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

Optimal. Leaf size=99 \[ a^2 A x+\frac{1}{4} a^2 D x^4+\frac{1}{5} b x^5 (2 a C+A b)+\frac{1}{3} a x^3 (a C+2 A b)+\frac{B \left (a+b x^2\right )^3}{6 b}+\frac{1}{3} a b D x^6+\frac{1}{7} b^2 C x^7+\frac{1}{8} b^2 D x^8 \]

[Out]

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

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Rubi [A]  time = 0.0723119, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {1582, 1810} \[ a^2 A x+\frac{1}{4} a^2 D x^4+\frac{1}{5} b x^5 (2 a C+A b)+\frac{1}{3} a x^3 (a C+2 A b)+\frac{B \left (a+b x^2\right )^3}{6 b}+\frac{1}{3} a b D x^6+\frac{1}{7} b^2 C x^7+\frac{1}{8} b^2 D x^8 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0357294, size = 88, normalized size = 0.89 \[ \frac{1}{840} \left (70 a^2 x (12 A+x (6 B+x (4 C+3 D x)))+28 a b x^3 (20 A+x (15 B+2 x (6 C+5 D x)))+b^2 x^5 (168 A+5 x (28 B+3 x (8 C+7 D x)))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(70*a^2*x*(12*A + x*(6*B + x*(4*C + 3*D*x))) + 28*a*b*x^3*(20*A + x*(15*B + 2*x*(6*C + 5*D*x))) + b^2*x^5*(168
*A + 5*x*(28*B + 3*x*(8*C + 7*D*x))))/840

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Maple [A]  time = 0.002, size = 99, normalized size = 1. \begin{align*}{\frac{{b}^{2}D{x}^{8}}{8}}+{\frac{{b}^{2}C{x}^{7}}{7}}+{\frac{ \left ({b}^{2}B+2\,abD \right ){x}^{6}}{6}}+{\frac{ \left ( A{b}^{2}+2\,abC \right ){x}^{5}}{5}}+{\frac{ \left ( 2\,Bba+{a}^{2}D \right ){x}^{4}}{4}}+{\frac{ \left ( 2\,Aab+{a}^{2}C \right ){x}^{3}}{3}}+{\frac{B{x}^{2}{a}^{2}}{2}}+{a}^{2}Ax \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.05777, size = 132, normalized size = 1.33 \begin{align*} \frac{1}{8} \, D b^{2} x^{8} + \frac{1}{7} \, C b^{2} x^{7} + \frac{1}{6} \,{\left (2 \, D a b + B b^{2}\right )} x^{6} + \frac{1}{5} \,{\left (2 \, C a b + A b^{2}\right )} x^{5} + \frac{1}{2} \, B a^{2} x^{2} + \frac{1}{4} \,{\left (D a^{2} + 2 \, B a b\right )} x^{4} + A a^{2} x + \frac{1}{3} \,{\left (C a^{2} + 2 \, A a b\right )} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.26653, size = 250, normalized size = 2.53 \begin{align*} \frac{1}{8} x^{8} b^{2} D + \frac{1}{7} x^{7} b^{2} C + \frac{1}{3} x^{6} b a D + \frac{1}{6} x^{6} b^{2} B + \frac{2}{5} x^{5} b a C + \frac{1}{5} x^{5} b^{2} A + \frac{1}{4} x^{4} a^{2} D + \frac{1}{2} x^{4} b a B + \frac{1}{3} x^{3} a^{2} C + \frac{2}{3} x^{3} b a A + \frac{1}{2} x^{2} a^{2} B + x a^{2} A \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.07495, size = 107, normalized size = 1.08 \begin{align*} A a^{2} x + \frac{B a^{2} x^{2}}{2} + \frac{C b^{2} x^{7}}{7} + \frac{D b^{2} x^{8}}{8} + x^{6} \left (\frac{B b^{2}}{6} + \frac{D a b}{3}\right ) + x^{5} \left (\frac{A b^{2}}{5} + \frac{2 C a b}{5}\right ) + x^{4} \left (\frac{B a b}{2} + \frac{D a^{2}}{4}\right ) + x^{3} \left (\frac{2 A a b}{3} + \frac{C a^{2}}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15732, size = 138, normalized size = 1.39 \begin{align*} \frac{1}{8} \, D b^{2} x^{8} + \frac{1}{7} \, C b^{2} x^{7} + \frac{1}{3} \, D a b x^{6} + \frac{1}{6} \, B b^{2} x^{6} + \frac{2}{5} \, C a b x^{5} + \frac{1}{5} \, A b^{2} x^{5} + \frac{1}{4} \, D a^{2} x^{4} + \frac{1}{2} \, B a b x^{4} + \frac{1}{3} \, C a^{2} x^{3} + \frac{2}{3} \, A a b x^{3} + \frac{1}{2} \, B a^{2} x^{2} + A a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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