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

Optimal. Leaf size=82 \[ \frac{1}{2} a^2 c x^2+\frac{1}{4} a^2 e x^4+\frac{2}{5} a b c x^5+\frac{d \left (a+b x^3\right )^3}{9 b}+\frac{2}{7} a b e x^7+\frac{1}{8} b^2 c x^8+\frac{1}{10} b^2 e x^{10} \]

[Out]

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

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Rubi [A]  time = 0.0510239, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {1582, 1850} \[ \frac{1}{2} a^2 c x^2+\frac{1}{4} a^2 e x^4+\frac{2}{5} a b c x^5+\frac{d \left (a+b x^3\right )^3}{9 b}+\frac{2}{7} a b e x^7+\frac{1}{8} b^2 c x^8+\frac{1}{10} b^2 e x^{10} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1582

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 1850

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

Mathematica [A]  time = 0.0030289, size = 97, normalized size = 1.18 \[ \frac{1}{2} a^2 c x^2+\frac{1}{3} a^2 d x^3+\frac{1}{4} a^2 e x^4+\frac{2}{5} a b c x^5+\frac{1}{3} a b d x^6+\frac{2}{7} a b e x^7+\frac{1}{8} b^2 c x^8+\frac{1}{9} b^2 d x^9+\frac{1}{10} b^2 e x^{10} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0., size = 80, normalized size = 1. \begin{align*}{\frac{{b}^{2}e{x}^{10}}{10}}+{\frac{{b}^{2}d{x}^{9}}{9}}+{\frac{{b}^{2}c{x}^{8}}{8}}+{\frac{2\,abe{x}^{7}}{7}}+{\frac{abd{x}^{6}}{3}}+{\frac{2\,abc{x}^{5}}{5}}+{\frac{{a}^{2}e{x}^{4}}{4}}+{\frac{{a}^{2}d{x}^{3}}{3}}+{\frac{{a}^{2}c{x}^{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.932844, size = 107, normalized size = 1.3 \begin{align*} \frac{1}{10} \, b^{2} e x^{10} + \frac{1}{9} \, b^{2} d x^{9} + \frac{1}{8} \, b^{2} c x^{8} + \frac{2}{7} \, a b e x^{7} + \frac{1}{3} \, a b d x^{6} + \frac{2}{5} \, a b c x^{5} + \frac{1}{4} \, a^{2} e x^{4} + \frac{1}{3} \, a^{2} d x^{3} + \frac{1}{2} \, a^{2} c x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.31605, size = 196, normalized size = 2.39 \begin{align*} \frac{1}{10} x^{10} e b^{2} + \frac{1}{9} x^{9} d b^{2} + \frac{1}{8} x^{8} c b^{2} + \frac{2}{7} x^{7} e b a + \frac{1}{3} x^{6} d b a + \frac{2}{5} x^{5} c b a + \frac{1}{4} x^{4} e a^{2} + \frac{1}{3} x^{3} d a^{2} + \frac{1}{2} x^{2} c a^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.071493, size = 94, normalized size = 1.15 \begin{align*} \frac{a^{2} c x^{2}}{2} + \frac{a^{2} d x^{3}}{3} + \frac{a^{2} e x^{4}}{4} + \frac{2 a b c x^{5}}{5} + \frac{a b d x^{6}}{3} + \frac{2 a b e x^{7}}{7} + \frac{b^{2} c x^{8}}{8} + \frac{b^{2} d x^{9}}{9} + \frac{b^{2} e x^{10}}{10} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.0604, size = 111, normalized size = 1.35 \begin{align*} \frac{1}{10} \, b^{2} x^{10} e + \frac{1}{9} \, b^{2} d x^{9} + \frac{1}{8} \, b^{2} c x^{8} + \frac{2}{7} \, a b x^{7} e + \frac{1}{3} \, a b d x^{6} + \frac{2}{5} \, a b c x^{5} + \frac{1}{4} \, a^{2} x^{4} e + \frac{1}{3} \, a^{2} d x^{3} + \frac{1}{2} \, a^{2} c x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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