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

Optimal. Leaf size=71 \[ -\frac{x (A b-a B)}{2 a^2 \left (a+b x^2\right )}-\frac{(3 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2} \sqrt{b}}-\frac{A}{a^2 x} \]

[Out]

-(A/(a^2*x)) - ((A*b - a*B)*x)/(2*a^2*(a + b*x^2)) - ((3*A*b - a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(5/2)*Sq
rt[b])

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Rubi [A]  time = 0.0511968, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {456, 453, 205} \[ -\frac{x (A b-a B)}{2 a^2 \left (a+b x^2\right )}-\frac{(3 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2} \sqrt{b}}-\frac{A}{a^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A/(a^2*x)) - ((A*b - a*B)*x)/(2*a^2*(a + b*x^2)) - ((3*A*b - a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(5/2)*Sq
rt[b])

Rule 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{A+B x^2}{x^2 \left (a+b x^2\right )^2} \, dx &=-\frac{(A b-a B) x}{2 a^2 \left (a+b x^2\right )}-\frac{1}{2} \int \frac{-\frac{2 A}{a}+\frac{(A b-a B) x^2}{a^2}}{x^2 \left (a+b x^2\right )} \, dx\\ &=-\frac{A}{a^2 x}-\frac{(A b-a B) x}{2 a^2 \left (a+b x^2\right )}-\frac{(3 A b-a B) \int \frac{1}{a+b x^2} \, dx}{2 a^2}\\ &=-\frac{A}{a^2 x}-\frac{(A b-a B) x}{2 a^2 \left (a+b x^2\right )}-\frac{(3 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2} \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.0323544, size = 70, normalized size = 0.99 \[ \frac{x (a B-A b)}{2 a^2 \left (a+b x^2\right )}+\frac{(a B-3 A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2} \sqrt{b}}-\frac{A}{a^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 85, normalized size = 1.2 \begin{align*} -{\frac{A}{{a}^{2}x}}-{\frac{Abx}{2\,{a}^{2} \left ( b{x}^{2}+a \right ) }}+{\frac{xB}{2\,a \left ( b{x}^{2}+a \right ) }}-{\frac{3\,Ab}{2\,{a}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{B}{2\,a}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.37724, size = 447, normalized size = 6.3 \begin{align*} \left [-\frac{4 \, A a^{2} b - 2 \,{\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{2} -{\left ({\left (B a b - 3 \, A b^{2}\right )} x^{3} +{\left (B a^{2} - 3 \, A a b\right )} x\right )} \sqrt{-a b} \log \left (\frac{b x^{2} + 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right )}{4 \,{\left (a^{3} b^{2} x^{3} + a^{4} b x\right )}}, -\frac{2 \, A a^{2} b -{\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{2} -{\left ({\left (B a b - 3 \, A b^{2}\right )} x^{3} +{\left (B a^{2} - 3 \, A a b\right )} x\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right )}{2 \,{\left (a^{3} b^{2} x^{3} + a^{4} b x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(4*A*a^2*b - 2*(B*a^2*b - 3*A*a*b^2)*x^2 - ((B*a*b - 3*A*b^2)*x^3 + (B*a^2 - 3*A*a*b)*x)*sqrt(-a*b)*log(
(b*x^2 + 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a^3*b^2*x^3 + a^4*b*x), -1/2*(2*A*a^2*b - (B*a^2*b - 3*A*a*b^2)*x^
2 - ((B*a*b - 3*A*b^2)*x^3 + (B*a^2 - 3*A*a*b)*x)*sqrt(a*b)*arctan(sqrt(a*b)*x/a))/(a^3*b^2*x^3 + a^4*b*x)]

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Sympy [A]  time = 0.632495, size = 114, normalized size = 1.61 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{5} b}} \left (- 3 A b + B a\right ) \log{\left (- a^{3} \sqrt{- \frac{1}{a^{5} b}} + x \right )}}{4} + \frac{\sqrt{- \frac{1}{a^{5} b}} \left (- 3 A b + B a\right ) \log{\left (a^{3} \sqrt{- \frac{1}{a^{5} b}} + x \right )}}{4} + \frac{- 2 A a + x^{2} \left (- 3 A b + B a\right )}{2 a^{3} x + 2 a^{2} b x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1373, size = 84, normalized size = 1.18 \begin{align*} \frac{{\left (B a - 3 \, A b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a^{2}} + \frac{B a x^{2} - 3 \, A b x^{2} - 2 \, A a}{2 \,{\left (b x^{3} + a x\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*a - 3*A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2) + 1/2*(B*a*x^2 - 3*A*b*x^2 - 2*A*a)/((b*x^3 + a*x)*a^2
)