3.62 \(\int \frac{x^8 (A+B x^2)}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=89 \[ -\frac{b x (b B-A c)}{2 c^3 \left (b+c x^2\right )}-\frac{x (2 b B-A c)}{c^3}+\frac{\sqrt{b} (5 b B-3 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 c^{7/2}}+\frac{B x^3}{3 c^2} \]

[Out]

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

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Rubi [A]  time = 0.0885159, antiderivative size = 89, 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, 455, 1153, 205} \[ -\frac{b x (b B-A c)}{2 c^3 \left (b+c x^2\right )}-\frac{x (2 b B-A c)}{c^3}+\frac{\sqrt{b} (5 b B-3 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 c^{7/2}}+\frac{B x^3}{3 c^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^8*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

-(((2*b*B - A*c)*x)/c^3) + (B*x^3)/(3*c^2) - (b*(b*B - A*c)*x)/(2*c^3*(b + c*x^2)) + (Sqrt[b]*(5*b*B - 3*A*c)*
ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*c^(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 455

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[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 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{x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac{x^4 \left (A+B x^2\right )}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac{b (b B-A c) x}{2 c^3 \left (b+c x^2\right )}-\frac{\int \frac{-b (b B-A c)+2 c (b B-A c) x^2-2 B c^2 x^4}{b+c x^2} \, dx}{2 c^3}\\ &=-\frac{b (b B-A c) x}{2 c^3 \left (b+c x^2\right )}-\frac{\int \left (2 (2 b B-A c)-2 B c x^2+\frac{-5 b^2 B+3 A b c}{b+c x^2}\right ) \, dx}{2 c^3}\\ &=-\frac{(2 b B-A c) x}{c^3}+\frac{B x^3}{3 c^2}-\frac{b (b B-A c) x}{2 c^3 \left (b+c x^2\right )}+\frac{(b (5 b B-3 A c)) \int \frac{1}{b+c x^2} \, dx}{2 c^3}\\ &=-\frac{(2 b B-A c) x}{c^3}+\frac{B x^3}{3 c^2}-\frac{b (b B-A c) x}{2 c^3 \left (b+c x^2\right )}+\frac{\sqrt{b} (5 b B-3 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0710759, size = 89, normalized size = 1. \[ \frac{x \left (A b c-b^2 B\right )}{2 c^3 \left (b+c x^2\right )}+\frac{x (A c-2 b B)}{c^3}+\frac{\sqrt{b} (5 b B-3 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 c^{7/2}}+\frac{B x^3}{3 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^8*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

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

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Maple [A]  time = 0.009, size = 105, normalized size = 1.2 \begin{align*}{\frac{B{x}^{3}}{3\,{c}^{2}}}+{\frac{Ax}{{c}^{2}}}-2\,{\frac{Bbx}{{c}^{3}}}+{\frac{Abx}{2\,{c}^{2} \left ( c{x}^{2}+b \right ) }}-{\frac{B{b}^{2}x}{2\,{c}^{3} \left ( c{x}^{2}+b \right ) }}-{\frac{3\,Ab}{2\,{c}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}+{\frac{5\,B{b}^{2}}{2\,{c}^{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(x^8*(B*x^2+A)/(c*x^4+b*x^2)^2,x)

[Out]

1/3*B*x^3/c^2+1/c^2*A*x-2/c^3*B*b*x+1/2*b/c^2*x/(c*x^2+b)*A-1/2*b^2/c^3*x/(c*x^2+b)*B-3/2*b/c^2/(b*c)^(1/2)*ar
ctan(x*c/(b*c)^(1/2))*A+5/2*b^2/c^3/(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(x^8*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.778381, size = 513, normalized size = 5.76 \begin{align*} \left [\frac{4 \, B c^{2} x^{5} - 4 \,{\left (5 \, B b c - 3 \, A c^{2}\right )} x^{3} - 3 \,{\left (5 \, B b^{2} - 3 \, A b c +{\left (5 \, B b c - 3 \, A c^{2}\right )} x^{2}\right )} \sqrt{-\frac{b}{c}} \log \left (\frac{c x^{2} - 2 \, c x \sqrt{-\frac{b}{c}} - b}{c x^{2} + b}\right ) - 6 \,{\left (5 \, B b^{2} - 3 \, A b c\right )} x}{12 \,{\left (c^{4} x^{2} + b c^{3}\right )}}, \frac{2 \, B c^{2} x^{5} - 2 \,{\left (5 \, B b c - 3 \, A c^{2}\right )} x^{3} + 3 \,{\left (5 \, B b^{2} - 3 \, A b c +{\left (5 \, B b c - 3 \, A c^{2}\right )} x^{2}\right )} \sqrt{\frac{b}{c}} \arctan \left (\frac{c x \sqrt{\frac{b}{c}}}{b}\right ) - 3 \,{\left (5 \, B b^{2} - 3 \, A b c\right )} x}{6 \,{\left (c^{4} x^{2} + b c^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.785215, size = 128, normalized size = 1.44 \begin{align*} \frac{B x^{3}}{3 c^{2}} - \frac{x \left (- A b c + B b^{2}\right )}{2 b c^{3} + 2 c^{4} x^{2}} - \frac{\sqrt{- \frac{b}{c^{7}}} \left (- 3 A c + 5 B b\right ) \log{\left (- c^{3} \sqrt{- \frac{b}{c^{7}}} + x \right )}}{4} + \frac{\sqrt{- \frac{b}{c^{7}}} \left (- 3 A c + 5 B b\right ) \log{\left (c^{3} \sqrt{- \frac{b}{c^{7}}} + x \right )}}{4} - \frac{x \left (- A c + 2 B b\right )}{c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(B*x**2+A)/(c*x**4+b*x**2)**2,x)

[Out]

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

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Giac [A]  time = 1.27138, size = 119, normalized size = 1.34 \begin{align*} \frac{{\left (5 \, B b^{2} - 3 \, A b c\right )} \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{2 \, \sqrt{b c} c^{3}} - \frac{B b^{2} x - A b c x}{2 \,{\left (c x^{2} + b\right )} c^{3}} + \frac{B c^{4} x^{3} - 6 \, B b c^{3} x + 3 \, A c^{4} x}{3 \, c^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

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