3.26.63 \(\int (d+e x)^3 (a+b x+c x^2)^p \, dx\) [2563]

3.26.63.1 Optimal result

Integrand size = 20, antiderivative size = 327 \[ \int (d+e x)^3 \left (a+b x+c x^2\right )^p \, dx=\frac {e (d+e x)^2 \left (a+b x+c x^2\right )^{1+p}}{2 c (2+p)}-\frac {e \left (b e (2 c d-b e) (2+p) (3+p)-2 c (3+2 p) \left (c d^2 (5+2 p)-e (a e+b d (2+p))\right )-2 c e (2 c d-b e) (1+p) (3+p) x\right ) \left (a+b x+c x^2\right )^{1+p}}{4 c^3 (1+p) (2+p) (3+2 p)}-\frac {2^{-1+p} (2 c d-b e) \left (b^2 e^2 (3+p)+2 c^2 d^2 (3+2 p)-2 c e (3 a e+b d (3+2 p))\right ) \left (-\frac {b-\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-1-p} \left (a+b x+c x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {b+\sqrt {b^2-4 a c}+2 c x}{2 \sqrt {b^2-4 a c}}\right )}{c^3 \sqrt {b^2-4 a c} (1+p) (3+2 p)} \]

output

1/2*e*(e*x+d)^2*(c*x^2+b*x+a)^(p+1)/c/(2+p)-1/4*e*(b*e*(-b*e+2*c*d)*(2+p)* 
(3+p)-2*c*(3+2*p)*(c*d^2*(5+2*p)-e*(a*e+b*d*(2+p)))-2*c*e*(-b*e+2*c*d)*(p+ 
1)*(3+p)*x)*(c*x^2+b*x+a)^(p+1)/c^3/(2+p)/(2*p^2+5*p+3)-2^(-1+p)*(-b*e+2*c 
*d)*(b^2*e^2*(3+p)+2*c^2*d^2*(3+2*p)-2*c*e*(3*a*e+b*d*(3+2*p)))*(c*x^2+b*x 
+a)^(p+1)*hypergeom([-p, p+1],[2+p],1/2*(b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a 
*c+b^2)^(1/2))*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(-1-p)/c 
^3/(p+1)/(3+2*p)/(-4*a*c+b^2)^(1/2)
 

3.26.63.2 Mathematica [C] (verified)

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

Time = 1.37 (sec) , antiderivative size = 558, normalized size of antiderivative = 1.71 \[ \int (d+e x)^3 \left (a+b x+c x^2\right )^p \, dx=\frac {1}{4} (a+x (b+c x))^p \left (6 d^2 e x^2 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {AppellF1}\left (2,-p,-p,3,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+4 d e^2 x^3 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {AppellF1}\left (3,-p,-p,4,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+e^3 x^4 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {AppellF1}\left (4,-p,-p,5,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+\frac {2^{1+p} d^3 \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {-b+\sqrt {b^2-4 a c}-2 c x}{2 \sqrt {b^2-4 a c}}\right )}{c (1+p)}\right ) \]

input

Integrate[(d + e*x)^3*(a + b*x + c*x^2)^p,x]
 

output

((a + x*(b + c*x))^p*((6*d^2*e*x^2*AppellF1[2, -p, -p, 3, (-2*c*x)/(b + Sq 
rt[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*d*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) + (e^3*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) + (2^(1 + p)*d^3*(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)))/4
 

3.26.63.3 Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1166, 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.

\(\displaystyle \int (d+e x)^3 \left (a+b x+c x^2\right )^p \, dx\)

\(\Big \downarrow \) 1166

\(\displaystyle \frac {\int (d+e x) \left (2 c (p+2) d^2-e (2 a e+b d (p+1))+e (2 c d-b e) (p+3) x\right ) \left (c x^2+b x+a\right )^pdx}{2 c (p+2)}+\frac {e (d+e x)^2 \left (a+b x+c x^2\right )^{p+1}}{2 c (p+2)}\)

\(\Big \downarrow \) 1225

\(\displaystyle \frac {\frac {(p+2) (2 c d-b e) \left (-2 c e (3 a e+b d (2 p+3))+b^2 e^2 (p+3)+2 c^2 d^2 (2 p+3)\right ) \int \left (c x^2+b x+a\right )^pdx}{2 c^2 (2 p+3)}-\frac {e \left (a+b x+c x^2\right )^{p+1} \left (-2 c (2 p+3) \left (c d^2 (2 p+5)-e (a e+b d (p+2))\right )-2 c e (p+1) (p+3) x (2 c d-b e)+b e (p+2) (p+3) (2 c d-b e)\right )}{2 c^2 (p+1) (2 p+3)}}{2 c (p+2)}+\frac {e (d+e x)^2 \left (a+b x+c x^2\right )^{p+1}}{2 c (p+2)}\)

\(\Big \downarrow \) 1096

\(\displaystyle \frac {-\frac {2^p (p+2) (2 c d-b e) \left (a+b x+c x^2\right )^{p+1} \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \left (-2 c e (3 a e+b d (2 p+3))+b^2 e^2 (p+3)+2 c^2 d^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{c^2 (p+1) (2 p+3) \sqrt {b^2-4 a c}}-\frac {e \left (a+b x+c x^2\right )^{p+1} \left (-2 c (2 p+3) \left (c d^2 (2 p+5)-e (a e+b d (p+2))\right )-2 c e (p+1) (p+3) x (2 c d-b e)+b e (p+2) (p+3) (2 c d-b e)\right )}{2 c^2 (p+1) (2 p+3)}}{2 c (p+2)}+\frac {e (d+e x)^2 \left (a+b x+c x^2\right )^{p+1}}{2 c (p+2)}\)

input

Int[(d + e*x)^3*(a + b*x + c*x^2)^p,x]
 

output

(e*(d + e*x)^2*(a + b*x + c*x^2)^(1 + p))/(2*c*(2 + p)) + (-1/2*(e*(b*e*(2 
*c*d - b*e)*(2 + p)*(3 + p) - 2*c*(3 + 2*p)*(c*d^2*(5 + 2*p) - e*(a*e + b* 
d*(2 + p))) - 2*c*e*(2*c*d - b*e)*(1 + p)*(3 + p)*x)*(a + b*x + c*x^2)^(1 
+ p))/(c^2*(1 + p)*(3 + 2*p)) - (2^p*(2*c*d - b*e)*(2 + p)*(b^2*e^2*(3 + p 
) + 2*c^2*d^2*(3 + 2*p) - 2*c*e*(3*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)*H 
ypergeometric2F1[-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))
 

3.26.63.3.1 Defintions of rubi rules used

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 1166

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]
 

rule 1225

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]
 

3.26.63.4 Maple [F]

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

input

int((e*x+d)^3*(c*x^2+b*x+a)^p,x)
 

output

int((e*x+d)^3*(c*x^2+b*x+a)^p,x)
 

3.26.63.5 Fricas [F]

\[ \int (d+e x)^3 \left (a+b x+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \]

input

integrate((e*x+d)^3*(c*x^2+b*x+a)^p,x, algorithm="fricas")
 

output

integral((e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*(c*x^2 + b*x + a)^p, x)
 

3.26.63.6 Sympy [F]

\[ \int (d+e x)^3 \left (a+b x+c x^2\right )^p \, dx=\int \left (d + e x\right )^{3} \left (a + b x + c x^{2}\right )^{p}\, dx \]

input

integrate((e*x+d)**3*(c*x**2+b*x+a)**p,x)
 

output

Integral((d + e*x)**3*(a + b*x + c*x**2)**p, x)
 

3.26.63.7 Maxima [F]

\[ \int (d+e x)^3 \left (a+b x+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \]

input

integrate((e*x+d)^3*(c*x^2+b*x+a)^p,x, algorithm="maxima")
 

output

integrate((e*x + d)^3*(c*x^2 + b*x + a)^p, x)
 

3.26.63.8 Giac [F]

\[ \int (d+e x)^3 \left (a+b x+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \]

input

integrate((e*x+d)^3*(c*x^2+b*x+a)^p,x, algorithm="giac")
 

output

integrate((e*x + d)^3*(c*x^2 + b*x + a)^p, x)
 

3.26.63.9 Mupad [F(-1)]

Timed out. \[ \int (d+e x)^3 \left (a+b x+c x^2\right )^p \, dx=\int {\left (d+e\,x\right )}^3\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \]

input

int((d + e*x)^3*(a + b*x + c*x^2)^p,x)
 

output

int((d + e*x)^3*(a + b*x + c*x^2)^p, x)