3.16 \(\int \frac{(a+b x^2)^2 (A+B x^2)}{x^3} \, dx\)

Optimal. Leaf size=51 \[ -\frac{a^2 A}{2 x^2}+\frac{1}{2} b x^2 (2 a B+A b)+a \log (x) (a B+2 A b)+\frac{1}{4} b^2 B x^4 \]

[Out]

-(a^2*A)/(2*x^2) + (b*(A*b + 2*a*B)*x^2)/2 + (b^2*B*x^4)/4 + a*(2*A*b + a*B)*Log[x]

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Rubi [A]  time = 0.0456132, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {446, 76} \[ -\frac{a^2 A}{2 x^2}+\frac{1}{2} b x^2 (2 a B+A b)+a \log (x) (a B+2 A b)+\frac{1}{4} b^2 B x^4 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*A)/(2*x^2) + (b*(A*b + 2*a*B)*x^2)/2 + (b^2*B*x^4)/4 + a*(2*A*b + a*B)*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 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

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

Mathematica [A]  time = 0.0234773, size = 49, normalized size = 0.96 \[ \frac{1}{4} \left (-\frac{2 a^2 A}{x^2}+2 b x^2 (2 a B+A b)+4 a \log (x) (a B+2 A b)+b^2 B x^4\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a^2*A)/x^2 + 2*b*(A*b + 2*a*B)*x^2 + b^2*B*x^4 + 4*a*(2*A*b + a*B)*Log[x])/4

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Maple [A]  time = 0.006, size = 50, normalized size = 1. \begin{align*}{\frac{{b}^{2}B{x}^{4}}{4}}+{\frac{A{x}^{2}{b}^{2}}{2}}+B{x}^{2}ab+2\,A\ln \left ( x \right ) ab+B\ln \left ( x \right ){a}^{2}-{\frac{A{a}^{2}}{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*B*x^4+1/2*A*x^2*b^2+B*x^2*a*b+2*A*ln(x)*a*b+B*ln(x)*a^2-1/2*a^2*A/x^2

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Maxima [A]  time = 0.986909, size = 70, normalized size = 1.37 \begin{align*} \frac{1}{4} \, B b^{2} x^{4} + \frac{1}{2} \,{\left (2 \, B a b + A b^{2}\right )} x^{2} + \frac{1}{2} \,{\left (B a^{2} + 2 \, A a b\right )} \log \left (x^{2}\right ) - \frac{A a^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*B*b^2*x^4 + 1/2*(2*B*a*b + A*b^2)*x^2 + 1/2*(B*a^2 + 2*A*a*b)*log(x^2) - 1/2*A*a^2/x^2

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Fricas [A]  time = 1.46967, size = 122, normalized size = 2.39 \begin{align*} \frac{B b^{2} x^{6} + 2 \,{\left (2 \, B a b + A b^{2}\right )} x^{4} + 4 \,{\left (B a^{2} + 2 \, A a b\right )} x^{2} \log \left (x\right ) - 2 \, A a^{2}}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(B*b^2*x^6 + 2*(2*B*a*b + A*b^2)*x^4 + 4*(B*a^2 + 2*A*a*b)*x^2*log(x) - 2*A*a^2)/x^2

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Sympy [A]  time = 0.361602, size = 48, normalized size = 0.94 \begin{align*} - \frac{A a^{2}}{2 x^{2}} + \frac{B b^{2} x^{4}}{4} + a \left (2 A b + B a\right ) \log{\left (x \right )} + x^{2} \left (\frac{A b^{2}}{2} + B a b\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**3,x)

[Out]

-A*a**2/(2*x**2) + B*b**2*x**4/4 + a*(2*A*b + B*a)*log(x) + x**2*(A*b**2/2 + B*a*b)

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Giac [A]  time = 1.62103, size = 95, normalized size = 1.86 \begin{align*} \frac{1}{4} \, B b^{2} x^{4} + B a b x^{2} + \frac{1}{2} \, A b^{2} x^{2} + \frac{1}{2} \,{\left (B a^{2} + 2 \, A a b\right )} \log \left (x^{2}\right ) - \frac{B a^{2} x^{2} + 2 \, A a b x^{2} + A a^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*B*b^2*x^4 + B*a*b*x^2 + 1/2*A*b^2*x^2 + 1/2*(B*a^2 + 2*A*a*b)*log(x^2) - 1/2*(B*a^2*x^2 + 2*A*a*b*x^2 + A*
a^2)/x^2