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

Optimal. Leaf size=78 \[ \frac{c^{3/2} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{b^{7/2}}-\frac{b B-A c}{3 b^2 x^3}+\frac{c (b B-A c)}{b^3 x}-\frac{A}{5 b x^5} \]

[Out]

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

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Rubi [A]  time = 0.0681362, antiderivative size = 78, 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, 453, 325, 205} \[ \frac{c^{3/2} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{b^{7/2}}-\frac{b B-A c}{3 b^2 x^3}+\frac{c (b B-A c)}{b^3 x}-\frac{A}{5 b x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 (b x^2+c x^4\right )} \, dx &=\int \frac{A+B x^2}{x^6 \left (b+c x^2\right )} \, dx\\ &=-\frac{A}{5 b x^5}-\frac{(-5 b B+5 A c) \int \frac{1}{x^4 \left (b+c x^2\right )} \, dx}{5 b}\\ &=-\frac{A}{5 b x^5}-\frac{b B-A c}{3 b^2 x^3}-\frac{(c (b B-A c)) \int \frac{1}{x^2 \left (b+c x^2\right )} \, dx}{b^2}\\ &=-\frac{A}{5 b x^5}-\frac{b B-A c}{3 b^2 x^3}+\frac{c (b B-A c)}{b^3 x}+\frac{\left (c^2 (b B-A c)\right ) \int \frac{1}{b+c x^2} \, dx}{b^3}\\ &=-\frac{A}{5 b x^5}-\frac{b B-A c}{3 b^2 x^3}+\frac{c (b B-A c)}{b^3 x}+\frac{c^{3/2} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0534885, size = 78, normalized size = 1. \[ \frac{c^{3/2} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{b^{7/2}}+\frac{A c-b B}{3 b^2 x^3}+\frac{c (b B-A c)}{b^3 x}-\frac{A}{5 b x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 96, normalized size = 1.2 \begin{align*} -{\frac{A}{5\,b{x}^{5}}}+{\frac{Ac}{3\,{b}^{2}{x}^{3}}}-{\frac{B}{3\,b{x}^{3}}}-{\frac{A{c}^{2}}{{b}^{3}x}}+{\frac{cB}{{b}^{2}x}}-{\frac{A{c}^{3}}{{b}^{3}}\arctan \left ({cx{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}+{\frac{{c}^{2}B}{{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)/x^4/(c*x^4+b*x^2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.834914, size = 398, normalized size = 5.1 \begin{align*} \left [-\frac{15 \,{\left (B b c - A c^{2}\right )} x^{5} \sqrt{-\frac{c}{b}} \log \left (\frac{c x^{2} - 2 \, b x \sqrt{-\frac{c}{b}} - b}{c x^{2} + b}\right ) - 30 \,{\left (B b c - A c^{2}\right )} x^{4} + 6 \, A b^{2} + 10 \,{\left (B b^{2} - A b c\right )} x^{2}}{30 \, b^{3} x^{5}}, \frac{15 \,{\left (B b c - A c^{2}\right )} x^{5} \sqrt{\frac{c}{b}} \arctan \left (x \sqrt{\frac{c}{b}}\right ) + 15 \,{\left (B b c - A c^{2}\right )} x^{4} - 3 \, A b^{2} - 5 \,{\left (B b^{2} - A b c\right )} x^{2}}{15 \, b^{3} x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/30*(15*(B*b*c - A*c^2)*x^5*sqrt(-c/b)*log((c*x^2 - 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)) - 30*(B*b*c - A*c^2)
*x^4 + 6*A*b^2 + 10*(B*b^2 - A*b*c)*x^2)/(b^3*x^5), 1/15*(15*(B*b*c - A*c^2)*x^5*sqrt(c/b)*arctan(x*sqrt(c/b))
+ 15*(B*b*c - A*c^2)*x^4 - 3*A*b^2 - 5*(B*b^2 - A*b*c)*x^2)/(b^3*x^5)]

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Sympy [B]  time = 0.677921, size = 163, normalized size = 2.09 \begin{align*} - \frac{\sqrt{- \frac{c^{3}}{b^{7}}} \left (- A c + B b\right ) \log{\left (- \frac{b^{4} \sqrt{- \frac{c^{3}}{b^{7}}} \left (- A c + B b\right )}{- A c^{3} + B b c^{2}} + x \right )}}{2} + \frac{\sqrt{- \frac{c^{3}}{b^{7}}} \left (- A c + B b\right ) \log{\left (\frac{b^{4} \sqrt{- \frac{c^{3}}{b^{7}}} \left (- A c + B b\right )}{- A c^{3} + B b c^{2}} + x \right )}}{2} + \frac{- 3 A b^{2} + x^{4} \left (- 15 A c^{2} + 15 B b c\right ) + x^{2} \left (5 A b c - 5 B b^{2}\right )}{15 b^{3} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.25137, size = 109, normalized size = 1.4 \begin{align*} \frac{{\left (B b c^{2} - A c^{3}\right )} \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{\sqrt{b c} b^{3}} + \frac{15 \, B b c x^{4} - 15 \, A c^{2} x^{4} - 5 \, B b^{2} x^{2} + 5 \, A b c x^{2} - 3 \, A b^{2}}{15 \, b^{3} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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