x + 24*(a + b*x + c*x**2)**p*a*b*c**2*e**2*f*p**2*x + 16*(a + b*x + c*x**2)...