3.145 \(\int (a+b x^2)^2 (c+d x^2) \, dx\)

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

[Out]

a^2*c*x + (a*(2*b*c + a*d)*x^3)/3 + (b*(b*c + 2*a*d)*x^5)/5 + (b^2*d*x^7)/7

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Rubi [A]  time = 0.0250865, 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} \[ a^2 c x+\frac{1}{5} b x^5 (2 a d+b c)+\frac{1}{3} a x^3 (a d+2 b c)+\frac{1}{7} b^2 d x^7 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2*(c + d*x^2),x]

[Out]

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

Mathematica [A]  time = 0.0074746, size = 50, normalized size = 1. \[ a^2 c x+\frac{1}{5} b x^5 (2 a d+b c)+\frac{1}{3} a x^3 (a d+2 b c)+\frac{1}{7} b^2 d x^7 \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2*(c + d*x^2),x]

[Out]

a^2*c*x + (a*(2*b*c + a*d)*x^3)/3 + (b*(b*c + 2*a*d)*x^5)/5 + (b^2*d*x^7)/7

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Maple [A]  time = 0., size = 49, normalized size = 1. \begin{align*}{\frac{{b}^{2}d{x}^{7}}{7}}+{\frac{ \left ( 2\,abd+{b}^{2}c \right ){x}^{5}}{5}}+{\frac{ \left ({a}^{2}d+2\,abc \right ){x}^{3}}{3}}+{a}^{2}cx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c),x)

[Out]

1/7*b^2*d*x^7+1/5*(2*a*b*d+b^2*c)*x^5+1/3*(a^2*d+2*a*b*c)*x^3+a^2*c*x

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Maxima [A]  time = 1.00811, size = 65, normalized size = 1.3 \begin{align*} \frac{1}{7} \, b^{2} d x^{7} + \frac{1}{5} \,{\left (b^{2} c + 2 \, a b d\right )} x^{5} + a^{2} c x + \frac{1}{3} \,{\left (2 \, a b c + a^{2} d\right )} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*b^2*d*x^7 + 1/5*(b^2*c + 2*a*b*d)*x^5 + a^2*c*x + 1/3*(2*a*b*c + a^2*d)*x^3

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Fricas [A]  time = 1.07312, size = 120, normalized size = 2.4 \begin{align*} \frac{1}{7} x^{7} d b^{2} + \frac{1}{5} x^{5} c b^{2} + \frac{2}{5} x^{5} d b a + \frac{2}{3} x^{3} c b a + \frac{1}{3} x^{3} d a^{2} + x c a^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*x^7*d*b^2 + 1/5*x^5*c*b^2 + 2/5*x^5*d*b*a + 2/3*x^3*c*b*a + 1/3*x^3*d*a^2 + x*c*a^2

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Sympy [A]  time = 0.067187, size = 53, normalized size = 1.06 \begin{align*} a^{2} c x + \frac{b^{2} d x^{7}}{7} + x^{5} \left (\frac{2 a b d}{5} + \frac{b^{2} c}{5}\right ) + x^{3} \left (\frac{a^{2} d}{3} + \frac{2 a b c}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c),x)

[Out]

a**2*c*x + b**2*d*x**7/7 + x**5*(2*a*b*d/5 + b**2*c/5) + x**3*(a**2*d/3 + 2*a*b*c/3)

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Giac [A]  time = 1.14705, size = 68, normalized size = 1.36 \begin{align*} \frac{1}{7} \, b^{2} d x^{7} + \frac{1}{5} \, b^{2} c x^{5} + \frac{2}{5} \, a b d x^{5} + \frac{2}{3} \, a b c x^{3} + \frac{1}{3} \, a^{2} d x^{3} + a^{2} c x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*b^2*d*x^7 + 1/5*b^2*c*x^5 + 2/5*a*b*d*x^5 + 2/3*a*b*c*x^3 + 1/3*a^2*d*x^3 + a^2*c*x