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

Optimal. Leaf size=158 \[ \frac{1}{4} a^2 d x^4+\frac{1}{5} a^2 e x^5+\frac{1}{6} a^2 f x^6+\frac{c \left (a+b x^3\right )^3}{9 b}+\frac{1}{10} b x^{10} (2 a g+b d)+\frac{1}{7} a x^7 (a g+2 b d)+\frac{1}{11} b x^{11} (2 a h+b e)+\frac{1}{8} a x^8 (a h+2 b e)+\frac{2}{9} a b f x^9+\frac{1}{12} b^2 f x^{12}+\frac{1}{13} b^2 g x^{13}+\frac{1}{14} b^2 h x^{14} \]

[Out]

(a^2*d*x^4)/4 + (a^2*e*x^5)/5 + (a^2*f*x^6)/6 + (a*(2*b*d + a*g)*x^7)/7 + (a*(2*b*e + a*h)*x^8)/8 + (2*a*b*f*x
^9)/9 + (b*(b*d + 2*a*g)*x^10)/10 + (b*(b*e + 2*a*h)*x^11)/11 + (b^2*f*x^12)/12 + (b^2*g*x^13)/13 + (b^2*h*x^1
4)/14 + (c*(a + b*x^3)^3)/(9*b)

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Rubi [A]  time = 0.125777, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {1582, 1850} \[ \frac{1}{4} a^2 d x^4+\frac{1}{5} a^2 e x^5+\frac{1}{6} a^2 f x^6+\frac{c \left (a+b x^3\right )^3}{9 b}+\frac{1}{10} b x^{10} (2 a g+b d)+\frac{1}{7} a x^7 (a g+2 b d)+\frac{1}{11} b x^{11} (2 a h+b e)+\frac{1}{8} a x^8 (a h+2 b e)+\frac{2}{9} a b f x^9+\frac{1}{12} b^2 f x^{12}+\frac{1}{13} b^2 g x^{13}+\frac{1}{14} b^2 h x^{14} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*d*x^4)/4 + (a^2*e*x^5)/5 + (a^2*f*x^6)/6 + (a*(2*b*d + a*g)*x^7)/7 + (a*(2*b*e + a*h)*x^8)/8 + (2*a*b*f*x
^9)/9 + (b*(b*d + 2*a*g)*x^10)/10 + (b*(b*e + 2*a*h)*x^11)/11 + (b^2*f*x^12)/12 + (b^2*g*x^13)/13 + (b^2*h*x^1
4)/14 + (c*(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^2 \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{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+f x^3+g x^4+h x^5\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+a^2 f x^5+a (2 b d+a g) x^6+a (2 b e+a h) x^7+2 a b f x^8+b (b d+2 a g) x^9+b (b e+2 a h) x^{10}+b^2 f x^{11}+b^2 g x^{12}+b^2 h x^{13}\right ) \, dx\\ &=\frac{1}{4} a^2 d x^4+\frac{1}{5} a^2 e x^5+\frac{1}{6} a^2 f x^6+\frac{1}{7} a (2 b d+a g) x^7+\frac{1}{8} a (2 b e+a h) x^8+\frac{2}{9} a b f x^9+\frac{1}{10} b (b d+2 a g) x^{10}+\frac{1}{11} b (b e+2 a h) x^{11}+\frac{1}{12} b^2 f x^{12}+\frac{1}{13} b^2 g x^{13}+\frac{1}{14} b^2 h x^{14}+\frac{c \left (a+b x^3\right )^3}{9 b}\\ \end{align*}

Mathematica [A]  time = 0.0759694, size = 150, normalized size = 0.95 \[ a^2 \left (\frac{c x^3}{3}+\frac{d x^4}{4}+\frac{e x^5}{5}+\frac{f x^6}{6}+\frac{g x^7}{7}+\frac{h x^8}{8}\right )+a b \left (\frac{c x^6}{3}+\frac{2 d x^7}{7}+\frac{e x^8}{4}+\frac{2 f x^9}{9}+\frac{g x^{10}}{5}+\frac{2 h x^{11}}{11}\right )+\frac{b^2 x^9 \left (20020 c+3 x \left (6006 d+5460 e x+55 x^2 \left (91 f+84 g x+78 h x^2\right )\right )\right )}{180180} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(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^3)/3 + (d*x^4)/4 + (e*x^5)/5 + (f*x^6)/6 + (g*x^7)/7 + (h*x^8)/8) + a*b*((c*x^6)/3 + (2*d*x^7)/7 + (
e*x^8)/4 + (2*f*x^9)/9 + (g*x^10)/5 + (2*h*x^11)/11) + (b^2*x^9*(20020*c + 3*x*(6006*d + 5460*e*x + 55*x^2*(91
*f + 84*g*x + 78*h*x^2))))/180180

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Maple [A]  time = 0.001, size = 152, normalized size = 1. \begin{align*}{\frac{{b}^{2}h{x}^{14}}{14}}+{\frac{{b}^{2}g{x}^{13}}{13}}+{\frac{{b}^{2}f{x}^{12}}{12}}+{\frac{ \left ( 2\,abh+{b}^{2}e \right ){x}^{11}}{11}}+{\frac{ \left ( 2\,abg+{b}^{2}d \right ){x}^{10}}{10}}+{\frac{ \left ( 2\,abf+{b}^{2}c \right ){x}^{9}}{9}}+{\frac{ \left ({a}^{2}h+2\,aeb \right ){x}^{8}}{8}}+{\frac{ \left ({a}^{2}g+2\,bda \right ){x}^{7}}{7}}+{\frac{ \left ({a}^{2}f+2\,abc \right ){x}^{6}}{6}}+{\frac{{a}^{2}e{x}^{5}}{5}}+{\frac{{a}^{2}d{x}^{4}}{4}}+{\frac{{a}^{2}c{x}^{3}}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/14*b^2*h*x^14+1/13*b^2*g*x^13+1/12*b^2*f*x^12+1/11*(2*a*b*h+b^2*e)*x^11+1/10*(2*a*b*g+b^2*d)*x^10+1/9*(2*a*b
*f+b^2*c)*x^9+1/8*(a^2*h+2*a*b*e)*x^8+1/7*(a^2*g+2*a*b*d)*x^7+1/6*(a^2*f+2*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

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Maxima [A]  time = 0.959115, size = 204, normalized size = 1.29 \begin{align*} \frac{1}{14} \, b^{2} h x^{14} + \frac{1}{13} \, b^{2} g x^{13} + \frac{1}{12} \, b^{2} f x^{12} + \frac{1}{11} \,{\left (b^{2} e + 2 \, a b h\right )} x^{11} + \frac{1}{10} \,{\left (b^{2} d + 2 \, a b g\right )} x^{10} + \frac{1}{9} \,{\left (b^{2} c + 2 \, a b f\right )} x^{9} + \frac{1}{8} \,{\left (2 \, a b e + a^{2} h\right )} x^{8} + \frac{1}{5} \, a^{2} e x^{5} + \frac{1}{7} \,{\left (2 \, a b d + a^{2} g\right )} x^{7} + \frac{1}{4} \, a^{2} d x^{4} + \frac{1}{6} \,{\left (2 \, a b c + a^{2} f\right )} x^{6} + \frac{1}{3} \, a^{2} c x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/14*b^2*h*x^14 + 1/13*b^2*g*x^13 + 1/12*b^2*f*x^12 + 1/11*(b^2*e + 2*a*b*h)*x^11 + 1/10*(b^2*d + 2*a*b*g)*x^1
0 + 1/9*(b^2*c + 2*a*b*f)*x^9 + 1/8*(2*a*b*e + a^2*h)*x^8 + 1/5*a^2*e*x^5 + 1/7*(2*a*b*d + a^2*g)*x^7 + 1/4*a^
2*d*x^4 + 1/6*(2*a*b*c + a^2*f)*x^6 + 1/3*a^2*c*x^3

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Fricas [A]  time = 0.861869, size = 405, normalized size = 2.56 \begin{align*} \frac{1}{14} x^{14} h b^{2} + \frac{1}{13} x^{13} g b^{2} + \frac{1}{12} x^{12} f b^{2} + \frac{1}{11} x^{11} e b^{2} + \frac{2}{11} x^{11} h b a + \frac{1}{10} x^{10} d b^{2} + \frac{1}{5} x^{10} g b a + \frac{1}{9} x^{9} c b^{2} + \frac{2}{9} x^{9} f b a + \frac{1}{4} x^{8} e b a + \frac{1}{8} x^{8} h a^{2} + \frac{2}{7} x^{7} d b a + \frac{1}{7} x^{7} g a^{2} + \frac{1}{3} x^{6} c b a + \frac{1}{6} x^{6} f a^{2} + \frac{1}{5} x^{5} e a^{2} + \frac{1}{4} x^{4} d a^{2} + \frac{1}{3} x^{3} c a^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/14*x^14*h*b^2 + 1/13*x^13*g*b^2 + 1/12*x^12*f*b^2 + 1/11*x^11*e*b^2 + 2/11*x^11*h*b*a + 1/10*x^10*d*b^2 + 1/
5*x^10*g*b*a + 1/9*x^9*c*b^2 + 2/9*x^9*f*b*a + 1/4*x^8*e*b*a + 1/8*x^8*h*a^2 + 2/7*x^7*d*b*a + 1/7*x^7*g*a^2 +
1/3*x^6*c*b*a + 1/6*x^6*f*a^2 + 1/5*x^5*e*a^2 + 1/4*x^4*d*a^2 + 1/3*x^3*c*a^2

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Sympy [A]  time = 0.100332, size = 167, normalized size = 1.06 \begin{align*} \frac{a^{2} c x^{3}}{3} + \frac{a^{2} d x^{4}}{4} + \frac{a^{2} e x^{5}}{5} + \frac{b^{2} f x^{12}}{12} + \frac{b^{2} g x^{13}}{13} + \frac{b^{2} h x^{14}}{14} + x^{11} \left (\frac{2 a b h}{11} + \frac{b^{2} e}{11}\right ) + x^{10} \left (\frac{a b g}{5} + \frac{b^{2} d}{10}\right ) + x^{9} \left (\frac{2 a b f}{9} + \frac{b^{2} c}{9}\right ) + x^{8} \left (\frac{a^{2} h}{8} + \frac{a b e}{4}\right ) + x^{7} \left (\frac{a^{2} g}{7} + \frac{2 a b d}{7}\right ) + x^{6} \left (\frac{a^{2} f}{6} + \frac{a b c}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(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**3/3 + a**2*d*x**4/4 + a**2*e*x**5/5 + b**2*f*x**12/12 + b**2*g*x**13/13 + b**2*h*x**14/14 + x**11*(2
*a*b*h/11 + b**2*e/11) + x**10*(a*b*g/5 + b**2*d/10) + x**9*(2*a*b*f/9 + b**2*c/9) + x**8*(a**2*h/8 + a*b*e/4)
+ x**7*(a**2*g/7 + 2*a*b*d/7) + x**6*(a**2*f/6 + a*b*c/3)

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Giac [A]  time = 1.07267, size = 216, normalized size = 1.37 \begin{align*} \frac{1}{14} \, b^{2} h x^{14} + \frac{1}{13} \, b^{2} g x^{13} + \frac{1}{12} \, b^{2} f x^{12} + \frac{2}{11} \, a b h x^{11} + \frac{1}{11} \, b^{2} x^{11} e + \frac{1}{10} \, b^{2} d x^{10} + \frac{1}{5} \, a b g x^{10} + \frac{1}{9} \, b^{2} c x^{9} + \frac{2}{9} \, a b f x^{9} + \frac{1}{8} \, a^{2} h x^{8} + \frac{1}{4} \, a b x^{8} e + \frac{2}{7} \, a b d x^{7} + \frac{1}{7} \, a^{2} g x^{7} + \frac{1}{3} \, a b c x^{6} + \frac{1}{6} \, a^{2} f x^{6} + \frac{1}{5} \, a^{2} x^{5} e + \frac{1}{4} \, a^{2} d x^{4} + \frac{1}{3} \, a^{2} c x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/14*b^2*h*x^14 + 1/13*b^2*g*x^13 + 1/12*b^2*f*x^12 + 2/11*a*b*h*x^11 + 1/11*b^2*x^11*e + 1/10*b^2*d*x^10 + 1/
5*a*b*g*x^10 + 1/9*b^2*c*x^9 + 2/9*a*b*f*x^9 + 1/8*a^2*h*x^8 + 1/4*a*b*x^8*e + 2/7*a*b*d*x^7 + 1/7*a^2*g*x^7 +
1/3*a*b*c*x^6 + 1/6*a^2*f*x^6 + 1/5*a^2*x^5*e + 1/4*a^2*d*x^4 + 1/3*a^2*c*x^3