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

Optimal. Leaf size=29 \[ \log (x) (a B+A b)-\frac{a A}{2 x^2}+\frac{1}{2} b B x^2 \]

[Out]

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

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Rubi [A]  time = 0.0202266, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {446, 76} \[ \log (x) (a B+A b)-\frac{a A}{2 x^2}+\frac{1}{2} b B x^2 \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*(A + B*x^2))/x^3,x]

[Out]

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

Mathematica [A]  time = 0.0112234, size = 29, normalized size = 1. \[ \log (x) (a B+A b)-\frac{a A}{2 x^2}+\frac{1}{2} b B x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 26, normalized size = 0.9 \begin{align*}{\frac{bB{x}^{2}}{2}}+A\ln \left ( x \right ) b+B\ln \left ( x \right ) a-{\frac{Aa}{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01205, size = 38, normalized size = 1.31 \begin{align*} \frac{1}{2} \, B b x^{2} + \frac{1}{2} \,{\left (B a + A b\right )} \log \left (x^{2}\right ) - \frac{A a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.47717, size = 70, normalized size = 2.41 \begin{align*} \frac{B b x^{4} + 2 \,{\left (B a + A b\right )} x^{2} \log \left (x\right ) - A a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.320552, size = 26, normalized size = 0.9 \begin{align*} - \frac{A a}{2 x^{2}} + \frac{B b x^{2}}{2} + \left (A b + B a\right ) \log{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14233, size = 57, normalized size = 1.97 \begin{align*} \frac{1}{2} \, B b x^{2} + \frac{1}{2} \,{\left (B a + A b\right )} \log \left (x^{2}\right ) - \frac{B a x^{2} + A b x^{2} + A a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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