\(\int (a+b x+c x^2)^p (A+B x+C x^2+D x^3) \, dx\) [28]

Optimal result

Integrand size = 28, antiderivative size = 357 \[ \int \left (a+b x+c x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=-\frac {\left (2 b c C (2+p)^2+2 a c D (3+2 p)-b^2 D \left (6+5 p+p^2\right )-2 B c^2 \left (6+7 p+2 p^2\right )\right ) \left (a+b x+c x^2\right )^{1+p}}{4 c^3 (1+p) (2+p) (3+2 p)}+\frac {(2 c C (2+p)-b D (3+p)) x \left (a+b x+c x^2\right )^{1+p}}{2 c^2 (2+p) (3+2 p)}+\frac {D x^2 \left (a+b x+c x^2\right )^{1+p}}{2 c (2+p)}-\frac {2^{-1+p} \left (2 b^2 c C (2+p)-b^3 D (3+p)-4 c^2 (a C-A c (3+2 p))+2 b c (3 a D-B c (3+2 p))\right ) \left (-\frac {b-\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-1-p} \left (a+b x+c x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {b+\sqrt {b^2-4 a c}+2 c x}{2 \sqrt {b^2-4 a c}}\right )}{c^3 \sqrt {b^2-4 a c} (1+p) (3+2 p)} \] Output:

-1/4*(2*b*c*C*(2+p)^2+2*a*c*D*(3+2*p)-b^2*D*(p^2+5*p+6)-2*B*c^2*(2*p^2+7*p 
+6))*(c*x^2+b*x+a)^(p+1)/c^3/(p+1)/(2+p)/(3+2*p)+1/2*(2*c*C*(2+p)-b*D*(3+p 
))*x*(c*x^2+b*x+a)^(p+1)/c^2/(2+p)/(3+2*p)+1/2*D*x^2*(c*x^2+b*x+a)^(p+1)/c 
/(2+p)-2^(-1+p)*(2*b^2*c*C*(2+p)-b^3*D*(3+p)-4*c^2*(C*a-A*c*(3+2*p))+2*b*c 
*(3*a*D-B*c*(3+2*p)))*(-(b-(-4*a*c+b^2)^(1/2)+2*c*x)/(-4*a*c+b^2)^(1/2))^( 
-1-p)*(c*x^2+b*x+a)^(p+1)*hypergeom([-p, p+1],[2+p],1/2*(b+(-4*a*c+b^2)^(1 
/2)+2*c*x)/(-4*a*c+b^2)^(1/2))/c^3/(-4*a*c+b^2)^(1/2)/(p+1)/(3+2*p)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 2.62 (sec) , antiderivative size = 550, normalized size of antiderivative = 1.54 \[ \int \left (a+b x+c x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{12} (a+x (b+c x))^p \left (6 B x^2 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {AppellF1}\left (2,-p,-p,3,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+4 C x^3 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {AppellF1}\left (3,-p,-p,4,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+3 D x^4 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {AppellF1}\left (4,-p,-p,5,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+\frac {3\ 2^{1+p} A \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {-b+\sqrt {b^2-4 a c}-2 c x}{2 \sqrt {b^2-4 a c}}\right )}{c (1+p)}\right ) \] Input:

Integrate[(a + b*x + c*x^2)^p*(A + B*x + C*x^2 + D*x^3),x]
 

Output:

((a + x*(b + c*x))^p*((6*B*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) + (4*C*x^3*AppellF1[3, -p, -p, 4, (-2*c*x)/(b + Sqr 
t[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*D*x^4*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])])/(((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^(1 + p)*A*(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*Sqrt[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)))/12
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2192, 2192, 25, 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.

\(\displaystyle \int \left (a+b x+c x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx\)

\(\Big \downarrow \) 2192

\(\displaystyle \frac {\int \left (c x^2+b x+a\right )^p \left ((2 c C (p+2)-b D (p+3)) x^2-2 (a D-B c (p+2)) x+2 A c (p+2)\right )dx}{2 c (p+2)}+\frac {D x^2 \left (a+b x+c x^2\right )^{p+1}}{2 c (p+2)}\)

\(\Big \downarrow \) 2192

\(\displaystyle \frac {\frac {\int -\left (\left (-2 A \left (2 p^2+7 p+6\right ) c^2+2 a C (p+2) c-a b D (p+3)+\left (-D \left (p^2+5 p+6\right ) b^2+2 c C (p+2)^2 b+2 a c D (2 p+3)-2 B c^2 \left (2 p^2+7 p+6\right )\right ) x\right ) \left (c x^2+b x+a\right )^p\right )dx}{c (2 p+3)}+\frac {x \left (a+b x+c x^2\right )^{p+1} (2 c C (p+2)-b D (p+3))}{c (2 p+3)}}{2 c (p+2)}+\frac {D x^2 \left (a+b x+c x^2\right )^{p+1}}{2 c (p+2)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {x \left (a+b x+c x^2\right )^{p+1} (2 c C (p+2)-b D (p+3))}{c (2 p+3)}-\frac {\int \left (-2 A \left (2 p^2+7 p+6\right ) c^2+2 a C (p+2) c-a b D (p+3)+\left (-D \left (p^2+5 p+6\right ) b^2+2 c C (p+2)^2 b+2 a c D (2 p+3)-2 B c^2 \left (2 p^2+7 p+6\right )\right ) x\right ) \left (c x^2+b x+a\right )^pdx}{c (2 p+3)}}{2 c (p+2)}+\frac {D x^2 \left (a+b x+c x^2\right )^{p+1}}{2 c (p+2)}\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {\frac {x \left (a+b x+c x^2\right )^{p+1} (2 c C (p+2)-b D (p+3))}{c (2 p+3)}-\frac {\frac {\left (a+b x+c x^2\right )^{p+1} \left (2 a c D (2 p+3)+b^2 (-D) \left (p^2+5 p+6\right )+2 b c C (p+2)^2-2 B c^2 \left (2 p^2+7 p+6\right )\right )}{2 c (p+1)}-\frac {(p+2) \left (-4 c^2 (a C-A c (2 p+3))+2 b c (3 a D-B c (2 p+3))+b^3 (-D) (p+3)+2 b^2 c C (p+2)\right ) \int \left (c x^2+b x+a\right )^pdx}{2 c}}{c (2 p+3)}}{2 c (p+2)}+\frac {D x^2 \left (a+b x+c x^2\right )^{p+1}}{2 c (p+2)}\)

\(\Big \downarrow \) 1096

\(\displaystyle \frac {\frac {x \left (a+b x+c x^2\right )^{p+1} (2 c C (p+2)-b D (p+3))}{c (2 p+3)}-\frac {\frac {2^p (p+2) \left (a+b x+c x^2\right )^{p+1} \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right ) \left (-4 c^2 (a C-A c (2 p+3))+2 b c (3 a D-B c (2 p+3))+b^3 (-D) (p+3)+2 b^2 c C (p+2)\right )}{c (p+1) \sqrt {b^2-4 a c}}+\frac {\left (a+b x+c x^2\right )^{p+1} \left (2 a c D (2 p+3)+b^2 (-D) \left (p^2+5 p+6\right )+2 b c C (p+2)^2-2 B c^2 \left (2 p^2+7 p+6\right )\right )}{2 c (p+1)}}{c (2 p+3)}}{2 c (p+2)}+\frac {D x^2 \left (a+b x+c x^2\right )^{p+1}}{2 c (p+2)}\)

Input:

Int[(a + b*x + c*x^2)^p*(A + B*x + C*x^2 + D*x^3),x]
 

Output:

(D*x^2*(a + b*x + c*x^2)^(1 + p))/(2*c*(2 + p)) + (((2*c*C*(2 + p) - b*D*( 
3 + p))*x*(a + b*x + c*x^2)^(1 + p))/(c*(3 + 2*p)) - (((2*b*c*C*(2 + p)^2 
+ 2*a*c*D*(3 + 2*p) - b^2*D*(6 + 5*p + p^2) - 2*B*c^2*(6 + 7*p + 2*p^2))*( 
a + b*x + c*x^2)^(1 + p))/(2*c*(1 + p)) + (2^p*(2 + p)*(2*b^2*c*C*(2 + p) 
- b^3*D*(3 + p) - 4*c^2*(a*C - A*c*(3 + 2*p)) + 2*b*c*(3*a*D - B*c*(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*Sqrt[b^2 - 4*a*c]*(1 + p)))/(c*( 
3 + 2*p)))/(2*c*(2 + p))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 1096

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]
 

rule 1160

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]
 

rule 2192

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = 
Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + 
 c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1))   Int[(a 
+ b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b 
*e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c 
, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]
 

Maple [F]

\[\int \left (c \,x^{2}+b x +a \right )^{p} \left (D x^{3}+C \,x^{2}+B x +A \right )d x\]

Input:

int((c*x^2+b*x+a)^p*(D*x^3+C*x^2+B*x+A),x)
 

Output:

int((c*x^2+b*x+a)^p*(D*x^3+C*x^2+B*x+A),x)
 

Fricas [F]

\[ \int \left (a+b x+c x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\int { {\left (D x^{3} + C x^{2} + B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \] Input:

integrate((c*x^2+b*x+a)^p*(D*x^3+C*x^2+B*x+A),x, algorithm="fricas")
 

Output:

integral((D*x^3 + C*x^2 + B*x + A)*(c*x^2 + b*x + a)^p, x)
 

Sympy [F]

\[ \int \left (a+b x+c x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\int \left (a + b x + c x^{2}\right )^{p} \left (A + B x + C x^{2} + D x^{3}\right )\, dx \] Input:

integrate((c*x**2+b*x+a)**p*(D*x**3+C*x**2+B*x+A),x)
 

Output:

Integral((a + b*x + c*x**2)**p*(A + B*x + C*x**2 + D*x**3), x)
 

Maxima [F]

\[ \int \left (a+b x+c x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\int { {\left (D x^{3} + C x^{2} + B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \] Input:

integrate((c*x^2+b*x+a)^p*(D*x^3+C*x^2+B*x+A),x, algorithm="maxima")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(c*x^2 + b*x + a)^p, x)
 

Giac [F]

\[ \int \left (a+b x+c x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\int { {\left (D x^{3} + C x^{2} + B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \] Input:

integrate((c*x^2+b*x+a)^p*(D*x^3+C*x^2+B*x+A),x, algorithm="giac")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(c*x^2 + b*x + a)^p, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b x+c x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\int {\left (c\,x^2+b\,x+a\right )}^p\,\left (A+B\,x+C\,x^2+x^3\,D\right ) \,d x \] Input:

int((a + b*x + c*x^2)^p*(A + B*x + C*x^2 + x^3*D),x)
 

Output:

int((a + b*x + c*x^2)^p*(A + B*x + C*x^2 + x^3*D), x)
 

Reduce [F]

\[ \int \left (a+b x+c x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\text {too large to display} \] Input:

int((c*x^2+b*x+a)^p*(D*x^3+C*x^2+B*x+A),x)
 

Output:

(4*(a + b*x + c*x**2)**p*a**2*b*c*d*p**2 + 20*(a + b*x + c*x**2)**p*a**2*b 
*c*d*p + 18*(a + b*x + c*x**2)**p*a**2*b*c*d + 16*(a + b*x + c*x**2)**p*a* 
*2*c**3*p**3 + 64*(a + b*x + c*x**2)**p*a**2*c**3*p**2 + 80*(a + b*x + c*x 
**2)**p*a**2*c**3*p + 32*(a + b*x + c*x**2)**p*a**2*c**3 - (a + b*x + c*x* 
*2)**p*a*b**3*d*p**2 - 5*(a + b*x + c*x**2)**p*a*b**3*d*p - 6*(a + b*x + c 
*x**2)**p*a*b**3*d - 2*(a + b*x + c*x**2)**p*a*b**2*c**2*p**2 - 6*(a + b*x 
 + c*x**2)**p*a*b**2*c**2*p - 4*(a + b*x + c*x**2)**p*a*b**2*c**2 - 4*(a + 
 b*x + c*x**2)**p*a*b**2*c*d*p**3*x - 20*(a + b*x + c*x**2)**p*a*b**2*c*d* 
p**2*x - 18*(a + b*x + c*x**2)**p*a*b**2*c*d*p*x + 16*(a + b*x + c*x**2)** 
p*a*b*c**3*p**3*x + 60*(a + b*x + c*x**2)**p*a*b*c**3*p**2*x + 68*(a + b*x 
 + c*x**2)**p*a*b*c**3*p*x + 24*(a + b*x + c*x**2)**p*a*b*c**3*x + 8*(a + 
b*x + c*x**2)**p*a*b*c**2*d*p**3*x**2 + 16*(a + b*x + c*x**2)**p*a*b*c**2* 
d*p**2*x**2 + 6*(a + b*x + c*x**2)**p*a*b*c**2*d*p*x**2 + (a + b*x + c*x** 
2)**p*b**4*d*p**3*x + 5*(a + b*x + c*x**2)**p*b**4*d*p**2*x + 6*(a + b*x + 
 c*x**2)**p*b**4*d*p*x + 2*(a + b*x + c*x**2)**p*b**3*c**2*p**3*x + 6*(a + 
 b*x + c*x**2)**p*b**3*c**2*p**2*x + 4*(a + b*x + c*x**2)**p*b**3*c**2*p*x 
 - 2*(a + b*x + c*x**2)**p*b**3*c*d*p**3*x**2 - 7*(a + b*x + c*x**2)**p*b* 
*3*c*d*p**2*x**2 - 3*(a + b*x + c*x**2)**p*b**3*c*d*p*x**2 + 12*(a + b*x + 
 c*x**2)**p*b**2*c**3*p**3*x**2 + 42*(a + b*x + c*x**2)**p*b**2*c**3*p**2* 
x**2 + 42*(a + b*x + c*x**2)**p*b**2*c**3*p*x**2 + 12*(a + b*x + c*x**2...