3.85 \(\int \frac{A+B x^2}{x^5 (a+b x^2)^2} \, dx\)

Optimal. Leaf size=97 \[ \frac{b (A b-a B)}{2 a^3 \left (a+b x^2\right )}+\frac{2 A b-a B}{2 a^3 x^2}-\frac{b (3 A b-2 a B) \log \left (a+b x^2\right )}{2 a^4}+\frac{b \log (x) (3 A b-2 a B)}{a^4}-\frac{A}{4 a^2 x^4} \]

[Out]

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

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Rubi [A]  time = 0.0950933, antiderivative size = 97, 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} \[ \frac{b (A b-a B)}{2 a^3 \left (a+b x^2\right )}+\frac{2 A b-a B}{2 a^3 x^2}-\frac{b (3 A b-2 a B) \log \left (a+b x^2\right )}{2 a^4}+\frac{b \log (x) (3 A b-2 a B)}{a^4}-\frac{A}{4 a^2 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^5*(a + b*x^2)^2),x]

[Out]

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

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^5 \left (a+b x^2\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{A+B x}{x^3 (a+b x)^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{A}{a^2 x^3}+\frac{-2 A b+a B}{a^3 x^2}-\frac{b (-3 A b+2 a B)}{a^4 x}+\frac{b^2 (-A b+a B)}{a^3 (a+b x)^2}+\frac{b^2 (-3 A b+2 a B)}{a^4 (a+b x)}\right ) \, dx,x,x^2\right )\\ &=-\frac{A}{4 a^2 x^4}+\frac{2 A b-a B}{2 a^3 x^2}+\frac{b (A b-a B)}{2 a^3 \left (a+b x^2\right )}+\frac{b (3 A b-2 a B) \log (x)}{a^4}-\frac{b (3 A b-2 a B) \log \left (a+b x^2\right )}{2 a^4}\\ \end{align*}

Mathematica [A]  time = 0.0971785, size = 85, normalized size = 0.88 \[ -\frac{\frac{a^2 A}{x^4}+\frac{2 a b (a B-A b)}{a+b x^2}+\frac{2 a (a B-2 A b)}{x^2}+2 b (3 A b-2 a B) \log \left (a+b x^2\right )-4 b \log (x) (3 A b-2 a B)}{4 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^5*(a + b*x^2)^2),x]

[Out]

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

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Maple [A]  time = 0.013, size = 114, normalized size = 1.2 \begin{align*} -{\frac{A}{4\,{a}^{2}{x}^{4}}}+{\frac{Ab}{{a}^{3}{x}^{2}}}-{\frac{B}{2\,{a}^{2}{x}^{2}}}+3\,{\frac{A\ln \left ( x \right ){b}^{2}}{{a}^{4}}}-2\,{\frac{bB\ln \left ( x \right ) }{{a}^{3}}}-{\frac{3\,{b}^{2}\ln \left ( b{x}^{2}+a \right ) A}{2\,{a}^{4}}}+{\frac{b\ln \left ( b{x}^{2}+a \right ) B}{{a}^{3}}}+{\frac{A{b}^{2}}{2\,{a}^{3} \left ( b{x}^{2}+a \right ) }}-{\frac{Bb}{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^5/(b*x^2+a)^2,x)

[Out]

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

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Maxima [A]  time = 1.03495, size = 143, normalized size = 1.47 \begin{align*} -\frac{2 \,{\left (2 \, B a b - 3 \, A b^{2}\right )} x^{4} + A a^{2} +{\left (2 \, B a^{2} - 3 \, A a b\right )} x^{2}}{4 \,{\left (a^{3} b x^{6} + a^{4} x^{4}\right )}} + \frac{{\left (2 \, B a b - 3 \, A b^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, a^{4}} - \frac{{\left (2 \, B a b - 3 \, A b^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.34228, size = 327, normalized size = 3.37 \begin{align*} -\frac{2 \,{\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{4} + A a^{3} +{\left (2 \, B a^{3} - 3 \, A a^{2} b\right )} x^{2} - 2 \,{\left ({\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} x^{6} +{\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{4}\right )} \log \left (b x^{2} + a\right ) + 4 \,{\left ({\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} x^{6} +{\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{4}\right )} \log \left (x\right )}{4 \,{\left (a^{4} b x^{6} + a^{5} x^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

-1/4*(2*(2*B*a^2*b - 3*A*a*b^2)*x^4 + A*a^3 + (2*B*a^3 - 3*A*a^2*b)*x^2 - 2*((2*B*a*b^2 - 3*A*b^3)*x^6 + (2*B*
a^2*b - 3*A*a*b^2)*x^4)*log(b*x^2 + a) + 4*((2*B*a*b^2 - 3*A*b^3)*x^6 + (2*B*a^2*b - 3*A*a*b^2)*x^4)*log(x))/(
a^4*b*x^6 + a^5*x^4)

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Sympy [A]  time = 1.3074, size = 100, normalized size = 1.03 \begin{align*} - \frac{A a^{2} + x^{4} \left (- 6 A b^{2} + 4 B a b\right ) + x^{2} \left (- 3 A a b + 2 B a^{2}\right )}{4 a^{4} x^{4} + 4 a^{3} b x^{6}} - \frac{b \left (- 3 A b + 2 B a\right ) \log{\left (x \right )}}{a^{4}} + \frac{b \left (- 3 A b + 2 B a\right ) \log{\left (\frac{a}{b} + x^{2} \right )}}{2 a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**5/(b*x**2+a)**2,x)

[Out]

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

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Giac [A]  time = 1.11282, size = 203, normalized size = 2.09 \begin{align*} -\frac{{\left (2 \, B a b - 3 \, A b^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{4}} + \frac{{\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{4} b} - \frac{2 \, B a b^{2} x^{2} - 3 \, A b^{3} x^{2} + 3 \, B a^{2} b - 4 \, A a b^{2}}{2 \,{\left (b x^{2} + a\right )} a^{4}} + \frac{6 \, B a b x^{4} - 9 \, A b^{2} x^{4} - 2 \, B a^{2} x^{2} + 4 \, A a b x^{2} - A a^{2}}{4 \, a^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(b*x^2+a)^2,x, algorithm="giac")

[Out]

-1/2*(2*B*a*b - 3*A*b^2)*log(x^2)/a^4 + 1/2*(2*B*a*b^2 - 3*A*b^3)*log(abs(b*x^2 + a))/(a^4*b) - 1/2*(2*B*a*b^2
*x^2 - 3*A*b^3*x^2 + 3*B*a^2*b - 4*A*a*b^2)/((b*x^2 + a)*a^4) + 1/4*(6*B*a*b*x^4 - 9*A*b^2*x^4 - 2*B*a^2*x^2 +
4*A*a*b*x^2 - A*a^2)/(a^4*x^4)