\(\int (b x+c x^2+d x^3)^p \, dx\) [172]

Optimal result

Integrand size = 16, antiderivative size = 132 \[ \int \left (b x+c x^2+d x^3\right )^p \, dx=\frac {x \left (1+\frac {2 d x}{c-\sqrt {c^2-4 b d}}\right )^{-p} \left (1+\frac {2 d x}{c+\sqrt {c^2-4 b d}}\right )^{-p} \left (b x+c x^2+d x^3\right )^p \operatorname {AppellF1}\left (1+p,-p,-p,2+p,-\frac {2 d x}{c-\sqrt {c^2-4 b d}},-\frac {2 d x}{c+\sqrt {c^2-4 b d}}\right )}{1+p} \] Output:

x*(d*x^3+c*x^2+b*x)^p*AppellF1(p+1,-p,-p,2+p,-2*d*x/(c-(-4*b*d+c^2)^(1/2)) 
,-2*d*x/(c+(-4*b*d+c^2)^(1/2)))/(p+1)/((1+2*d*x/(c-(-4*b*d+c^2)^(1/2)))^p) 
/((1+2*d*x/(c+(-4*b*d+c^2)^(1/2)))^p)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.98 \[ \int \left (b x+c x^2+d x^3\right )^p \, dx=\frac {x \left (1+\frac {2 d x}{c-\sqrt {c^2-4 b d}}\right )^{-p} \left (1+\frac {2 d x}{c+\sqrt {c^2-4 b d}}\right )^{-p} (x (b+x (c+d x)))^p \operatorname {AppellF1}\left (1+p,-p,-p,2+p,-\frac {2 d x}{c+\sqrt {c^2-4 b d}},\frac {2 d x}{-c+\sqrt {c^2-4 b d}}\right )}{1+p} \] Input:

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

Output:

(x*(x*(b + x*(c + d*x)))^p*AppellF1[1 + p, -p, -p, 2 + p, (-2*d*x)/(c + Sq 
rt[c^2 - 4*b*d]), (2*d*x)/(-c + Sqrt[c^2 - 4*b*d])])/((1 + p)*(1 + (2*d*x) 
/(c - Sqrt[c^2 - 4*b*d]))^p*(1 + (2*d*x)/(c + Sqrt[c^2 - 4*b*d]))^p)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1955, 1179, 150}

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.

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

\(\Big \downarrow \) 1955

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

\(\Big \downarrow \) 1179

\(\displaystyle x^{-p} \left (\frac {2 d x}{c-\sqrt {c^2-4 b d}}+1\right )^{-p} \left (\frac {2 d x}{\sqrt {c^2-4 b d}+c}+1\right )^{-p} \left (b x+c x^2+d x^3\right )^p \int x^p \left (\frac {2 d x}{c-\sqrt {c^2-4 b d}}+1\right )^p \left (\frac {2 d x}{c+\sqrt {c^2-4 b d}}+1\right )^pdx\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {x \left (\frac {2 d x}{c-\sqrt {c^2-4 b d}}+1\right )^{-p} \left (\frac {2 d x}{\sqrt {c^2-4 b d}+c}+1\right )^{-p} \left (b x+c x^2+d x^3\right )^p \operatorname {AppellF1}\left (p+1,-p,-p,p+2,-\frac {2 d x}{c-\sqrt {c^2-4 b d}},-\frac {2 d x}{c+\sqrt {c^2-4 b d}}\right )}{p+1}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 1179

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(a + b*x + c*x^2)^p/(e*(1 - ( 
d + e*x)/(d - e*((b - q)/(2*c))))^p*(1 - (d + e*x)/(d - e*((b + q)/(2*c)))) 
^p)   Subst[Int[x^m*Simp[1 - x/(d - e*((b - q)/(2*c))), x]^p*Simp[1 - x/(d 
- e*((b + q)/(2*c))), x]^p, x], x, d + e*x], x]] /; FreeQ[{a, b, c, d, e, m 
, p}, x]
 

rule 1955

Int[((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol 
] :> Simp[(a*x^q + b*x^n + c*x^(2*n - q))^p/(x^(p*q)*(a + b*x^(n - q) + c*x 
^(2*(n - q)))^p)   Int[x^(p*q)*(a + b*x^(n - q) + c*x^(2*(n - q)))^p, x], x 
] /; FreeQ[{a, b, c, n, p, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] &&  !In 
tegerQ[p]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \left (b x+c x^2+d x^3\right )^p \, dx=\frac {\left (d \,x^{3}+c \,x^{2}+b x \right )^{p} b +\left (d \,x^{3}+c \,x^{2}+b x \right )^{p} c x -3 \left (\int \frac {\left (d \,x^{3}+c \,x^{2}+b x \right )^{p}}{3 d p \,x^{3}+3 c p \,x^{2}+d \,x^{3}+3 b p x +c \,x^{2}+b x}d x \right ) b^{2} p^{2}-\left (\int \frac {\left (d \,x^{3}+c \,x^{2}+b x \right )^{p}}{3 d p \,x^{3}+3 c p \,x^{2}+d \,x^{3}+3 b p x +c \,x^{2}+b x}d x \right ) b^{2} p -9 \left (\int \frac {\left (d \,x^{3}+c \,x^{2}+b x \right )^{p} x}{3 d p \,x^{2}+3 c p x +d \,x^{2}+3 b p +c x +b}d x \right ) b d \,p^{2}-3 \left (\int \frac {\left (d \,x^{3}+c \,x^{2}+b x \right )^{p} x}{3 d p \,x^{2}+3 c p x +d \,x^{2}+3 b p +c x +b}d x \right ) b d p +3 \left (\int \frac {\left (d \,x^{3}+c \,x^{2}+b x \right )^{p} x}{3 d p \,x^{2}+3 c p x +d \,x^{2}+3 b p +c x +b}d x \right ) c^{2} p^{2}+\left (\int \frac {\left (d \,x^{3}+c \,x^{2}+b x \right )^{p} x}{3 d p \,x^{2}+3 c p x +d \,x^{2}+3 b p +c x +b}d x \right ) c^{2} p}{c \left (3 p +1\right )} \] Input:

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

Output:

((b*x + c*x**2 + d*x**3)**p*b + (b*x + c*x**2 + d*x**3)**p*c*x - 3*int((b* 
x + c*x**2 + d*x**3)**p/(3*b*p*x + b*x + 3*c*p*x**2 + c*x**2 + 3*d*p*x**3 
+ d*x**3),x)*b**2*p**2 - int((b*x + c*x**2 + d*x**3)**p/(3*b*p*x + b*x + 3 
*c*p*x**2 + c*x**2 + 3*d*p*x**3 + d*x**3),x)*b**2*p - 9*int(((b*x + c*x**2 
 + d*x**3)**p*x)/(3*b*p + b + 3*c*p*x + c*x + 3*d*p*x**2 + d*x**2),x)*b*d* 
p**2 - 3*int(((b*x + c*x**2 + d*x**3)**p*x)/(3*b*p + b + 3*c*p*x + c*x + 3 
*d*p*x**2 + d*x**2),x)*b*d*p + 3*int(((b*x + c*x**2 + d*x**3)**p*x)/(3*b*p 
 + b + 3*c*p*x + c*x + 3*d*p*x**2 + d*x**2),x)*c**2*p**2 + int(((b*x + c*x 
**2 + d*x**3)**p*x)/(3*b*p + b + 3*c*p*x + c*x + 3*d*p*x**2 + d*x**2),x)*c 
**2*p)/(c*(3*p + 1))