Integrand size = 20, antiderivative size = 187 \[ \int (d+e x)^2 \left (a+b x+c x^2\right )^p \, dx=-\frac {e (b e (2+p)-2 c d (3+2 p)-2 c e (1+p) x) \left (a+b x+c x^2\right )^{1+p}}{2 c^2 (1+p) (3+2 p)}+\frac {2^{-2 (1+p)} \left (2 c d (c d-b e) (3+2 p)-e^2 \left (2 a c-b^2 (2+p)\right )\right ) (b+2 c x) \left (a+b x+c x^2\right )^p \left (-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{c^3 (3+2 p)} \] Output:
-1/2*e*(b*e*(2+p)-2*c*d*(3+2*p)-2*c*e*(p+1)*x)*(c*x^2+b*x+a)^(p+1)/c^2/(p+ 1)/(3+2*p)+(2*c*d*(-b*e+c*d)*(3+2*p)-e^2*(2*a*c-b^2*(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))/(2^(2*p+2) )/c^3/(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 = 1.23 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.21 \[ \int (d+e x)^2 \left (a+b x+c x^2\right )^p \, dx=\frac {1}{6} (a+x (b+c x))^p \left (6 d e x^2 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {AppellF1}\left (2,-p,-p,3,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+2 e^2 x^3 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {AppellF1}\left (3,-p,-p,4,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+\frac {3\ 2^p d^2 \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {-b+\sqrt {b^2-4 a c}-2 c x}{2 \sqrt {b^2-4 a c}}\right )}{c (1+p)}\right ) \] Input:
Integrate[(d + e*x)^2*(a + b*x + c*x^2)^p,x]
Output:
((a + x*(b + c*x))^p*((6*d*e*x^2*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])])/(((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) + (2*e^2*x^3*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])])/(((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) + (3*2^p*d^2*(b - Sqrt[b^2 - 4*a*c] + 2*c*x) *Hypergeometric2F1[-p, 1 + p, 2 + p, (-b + Sqrt[b^2 - 4*a*c] - 2*c*x)/(2*S qrt[b^2 - 4*a*c])])/(c*(1 + p)*((b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c])^p)))/6
Time = 0.39 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.32, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1166, 1160, 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.
| \(\displaystyle \int (d+e x)^2 \left (a+b x+c x^2\right )^p \, dx\) |
| \(\Big \downarrow \) 1166 |
| \(\displaystyle \frac {\int \left (c (2 p+3) d^2-e (a e+b d (p+1))+e (2 c d-b e) (p+2) x\right ) \left (c x^2+b x+a\right )^pdx}{c (2 p+3)}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{p+1}}{c (2 p+3)}\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {\frac {\left (-2 c e (a e+b d (2 p+3))+b^2 e^2 (p+2)+2 c^2 d^2 (2 p+3)\right ) \int \left (c x^2+b x+a\right )^pdx}{2 c}+\frac {e (p+2) (2 c d-b e) \left (a+b x+c x^2\right )^{p+1}}{2 c (p+1)}}{c (2 p+3)}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{p+1}}{c (2 p+3)}\) |
| \(\Big \downarrow \) 1096 |
| \(\displaystyle \frac {\frac {e (p+2) (2 c d-b e) \left (a+b x+c x^2\right )^{p+1}}{2 c (p+1)}-\frac {2^p \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \left (a+b x+c x^2\right )^{p+1} \left (-2 c e (a e+b d (2 p+3))+b^2 e^2 (p+2)+2 c^2 d^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{c (p+1) \sqrt {b^2-4 a c}}}{c (2 p+3)}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{p+1}}{c (2 p+3)}\) |
Input:
Int[(d + e*x)^2*(a + b*x + c*x^2)^p,x]
Output:
(e*(d + e*x)*(a + b*x + c*x^2)^(1 + p))/(c*(3 + 2*p)) + ((e*(2*c*d - b*e)* (2 + p)*(a + b*x + c*x^2)^(1 + p))/(2*c*(1 + p)) - (2^p*(b^2*e^2*(2 + p) + 2*c^2*d^2*(3 + 2*p) - 2*c*e*(a*e + b*d*(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)*Hyperg eometric2F1[-p, 1 + p, 2 + p, (b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(2*Sqrt[b^2 - 4*a*c])])/(c*Sqrt[b^2 - 4*a*c]*(1 + p)))/(c*(3 + 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_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[1/(c*(m + 2*p + 1)) Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* (a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat icQ[a, b, c, d, e, m, p, x]
\[\int \left (e x +d \right )^{2} \left (c \,x^{2}+b x +a \right )^{p}d x\]
Input:
int((e*x+d)^2*(c*x^2+b*x+a)^p,x)
Output:
int((e*x+d)^2*(c*x^2+b*x+a)^p,x)
\[ \int (d+e x)^2 \left (a+b x+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \] Input:
integrate((e*x+d)^2*(c*x^2+b*x+a)^p,x, algorithm="fricas")
Output:
integral((e^2*x^2 + 2*d*e*x + d^2)*(c*x^2 + b*x + a)^p, x)
\[ \int (d+e x)^2 \left (a+b x+c x^2\right )^p \, dx=\int \left (d + e x\right )^{2} \left (a + b x + c x^{2}\right )^{p}\, dx \] Input:
integrate((e*x+d)**2*(c*x**2+b*x+a)**p,x)
Output:
Integral((d + e*x)**2*(a + b*x + c*x**2)**p, x)
\[ \int (d+e x)^2 \left (a+b x+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \] Input:
integrate((e*x+d)^2*(c*x^2+b*x+a)^p,x, algorithm="maxima")
Output:
integrate((e*x + d)^2*(c*x^2 + b*x + a)^p, x)
\[ \int (d+e x)^2 \left (a+b x+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \] Input:
integrate((e*x+d)^2*(c*x^2+b*x+a)^p,x, algorithm="giac")
Output:
integrate((e*x + d)^2*(c*x^2 + b*x + a)^p, x)
Timed out. \[ \int (d+e x)^2 \left (a+b x+c x^2\right )^p \, dx=\int {\left (d+e\,x\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \] Input:
int((d + e*x)^2*(a + b*x + c*x^2)^p,x)
Output:
int((d + e*x)^2*(a + b*x + c*x^2)^p, x)
\[ \int (d+e x)^2 \left (a+b x+c x^2\right )^p \, dx=\text {too large to display} \] Input:
int((e*x+d)^2*(c*x^2+b*x+a)^p,x)
Output:
( - 4*(a + b*x + c*x**2)**p*a**2*c*e**2*p - 4*(a + b*x + c*x**2)**p*a**2*c *e**2 + (a + b*x + c*x**2)**p*a*b**2*e**2*p + 2*(a + b*x + c*x**2)**p*a*b* *2*e**2 - 4*(a + b*x + c*x**2)**p*a*b*c*d*e*p - 6*(a + b*x + c*x**2)**p*a* b*c*d*e + 4*(a + b*x + c*x**2)**p*a*b*c*e**2*p**2*x + 4*(a + b*x + c*x**2) **p*a*b*c*e**2*p*x + 8*(a + b*x + c*x**2)**p*a*c**2*d**2*p**2 + 20*(a + b* x + c*x**2)**p*a*c**2*d**2*p + 12*(a + b*x + c*x**2)**p*a*c**2*d**2 - (a + b*x + c*x**2)**p*b**3*e**2*p**2*x - 2*(a + b*x + c*x**2)**p*b**3*e**2*p*x + 4*(a + b*x + c*x**2)**p*b**2*c*d*e*p**2*x + 6*(a + b*x + c*x**2)**p*b** 2*c*d*e*p*x + 2*(a + b*x + c*x**2)**p*b**2*c*e**2*p**2*x**2 + (a + b*x + c *x**2)**p*b**2*c*e**2*p*x**2 + 4*(a + b*x + c*x**2)**p*b*c**2*d**2*p**2*x + 10*(a + b*x + c*x**2)**p*b*c**2*d**2*p*x + 6*(a + b*x + c*x**2)**p*b*c** 2*d**2*x + 8*(a + b*x + c*x**2)**p*b*c**2*d*e*p**2*x**2 + 16*(a + b*x + c* x**2)**p*b*c**2*d*e*p*x**2 + 6*(a + b*x + c*x**2)**p*b*c**2*d*e*x**2 + 4*( a + b*x + c*x**2)**p*b*c**2*e**2*p**2*x**3 + 6*(a + b*x + c*x**2)**p*b*c** 2*e**2*p*x**3 + 2*(a + b*x + c*x**2)**p*b*c**2*e**2*x**3 + 32*int(((a + b* x + c*x**2)**p*x)/(4*a*p**2 + 8*a*p + 3*a + 4*b*p**2*x + 8*b*p*x + 3*b*x + 4*c*p**2*x**2 + 8*c*p*x**2 + 3*c*x**2),x)*a**2*c**2*e**2*p**4 + 96*int((( a + b*x + c*x**2)**p*x)/(4*a*p**2 + 8*a*p + 3*a + 4*b*p**2*x + 8*b*p*x + 3 *b*x + 4*c*p**2*x**2 + 8*c*p*x**2 + 3*c*x**2),x)*a**2*c**2*e**2*p**3 + 88* int(((a + b*x + c*x**2)**p*x)/(4*a*p**2 + 8*a*p + 3*a + 4*b*p**2*x + 8*...