3.387 \(\int (a+b x^3)^2 (c+d x+e x^2+f x^3+g x^4+h x^5) \, dx\)

Optimal. Leaf size=153 \[ a^2 c x+\frac{1}{2} a^2 d x^2+\frac{1}{6} a^2 h x^6+\frac{1}{7} b x^7 (2 a f+b c)+\frac{1}{4} a x^4 (a f+2 b c)+\frac{1}{8} b x^8 (2 a g+b d)+\frac{1}{5} a x^5 (a g+2 b d)+\frac{e \left (a+b x^3\right )^3}{9 b}+\frac{2}{9} a b h x^9+\frac{1}{10} b^2 f x^{10}+\frac{1}{11} b^2 g x^{11}+\frac{1}{12} b^2 h x^{12} \]

[Out]

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

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Rubi [A]  time = 0.12658, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.057, Rules used = {1582, 1850} \[ a^2 c x+\frac{1}{2} a^2 d x^2+\frac{1}{6} a^2 h x^6+\frac{1}{7} b x^7 (2 a f+b c)+\frac{1}{4} a x^4 (a f+2 b c)+\frac{1}{8} b x^8 (2 a g+b d)+\frac{1}{5} a x^5 (a g+2 b d)+\frac{e \left (a+b x^3\right )^3}{9 b}+\frac{2}{9} a b h x^9+\frac{1}{10} b^2 f x^{10}+\frac{1}{11} b^2 g x^{11}+\frac{1}{12} b^2 h x^{12} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^2*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5),x]

[Out]

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

Mathematica [A]  time = 0.0742271, size = 125, normalized size = 0.82 \[ \frac{462 a^2 x \left (60 c+x \left (30 d+x \left (20 e+15 f x+12 g x^2+10 h x^3\right )\right )\right )+22 a b x^4 (630 c+x (504 d+5 x (84 e+x (72 f+7 x (9 g+8 h x)))))+b^2 x^7 \left (3960 c+7 x \left (495 d+440 e x+6 x^2 \left (66 f+60 g x+55 h x^2\right )\right )\right )}{27720} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^2*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5),x]

[Out]

(b^2*x^7*(3960*c + 7*x*(495*d + 440*e*x + 6*x^2*(66*f + 60*g*x + 55*h*x^2))) + 462*a^2*x*(60*c + x*(30*d + x*(
20*e + 15*f*x + 12*g*x^2 + 10*h*x^3))) + 22*a*b*x^4*(630*c + x*(504*d + 5*x*(84*e + x*(72*f + 7*x*(9*g + 8*h*x
))))))/27720

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Maple [A]  time = 0.002, size = 149, normalized size = 1. \begin{align*}{\frac{{b}^{2}h{x}^{12}}{12}}+{\frac{{b}^{2}g{x}^{11}}{11}}+{\frac{{b}^{2}f{x}^{10}}{10}}+{\frac{ \left ( 2\,abh+{b}^{2}e \right ){x}^{9}}{9}}+{\frac{ \left ( 2\,abg+{b}^{2}d \right ){x}^{8}}{8}}+{\frac{ \left ( 2\,abf+{b}^{2}c \right ){x}^{7}}{7}}+{\frac{ \left ({a}^{2}h+2\,aeb \right ){x}^{6}}{6}}+{\frac{ \left ({a}^{2}g+2\,bda \right ){x}^{5}}{5}}+{\frac{ \left ({a}^{2}f+2\,abc \right ){x}^{4}}{4}}+{\frac{{a}^{2}e{x}^{3}}{3}}+{\frac{{a}^{2}d{x}^{2}}{2}}+{a}^{2}cx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^2*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c),x)

[Out]

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

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Maxima [A]  time = 0.947005, size = 200, normalized size = 1.31 \begin{align*} \frac{1}{12} \, b^{2} h x^{12} + \frac{1}{11} \, b^{2} g x^{11} + \frac{1}{10} \, b^{2} f x^{10} + \frac{1}{9} \,{\left (b^{2} e + 2 \, a b h\right )} x^{9} + \frac{1}{8} \,{\left (b^{2} d + 2 \, a b g\right )} x^{8} + \frac{1}{7} \,{\left (b^{2} c + 2 \, a b f\right )} x^{7} + \frac{1}{6} \,{\left (2 \, a b e + a^{2} h\right )} x^{6} + \frac{1}{3} \, a^{2} e x^{3} + \frac{1}{5} \,{\left (2 \, a b d + a^{2} g\right )} x^{5} + \frac{1}{2} \, a^{2} d x^{2} + \frac{1}{4} \,{\left (2 \, a b c + a^{2} f\right )} x^{4} + a^{2} c x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.900123, size = 387, normalized size = 2.53 \begin{align*} \frac{1}{12} x^{12} h b^{2} + \frac{1}{11} x^{11} g b^{2} + \frac{1}{10} x^{10} f b^{2} + \frac{1}{9} x^{9} e b^{2} + \frac{2}{9} x^{9} h b a + \frac{1}{8} x^{8} d b^{2} + \frac{1}{4} x^{8} g b a + \frac{1}{7} x^{7} c b^{2} + \frac{2}{7} x^{7} f b a + \frac{1}{3} x^{6} e b a + \frac{1}{6} x^{6} h a^{2} + \frac{2}{5} x^{5} d b a + \frac{1}{5} x^{5} g a^{2} + \frac{1}{2} x^{4} c b a + \frac{1}{4} x^{4} f a^{2} + \frac{1}{3} x^{3} e a^{2} + \frac{1}{2} x^{2} d a^{2} + x c a^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.087431, size = 163, normalized size = 1.07 \begin{align*} a^{2} c x + \frac{a^{2} d x^{2}}{2} + \frac{a^{2} e x^{3}}{3} + \frac{b^{2} f x^{10}}{10} + \frac{b^{2} g x^{11}}{11} + \frac{b^{2} h x^{12}}{12} + x^{9} \left (\frac{2 a b h}{9} + \frac{b^{2} e}{9}\right ) + x^{8} \left (\frac{a b g}{4} + \frac{b^{2} d}{8}\right ) + x^{7} \left (\frac{2 a b f}{7} + \frac{b^{2} c}{7}\right ) + x^{6} \left (\frac{a^{2} h}{6} + \frac{a b e}{3}\right ) + x^{5} \left (\frac{a^{2} g}{5} + \frac{2 a b d}{5}\right ) + x^{4} \left (\frac{a^{2} f}{4} + \frac{a b c}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**2*(h*x**5+g*x**4+f*x**3+e*x**2+d*x+c),x)

[Out]

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

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Giac [A]  time = 1.05691, size = 212, normalized size = 1.39 \begin{align*} \frac{1}{12} \, b^{2} h x^{12} + \frac{1}{11} \, b^{2} g x^{11} + \frac{1}{10} \, b^{2} f x^{10} + \frac{2}{9} \, a b h x^{9} + \frac{1}{9} \, b^{2} x^{9} e + \frac{1}{8} \, b^{2} d x^{8} + \frac{1}{4} \, a b g x^{8} + \frac{1}{7} \, b^{2} c x^{7} + \frac{2}{7} \, a b f x^{7} + \frac{1}{6} \, a^{2} h x^{6} + \frac{1}{3} \, a b x^{6} e + \frac{2}{5} \, a b d x^{5} + \frac{1}{5} \, a^{2} g x^{5} + \frac{1}{2} \, a b c x^{4} + \frac{1}{4} \, a^{2} f x^{4} + \frac{1}{3} \, a^{2} x^{3} e + \frac{1}{2} \, a^{2} d x^{2} + a^{2} c x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c),x, algorithm="giac")

[Out]

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