3.8.92 \(\int x^2 (A+B x) (a+b x+c x^2)^3 \, dx\)

Optimal. Leaf size=166 \[ \frac {1}{3} a^3 A x^3+\frac {1}{4} a^2 x^4 (a B+3 A b)+\frac {3}{8} c x^8 \left (a B c+A b c+b^2 B\right )+\frac {3}{5} a x^5 \left (A \left (a c+b^2\right )+a b B\right )+\frac {1}{7} x^7 \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {1}{6} x^6 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {1}{9} c^2 x^9 (A c+3 b B)+\frac {1}{10} B c^3 x^{10} \]

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Rubi [A]  time = 0.26, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {765} \begin {gather*} \frac {1}{4} a^2 x^4 (a B+3 A b)+\frac {1}{3} a^3 A x^3+\frac {1}{7} x^7 \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {3}{8} c x^8 \left (a B c+A b c+b^2 B\right )+\frac {1}{6} x^6 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {3}{5} a x^5 \left (A \left (a c+b^2\right )+a b B\right )+\frac {1}{9} c^2 x^9 (A c+3 b B)+\frac {1}{10} B c^3 x^{10} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(A + B*x)*(a + b*x + c*x^2)^3,x]

[Out]

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

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 166, normalized size = 1.00 \begin {gather*} \frac {1}{3} a^3 A x^3+\frac {1}{4} a^2 x^4 (a B+3 A b)+\frac {3}{8} c x^8 \left (a B c+A b c+b^2 B\right )+\frac {3}{5} a x^5 \left (A \left (a c+b^2\right )+a b B\right )+\frac {1}{7} x^7 \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {1}{6} x^6 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {1}{9} c^2 x^9 (A c+3 b B)+\frac {1}{10} B c^3 x^{10} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(A + B*x)*(a + b*x + c*x^2)^3,x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^2 (A+B x) \left (a+b x+c x^2\right )^3 \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^2*(A + B*x)*(a + b*x + c*x^2)^3,x]

[Out]

IntegrateAlgebraic[x^2*(A + B*x)*(a + b*x + c*x^2)^3, x]

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fricas [A]  time = 0.36, size = 192, normalized size = 1.16 \begin {gather*} \frac {1}{10} x^{10} c^{3} B + \frac {1}{3} x^{9} c^{2} b B + \frac {1}{9} x^{9} c^{3} A + \frac {3}{8} x^{8} c b^{2} B + \frac {3}{8} x^{8} c^{2} a B + \frac {3}{8} x^{8} c^{2} b A + \frac {1}{7} x^{7} b^{3} B + \frac {6}{7} x^{7} c b a B + \frac {3}{7} x^{7} c b^{2} A + \frac {3}{7} x^{7} c^{2} a A + \frac {1}{2} x^{6} b^{2} a B + \frac {1}{2} x^{6} c a^{2} B + \frac {1}{6} x^{6} b^{3} A + x^{6} c b a A + \frac {3}{5} x^{5} b a^{2} B + \frac {3}{5} x^{5} b^{2} a A + \frac {3}{5} x^{5} c a^{2} A + \frac {1}{4} x^{4} a^{3} B + \frac {3}{4} x^{4} b a^{2} A + \frac {1}{3} x^{3} a^{3} A \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 192, normalized size = 1.16 \begin {gather*} \frac {1}{10} \, B c^{3} x^{10} + \frac {1}{3} \, B b c^{2} x^{9} + \frac {1}{9} \, A c^{3} x^{9} + \frac {3}{8} \, B b^{2} c x^{8} + \frac {3}{8} \, B a c^{2} x^{8} + \frac {3}{8} \, A b c^{2} x^{8} + \frac {1}{7} \, B b^{3} x^{7} + \frac {6}{7} \, B a b c x^{7} + \frac {3}{7} \, A b^{2} c x^{7} + \frac {3}{7} \, A a c^{2} x^{7} + \frac {1}{2} \, B a b^{2} x^{6} + \frac {1}{6} \, A b^{3} x^{6} + \frac {1}{2} \, B a^{2} c x^{6} + A a b c x^{6} + \frac {3}{5} \, B a^{2} b x^{5} + \frac {3}{5} \, A a b^{2} x^{5} + \frac {3}{5} \, A a^{2} c x^{5} + \frac {1}{4} \, B a^{3} x^{4} + \frac {3}{4} \, A a^{2} b x^{4} + \frac {1}{3} \, A a^{3} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 226, normalized size = 1.36 \begin {gather*} \frac {B \,c^{3} x^{10}}{10}+\frac {\left (A \,c^{3}+3 B b \,c^{2}\right ) x^{9}}{9}+\frac {\left (3 A b \,c^{2}+\left (a \,c^{2}+2 b^{2} c +\left (2 a c +b^{2}\right ) c \right ) B \right ) x^{8}}{8}+\frac {A \,a^{3} x^{3}}{3}+\frac {\left (\left (a \,c^{2}+2 b^{2} c +\left (2 a c +b^{2}\right ) c \right ) A +\left (4 a b c +\left (2 a c +b^{2}\right ) b \right ) B \right ) x^{7}}{7}+\frac {\left (\left (4 a b c +\left (2 a c +b^{2}\right ) b \right ) A +\left (a^{2} c +2 a \,b^{2}+\left (2 a c +b^{2}\right ) a \right ) B \right ) x^{6}}{6}+\frac {\left (3 B \,a^{2} b +\left (a^{2} c +2 a \,b^{2}+\left (2 a c +b^{2}\right ) a \right ) A \right ) x^{5}}{5}+\frac {\left (3 A \,a^{2} b +B \,a^{3}\right ) x^{4}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(B*x+A)*(c*x^2+b*x+a)^3,x)

[Out]

1/10*B*c^3*x^10+1/9*(A*c^3+3*B*b*c^2)*x^9+1/8*(3*A*b*c^2+B*(a*c^2+2*b^2*c+c*(2*a*c+b^2)))*x^8+1/7*(A*(a*c^2+2*
b^2*c+c*(2*a*c+b^2))+B*(4*a*b*c+b*(2*a*c+b^2)))*x^7+1/6*(A*(4*a*b*c+b*(2*a*c+b^2))+B*(a*(2*a*c+b^2)+2*a*b^2+c*
a^2))*x^6+1/5*(A*(a*(2*a*c+b^2)+2*a*b^2+c*a^2)+3*B*a^2*b)*x^5+1/4*(3*A*a^2*b+B*a^3)*x^4+1/3*A*a^3*x^3

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maxima [A]  time = 0.71, size = 166, normalized size = 1.00 \begin {gather*} \frac {1}{10} \, B c^{3} x^{10} + \frac {1}{9} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{9} + \frac {3}{8} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{8} + \frac {1}{7} \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{7} + \frac {1}{3} \, A a^{3} x^{3} + \frac {1}{6} \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{6} + \frac {3}{5} \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{5} + \frac {1}{4} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*B*c^3*x^10 + 1/9*(3*B*b*c^2 + A*c^3)*x^9 + 3/8*(B*b^2*c + (B*a + A*b)*c^2)*x^8 + 1/7*(B*b^3 + 3*A*a*c^2 +
3*(2*B*a*b + A*b^2)*c)*x^7 + 1/3*A*a^3*x^3 + 1/6*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^6 + 3/5*(B*a^2
*b + A*a*b^2 + A*a^2*c)*x^5 + 1/4*(B*a^3 + 3*A*a^2*b)*x^4

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mupad [B]  time = 0.07, size = 168, normalized size = 1.01 \begin {gather*} x^6\,\left (\frac {B\,c\,a^2}{2}+\frac {B\,a\,b^2}{2}+A\,c\,a\,b+\frac {A\,b^3}{6}\right )+x^7\,\left (\frac {B\,b^3}{7}+\frac {3\,A\,b^2\,c}{7}+\frac {6\,B\,a\,b\,c}{7}+\frac {3\,A\,a\,c^2}{7}\right )+x^4\,\left (\frac {B\,a^3}{4}+\frac {3\,A\,b\,a^2}{4}\right )+x^9\,\left (\frac {A\,c^3}{9}+\frac {B\,b\,c^2}{3}\right )+x^5\,\left (\frac {3\,B\,a^2\,b}{5}+\frac {3\,A\,c\,a^2}{5}+\frac {3\,A\,a\,b^2}{5}\right )+x^8\,\left (\frac {3\,B\,b^2\,c}{8}+\frac {3\,A\,b\,c^2}{8}+\frac {3\,B\,a\,c^2}{8}\right )+\frac {A\,a^3\,x^3}{3}+\frac {B\,c^3\,x^{10}}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(A + B*x)*(a + b*x + c*x^2)^3,x)

[Out]

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

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sympy [A]  time = 0.11, size = 201, normalized size = 1.21 \begin {gather*} \frac {A a^{3} x^{3}}{3} + \frac {B c^{3} x^{10}}{10} + x^{9} \left (\frac {A c^{3}}{9} + \frac {B b c^{2}}{3}\right ) + x^{8} \left (\frac {3 A b c^{2}}{8} + \frac {3 B a c^{2}}{8} + \frac {3 B b^{2} c}{8}\right ) + x^{7} \left (\frac {3 A a c^{2}}{7} + \frac {3 A b^{2} c}{7} + \frac {6 B a b c}{7} + \frac {B b^{3}}{7}\right ) + x^{6} \left (A a b c + \frac {A b^{3}}{6} + \frac {B a^{2} c}{2} + \frac {B a b^{2}}{2}\right ) + x^{5} \left (\frac {3 A a^{2} c}{5} + \frac {3 A a b^{2}}{5} + \frac {3 B a^{2} b}{5}\right ) + x^{4} \left (\frac {3 A a^{2} b}{4} + \frac {B a^{3}}{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(B*x+A)*(c*x**2+b*x+a)**3,x)

[Out]

A*a**3*x**3/3 + B*c**3*x**10/10 + x**9*(A*c**3/9 + B*b*c**2/3) + x**8*(3*A*b*c**2/8 + 3*B*a*c**2/8 + 3*B*b**2*
c/8) + x**7*(3*A*a*c**2/7 + 3*A*b**2*c/7 + 6*B*a*b*c/7 + B*b**3/7) + x**6*(A*a*b*c + A*b**3/6 + B*a**2*c/2 + B
*a*b**2/2) + x**5*(3*A*a**2*c/5 + 3*A*a*b**2/5 + 3*B*a**2*b/5) + x**4*(3*A*a**2*b/4 + B*a**3/4)

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