3.155 \(\int \frac{(a+b x^2)^2 (c+d x^2)^2}{x} \, dx\)

Optimal. Leaf size=80 \[ \frac{1}{4} x^4 \left (a^2 d^2+4 a b c d+b^2 c^2\right )+a^2 c^2 \log (x)+\frac{1}{3} b d x^6 (a d+b c)+a c x^2 (a d+b c)+\frac{1}{8} b^2 d^2 x^8 \]

[Out]

a*c*(b*c + a*d)*x^2 + ((b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^4)/4 + (b*d*(b*c + a*d)*x^6)/3 + (b^2*d^2*x^8)/8 + a^
2*c^2*Log[x]

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Rubi [A]  time = 0.0750831, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {446, 88} \[ \frac{1}{4} x^4 \left (a^2 d^2+4 a b c d+b^2 c^2\right )+a^2 c^2 \log (x)+\frac{1}{3} b d x^6 (a d+b c)+a c x^2 (a d+b c)+\frac{1}{8} b^2 d^2 x^8 \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*c*(b*c + a*d)*x^2 + ((b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^4)/4 + (b*d*(b*c + a*d)*x^6)/3 + (b^2*d^2*x^8)/8 + a^
2*c^2*Log[x]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2}{x} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^2 (c+d x)^2}{x} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (2 a c (b c+a d)+\frac{a^2 c^2}{x}+\left (b^2 c^2+4 a b c d+a^2 d^2\right ) x+2 b d (b c+a d) x^2+b^2 d^2 x^3\right ) \, dx,x,x^2\right )\\ &=a c (b c+a d) x^2+\frac{1}{4} \left (b^2 c^2+4 a b c d+a^2 d^2\right ) x^4+\frac{1}{3} b d (b c+a d) x^6+\frac{1}{8} b^2 d^2 x^8+a^2 c^2 \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0240972, size = 80, normalized size = 1. \[ \frac{1}{4} x^4 \left (a^2 d^2+4 a b c d+b^2 c^2\right )+a^2 c^2 \log (x)+\frac{1}{3} b d x^6 (a d+b c)+a c x^2 (a d+b c)+\frac{1}{8} b^2 d^2 x^8 \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*c*(b*c + a*d)*x^2 + ((b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^4)/4 + (b*d*(b*c + a*d)*x^6)/3 + (b^2*d^2*x^8)/8 + a^
2*c^2*Log[x]

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Maple [A]  time = 0.002, size = 90, normalized size = 1.1 \begin{align*}{\frac{{b}^{2}{d}^{2}{x}^{8}}{8}}+{\frac{{x}^{6}ab{d}^{2}}{3}}+{\frac{{x}^{6}{b}^{2}cd}{3}}+{\frac{{x}^{4}{a}^{2}{d}^{2}}{4}}+{x}^{4}abcd+{\frac{{x}^{4}{b}^{2}{c}^{2}}{4}}+{x}^{2}{a}^{2}cd+a{c}^{2}b{x}^{2}+{a}^{2}{c}^{2}\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^2*d^2*x^8+1/3*x^6*a*b*d^2+1/3*x^6*b^2*c*d+1/4*x^4*a^2*d^2+x^4*a*b*c*d+1/4*x^4*b^2*c^2+x^2*a^2*c*d+a*c^2*
b*x^2+a^2*c^2*ln(x)

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Maxima [A]  time = 0.975723, size = 115, normalized size = 1.44 \begin{align*} \frac{1}{8} \, b^{2} d^{2} x^{8} + \frac{1}{3} \,{\left (b^{2} c d + a b d^{2}\right )} x^{6} + \frac{1}{4} \,{\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} x^{4} + \frac{1}{2} \, a^{2} c^{2} \log \left (x^{2}\right ) +{\left (a b c^{2} + a^{2} c d\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^2*d^2*x^8 + 1/3*(b^2*c*d + a*b*d^2)*x^6 + 1/4*(b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^4 + 1/2*a^2*c^2*log(x^2)
+ (a*b*c^2 + a^2*c*d)*x^2

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Fricas [A]  time = 1.28162, size = 178, normalized size = 2.22 \begin{align*} \frac{1}{8} \, b^{2} d^{2} x^{8} + \frac{1}{3} \,{\left (b^{2} c d + a b d^{2}\right )} x^{6} + \frac{1}{4} \,{\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} \log \left (x\right ) +{\left (a b c^{2} + a^{2} c d\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^2*d^2*x^8 + 1/3*(b^2*c*d + a*b*d^2)*x^6 + 1/4*(b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^4 + a^2*c^2*log(x) + (a*
b*c^2 + a^2*c*d)*x^2

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Sympy [A]  time = 0.326519, size = 85, normalized size = 1.06 \begin{align*} a^{2} c^{2} \log{\left (x \right )} + \frac{b^{2} d^{2} x^{8}}{8} + x^{6} \left (\frac{a b d^{2}}{3} + \frac{b^{2} c d}{3}\right ) + x^{4} \left (\frac{a^{2} d^{2}}{4} + a b c d + \frac{b^{2} c^{2}}{4}\right ) + x^{2} \left (a^{2} c d + a b c^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*c**2*log(x) + b**2*d**2*x**8/8 + x**6*(a*b*d**2/3 + b**2*c*d/3) + x**4*(a**2*d**2/4 + a*b*c*d + b**2*c**2
/4) + x**2*(a**2*c*d + a*b*c**2)

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Giac [A]  time = 1.17909, size = 124, normalized size = 1.55 \begin{align*} \frac{1}{8} \, b^{2} d^{2} x^{8} + \frac{1}{3} \, b^{2} c d x^{6} + \frac{1}{3} \, a b d^{2} x^{6} + \frac{1}{4} \, b^{2} c^{2} x^{4} + a b c d x^{4} + \frac{1}{4} \, a^{2} d^{2} x^{4} + a b c^{2} x^{2} + a^{2} c d x^{2} + \frac{1}{2} \, a^{2} c^{2} \log \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^2*d^2*x^8 + 1/3*b^2*c*d*x^6 + 1/3*a*b*d^2*x^6 + 1/4*b^2*c^2*x^4 + a*b*c*d*x^4 + 1/4*a^2*d^2*x^4 + a*b*c^
2*x^2 + a^2*c*d*x^2 + 1/2*a^2*c^2*log(x^2)