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

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

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1582, 1850} \[ \frac {1}{2} a^2 c x^2+\frac {1}{4} a^2 e x^4+\frac {1}{6} a^2 g x^6+\frac {1}{8} b x^8 (2 a f+b c)+\frac {1}{5} a x^5 (a f+2 b c)+\frac {d \left (a+b x^3\right )^3}{9 b}+\frac {1}{10} b x^{10} (2 a h+b e)+\frac {1}{7} a x^7 (a h+2 b e)+\frac {2}{9} a b g x^9+\frac {1}{11} b^2 f x^{11}+\frac {1}{12} b^2 g x^{12}+\frac {1}{13} b^2 h x^{13} \]

Antiderivative was successfully verified.

[In]

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

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Mathematica [A]  time = 0.03, size = 163, normalized size = 1.03 \[ \frac {1}{2} a^2 c x^2+\frac {1}{3} a^2 d x^3+\frac {1}{4} a^2 e x^4+\frac {1}{8} b x^8 (2 a f+b c)+\frac {1}{5} a x^5 (a f+2 b c)+\frac {1}{9} b x^9 (2 a g+b d)+\frac {1}{6} a x^6 (a g+2 b d)+\frac {1}{10} b x^{10} (2 a h+b e)+\frac {1}{7} a x^7 (a h+2 b e)+\frac {1}{11} b^2 f x^{11}+\frac {1}{12} b^2 g x^{12}+\frac {1}{13} b^2 h x^{13} \]

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 0.36, size = 157, normalized size = 0.99 \[ \frac {1}{13} x^{13} h b^{2} + \frac {1}{12} x^{12} g b^{2} + \frac {1}{11} x^{11} f b^{2} + \frac {1}{10} x^{10} e b^{2} + \frac {1}{5} x^{10} h b a + \frac {1}{9} x^{9} d b^{2} + \frac {2}{9} x^{9} g b a + \frac {1}{8} x^{8} c b^{2} + \frac {1}{4} x^{8} f b a + \frac {2}{7} x^{7} e b a + \frac {1}{7} x^{7} h a^{2} + \frac {1}{3} x^{6} d b a + \frac {1}{6} x^{6} g a^{2} + \frac {2}{5} x^{5} c b a + \frac {1}{5} x^{5} f a^{2} + \frac {1}{4} x^{4} e a^{2} + \frac {1}{3} x^{3} d a^{2} + \frac {1}{2} x^{2} c a^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13*x^13*h*b^2 + 1/12*x^12*g*b^2 + 1/11*x^11*f*b^2 + 1/10*x^10*e*b^2 + 1/5*x^10*h*b*a + 1/9*x^9*d*b^2 + 2/9*x
^9*g*b*a + 1/8*x^8*c*b^2 + 1/4*x^8*f*b*a + 2/7*x^7*e*b*a + 1/7*x^7*h*a^2 + 1/3*x^6*d*b*a + 1/6*x^6*g*a^2 + 2/5
*x^5*c*b*a + 1/5*x^5*f*a^2 + 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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giac [A]  time = 0.19, size = 160, normalized size = 1.01 \[ \frac {1}{13} \, b^{2} h x^{13} + \frac {1}{12} \, b^{2} g x^{12} + \frac {1}{11} \, b^{2} f x^{11} + \frac {1}{5} \, a b h x^{10} + \frac {1}{10} \, b^{2} x^{10} e + \frac {1}{9} \, b^{2} d x^{9} + \frac {2}{9} \, a b g x^{9} + \frac {1}{8} \, b^{2} c x^{8} + \frac {1}{4} \, a b f x^{8} + \frac {1}{7} \, a^{2} h x^{7} + \frac {2}{7} \, a b x^{7} e + \frac {1}{3} \, a b d x^{6} + \frac {1}{6} \, a^{2} g x^{6} + \frac {2}{5} \, a b c x^{5} + \frac {1}{5} \, a^{2} f x^{5} + \frac {1}{4} \, a^{2} x^{4} e + \frac {1}{3} \, a^{2} d x^{3} + \frac {1}{2} \, a^{2} c x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 152, normalized size = 0.96 \[ \frac {b^{2} h \,x^{13}}{13}+\frac {b^{2} g \,x^{12}}{12}+\frac {b^{2} f \,x^{11}}{11}+\frac {\left (2 a b h +b^{2} e \right ) x^{10}}{10}+\frac {\left (2 a b g +b^{2} d \right ) x^{9}}{9}+\frac {\left (2 a b f +c \,b^{2}\right ) x^{8}}{8}+\frac {a^{2} e \,x^{4}}{4}+\frac {\left (a^{2} h +2 b e a \right ) x^{7}}{7}+\frac {a^{2} d \,x^{3}}{3}+\frac {\left (a^{2} g +2 b d a \right ) x^{6}}{6}+\frac {a^{2} c \,x^{2}}{2}+\frac {\left (a^{2} f +2 a b c \right ) x^{5}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13*b^2*h*x^13+1/12*b^2*g*x^12+1/11*b^2*f*x^11+1/10*(2*a*b*h+b^2*e)*x^10+1/9*(2*a*b*g+b^2*d)*x^9+1/8*(2*a*b*f
+b^2*c)*x^8+1/7*(a^2*h+2*a*b*e)*x^7+1/6*(a^2*g+2*a*b*d)*x^6+1/5*(a^2*f+2*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 = 1.28, size = 151, normalized size = 0.96 \[ \frac {1}{13} \, b^{2} h x^{13} + \frac {1}{12} \, b^{2} g x^{12} + \frac {1}{11} \, b^{2} f x^{11} + \frac {1}{10} \, {\left (b^{2} e + 2 \, a b h\right )} x^{10} + \frac {1}{9} \, {\left (b^{2} d + 2 \, a b g\right )} x^{9} + \frac {1}{8} \, {\left (b^{2} c + 2 \, a b f\right )} x^{8} + \frac {1}{7} \, {\left (2 \, a b e + a^{2} h\right )} x^{7} + \frac {1}{4} \, a^{2} e x^{4} + \frac {1}{6} \, {\left (2 \, a b d + a^{2} g\right )} x^{6} + \frac {1}{3} \, a^{2} d x^{3} + \frac {1}{5} \, {\left (2 \, a b c + a^{2} f\right )} x^{5} + \frac {1}{2} \, a^{2} c x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.09, size = 151, normalized size = 0.96 \[ x^5\,\left (\frac {f\,a^2}{5}+\frac {2\,b\,c\,a}{5}\right )+x^8\,\left (\frac {c\,b^2}{8}+\frac {a\,f\,b}{4}\right )+x^6\,\left (\frac {g\,a^2}{6}+\frac {b\,d\,a}{3}\right )+x^9\,\left (\frac {d\,b^2}{9}+\frac {2\,a\,g\,b}{9}\right )+x^7\,\left (\frac {h\,a^2}{7}+\frac {2\,b\,e\,a}{7}\right )+x^{10}\,\left (\frac {e\,b^2}{10}+\frac {a\,h\,b}{5}\right )+\frac {a^2\,c\,x^2}{2}+\frac {a^2\,d\,x^3}{3}+\frac {a^2\,e\,x^4}{4}+\frac {b^2\,f\,x^{11}}{11}+\frac {b^2\,g\,x^{12}}{12}+\frac {b^2\,h\,x^{13}}{13} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.10, size = 167, normalized size = 1.06 \[ \frac {a^{2} c x^{2}}{2} + \frac {a^{2} d x^{3}}{3} + \frac {a^{2} e x^{4}}{4} + \frac {b^{2} f x^{11}}{11} + \frac {b^{2} g x^{12}}{12} + \frac {b^{2} h x^{13}}{13} + x^{10} \left (\frac {a b h}{5} + \frac {b^{2} e}{10}\right ) + x^{9} \left (\frac {2 a b g}{9} + \frac {b^{2} d}{9}\right ) + x^{8} \left (\frac {a b f}{4} + \frac {b^{2} c}{8}\right ) + x^{7} \left (\frac {a^{2} h}{7} + \frac {2 a b e}{7}\right ) + x^{6} \left (\frac {a^{2} g}{6} + \frac {a b d}{3}\right ) + x^{5} \left (\frac {a^{2} f}{5} + \frac {2 a b c}{5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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