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

Optimal. Leaf size=111 \[ \frac{c^2 x (b B-A c)}{2 b^4 \left (b+c x^2\right )}+\frac{c^{3/2} (5 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{9/2}}-\frac{b B-2 A c}{3 b^3 x^3}+\frac{c (2 b B-3 A c)}{b^4 x}-\frac{A}{5 b^2 x^5} \]

[Out]

-A/(5*b^2*x^5) - (b*B - 2*A*c)/(3*b^3*x^3) + (c*(2*b*B - 3*A*c))/(b^4*x) + (c^2*(b*B - A*c)*x)/(2*b^4*(b + c*x
^2)) + (c^(3/2)*(5*b*B - 7*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.198159, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1584, 456, 1802, 205} \[ \frac{c^2 x (b B-A c)}{2 b^4 \left (b+c x^2\right )}+\frac{c^{3/2} (5 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{9/2}}-\frac{b B-2 A c}{3 b^3 x^3}+\frac{c (2 b B-3 A c)}{b^4 x}-\frac{A}{5 b^2 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(5*b^2*x^5) - (b*B - 2*A*c)/(3*b^3*x^3) + (c*(2*b*B - 3*A*c))/(b^4*x) + (c^2*(b*B - A*c)*x)/(2*b^4*(b + c*x
^2)) + (c^(3/2)*(5*b*B - 7*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(9/2))

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, 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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -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^2 \left (b x^2+c x^4\right )^2} \, dx &=\int \frac{A+B x^2}{x^6 \left (b+c x^2\right )^2} \, dx\\ &=\frac{c^2 (b B-A c) x}{2 b^4 \left (b+c x^2\right )}-\frac{1}{2} c^2 \int \frac{-\frac{2 A}{b c^2}-\frac{2 (b B-A c) x^2}{b^2 c^2}+\frac{2 (b B-A c) x^4}{b^3 c}-\frac{(b B-A c) x^6}{b^4}}{x^6 \left (b+c x^2\right )} \, dx\\ &=\frac{c^2 (b B-A c) x}{2 b^4 \left (b+c x^2\right )}-\frac{1}{2} c^2 \int \left (-\frac{2 A}{b^2 c^2 x^6}-\frac{2 (b B-2 A c)}{b^3 c^2 x^4}+\frac{2 (2 b B-3 A c)}{b^4 c x^2}+\frac{-5 b B+7 A c}{b^4 \left (b+c x^2\right )}\right ) \, dx\\ &=-\frac{A}{5 b^2 x^5}-\frac{b B-2 A c}{3 b^3 x^3}+\frac{c (2 b B-3 A c)}{b^4 x}+\frac{c^2 (b B-A c) x}{2 b^4 \left (b+c x^2\right )}+\frac{\left (c^2 (5 b B-7 A c)\right ) \int \frac{1}{b+c x^2} \, dx}{2 b^4}\\ &=-\frac{A}{5 b^2 x^5}-\frac{b B-2 A c}{3 b^3 x^3}+\frac{c (2 b B-3 A c)}{b^4 x}+\frac{c^2 (b B-A c) x}{2 b^4 \left (b+c x^2\right )}+\frac{c^{3/2} (5 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0899056, size = 112, normalized size = 1.01 \[ \frac{c^2 x (b B-A c)}{2 b^4 \left (b+c x^2\right )}+\frac{c^{3/2} (5 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{9/2}}+\frac{2 A c-b B}{3 b^3 x^3}+\frac{c (2 b B-3 A c)}{b^4 x}-\frac{A}{5 b^2 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(5*b^2*x^5) + (-(b*B) + 2*A*c)/(3*b^3*x^3) + (c*(2*b*B - 3*A*c))/(b^4*x) + (c^2*(b*B - A*c)*x)/(2*b^4*(b +
c*x^2)) + (c^(3/2)*(5*b*B - 7*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(9/2))

________________________________________________________________________________________

Maple [A]  time = 0.013, size = 136, normalized size = 1.2 \begin{align*} -{\frac{A}{5\,{b}^{2}{x}^{5}}}+{\frac{2\,Ac}{3\,{b}^{3}{x}^{3}}}-{\frac{B}{3\,{b}^{2}{x}^{3}}}-3\,{\frac{A{c}^{2}}{{b}^{4}x}}+2\,{\frac{Bc}{{b}^{3}x}}-{\frac{A{c}^{3}x}{2\,{b}^{4} \left ( c{x}^{2}+b \right ) }}+{\frac{B{c}^{2}x}{2\,{b}^{3} \left ( c{x}^{2}+b \right ) }}-{\frac{7\,A{c}^{3}}{2\,{b}^{4}}\arctan \left ({cx{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}+{\frac{5\,B{c}^{2}}{2\,{b}^{3}}\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)/x^2/(c*x^4+b*x^2)^2,x)

[Out]

-1/5*A/b^2/x^5+2/3/b^3/x^3*A*c-1/3/b^2/x^3*B-3*c^2/b^4/x*A+2*c/b^3/x*B-1/2/b^4*c^3*x/(c*x^2+b)*A+1/2/b^3*c^2*x
/(c*x^2+b)*B-7/2/b^4*c^3/(b*c)^(1/2)*arctan(x*c/(b*c)^(1/2))*A+5/2/b^3*c^2/(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)/x^2/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 0.847408, size = 653, normalized size = 5.88 \begin{align*} \left [\frac{30 \,{\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{6} + 20 \,{\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{4} - 12 \, A b^{3} - 4 \,{\left (5 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2} - 15 \,{\left ({\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{7} +{\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{5}\right )} \sqrt{-\frac{c}{b}} \log \left (\frac{c x^{2} - 2 \, b x \sqrt{-\frac{c}{b}} - b}{c x^{2} + b}\right )}{60 \,{\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}, \frac{15 \,{\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{6} + 10 \,{\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{4} - 6 \, A b^{3} - 2 \,{\left (5 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2} + 15 \,{\left ({\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{7} +{\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{5}\right )} \sqrt{\frac{c}{b}} \arctan \left (x \sqrt{\frac{c}{b}}\right )}{30 \,{\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/60*(30*(5*B*b*c^2 - 7*A*c^3)*x^6 + 20*(5*B*b^2*c - 7*A*b*c^2)*x^4 - 12*A*b^3 - 4*(5*B*b^3 - 7*A*b^2*c)*x^2
- 15*((5*B*b*c^2 - 7*A*c^3)*x^7 + (5*B*b^2*c - 7*A*b*c^2)*x^5)*sqrt(-c/b)*log((c*x^2 - 2*b*x*sqrt(-c/b) - b)/(
c*x^2 + b)))/(b^4*c*x^7 + b^5*x^5), 1/30*(15*(5*B*b*c^2 - 7*A*c^3)*x^6 + 10*(5*B*b^2*c - 7*A*b*c^2)*x^4 - 6*A*
b^3 - 2*(5*B*b^3 - 7*A*b^2*c)*x^2 + 15*((5*B*b*c^2 - 7*A*c^3)*x^7 + (5*B*b^2*c - 7*A*b*c^2)*x^5)*sqrt(c/b)*arc
tan(x*sqrt(c/b)))/(b^4*c*x^7 + b^5*x^5)]

________________________________________________________________________________________

Sympy [B]  time = 1.02174, size = 218, normalized size = 1.96 \begin{align*} - \frac{\sqrt{- \frac{c^{3}}{b^{9}}} \left (- 7 A c + 5 B b\right ) \log{\left (- \frac{b^{5} \sqrt{- \frac{c^{3}}{b^{9}}} \left (- 7 A c + 5 B b\right )}{- 7 A c^{3} + 5 B b c^{2}} + x \right )}}{4} + \frac{\sqrt{- \frac{c^{3}}{b^{9}}} \left (- 7 A c + 5 B b\right ) \log{\left (\frac{b^{5} \sqrt{- \frac{c^{3}}{b^{9}}} \left (- 7 A c + 5 B b\right )}{- 7 A c^{3} + 5 B b c^{2}} + x \right )}}{4} + \frac{- 6 A b^{3} + x^{6} \left (- 105 A c^{3} + 75 B b c^{2}\right ) + x^{4} \left (- 70 A b c^{2} + 50 B b^{2} c\right ) + x^{2} \left (14 A b^{2} c - 10 B b^{3}\right )}{30 b^{5} x^{5} + 30 b^{4} c x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-c**3/b**9)*(-7*A*c + 5*B*b)*log(-b**5*sqrt(-c**3/b**9)*(-7*A*c + 5*B*b)/(-7*A*c**3 + 5*B*b*c**2) + x)/4
+ sqrt(-c**3/b**9)*(-7*A*c + 5*B*b)*log(b**5*sqrt(-c**3/b**9)*(-7*A*c + 5*B*b)/(-7*A*c**3 + 5*B*b*c**2) + x)/
4 + (-6*A*b**3 + x**6*(-105*A*c**3 + 75*B*b*c**2) + x**4*(-70*A*b*c**2 + 50*B*b**2*c) + x**2*(14*A*b**2*c - 10
*B*b**3))/(30*b**5*x**5 + 30*b**4*c*x**7)

________________________________________________________________________________________

Giac [A]  time = 1.1418, size = 151, normalized size = 1.36 \begin{align*} \frac{{\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{2 \, \sqrt{b c} b^{4}} + \frac{B b c^{2} x - A c^{3} x}{2 \,{\left (c x^{2} + b\right )} b^{4}} + \frac{30 \, B b c x^{4} - 45 \, A c^{2} x^{4} - 5 \, B b^{2} x^{2} + 10 \, A b c x^{2} - 3 \, A b^{2}}{15 \, b^{4} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(5*B*b*c^2 - 7*A*c^3)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^4) + 1/2*(B*b*c^2*x - A*c^3*x)/((c*x^2 + b)*b^4)
+ 1/15*(30*B*b*c*x^4 - 45*A*c^2*x^4 - 5*B*b^2*x^2 + 10*A*b*c*x^2 - 3*A*b^2)/(b^4*x^5)