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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.117439, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {1593, 456, 1261, 205} \[ -\frac{c x (b B-A c)}{2 b^3 \left (b+c x^2\right )}-\frac{b B-2 A c}{b^3 x}-\frac{\sqrt{c} (3 b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{7/2}}-\frac{A}{3 b^2 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(b*x^2 + c*x^4)^2,x]

[Out]

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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

Mathematica [A]  time = 0.0715265, size = 90, normalized size = 1. \[ -\frac{c x (b B-A c)}{2 b^3 \left (b+c x^2\right )}+\frac{2 A c-b B}{b^3 x}-\frac{\sqrt{c} (3 b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{7/2}}-\frac{A}{3 b^2 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.013, size = 110, normalized size = 1.2 \begin{align*} -{\frac{A}{3\,{b}^{2}{x}^{3}}}+2\,{\frac{Ac}{{b}^{3}x}}-{\frac{B}{{b}^{2}x}}+{\frac{A{c}^{2}x}{2\,{b}^{3} \left ( c{x}^{2}+b \right ) }}-{\frac{Bcx}{2\,{b}^{2} \left ( c{x}^{2}+b \right ) }}+{\frac{5\,A{c}^{2}}{2\,{b}^{3}}\arctan \left ({cx{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}-{\frac{3\,Bc}{2\,{b}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(c*x^4+b*x^2)^2,x)

[Out]

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

________________________________________________________________________________________

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)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 0.846885, size = 532, normalized size = 5.91 \begin{align*} \left [-\frac{6 \,{\left (3 \, B b c - 5 \, A c^{2}\right )} x^{4} + 4 \, A b^{2} + 4 \,{\left (3 \, B b^{2} - 5 \, A b c\right )} x^{2} + 3 \,{\left ({\left (3 \, B b c - 5 \, A c^{2}\right )} x^{5} +{\left (3 \, B b^{2} - 5 \, A b c\right )} x^{3}\right )} \sqrt{-\frac{c}{b}} \log \left (\frac{c x^{2} + 2 \, b x \sqrt{-\frac{c}{b}} - b}{c x^{2} + b}\right )}{12 \,{\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}, -\frac{3 \,{\left (3 \, B b c - 5 \, A c^{2}\right )} x^{4} + 2 \, A b^{2} + 2 \,{\left (3 \, B b^{2} - 5 \, A b c\right )} x^{2} + 3 \,{\left ({\left (3 \, B b c - 5 \, A c^{2}\right )} x^{5} +{\left (3 \, B b^{2} - 5 \, A b c\right )} x^{3}\right )} \sqrt{\frac{c}{b}} \arctan \left (x \sqrt{\frac{c}{b}}\right )}{6 \,{\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [B]  time = 0.827283, size = 184, normalized size = 2.04 \begin{align*} \frac{\sqrt{- \frac{c}{b^{7}}} \left (- 5 A c + 3 B b\right ) \log{\left (- \frac{b^{4} \sqrt{- \frac{c}{b^{7}}} \left (- 5 A c + 3 B b\right )}{- 5 A c^{2} + 3 B b c} + x \right )}}{4} - \frac{\sqrt{- \frac{c}{b^{7}}} \left (- 5 A c + 3 B b\right ) \log{\left (\frac{b^{4} \sqrt{- \frac{c}{b^{7}}} \left (- 5 A c + 3 B b\right )}{- 5 A c^{2} + 3 B b c} + x \right )}}{4} - \frac{2 A b^{2} + x^{4} \left (- 15 A c^{2} + 9 B b c\right ) + x^{2} \left (- 10 A b c + 6 B b^{2}\right )}{6 b^{4} x^{3} + 6 b^{3} c x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.15127, size = 115, normalized size = 1.28 \begin{align*} -\frac{{\left (3 \, B b c - 5 \, A c^{2}\right )} \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{2 \, \sqrt{b c} b^{3}} - \frac{B b c x - A c^{2} x}{2 \,{\left (c x^{2} + b\right )} b^{3}} - \frac{3 \, B b x^{2} - 6 \, A c x^{2} + A b}{3 \, b^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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