\(\int x (c+d x+e x^2) (a+b x^3)^3 \, dx\) [326]

Optimal result

Integrand size = 21, antiderivative size = 110 \[ \int x \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=\frac {1}{2} a^3 c x^2+\frac {1}{4} a^3 e x^4+\frac {3}{5} a^2 b c x^5+\frac {3}{7} a^2 b e x^7+\frac {3}{8} a b^2 c x^8+\frac {3}{10} a b^2 e x^{10}+\frac {1}{11} b^3 c x^{11}+\frac {1}{13} b^3 e x^{13}+\frac {d \left (a+b x^3\right )^4}{12 b} \]

[Out]

1/2*a^3*c*x^2+1/4*a^3*e*x^4+3/5*a^2*b*c*x^5+3/7*a^2*b*e*x^7+3/8*a*b^2*c*x^8+3/10*a*b^2*e*x^10+1/11*b^3*c*x^11+
1/13*b^3*e*x^13+1/12*d*(b*x^3+a)^4/b

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1596, 1864} \[ \int x \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=\frac {1}{2} a^3 c x^2+\frac {1}{4} a^3 e x^4+\frac {3}{5} a^2 b c x^5+\frac {3}{7} a^2 b e x^7+\frac {3}{8} a b^2 c x^8+\frac {3}{10} a b^2 e x^{10}+\frac {d \left (a+b x^3\right )^4}{12 b}+\frac {1}{11} b^3 c x^{11}+\frac {1}{13} b^3 e x^{13} \]

[In]

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

[Out]

(a^3*c*x^2)/2 + (a^3*e*x^4)/4 + (3*a^2*b*c*x^5)/5 + (3*a^2*b*e*x^7)/7 + (3*a*b^2*c*x^8)/8 + (3*a*b^2*e*x^10)/1
0 + (b^3*c*x^11)/11 + (b^3*e*x^13)/13 + (d*(a + b*x^3)^4)/(12*b)

Rule 1596

Int[(Px_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[Coeff[Px, x, n - 1]*((a + b*x^n)^(p + 1)/(b*n*(p +
1))), x] + Int[(Px - Coeff[Px, x, n - 1]*x^(n - 1))*(a + b*x^n)^p, x] /; FreeQ[{a, b}, x] && PolyQ[Px, x] && I
GtQ[p, 1] && IGtQ[n, 1] && NeQ[Coeff[Px, x, n - 1], 0] && NeQ[Px, Coeff[Px, x, n - 1]*x^(n - 1)] &&  !MatchQ[P
x, (Qx_.)*((c_) + (d_.)*x^(m_))^(q_) /; FreeQ[{c, d}, x] && PolyQ[Qx, x] && IGtQ[q, 1] && IGtQ[m, 1] && NeQ[Co
eff[Qx*(a + b*x^n)^p, x, m - 1], 0] && GtQ[m*q, n*p]]

Rule 1864

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = \frac {d \left (a+b x^3\right )^4}{12 b}+\int \left (a+b x^3\right )^3 \left (-d x^2+x \left (c+d x+e x^2\right )\right ) \, dx \\ & = \frac {d \left (a+b x^3\right )^4}{12 b}+\int \left (a^3 c x+a^3 e x^3+3 a^2 b c x^4+3 a^2 b e x^6+3 a b^2 c x^7+3 a b^2 e x^9+b^3 c x^{10}+b^3 e x^{12}\right ) \, dx \\ & = \frac {1}{2} a^3 c x^2+\frac {1}{4} a^3 e x^4+\frac {3}{5} a^2 b c x^5+\frac {3}{7} a^2 b e x^7+\frac {3}{8} a b^2 c x^8+\frac {3}{10} a b^2 e x^{10}+\frac {1}{11} b^3 c x^{11}+\frac {1}{13} b^3 e x^{13}+\frac {d \left (a+b x^3\right )^4}{12 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.26 \[ \int x \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=\frac {1}{2} a^3 c x^2+\frac {1}{3} a^3 d x^3+\frac {1}{4} a^3 e x^4+\frac {3}{5} a^2 b c x^5+\frac {1}{2} a^2 b d x^6+\frac {3}{7} a^2 b e x^7+\frac {3}{8} a b^2 c x^8+\frac {1}{3} a b^2 d x^9+\frac {3}{10} a b^2 e x^{10}+\frac {1}{11} b^3 c x^{11}+\frac {1}{12} b^3 d x^{12}+\frac {1}{13} b^3 e x^{13} \]

[In]

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

[Out]

(a^3*c*x^2)/2 + (a^3*d*x^3)/3 + (a^3*e*x^4)/4 + (3*a^2*b*c*x^5)/5 + (a^2*b*d*x^6)/2 + (3*a^2*b*e*x^7)/7 + (3*a
*b^2*c*x^8)/8 + (a*b^2*d*x^9)/3 + (3*a*b^2*e*x^10)/10 + (b^3*c*x^11)/11 + (b^3*d*x^12)/12 + (b^3*e*x^13)/13

Maple [A] (verified)

Time = 1.58 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05

method	result	size
gosper	\(\frac {1}{13} b^{3} e \,x^{13}+\frac {1}{12} b^{3} d \,x^{12}+\frac {1}{11} b^{3} c \,x^{11}+\frac {3}{10} a \,b^{2} e \,x^{10}+\frac {1}{3} x^{9} a \,b^{2} d +\frac {3}{8} a \,b^{2} c \,x^{8}+\frac {3}{7} a^{2} b e \,x^{7}+\frac {1}{2} a^{2} b d \,x^{6}+\frac {3}{5} a^{2} b c \,x^{5}+\frac {1}{4} a^{3} e \,x^{4}+\frac {1}{3} a^{3} d \,x^{3}+\frac {1}{2} a^{3} c \,x^{2}\)	\(116\)
default	\(\frac {1}{13} b^{3} e \,x^{13}+\frac {1}{12} b^{3} d \,x^{12}+\frac {1}{11} b^{3} c \,x^{11}+\frac {3}{10} a \,b^{2} e \,x^{10}+\frac {1}{3} x^{9} a \,b^{2} d +\frac {3}{8} a \,b^{2} c \,x^{8}+\frac {3}{7} a^{2} b e \,x^{7}+\frac {1}{2} a^{2} b d \,x^{6}+\frac {3}{5} a^{2} b c \,x^{5}+\frac {1}{4} a^{3} e \,x^{4}+\frac {1}{3} a^{3} d \,x^{3}+\frac {1}{2} a^{3} c \,x^{2}\)	\(116\)
norman	\(\frac {1}{13} b^{3} e \,x^{13}+\frac {1}{12} b^{3} d \,x^{12}+\frac {1}{11} b^{3} c \,x^{11}+\frac {3}{10} a \,b^{2} e \,x^{10}+\frac {1}{3} x^{9} a \,b^{2} d +\frac {3}{8} a \,b^{2} c \,x^{8}+\frac {3}{7} a^{2} b e \,x^{7}+\frac {1}{2} a^{2} b d \,x^{6}+\frac {3}{5} a^{2} b c \,x^{5}+\frac {1}{4} a^{3} e \,x^{4}+\frac {1}{3} a^{3} d \,x^{3}+\frac {1}{2} a^{3} c \,x^{2}\)	\(116\)
risch	\(\frac {1}{13} b^{3} e \,x^{13}+\frac {1}{12} b^{3} d \,x^{12}+\frac {1}{11} b^{3} c \,x^{11}+\frac {3}{10} a \,b^{2} e \,x^{10}+\frac {1}{3} x^{9} a \,b^{2} d +\frac {3}{8} a \,b^{2} c \,x^{8}+\frac {3}{7} a^{2} b e \,x^{7}+\frac {1}{2} a^{2} b d \,x^{6}+\frac {3}{5} a^{2} b c \,x^{5}+\frac {1}{4} a^{3} e \,x^{4}+\frac {1}{3} a^{3} d \,x^{3}+\frac {1}{2} a^{3} c \,x^{2}\)	\(116\)
parallelrisch	\(\frac {1}{13} b^{3} e \,x^{13}+\frac {1}{12} b^{3} d \,x^{12}+\frac {1}{11} b^{3} c \,x^{11}+\frac {3}{10} a \,b^{2} e \,x^{10}+\frac {1}{3} x^{9} a \,b^{2} d +\frac {3}{8} a \,b^{2} c \,x^{8}+\frac {3}{7} a^{2} b e \,x^{7}+\frac {1}{2} a^{2} b d \,x^{6}+\frac {3}{5} a^{2} b c \,x^{5}+\frac {1}{4} a^{3} e \,x^{4}+\frac {1}{3} a^{3} d \,x^{3}+\frac {1}{2} a^{3} c \,x^{2}\)	\(116\)

[In]

int(x*(e*x^2+d*x+c)*(b*x^3+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/13*b^3*e*x^13+1/12*b^3*d*x^12+1/11*b^3*c*x^11+3/10*a*b^2*e*x^10+1/3*x^9*a*b^2*d+3/8*a*b^2*c*x^8+3/7*a^2*b*e*
x^7+1/2*a^2*b*d*x^6+3/5*a^2*b*c*x^5+1/4*a^3*e*x^4+1/3*a^3*d*x^3+1/2*a^3*c*x^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.05 \[ \int x \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=\frac {1}{13} \, b^{3} e x^{13} + \frac {1}{12} \, b^{3} d x^{12} + \frac {1}{11} \, b^{3} c x^{11} + \frac {3}{10} \, a b^{2} e x^{10} + \frac {1}{3} \, a b^{2} d x^{9} + \frac {3}{8} \, a b^{2} c x^{8} + \frac {3}{7} \, a^{2} b e x^{7} + \frac {1}{2} \, a^{2} b d x^{6} + \frac {3}{5} \, a^{2} b c x^{5} + \frac {1}{4} \, a^{3} e x^{4} + \frac {1}{3} \, a^{3} d x^{3} + \frac {1}{2} \, a^{3} c x^{2} \]

[In]

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

[Out]

1/13*b^3*e*x^13 + 1/12*b^3*d*x^12 + 1/11*b^3*c*x^11 + 3/10*a*b^2*e*x^10 + 1/3*a*b^2*d*x^9 + 3/8*a*b^2*c*x^8 +
3/7*a^2*b*e*x^7 + 1/2*a^2*b*d*x^6 + 3/5*a^2*b*c*x^5 + 1/4*a^3*e*x^4 + 1/3*a^3*d*x^3 + 1/2*a^3*c*x^2

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.25 \[ \int x \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=\frac {a^{3} c x^{2}}{2} + \frac {a^{3} d x^{3}}{3} + \frac {a^{3} e x^{4}}{4} + \frac {3 a^{2} b c x^{5}}{5} + \frac {a^{2} b d x^{6}}{2} + \frac {3 a^{2} b e x^{7}}{7} + \frac {3 a b^{2} c x^{8}}{8} + \frac {a b^{2} d x^{9}}{3} + \frac {3 a b^{2} e x^{10}}{10} + \frac {b^{3} c x^{11}}{11} + \frac {b^{3} d x^{12}}{12} + \frac {b^{3} e x^{13}}{13} \]

[In]

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

[Out]

a**3*c*x**2/2 + a**3*d*x**3/3 + a**3*e*x**4/4 + 3*a**2*b*c*x**5/5 + a**2*b*d*x**6/2 + 3*a**2*b*e*x**7/7 + 3*a*
b**2*c*x**8/8 + a*b**2*d*x**9/3 + 3*a*b**2*e*x**10/10 + b**3*c*x**11/11 + b**3*d*x**12/12 + b**3*e*x**13/13

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.05 \[ \int x \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=\frac {1}{13} \, b^{3} e x^{13} + \frac {1}{12} \, b^{3} d x^{12} + \frac {1}{11} \, b^{3} c x^{11} + \frac {3}{10} \, a b^{2} e x^{10} + \frac {1}{3} \, a b^{2} d x^{9} + \frac {3}{8} \, a b^{2} c x^{8} + \frac {3}{7} \, a^{2} b e x^{7} + \frac {1}{2} \, a^{2} b d x^{6} + \frac {3}{5} \, a^{2} b c x^{5} + \frac {1}{4} \, a^{3} e x^{4} + \frac {1}{3} \, a^{3} d x^{3} + \frac {1}{2} \, a^{3} c x^{2} \]

[In]

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

[Out]

1/13*b^3*e*x^13 + 1/12*b^3*d*x^12 + 1/11*b^3*c*x^11 + 3/10*a*b^2*e*x^10 + 1/3*a*b^2*d*x^9 + 3/8*a*b^2*c*x^8 +
3/7*a^2*b*e*x^7 + 1/2*a^2*b*d*x^6 + 3/5*a^2*b*c*x^5 + 1/4*a^3*e*x^4 + 1/3*a^3*d*x^3 + 1/2*a^3*c*x^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.05 \[ \int x \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=\frac {1}{13} \, b^{3} e x^{13} + \frac {1}{12} \, b^{3} d x^{12} + \frac {1}{11} \, b^{3} c x^{11} + \frac {3}{10} \, a b^{2} e x^{10} + \frac {1}{3} \, a b^{2} d x^{9} + \frac {3}{8} \, a b^{2} c x^{8} + \frac {3}{7} \, a^{2} b e x^{7} + \frac {1}{2} \, a^{2} b d x^{6} + \frac {3}{5} \, a^{2} b c x^{5} + \frac {1}{4} \, a^{3} e x^{4} + \frac {1}{3} \, a^{3} d x^{3} + \frac {1}{2} \, a^{3} c x^{2} \]

[In]

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

[Out]

1/13*b^3*e*x^13 + 1/12*b^3*d*x^12 + 1/11*b^3*c*x^11 + 3/10*a*b^2*e*x^10 + 1/3*a*b^2*d*x^9 + 3/8*a*b^2*c*x^8 +
3/7*a^2*b*e*x^7 + 1/2*a^2*b*d*x^6 + 3/5*a^2*b*c*x^5 + 1/4*a^3*e*x^4 + 1/3*a^3*d*x^3 + 1/2*a^3*c*x^2

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.05 \[ \int x \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=\frac {e\,a^3\,x^4}{4}+\frac {d\,a^3\,x^3}{3}+\frac {c\,a^3\,x^2}{2}+\frac {3\,e\,a^2\,b\,x^7}{7}+\frac {d\,a^2\,b\,x^6}{2}+\frac {3\,c\,a^2\,b\,x^5}{5}+\frac {3\,e\,a\,b^2\,x^{10}}{10}+\frac {d\,a\,b^2\,x^9}{3}+\frac {3\,c\,a\,b^2\,x^8}{8}+\frac {e\,b^3\,x^{13}}{13}+\frac {d\,b^3\,x^{12}}{12}+\frac {c\,b^3\,x^{11}}{11} \]

[In]

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

[Out]

(a^3*c*x^2)/2 + (a^3*d*x^3)/3 + (b^3*c*x^11)/11 + (a^3*e*x^4)/4 + (b^3*d*x^12)/12 + (b^3*e*x^13)/13 + (3*a^2*b
*c*x^5)/5 + (3*a*b^2*c*x^8)/8 + (a^2*b*d*x^6)/2 + (a*b^2*d*x^9)/3 + (3*a^2*b*e*x^7)/7 + (3*a*b^2*e*x^10)/10