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

Optimal. Leaf size=68 \[ \frac{A}{2 a^2 \left (a+b x^2\right )}-\frac{A \log \left (a+b x^2\right )}{2 a^3}+\frac{A \log (x)}{a^3}+\frac{A b-a B}{4 a b \left (a+b x^2\right )^2} \]

[Out]

(A*b - a*B)/(4*a*b*(a + b*x^2)^2) + A/(2*a^2*(a + b*x^2)) + (A*Log[x])/a^3 - (A*Log[a + b*x^2])/(2*a^3)

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Rubi [A]  time = 0.0604057, antiderivative size = 68, 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, 77} \[ \frac{A}{2 a^2 \left (a+b x^2\right )}-\frac{A \log \left (a+b x^2\right )}{2 a^3}+\frac{A \log (x)}{a^3}+\frac{A b-a B}{4 a b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x*(a + b*x^2)^3),x]

[Out]

(A*b - a*B)/(4*a*b*(a + b*x^2)^2) + A/(2*a^2*(a + b*x^2)) + (A*Log[x])/a^3 - (A*Log[a + b*x^2])/(2*a^3)

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.0466887, size = 59, normalized size = 0.87 \[ \frac{\frac{a \left (a^2 (-B)+3 a A b+2 A b^2 x^2\right )}{b \left (a+b x^2\right )^2}-2 A \log \left (a+b x^2\right )+4 A \log (x)}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(3*a*A*b - a^2*B + 2*A*b^2*x^2))/(b*(a + b*x^2)^2) + 4*A*Log[x] - 2*A*Log[a + b*x^2])/(4*a^3)

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Maple [A]  time = 0.011, size = 68, normalized size = 1. \begin{align*}{\frac{A\ln \left ( x \right ) }{{a}^{3}}}+{\frac{A}{4\,a \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{B}{4\,b \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{A\ln \left ( b{x}^{2}+a \right ) }{2\,{a}^{3}}}+{\frac{A}{2\,{a}^{2} \left ( b{x}^{2}+a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*ln(x)/a^3+1/4/a/(b*x^2+a)^2*A-1/4/b/(b*x^2+a)^2*B-1/2*A*ln(b*x^2+a)/a^3+1/2*A/a^2/(b*x^2+a)

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Maxima [A]  time = 1.01256, size = 104, normalized size = 1.53 \begin{align*} \frac{2 \, A b^{2} x^{2} - B a^{2} + 3 \, A a b}{4 \,{\left (a^{2} b^{3} x^{4} + 2 \, a^{3} b^{2} x^{2} + a^{4} b\right )}} - \frac{A \log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac{A \log \left (x^{2}\right )}{2 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*A*b^2*x^2 - B*a^2 + 3*A*a*b)/(a^2*b^3*x^4 + 2*a^3*b^2*x^2 + a^4*b) - 1/2*A*log(b*x^2 + a)/a^3 + 1/2*A*l
og(x^2)/a^3

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Fricas [A]  time = 1.23087, size = 250, normalized size = 3.68 \begin{align*} \frac{2 \, A a b^{2} x^{2} - B a^{3} + 3 \, A a^{2} b - 2 \,{\left (A b^{3} x^{4} + 2 \, A a b^{2} x^{2} + A a^{2} b\right )} \log \left (b x^{2} + a\right ) + 4 \,{\left (A b^{3} x^{4} + 2 \, A a b^{2} x^{2} + A a^{2} b\right )} \log \left (x\right )}{4 \,{\left (a^{3} b^{3} x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{5} b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*A*a*b^2*x^2 - B*a^3 + 3*A*a^2*b - 2*(A*b^3*x^4 + 2*A*a*b^2*x^2 + A*a^2*b)*log(b*x^2 + a) + 4*(A*b^3*x^4
+ 2*A*a*b^2*x^2 + A*a^2*b)*log(x))/(a^3*b^3*x^4 + 2*a^4*b^2*x^2 + a^5*b)

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Sympy [A]  time = 0.743284, size = 75, normalized size = 1.1 \begin{align*} \frac{A \log{\left (x \right )}}{a^{3}} - \frac{A \log{\left (\frac{a}{b} + x^{2} \right )}}{2 a^{3}} + \frac{3 A a b + 2 A b^{2} x^{2} - B a^{2}}{4 a^{4} b + 8 a^{3} b^{2} x^{2} + 4 a^{2} b^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*log(x)/a**3 - A*log(a/b + x**2)/(2*a**3) + (3*A*a*b + 2*A*b**2*x**2 - B*a**2)/(4*a**4*b + 8*a**3*b**2*x**2 +
4*a**2*b**3*x**4)

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Giac [A]  time = 1.15929, size = 103, normalized size = 1.51 \begin{align*} \frac{A \log \left (x^{2}\right )}{2 \, a^{3}} - \frac{A \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{3}} + \frac{3 \, A b^{3} x^{4} + 8 \, A a b^{2} x^{2} - B a^{3} + 6 \, A a^{2} b}{4 \,{\left (b x^{2} + a\right )}^{2} a^{3} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*A*log(x^2)/a^3 - 1/2*A*log(abs(b*x^2 + a))/a^3 + 1/4*(3*A*b^3*x^4 + 8*A*a*b^2*x^2 - B*a^3 + 6*A*a^2*b)/((b
*x^2 + a)^2*a^3*b)