3.84 \(\int \frac{A+B x^2}{x^4 (a+b x^2)^2} \, dx\)

Optimal. Leaf size=90 \[ \frac{b x (A b-a B)}{2 a^3 \left (a+b x^2\right )}+\frac{2 A b-a B}{a^3 x}+\frac{\sqrt{b} (5 A b-3 a B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{7/2}}-\frac{A}{3 a^2 x^3} \]

[Out]

-A/(3*a^2*x^3) + (2*A*b - a*B)/(a^3*x) + (b*(A*b - a*B)*x)/(2*a^3*(a + b*x^2)) + (Sqrt[b]*(5*A*b - 3*a*B)*ArcT
an[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(7/2))

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Rubi [A]  time = 0.105757, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {456, 1261, 205} \[ \frac{b x (A b-a B)}{2 a^3 \left (a+b x^2\right )}+\frac{2 A b-a B}{a^3 x}+\frac{\sqrt{b} (5 A b-3 a B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{7/2}}-\frac{A}{3 a^2 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^4*(a + b*x^2)^2),x]

[Out]

-A/(3*a^2*x^3) + (2*A*b - a*B)/(a^3*x) + (b*(A*b - a*B)*x)/(2*a^3*(a + b*x^2)) + (Sqrt[b]*(5*A*b - 3*a*B)*ArcT
an[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(7/2))

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 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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^4 \left (a+b x^2\right )^2} \, dx &=\frac{b (A b-a B) x}{2 a^3 \left (a+b x^2\right )}-\frac{1}{2} b \int \frac{-\frac{2 A}{a b}+\frac{2 (A b-a B) x^2}{a^2 b}-\frac{(A b-a B) x^4}{a^3}}{x^4 \left (a+b x^2\right )} \, dx\\ &=\frac{b (A b-a B) x}{2 a^3 \left (a+b x^2\right )}-\frac{1}{2} b \int \left (-\frac{2 A}{a^2 b x^4}-\frac{2 (-2 A b+a B)}{a^3 b x^2}+\frac{-5 A b+3 a B}{a^3 \left (a+b x^2\right )}\right ) \, dx\\ &=-\frac{A}{3 a^2 x^3}+\frac{2 A b-a B}{a^3 x}+\frac{b (A b-a B) x}{2 a^3 \left (a+b x^2\right )}+\frac{(b (5 A b-3 a B)) \int \frac{1}{a+b x^2} \, dx}{2 a^3}\\ &=-\frac{A}{3 a^2 x^3}+\frac{2 A b-a B}{a^3 x}+\frac{b (A b-a B) x}{2 a^3 \left (a+b x^2\right )}+\frac{\sqrt{b} (5 A b-3 a B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0732672, size = 90, normalized size = 1. \[ -\frac{b x (a B-A b)}{2 a^3 \left (a+b x^2\right )}+\frac{2 A b-a B}{a^3 x}-\frac{\sqrt{b} (3 a B-5 A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{7/2}}-\frac{A}{3 a^2 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^4*(a + b*x^2)^2),x]

[Out]

-A/(3*a^2*x^3) + (2*A*b - a*B)/(a^3*x) - (b*(-(A*b) + a*B)*x)/(2*a^3*(a + b*x^2)) - (Sqrt[b]*(-5*A*b + 3*a*B)*
ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(7/2))

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Maple [A]  time = 0.01, size = 110, normalized size = 1.2 \begin{align*} -{\frac{A}{3\,{a}^{2}{x}^{3}}}+2\,{\frac{Ab}{{a}^{3}x}}-{\frac{B}{{a}^{2}x}}+{\frac{{b}^{2}Ax}{2\,{a}^{3} \left ( b{x}^{2}+a \right ) }}-{\frac{bBx}{2\,{a}^{2} \left ( b{x}^{2}+a \right ) }}+{\frac{5\,{b}^{2}A}{2\,{a}^{3}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{3\,bB}{2\,{a}^{2}}\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^4/(b*x^2+a)^2,x)

[Out]

-1/3*A/a^2/x^3+2/a^3/x*A*b-1/a^2/x*B+1/2/a^3*b^2*x/(b*x^2+a)*A-1/2/a^2*b*x/(b*x^2+a)*B+5/2/a^3*b^2/(a*b)^(1/2)
*arctan(b*x/(a*b)^(1/2))*A-3/2/a^2*b/(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^4/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.32063, size = 532, normalized size = 5.91 \begin{align*} \left [-\frac{6 \,{\left (3 \, B a b - 5 \, A b^{2}\right )} x^{4} + 4 \, A a^{2} + 4 \,{\left (3 \, B a^{2} - 5 \, A a b\right )} x^{2} + 3 \,{\left ({\left (3 \, B a b - 5 \, A b^{2}\right )} x^{5} +{\left (3 \, B a^{2} - 5 \, A a b\right )} x^{3}\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{b x^{2} + 2 \, a x \sqrt{-\frac{b}{a}} - a}{b x^{2} + a}\right )}{12 \,{\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}, -\frac{3 \,{\left (3 \, B a b - 5 \, A b^{2}\right )} x^{4} + 2 \, A a^{2} + 2 \,{\left (3 \, B a^{2} - 5 \, A a b\right )} x^{2} + 3 \,{\left ({\left (3 \, B a b - 5 \, A b^{2}\right )} x^{5} +{\left (3 \, B a^{2} - 5 \, A a b\right )} x^{3}\right )} \sqrt{\frac{b}{a}} \arctan \left (x \sqrt{\frac{b}{a}}\right )}{6 \,{\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^4/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

[-1/12*(6*(3*B*a*b - 5*A*b^2)*x^4 + 4*A*a^2 + 4*(3*B*a^2 - 5*A*a*b)*x^2 + 3*((3*B*a*b - 5*A*b^2)*x^5 + (3*B*a^
2 - 5*A*a*b)*x^3)*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^3*b*x^5 + a^4*x^3), -1/6*(3*(
3*B*a*b - 5*A*b^2)*x^4 + 2*A*a^2 + 2*(3*B*a^2 - 5*A*a*b)*x^2 + 3*((3*B*a*b - 5*A*b^2)*x^5 + (3*B*a^2 - 5*A*a*b
)*x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)))/(a^3*b*x^5 + a^4*x^3)]

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Sympy [B]  time = 0.803905, size = 184, normalized size = 2.04 \begin{align*} \frac{\sqrt{- \frac{b}{a^{7}}} \left (- 5 A b + 3 B a\right ) \log{\left (- \frac{a^{4} \sqrt{- \frac{b}{a^{7}}} \left (- 5 A b + 3 B a\right )}{- 5 A b^{2} + 3 B a b} + x \right )}}{4} - \frac{\sqrt{- \frac{b}{a^{7}}} \left (- 5 A b + 3 B a\right ) \log{\left (\frac{a^{4} \sqrt{- \frac{b}{a^{7}}} \left (- 5 A b + 3 B a\right )}{- 5 A b^{2} + 3 B a b} + x \right )}}{4} - \frac{2 A a^{2} + x^{4} \left (- 15 A b^{2} + 9 B a b\right ) + x^{2} \left (- 10 A a b + 6 B a^{2}\right )}{6 a^{4} x^{3} + 6 a^{3} b x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**4/(b*x**2+a)**2,x)

[Out]

sqrt(-b/a**7)*(-5*A*b + 3*B*a)*log(-a**4*sqrt(-b/a**7)*(-5*A*b + 3*B*a)/(-5*A*b**2 + 3*B*a*b) + x)/4 - sqrt(-b
/a**7)*(-5*A*b + 3*B*a)*log(a**4*sqrt(-b/a**7)*(-5*A*b + 3*B*a)/(-5*A*b**2 + 3*B*a*b) + x)/4 - (2*A*a**2 + x**
4*(-15*A*b**2 + 9*B*a*b) + x**2*(-10*A*a*b + 6*B*a**2))/(6*a**4*x**3 + 6*a**3*b*x**5)

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Giac [A]  time = 1.14811, size = 115, normalized size = 1.28 \begin{align*} -\frac{{\left (3 \, B a b - 5 \, A b^{2}\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a^{3}} - \frac{B a b x - A b^{2} x}{2 \,{\left (b x^{2} + a\right )} a^{3}} - \frac{3 \, B a x^{2} - 6 \, A b x^{2} + A a}{3 \, a^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^4/(b*x^2+a)^2,x, algorithm="giac")

[Out]

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