3.80 \(\int x (a+b x^2)^3 (A+B x+C x^2+D x^3) \, dx\)

Optimal. Leaf size=138 \[ \frac{1}{5} a^2 x^5 (a D+3 b B)+\frac{1}{2} a^2 b C x^6+\frac{1}{3} a^3 B x^3+\frac{1}{4} a^3 C x^4+\frac{A \left (a+b x^2\right )^4}{8 b}+\frac{1}{9} b^2 x^9 (3 a D+b B)+\frac{3}{8} a b^2 C x^8+\frac{3}{7} a b x^7 (a D+b B)+\frac{1}{10} b^3 C x^{10}+\frac{1}{11} b^3 D x^{11} \]

[Out]

(a^3*B*x^3)/3 + (a^3*C*x^4)/4 + (a^2*(3*b*B + a*D)*x^5)/5 + (a^2*b*C*x^6)/2 + (3*a*b*(b*B + a*D)*x^7)/7 + (3*a
*b^2*C*x^8)/8 + (b^2*(b*B + 3*a*D)*x^9)/9 + (b^3*C*x^10)/10 + (b^3*D*x^11)/11 + (A*(a + b*x^2)^4)/(8*b)

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Rubi [A]  time = 0.0947486, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {1582, 1810} \[ \frac{1}{5} a^2 x^5 (a D+3 b B)+\frac{1}{2} a^2 b C x^6+\frac{1}{3} a^3 B x^3+\frac{1}{4} a^3 C x^4+\frac{A \left (a+b x^2\right )^4}{8 b}+\frac{1}{9} b^2 x^9 (3 a D+b B)+\frac{3}{8} a b^2 C x^8+\frac{3}{7} a b x^7 (a D+b B)+\frac{1}{10} b^3 C x^{10}+\frac{1}{11} b^3 D x^{11} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*B*x^3)/3 + (a^3*C*x^4)/4 + (a^2*(3*b*B + a*D)*x^5)/5 + (a^2*b*C*x^6)/2 + (3*a*b*(b*B + a*D)*x^7)/7 + (3*a
*b^2*C*x^8)/8 + (b^2*(b*B + 3*a*D)*x^9)/9 + (b^3*C*x^10)/10 + (b^3*D*x^11)/11 + (A*(a + b*x^2)^4)/(8*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 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin{align*} \int x \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx &=\frac{A \left (a+b x^2\right )^4}{8 b}+\int \left (a+b x^2\right )^3 \left (-A x+x \left (A+B x+C x^2+D x^3\right )\right ) \, dx\\ &=\frac{A \left (a+b x^2\right )^4}{8 b}+\int \left (a^3 B x^2+a^3 C x^3+a^2 (3 b B+a D) x^4+3 a^2 b C x^5+3 a b (b B+a D) x^6+3 a b^2 C x^7+b^2 (b B+3 a D) x^8+b^3 C x^9+b^3 D x^{10}\right ) \, dx\\ &=\frac{1}{3} a^3 B x^3+\frac{1}{4} a^3 C x^4+\frac{1}{5} a^2 (3 b B+a D) x^5+\frac{1}{2} a^2 b C x^6+\frac{3}{7} a b (b B+a D) x^7+\frac{3}{8} a b^2 C x^8+\frac{1}{9} b^2 (b B+3 a D) x^9+\frac{1}{10} b^3 C x^{10}+\frac{1}{11} b^3 D x^{11}+\frac{A \left (a+b x^2\right )^4}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.0499018, size = 124, normalized size = 0.9 \[ \frac{198 a^2 b x^4 (105 A+2 x (42 B+5 x (7 C+6 D x)))+462 a^3 x^2 (30 A+x (20 B+3 x (5 C+4 D x)))+165 a b^2 x^6 (84 A+x (72 B+7 x (9 C+8 D x)))+7 b^3 x^8 \left (495 A+4 x \left (110 B+99 C x+90 D x^2\right )\right )}{27720} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*b^3*x^8*(495*A + 4*x*(110*B + 99*C*x + 90*D*x^2)) + 462*a^3*x^2*(30*A + x*(20*B + 3*x*(5*C + 4*D*x))) + 198
*a^2*b*x^4*(105*A + 2*x*(42*B + 5*x*(7*C + 6*D*x))) + 165*a*b^2*x^6*(84*A + x*(72*B + 7*x*(9*C + 8*D*x))))/277
20

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Maple [A]  time = 0.001, size = 150, normalized size = 1.1 \begin{align*}{\frac{{b}^{3}D{x}^{11}}{11}}+{\frac{{b}^{3}C{x}^{10}}{10}}+{\frac{ \left ({b}^{3}B+3\,a{b}^{2}D \right ){x}^{9}}{9}}+{\frac{ \left ( A{b}^{3}+3\,a{b}^{2}C \right ){x}^{8}}{8}}+{\frac{ \left ( 3\,a{b}^{2}B+3\,{a}^{2}bD \right ){x}^{7}}{7}}+{\frac{ \left ( 3\,a{b}^{2}A+3\,{a}^{2}bC \right ){x}^{6}}{6}}+{\frac{ \left ( 3\,{a}^{2}bB+{a}^{3}D \right ){x}^{5}}{5}}+{\frac{ \left ( 3\,A{a}^{2}b+{a}^{3}C \right ){x}^{4}}{4}}+{\frac{{a}^{3}B{x}^{3}}{3}}+{\frac{{a}^{3}A{x}^{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*b^3*D*x^11+1/10*b^3*C*x^10+1/9*(B*b^3+3*D*a*b^2)*x^9+1/8*(A*b^3+3*C*a*b^2)*x^8+1/7*(3*B*a*b^2+3*D*a^2*b)*
x^7+1/6*(3*A*a*b^2+3*C*a^2*b)*x^6+1/5*(3*B*a^2*b+D*a^3)*x^5+1/4*(3*A*a^2*b+C*a^3)*x^4+1/3*a^3*B*x^3+1/2*a^3*A*
x^2

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Maxima [A]  time = 1.0011, size = 196, normalized size = 1.42 \begin{align*} \frac{1}{11} \, D b^{3} x^{11} + \frac{1}{10} \, C b^{3} x^{10} + \frac{1}{9} \,{\left (3 \, D a b^{2} + B b^{3}\right )} x^{9} + \frac{1}{8} \,{\left (3 \, C a b^{2} + A b^{3}\right )} x^{8} + \frac{3}{7} \,{\left (D a^{2} b + B a b^{2}\right )} x^{7} + \frac{1}{3} \, B a^{3} x^{3} + \frac{1}{2} \,{\left (C a^{2} b + A a b^{2}\right )} x^{6} + \frac{1}{2} \, A a^{3} x^{2} + \frac{1}{5} \,{\left (D a^{3} + 3 \, B a^{2} b\right )} x^{5} + \frac{1}{4} \,{\left (C a^{3} + 3 \, A a^{2} b\right )} x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*D*b^3*x^11 + 1/10*C*b^3*x^10 + 1/9*(3*D*a*b^2 + B*b^3)*x^9 + 1/8*(3*C*a*b^2 + A*b^3)*x^8 + 3/7*(D*a^2*b +
B*a*b^2)*x^7 + 1/3*B*a^3*x^3 + 1/2*(C*a^2*b + A*a*b^2)*x^6 + 1/2*A*a^3*x^2 + 1/5*(D*a^3 + 3*B*a^2*b)*x^5 + 1/
4*(C*a^3 + 3*A*a^2*b)*x^4

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Fricas [A]  time = 1.28047, size = 371, normalized size = 2.69 \begin{align*} \frac{1}{11} x^{11} b^{3} D + \frac{1}{10} x^{10} b^{3} C + \frac{1}{3} x^{9} b^{2} a D + \frac{1}{9} x^{9} b^{3} B + \frac{3}{8} x^{8} b^{2} a C + \frac{1}{8} x^{8} b^{3} A + \frac{3}{7} x^{7} b a^{2} D + \frac{3}{7} x^{7} b^{2} a B + \frac{1}{2} x^{6} b a^{2} C + \frac{1}{2} x^{6} b^{2} a A + \frac{1}{5} x^{5} a^{3} D + \frac{3}{5} x^{5} b a^{2} B + \frac{1}{4} x^{4} a^{3} C + \frac{3}{4} x^{4} b a^{2} A + \frac{1}{3} x^{3} a^{3} B + \frac{1}{2} x^{2} a^{3} A \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*x^11*b^3*D + 1/10*x^10*b^3*C + 1/3*x^9*b^2*a*D + 1/9*x^9*b^3*B + 3/8*x^8*b^2*a*C + 1/8*x^8*b^3*A + 3/7*x^
7*b*a^2*D + 3/7*x^7*b^2*a*B + 1/2*x^6*b*a^2*C + 1/2*x^6*b^2*a*A + 1/5*x^5*a^3*D + 3/5*x^5*b*a^2*B + 1/4*x^4*a^
3*C + 3/4*x^4*b*a^2*A + 1/3*x^3*a^3*B + 1/2*x^2*a^3*A

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Sympy [A]  time = 0.084875, size = 163, normalized size = 1.18 \begin{align*} \frac{A a^{3} x^{2}}{2} + \frac{B a^{3} x^{3}}{3} + \frac{C b^{3} x^{10}}{10} + \frac{D b^{3} x^{11}}{11} + x^{9} \left (\frac{B b^{3}}{9} + \frac{D a b^{2}}{3}\right ) + x^{8} \left (\frac{A b^{3}}{8} + \frac{3 C a b^{2}}{8}\right ) + x^{7} \left (\frac{3 B a b^{2}}{7} + \frac{3 D a^{2} b}{7}\right ) + x^{6} \left (\frac{A a b^{2}}{2} + \frac{C a^{2} b}{2}\right ) + x^{5} \left (\frac{3 B a^{2} b}{5} + \frac{D a^{3}}{5}\right ) + x^{4} \left (\frac{3 A a^{2} b}{4} + \frac{C a^{3}}{4}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**3*x**2/2 + B*a**3*x**3/3 + C*b**3*x**10/10 + D*b**3*x**11/11 + x**9*(B*b**3/9 + D*a*b**2/3) + x**8*(A*b**
3/8 + 3*C*a*b**2/8) + x**7*(3*B*a*b**2/7 + 3*D*a**2*b/7) + x**6*(A*a*b**2/2 + C*a**2*b/2) + x**5*(3*B*a**2*b/5
+ D*a**3/5) + x**4*(3*A*a**2*b/4 + C*a**3/4)

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Giac [A]  time = 1.2291, size = 207, normalized size = 1.5 \begin{align*} \frac{1}{11} \, D b^{3} x^{11} + \frac{1}{10} \, C b^{3} x^{10} + \frac{1}{3} \, D a b^{2} x^{9} + \frac{1}{9} \, B b^{3} x^{9} + \frac{3}{8} \, C a b^{2} x^{8} + \frac{1}{8} \, A b^{3} x^{8} + \frac{3}{7} \, D a^{2} b x^{7} + \frac{3}{7} \, B a b^{2} x^{7} + \frac{1}{2} \, C a^{2} b x^{6} + \frac{1}{2} \, A a b^{2} x^{6} + \frac{1}{5} \, D a^{3} x^{5} + \frac{3}{5} \, B a^{2} b x^{5} + \frac{1}{4} \, C a^{3} x^{4} + \frac{3}{4} \, A a^{2} b x^{4} + \frac{1}{3} \, B a^{3} x^{3} + \frac{1}{2} \, A a^{3} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*D*b^3*x^11 + 1/10*C*b^3*x^10 + 1/3*D*a*b^2*x^9 + 1/9*B*b^3*x^9 + 3/8*C*a*b^2*x^8 + 1/8*A*b^3*x^8 + 3/7*D*
a^2*b*x^7 + 3/7*B*a*b^2*x^7 + 1/2*C*a^2*b*x^6 + 1/2*A*a*b^2*x^6 + 1/5*D*a^3*x^5 + 3/5*B*a^2*b*x^5 + 1/4*C*a^3*
x^4 + 3/4*A*a^2*b*x^4 + 1/3*B*a^3*x^3 + 1/2*A*a^3*x^2