3.2.98 \(\int \frac {x^4}{(b x^2+c x^4)^2} \, dx\) [198]

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

[Out]

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

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Rubi [A]
time = 0.01, 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 = {1598, 205, 211} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{3/2} \sqrt {c}}+\frac {x}{2 b \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/(b*x^2 + c*x^4)^2,x]

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 1598

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

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

Antiderivative was successfully verified.

[In]

Integrate[x^4/(b*x^2 + c*x^4)^2,x]

[Out]

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

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Maple [A]
time = 0.09, size = 36, normalized size = 0.80

method	result	size
default	\(\frac {x}{2 b \left (c \,x^{2}+b \right )}+\frac {\arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 b \sqrt {b c}}\)	\(36\)
risch	\(\frac {x}{2 b \left (c \,x^{2}+b \right )}-\frac {\ln \left (c x +\sqrt {-b c}\right )}{4 \sqrt {-b c}\, b}+\frac {\ln \left (-c x +\sqrt {-b c}\right )}{4 \sqrt {-b c}\, b}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(c*x^4+b*x^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 35, normalized size = 0.78 \begin {gather*} \frac {x}{2 \, {\left (b c x^{2} + b^{2}\right )}} + \frac {\arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.40, 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^{2} c^{2} x^{2} + b^{3} c\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^{2} c^{2} x^{2} + b^{3} c\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (36) = 72\).
time = 0.09, size = 78, normalized size = 1.73 \begin {gather*} \frac {x}{2 b^{2} + 2 b c x^{2}} - \frac {\sqrt {- \frac {1}{b^{3} c}} \log {\left (- b^{2} \sqrt {- \frac {1}{b^{3} c}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{b^{3} c}} \log {\left (b^{2} \sqrt {- \frac {1}{b^{3} c}} + x \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(2*b**2 + 2*b*c*x**2) - sqrt(-1/(b**3*c))*log(-b**2*sqrt(-1/(b**3*c)) + x)/4 + sqrt(-1/(b**3*c))*log(b**2*sq
rt(-1/(b**3*c)) + x)/4

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Giac [A]
time = 5.92, size = 35, normalized size = 0.78 \begin {gather*} \frac {\arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b} + \frac {x}{2 \, {\left (c x^{2} + b\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.04, size = 33, normalized size = 0.73 \begin {gather*} \frac {x}{2\,b\,\left (c\,x^2+b\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{2\,b^{3/2}\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x^2 + c*x^4)^2,x)

[Out]

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

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