\(\int (c+d x+e x^2) (a+b x^3)^3 \, dx\) [327]

Optimal result

Integrand size = 20, antiderivative size = 105 \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=a^3 c x+\frac {1}{2} a^3 d x^2+\frac {3}{4} a^2 b c x^4+\frac {3}{5} a^2 b d x^5+\frac {3}{7} a b^2 c x^7+\frac {3}{8} a b^2 d x^8+\frac {1}{10} b^3 c x^{10}+\frac {1}{11} b^3 d x^{11}+\frac {e \left (a+b x^3\right )^4}{12 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1596, 1864} \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=a^3 c x+\frac {1}{2} a^3 d x^2+\frac {3}{4} a^2 b c x^4+\frac {3}{5} a^2 b d x^5+\frac {3}{7} a b^2 c x^7+\frac {3}{8} a b^2 d x^8+\frac {e \left (a+b x^3\right )^4}{12 b}+\frac {1}{10} b^3 c x^{10}+\frac {1}{11} b^3 d x^{11} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.28 \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=a^3 c x+\frac {1}{2} a^3 d x^2+\frac {1}{3} a^3 e x^3+\frac {3}{4} a^2 b c x^4+\frac {3}{5} a^2 b d x^5+\frac {1}{2} a^2 b e x^6+\frac {3}{7} a b^2 c x^7+\frac {3}{8} a b^2 d x^8+\frac {1}{3} a b^2 e x^9+\frac {1}{10} b^3 c x^{10}+\frac {1}{11} b^3 d x^{11}+\frac {1}{12} b^3 e x^{12} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.53 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.08

method	result	size
gosper	\(\frac {1}{12} b^{3} e \,x^{12}+\frac {1}{11} b^{3} d \,x^{11}+\frac {1}{10} b^{3} c \,x^{10}+\frac {1}{3} a \,b^{2} e \,x^{9}+\frac {3}{8} x^{8} b^{2} d a +\frac {3}{7} a \,b^{2} c \,x^{7}+\frac {1}{2} a^{2} b e \,x^{6}+\frac {3}{5} x^{5} b d \,a^{2}+\frac {3}{4} a^{2} b c \,x^{4}+\frac {1}{3} a^{3} e \,x^{3}+\frac {1}{2} a^{3} d \,x^{2}+a^{3} c x\)	\(113\)
default	\(\frac {1}{12} b^{3} e \,x^{12}+\frac {1}{11} b^{3} d \,x^{11}+\frac {1}{10} b^{3} c \,x^{10}+\frac {1}{3} a \,b^{2} e \,x^{9}+\frac {3}{8} x^{8} b^{2} d a +\frac {3}{7} a \,b^{2} c \,x^{7}+\frac {1}{2} a^{2} b e \,x^{6}+\frac {3}{5} x^{5} b d \,a^{2}+\frac {3}{4} a^{2} b c \,x^{4}+\frac {1}{3} a^{3} e \,x^{3}+\frac {1}{2} a^{3} d \,x^{2}+a^{3} c x\)	\(113\)
norman	\(\frac {1}{12} b^{3} e \,x^{12}+\frac {1}{11} b^{3} d \,x^{11}+\frac {1}{10} b^{3} c \,x^{10}+\frac {1}{3} a \,b^{2} e \,x^{9}+\frac {3}{8} x^{8} b^{2} d a +\frac {3}{7} a \,b^{2} c \,x^{7}+\frac {1}{2} a^{2} b e \,x^{6}+\frac {3}{5} x^{5} b d \,a^{2}+\frac {3}{4} a^{2} b c \,x^{4}+\frac {1}{3} a^{3} e \,x^{3}+\frac {1}{2} a^{3} d \,x^{2}+a^{3} c x\)	\(113\)
risch	\(\frac {1}{12} b^{3} e \,x^{12}+\frac {1}{11} b^{3} d \,x^{11}+\frac {1}{10} b^{3} c \,x^{10}+\frac {1}{3} a \,b^{2} e \,x^{9}+\frac {3}{8} x^{8} b^{2} d a +\frac {3}{7} a \,b^{2} c \,x^{7}+\frac {1}{2} a^{2} b e \,x^{6}+\frac {3}{5} x^{5} b d \,a^{2}+\frac {3}{4} a^{2} b c \,x^{4}+\frac {1}{3} a^{3} e \,x^{3}+\frac {1}{2} a^{3} d \,x^{2}+a^{3} c x\)	\(113\)
parallelrisch	\(\frac {1}{12} b^{3} e \,x^{12}+\frac {1}{11} b^{3} d \,x^{11}+\frac {1}{10} b^{3} c \,x^{10}+\frac {1}{3} a \,b^{2} e \,x^{9}+\frac {3}{8} x^{8} b^{2} d a +\frac {3}{7} a \,b^{2} c \,x^{7}+\frac {1}{2} a^{2} b e \,x^{6}+\frac {3}{5} x^{5} b d \,a^{2}+\frac {3}{4} a^{2} b c \,x^{4}+\frac {1}{3} a^{3} e \,x^{3}+\frac {1}{2} a^{3} d \,x^{2}+a^{3} c x\)	\(113\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.07 \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=\frac {1}{12} \, b^{3} e x^{12} + \frac {1}{11} \, b^{3} d x^{11} + \frac {1}{10} \, b^{3} c x^{10} + \frac {1}{3} \, a b^{2} e x^{9} + \frac {3}{8} \, a b^{2} d x^{8} + \frac {3}{7} \, a b^{2} c x^{7} + \frac {1}{2} \, a^{2} b e x^{6} + \frac {3}{5} \, a^{2} b d x^{5} + \frac {3}{4} \, a^{2} b c x^{4} + \frac {1}{3} \, a^{3} e x^{3} + \frac {1}{2} \, a^{3} d x^{2} + a^{3} c x \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.28 \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=a^{3} c x + \frac {a^{3} d x^{2}}{2} + \frac {a^{3} e x^{3}}{3} + \frac {3 a^{2} b c x^{4}}{4} + \frac {3 a^{2} b d x^{5}}{5} + \frac {a^{2} b e x^{6}}{2} + \frac {3 a b^{2} c x^{7}}{7} + \frac {3 a b^{2} d x^{8}}{8} + \frac {a b^{2} e x^{9}}{3} + \frac {b^{3} c x^{10}}{10} + \frac {b^{3} d x^{11}}{11} + \frac {b^{3} e x^{12}}{12} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.07 \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=\frac {1}{12} \, b^{3} e x^{12} + \frac {1}{11} \, b^{3} d x^{11} + \frac {1}{10} \, b^{3} c x^{10} + \frac {1}{3} \, a b^{2} e x^{9} + \frac {3}{8} \, a b^{2} d x^{8} + \frac {3}{7} \, a b^{2} c x^{7} + \frac {1}{2} \, a^{2} b e x^{6} + \frac {3}{5} \, a^{2} b d x^{5} + \frac {3}{4} \, a^{2} b c x^{4} + \frac {1}{3} \, a^{3} e x^{3} + \frac {1}{2} \, a^{3} d x^{2} + a^{3} c x \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.07 \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=\frac {1}{12} \, b^{3} e x^{12} + \frac {1}{11} \, b^{3} d x^{11} + \frac {1}{10} \, b^{3} c x^{10} + \frac {1}{3} \, a b^{2} e x^{9} + \frac {3}{8} \, a b^{2} d x^{8} + \frac {3}{7} \, a b^{2} c x^{7} + \frac {1}{2} \, a^{2} b e x^{6} + \frac {3}{5} \, a^{2} b d x^{5} + \frac {3}{4} \, a^{2} b c x^{4} + \frac {1}{3} \, a^{3} e x^{3} + \frac {1}{2} \, a^{3} d x^{2} + a^{3} c x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.07 \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^3 \, dx=\frac {e\,a^3\,x^3}{3}+\frac {d\,a^3\,x^2}{2}+c\,a^3\,x+\frac {e\,a^2\,b\,x^6}{2}+\frac {3\,d\,a^2\,b\,x^5}{5}+\frac {3\,c\,a^2\,b\,x^4}{4}+\frac {e\,a\,b^2\,x^9}{3}+\frac {3\,d\,a\,b^2\,x^8}{8}+\frac {3\,c\,a\,b^2\,x^7}{7}+\frac {e\,b^3\,x^{12}}{12}+\frac {d\,b^3\,x^{11}}{11}+\frac {c\,b^3\,x^{10}}{10} \]

[In]

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

[Out]

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