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

Optimal. Leaf size=55 \[ -\frac {3 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{5/2}}-\frac {x^3}{2 c \left (b+c x^2\right )}+\frac {3 x}{2 c^2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1584, 288, 321, 205} \[ -\frac {3 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{5/2}}-\frac {x^3}{2 c \left (b+c x^2\right )}+\frac {3 x}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x^8/(b*x^2 + c*x^4)^2,x]

[Out]

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

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]

Rule 288

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

Rule 321

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

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]

Rubi steps

\begin {align*} \int \frac {x^8}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {x^4}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac {x^3}{2 c \left (b+c x^2\right )}+\frac {3 \int \frac {x^2}{b+c x^2} \, dx}{2 c}\\ &=\frac {3 x}{2 c^2}-\frac {x^3}{2 c \left (b+c x^2\right )}-\frac {(3 b) \int \frac {1}{b+c x^2} \, dx}{2 c^2}\\ &=\frac {3 x}{2 c^2}-\frac {x^3}{2 c \left (b+c x^2\right )}-\frac {3 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 51, normalized size = 0.93 \[ -\frac {3 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{5/2}}+\frac {b x}{2 c^2 \left (b+c x^2\right )}+\frac {x}{c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/(b*x^2 + c*x^4)^2,x]

[Out]

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

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fricas [A]  time = 0.62, size = 136, normalized size = 2.47 \[ \left [\frac {4 \, c x^{3} + 3 \, {\left (c x^{2} + b\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x^{2} - 2 \, c x \sqrt {-\frac {b}{c}} - b}{c x^{2} + b}\right ) + 6 \, b x}{4 \, {\left (c^{3} x^{2} + b c^{2}\right )}}, \frac {2 \, c x^{3} - 3 \, {\left (c x^{2} + b\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c x \sqrt {\frac {b}{c}}}{b}\right ) + 3 \, b x}{2 \, {\left (c^{3} x^{2} + b c^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 42, normalized size = 0.76 \[ -\frac {3 \, b \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c^{2}} + \frac {b x}{2 \, {\left (c x^{2} + b\right )} c^{2}} + \frac {x}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*b*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^2) + 1/2*b*x/((c*x^2 + b)*c^2) + x/c^2

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maple [A]  time = 0.01, size = 43, normalized size = 0.78 \[ \frac {b x}{2 \left (c \,x^{2}+b \right ) c^{2}}-\frac {3 b \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}\, c^{2}}+\frac {x}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(c*x^4+b*x^2)^2,x)

[Out]

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

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maxima [A]  time = 2.94, size = 45, normalized size = 0.82 \[ \frac {b x}{2 \, {\left (c^{3} x^{2} + b c^{2}\right )}} - \frac {3 \, b \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c^{2}} + \frac {x}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*x/(c^3*x^2 + b*c^2) - 3/2*b*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^2) + x/c^2

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mupad [B]  time = 4.17, size = 43, normalized size = 0.78 \[ \frac {x}{c^2}+\frac {b\,x}{2\,\left (c^3\,x^2+b\,c^2\right )}-\frac {3\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{2\,c^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(b*x^2 + c*x^4)^2,x)

[Out]

x/c^2 + (b*x)/(2*(b*c^2 + c^3*x^2)) - (3*b^(1/2)*atan((c^(1/2)*x)/b^(1/2)))/(2*c^(5/2))

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sympy [A]  time = 0.28, size = 83, normalized size = 1.51 \[ \frac {b x}{2 b c^{2} + 2 c^{3} x^{2}} + \frac {3 \sqrt {- \frac {b}{c^{5}}} \log {\left (- c^{2} \sqrt {- \frac {b}{c^{5}}} + x \right )}}{4} - \frac {3 \sqrt {- \frac {b}{c^{5}}} \log {\left (c^{2} \sqrt {- \frac {b}{c^{5}}} + x \right )}}{4} + \frac {x}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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