3.13 \(\int (a+b x^2)^2 (A+B x^2) \, dx\)

Optimal. Leaf size=50 \[ a^2 A x+\frac{1}{5} b x^5 (2 a B+A b)+\frac{1}{3} a x^3 (a B+2 A b)+\frac{1}{7} b^2 B x^7 \]

[Out]

a^2*A*x + (a*(2*A*b + a*B)*x^3)/3 + (b*(A*b + 2*a*B)*x^5)/5 + (b^2*B*x^7)/7

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Rubi [A]  time = 0.0223412, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {373} \[ a^2 A x+\frac{1}{5} b x^5 (2 a B+A b)+\frac{1}{3} a x^3 (a B+2 A b)+\frac{1}{7} b^2 B x^7 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2*(A + B*x^2),x]

[Out]

a^2*A*x + (a*(2*A*b + a*B)*x^3)/3 + (b*(A*b + 2*a*B)*x^5)/5 + (b^2*B*x^7)/7

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

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

Mathematica [A]  time = 0.0073365, size = 50, normalized size = 1. \[ a^2 A x+\frac{1}{5} b x^5 (2 a B+A b)+\frac{1}{3} a x^3 (a B+2 A b)+\frac{1}{7} b^2 B x^7 \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*A*x + (a*(2*A*b + a*B)*x^3)/3 + (b*(A*b + 2*a*B)*x^5)/5 + (b^2*B*x^7)/7

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Maple [A]  time = 0.001, size = 49, normalized size = 1. \begin{align*}{\frac{{b}^{2}B{x}^{7}}{7}}+{\frac{ \left ({b}^{2}A+2\,abB \right ){x}^{5}}{5}}+{\frac{ \left ( 2\,abA+{a}^{2}B \right ){x}^{3}}{3}}+{a}^{2}Ax \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*b^2*B*x^7+1/5*(A*b^2+2*B*a*b)*x^5+1/3*(2*A*a*b+B*a^2)*x^3+a^2*A*x

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Maxima [A]  time = 1.00006, size = 65, normalized size = 1.3 \begin{align*} \frac{1}{7} \, B b^{2} x^{7} + \frac{1}{5} \,{\left (2 \, B a b + A b^{2}\right )} x^{5} + A a^{2} x + \frac{1}{3} \,{\left (B 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*(B*x^2+A),x, algorithm="maxima")

[Out]

1/7*B*b^2*x^7 + 1/5*(2*B*a*b + A*b^2)*x^5 + A*a^2*x + 1/3*(B*a^2 + 2*A*a*b)*x^3

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Fricas [A]  time = 1.28013, size = 120, normalized size = 2.4 \begin{align*} \frac{1}{7} x^{7} b^{2} B + \frac{2}{5} x^{5} b a B + \frac{1}{5} x^{5} b^{2} A + \frac{1}{3} x^{3} a^{2} B + \frac{2}{3} x^{3} b a A + x a^{2} A \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*x^7*b^2*B + 2/5*x^5*b*a*B + 1/5*x^5*b^2*A + 1/3*x^3*a^2*B + 2/3*x^3*b*a*A + x*a^2*A

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Sympy [A]  time = 0.065136, size = 53, normalized size = 1.06 \begin{align*} A a^{2} x + \frac{B b^{2} x^{7}}{7} + x^{5} \left (\frac{A b^{2}}{5} + \frac{2 B a b}{5}\right ) + x^{3} \left (\frac{2 A a b}{3} + \frac{B a^{2}}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**2*x + B*b**2*x**7/7 + x**5*(A*b**2/5 + 2*B*a*b/5) + x**3*(2*A*a*b/3 + B*a**2/3)

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Giac [A]  time = 1.11558, size = 68, normalized size = 1.36 \begin{align*} \frac{1}{7} \, B b^{2} x^{7} + \frac{2}{5} \, B a b x^{5} + \frac{1}{5} \, A b^{2} x^{5} + \frac{1}{3} \, B a^{2} x^{3} + \frac{2}{3} \, A a b x^{3} + A a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*B*b^2*x^7 + 2/5*B*a*b*x^5 + 1/5*A*b^2*x^5 + 1/3*B*a^2*x^3 + 2/3*A*a*b*x^3 + A*a^2*x