3.96 \(\int \frac{A+B x^2}{x^5 (a+b x^2)^3} \, dx\)

Optimal. Leaf size=124 \[ \frac{b (3 A b-2 a B)}{2 a^4 \left (a+b x^2\right )}+\frac{3 A b-a B}{2 a^4 x^2}+\frac{b (A b-a B)}{4 a^3 \left (a+b x^2\right )^2}-\frac{3 b (2 A b-a B) \log \left (a+b x^2\right )}{2 a^5}+\frac{3 b \log (x) (2 A b-a B)}{a^5}-\frac{A}{4 a^3 x^4} \]

[Out]

-A/(4*a^3*x^4) + (3*A*b - a*B)/(2*a^4*x^2) + (b*(A*b - a*B))/(4*a^3*(a + b*x^2)^2) + (b*(3*A*b - 2*a*B))/(2*a^
4*(a + b*x^2)) + (3*b*(2*A*b - a*B)*Log[x])/a^5 - (3*b*(2*A*b - a*B)*Log[a + b*x^2])/(2*a^5)

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Rubi [A]  time = 0.129733, antiderivative size = 124, 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} \[ \frac{b (3 A b-2 a B)}{2 a^4 \left (a+b x^2\right )}+\frac{3 A b-a B}{2 a^4 x^2}+\frac{b (A b-a B)}{4 a^3 \left (a+b x^2\right )^2}-\frac{3 b (2 A b-a B) \log \left (a+b x^2\right )}{2 a^5}+\frac{3 b \log (x) (2 A b-a B)}{a^5}-\frac{A}{4 a^3 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^5*(a + b*x^2)^3),x]

[Out]

-A/(4*a^3*x^4) + (3*A*b - a*B)/(2*a^4*x^2) + (b*(A*b - a*B))/(4*a^3*(a + b*x^2)^2) + (b*(3*A*b - 2*a*B))/(2*a^
4*(a + b*x^2)) + (3*b*(2*A*b - a*B)*Log[x])/a^5 - (3*b*(2*A*b - a*B)*Log[a + b*x^2])/(2*a^5)

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

Mathematica [A]  time = 0.0785861, size = 108, normalized size = 0.87 \[ \frac{\frac{a^2 b (A b-a B)}{\left (a+b x^2\right )^2}-\frac{a^2 A}{x^4}+\frac{2 a b (3 A b-2 a B)}{a+b x^2}-\frac{2 a (a B-3 A b)}{x^2}+6 b (a B-2 A b) \log \left (a+b x^2\right )+12 b \log (x) (2 A b-a B)}{4 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^5*(a + b*x^2)^3),x]

[Out]

(-((a^2*A)/x^4) - (2*a*(-3*A*b + a*B))/x^2 + (a^2*b*(A*b - a*B))/(a + b*x^2)^2 + (2*a*b*(3*A*b - 2*a*B))/(a +
b*x^2) + 12*b*(2*A*b - a*B)*Log[x] + 6*b*(-2*A*b + a*B)*Log[a + b*x^2])/(4*a^5)

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Maple [A]  time = 0.017, size = 150, normalized size = 1.2 \begin{align*} -{\frac{A}{4\,{a}^{3}{x}^{4}}}+{\frac{3\,Ab}{2\,{a}^{4}{x}^{2}}}-{\frac{B}{2\,{a}^{3}{x}^{2}}}+6\,{\frac{A\ln \left ( x \right ){b}^{2}}{{a}^{5}}}-3\,{\frac{bB\ln \left ( x \right ) }{{a}^{4}}}+{\frac{A{b}^{2}}{4\,{a}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{Bb}{4\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}-3\,{\frac{{b}^{2}\ln \left ( b{x}^{2}+a \right ) A}{{a}^{5}}}+{\frac{3\,b\ln \left ( b{x}^{2}+a \right ) B}{2\,{a}^{4}}}+{\frac{3\,A{b}^{2}}{2\,{a}^{4} \left ( b{x}^{2}+a \right ) }}-{\frac{Bb}{{a}^{3} \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)^3,x)

[Out]

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

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Maxima [A]  time = 1.01524, size = 185, normalized size = 1.49 \begin{align*} -\frac{6 \,{\left (B a b^{2} - 2 \, A b^{3}\right )} x^{6} + 9 \,{\left (B a^{2} b - 2 \, A a b^{2}\right )} x^{4} + A a^{3} + 2 \,{\left (B a^{3} - 2 \, A a^{2} b\right )} x^{2}}{4 \,{\left (a^{4} b^{2} x^{8} + 2 \, a^{5} b x^{6} + a^{6} x^{4}\right )}} + \frac{3 \,{\left (B a b - 2 \, A b^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, a^{5}} - \frac{3 \,{\left (B a b - 2 \, A b^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

-1/4*(6*(B*a*b^2 - 2*A*b^3)*x^6 + 9*(B*a^2*b - 2*A*a*b^2)*x^4 + A*a^3 + 2*(B*a^3 - 2*A*a^2*b)*x^2)/(a^4*b^2*x^
8 + 2*a^5*b*x^6 + a^6*x^4) + 3/2*(B*a*b - 2*A*b^2)*log(b*x^2 + a)/a^5 - 3/2*(B*a*b - 2*A*b^2)*log(x^2)/a^5

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Fricas [B]  time = 1.30121, size = 474, normalized size = 3.82 \begin{align*} -\frac{6 \,{\left (B a^{2} b^{2} - 2 \, A a b^{3}\right )} x^{6} + A a^{4} + 9 \,{\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x^{4} + 2 \,{\left (B a^{4} - 2 \, A a^{3} b\right )} x^{2} - 6 \,{\left ({\left (B a b^{3} - 2 \, A b^{4}\right )} x^{8} + 2 \,{\left (B a^{2} b^{2} - 2 \, A a b^{3}\right )} x^{6} +{\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x^{4}\right )} \log \left (b x^{2} + a\right ) + 12 \,{\left ({\left (B a b^{3} - 2 \, A b^{4}\right )} x^{8} + 2 \,{\left (B a^{2} b^{2} - 2 \, A a b^{3}\right )} x^{6} +{\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x^{4}\right )} \log \left (x\right )}{4 \,{\left (a^{5} b^{2} x^{8} + 2 \, a^{6} b x^{6} + a^{7} 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)^3,x, algorithm="fricas")

[Out]

-1/4*(6*(B*a^2*b^2 - 2*A*a*b^3)*x^6 + A*a^4 + 9*(B*a^3*b - 2*A*a^2*b^2)*x^4 + 2*(B*a^4 - 2*A*a^3*b)*x^2 - 6*((
B*a*b^3 - 2*A*b^4)*x^8 + 2*(B*a^2*b^2 - 2*A*a*b^3)*x^6 + (B*a^3*b - 2*A*a^2*b^2)*x^4)*log(b*x^2 + a) + 12*((B*
a*b^3 - 2*A*b^4)*x^8 + 2*(B*a^2*b^2 - 2*A*a*b^3)*x^6 + (B*a^3*b - 2*A*a^2*b^2)*x^4)*log(x))/(a^5*b^2*x^8 + 2*a
^6*b*x^6 + a^7*x^4)

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Sympy [A]  time = 1.82531, size = 136, normalized size = 1.1 \begin{align*} - \frac{A a^{3} + x^{6} \left (- 12 A b^{3} + 6 B a b^{2}\right ) + x^{4} \left (- 18 A a b^{2} + 9 B a^{2} b\right ) + x^{2} \left (- 4 A a^{2} b + 2 B a^{3}\right )}{4 a^{6} x^{4} + 8 a^{5} b x^{6} + 4 a^{4} b^{2} x^{8}} - \frac{3 b \left (- 2 A b + B a\right ) \log{\left (x \right )}}{a^{5}} + \frac{3 b \left (- 2 A b + B a\right ) \log{\left (\frac{a}{b} + x^{2} \right )}}{2 a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**5/(b*x**2+a)**3,x)

[Out]

-(A*a**3 + x**6*(-12*A*b**3 + 6*B*a*b**2) + x**4*(-18*A*a*b**2 + 9*B*a**2*b) + x**2*(-4*A*a**2*b + 2*B*a**3))/
(4*a**6*x**4 + 8*a**5*b*x**6 + 4*a**4*b**2*x**8) - 3*b*(-2*A*b + B*a)*log(x)/a**5 + 3*b*(-2*A*b + B*a)*log(a/b
+ x**2)/(2*a**5)

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Giac [A]  time = 1.15009, size = 180, normalized size = 1.45 \begin{align*} -\frac{3 \,{\left (B a b - 2 \, A b^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{5}} + \frac{3 \,{\left (B a b^{2} - 2 \, A b^{3}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{5} b} - \frac{6 \, B a b^{2} x^{6} - 12 \, A b^{3} x^{6} + 9 \, B a^{2} b x^{4} - 18 \, A a b^{2} x^{4} + 2 \, B a^{3} x^{2} - 4 \, A a^{2} b x^{2} + A a^{3}}{4 \,{\left (b x^{4} + a 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)^3,x, algorithm="giac")

[Out]

-3/2*(B*a*b - 2*A*b^2)*log(x^2)/a^5 + 3/2*(B*a*b^2 - 2*A*b^3)*log(abs(b*x^2 + a))/(a^5*b) - 1/4*(6*B*a*b^2*x^6
- 12*A*b^3*x^6 + 9*B*a^2*b*x^4 - 18*A*a*b^2*x^4 + 2*B*a^3*x^2 - 4*A*a^2*b*x^2 + A*a^3)/((b*x^4 + a*x^2)^2*a^4
)