3.1.78 \(\int \frac {x^6}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=45 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 \sqrt {b} c^{3/2}}-\frac {x}{2 c \left (b+c x^2\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1584, 288, 205} \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 \sqrt {b} c^{3/2}}-\frac {x}{2 c \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^6/(b*x^2 + c*x^4)^2,x]

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 45, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 \sqrt {b} c^{3/2}}-\frac {x}{2 c \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^6/(b*x^2 + c*x^4)^2,x]

[Out]

-1/2*x/(c*(b + c*x^2)) + ArcTan[(Sqrt[c]*x)/Sqrt[b]]/(2*Sqrt[b]*c^(3/2))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^6}{\left (b x^2+c x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^6/(b*x^2 + c*x^4)^2,x]

[Out]

IntegrateAlgebraic[x^6/(b*x^2 + c*x^4)^2, x]

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fricas [A]  time = 0.51, size = 120, normalized size = 2.67 \begin {gather*} \left [-\frac {2 \, b c x + {\left (c x^{2} + b\right )} \sqrt {-b c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-b c} x - b}{c x^{2} + b}\right )}{4 \, {\left (b c^{3} x^{2} + b^{2} c^{2}\right )}}, -\frac {b c x - {\left (c x^{2} + b\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c} x}{b}\right )}{2 \, {\left (b c^{3} x^{2} + b^{2} c^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 35, normalized size = 0.78 \begin {gather*} \frac {\arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c} - \frac {x}{2 \, {\left (c x^{2} + b\right )} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 36, normalized size = 0.80 \begin {gather*} -\frac {x}{2 \left (c \,x^{2}+b \right ) c}+\frac {\arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}\, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(c*x^4+b*x^2)^2,x)

[Out]

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

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maxima [A]  time = 3.02, size = 36, normalized size = 0.80 \begin {gather*} -\frac {x}{2 \, {\left (c^{2} x^{2} + b c\right )}} + \frac {\arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.15, size = 33, normalized size = 0.73 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{2\,\sqrt {b}\,c^{3/2}}-\frac {x}{2\,c\,\left (c\,x^2+b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(b*x^2 + c*x^4)^2,x)

[Out]

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

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sympy [B]  time = 0.23, size = 78, normalized size = 1.73 \begin {gather*} - \frac {x}{2 b c + 2 c^{2} x^{2}} - \frac {\sqrt {- \frac {1}{b c^{3}}} \log {\left (- b c \sqrt {- \frac {1}{b c^{3}}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{b c^{3}}} \log {\left (b c \sqrt {- \frac {1}{b c^{3}}} + x \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(c*x**4+b*x**2)**2,x)

[Out]

-x/(2*b*c + 2*c**2*x**2) - sqrt(-1/(b*c**3))*log(-b*c*sqrt(-1/(b*c**3)) + x)/4 + sqrt(-1/(b*c**3))*log(b*c*sqr
t(-1/(b*c**3)) + x)/4

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