\(\int x^2 (c+d x+e x^2) (a+b x^3)^2 \, dx\) [319]

Optimal result

Integrand size = 23, antiderivative size = 82 \[ \int x^2 \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=\frac {1}{4} a^2 d x^4+\frac {1}{5} a^2 e x^5+\frac {2}{7} a b d x^7+\frac {1}{4} a b e x^8+\frac {1}{10} b^2 d x^{10}+\frac {1}{11} b^2 e x^{11}+\frac {c \left (a+b x^3\right )^3}{9 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1596, 1864} \[ \int x^2 \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=\frac {1}{4} a^2 d x^4+\frac {1}{5} a^2 e x^5+\frac {c \left (a+b x^3\right )^3}{9 b}+\frac {2}{7} a b d x^7+\frac {1}{4} a b e x^8+\frac {1}{10} b^2 d x^{10}+\frac {1}{11} b^2 e x^{11} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.18 \[ \int x^2 \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=\frac {1}{3} a^2 c x^3+\frac {1}{4} a^2 d x^4+\frac {1}{5} a^2 e x^5+\frac {1}{3} a b c x^6+\frac {2}{7} a b d x^7+\frac {1}{4} a b e x^8+\frac {1}{9} b^2 c x^9+\frac {1}{10} b^2 d x^{10}+\frac {1}{11} b^2 e x^{11} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.53 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98

method	result	size
gosper	\(\frac {1}{11} b^{2} e \,x^{11}+\frac {1}{10} b^{2} d \,x^{10}+\frac {1}{9} b^{2} c \,x^{9}+\frac {1}{4} a b e \,x^{8}+\frac {2}{7} a d \,x^{7} b +\frac {1}{3} a b c \,x^{6}+\frac {1}{5} a^{2} e \,x^{5}+\frac {1}{4} a^{2} d \,x^{4}+\frac {1}{3} a^{2} c \,x^{3}\)	\(80\)
default	\(\frac {1}{11} b^{2} e \,x^{11}+\frac {1}{10} b^{2} d \,x^{10}+\frac {1}{9} b^{2} c \,x^{9}+\frac {1}{4} a b e \,x^{8}+\frac {2}{7} a d \,x^{7} b +\frac {1}{3} a b c \,x^{6}+\frac {1}{5} a^{2} e \,x^{5}+\frac {1}{4} a^{2} d \,x^{4}+\frac {1}{3} a^{2} c \,x^{3}\)	\(80\)
norman	\(\frac {1}{11} b^{2} e \,x^{11}+\frac {1}{10} b^{2} d \,x^{10}+\frac {1}{9} b^{2} c \,x^{9}+\frac {1}{4} a b e \,x^{8}+\frac {2}{7} a d \,x^{7} b +\frac {1}{3} a b c \,x^{6}+\frac {1}{5} a^{2} e \,x^{5}+\frac {1}{4} a^{2} d \,x^{4}+\frac {1}{3} a^{2} c \,x^{3}\)	\(80\)
risch	\(\frac {1}{11} b^{2} e \,x^{11}+\frac {1}{10} b^{2} d \,x^{10}+\frac {1}{9} b^{2} c \,x^{9}+\frac {1}{4} a b e \,x^{8}+\frac {2}{7} a d \,x^{7} b +\frac {1}{3} a b c \,x^{6}+\frac {1}{5} a^{2} e \,x^{5}+\frac {1}{4} a^{2} d \,x^{4}+\frac {1}{3} a^{2} c \,x^{3}\)	\(80\)
parallelrisch	\(\frac {1}{11} b^{2} e \,x^{11}+\frac {1}{10} b^{2} d \,x^{10}+\frac {1}{9} b^{2} c \,x^{9}+\frac {1}{4} a b e \,x^{8}+\frac {2}{7} a d \,x^{7} b +\frac {1}{3} a b c \,x^{6}+\frac {1}{5} a^{2} e \,x^{5}+\frac {1}{4} a^{2} d \,x^{4}+\frac {1}{3} a^{2} c \,x^{3}\)	\(80\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.96 \[ \int x^2 \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=\frac {1}{11} \, b^{2} e x^{11} + \frac {1}{10} \, b^{2} d x^{10} + \frac {1}{9} \, b^{2} c x^{9} + \frac {1}{4} \, a b e x^{8} + \frac {2}{7} \, a b d x^{7} + \frac {1}{3} \, a b c x^{6} + \frac {1}{5} \, a^{2} e x^{5} + \frac {1}{4} \, a^{2} d x^{4} + \frac {1}{3} \, a^{2} c x^{3} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.12 \[ \int x^2 \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=\frac {a^{2} c x^{3}}{3} + \frac {a^{2} d x^{4}}{4} + \frac {a^{2} e x^{5}}{5} + \frac {a b c x^{6}}{3} + \frac {2 a b d x^{7}}{7} + \frac {a b e x^{8}}{4} + \frac {b^{2} c x^{9}}{9} + \frac {b^{2} d x^{10}}{10} + \frac {b^{2} e x^{11}}{11} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.96 \[ \int x^2 \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=\frac {1}{11} \, b^{2} e x^{11} + \frac {1}{10} \, b^{2} d x^{10} + \frac {1}{9} \, b^{2} c x^{9} + \frac {1}{4} \, a b e x^{8} + \frac {2}{7} \, a b d x^{7} + \frac {1}{3} \, a b c x^{6} + \frac {1}{5} \, a^{2} e x^{5} + \frac {1}{4} \, a^{2} d x^{4} + \frac {1}{3} \, a^{2} c x^{3} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.96 \[ \int x^2 \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=\frac {1}{11} \, b^{2} e x^{11} + \frac {1}{10} \, b^{2} d x^{10} + \frac {1}{9} \, b^{2} c x^{9} + \frac {1}{4} \, a b e x^{8} + \frac {2}{7} \, a b d x^{7} + \frac {1}{3} \, a b c x^{6} + \frac {1}{5} \, a^{2} e x^{5} + \frac {1}{4} \, a^{2} d x^{4} + \frac {1}{3} \, a^{2} c x^{3} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.96 \[ \int x^2 \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=\frac {e\,a^2\,x^5}{5}+\frac {d\,a^2\,x^4}{4}+\frac {c\,a^2\,x^3}{3}+\frac {e\,a\,b\,x^8}{4}+\frac {2\,d\,a\,b\,x^7}{7}+\frac {c\,a\,b\,x^6}{3}+\frac {e\,b^2\,x^{11}}{11}+\frac {d\,b^2\,x^{10}}{10}+\frac {c\,b^2\,x^9}{9} \]

[In]

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

[Out]

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