3.2.100 \(\int \frac {x^2}{(b x^2+c x^4)^2} \, dx\) [200]

Optimal. Leaf size=57 \[ -\frac {3}{2 b^2 x}+\frac {1}{2 b x \left (b+c x^2\right )}-\frac {3 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 57, 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 = {1598, 296, 331, 211} \begin {gather*} -\frac {3 \sqrt {c} \text {ArcTan}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{5/2}}-\frac {3}{2 b^2 x}+\frac {1}{2 b x \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(b*x^2 + c*x^4)^2,x]

[Out]

-3/(2*b^2*x) + 1/(2*b*x*(b + c*x^2)) - (3*Sqrt[c]*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(5/2))

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 296

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

Rule 331

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

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Mathematica [A]
time = 0.03, size = 54, normalized size = 0.95 \begin {gather*} -\frac {1}{b^2 x}-\frac {c x}{2 b^2 \left (b+c x^2\right )}-\frac {3 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(b*x^2 + c*x^4)^2,x]

[Out]

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

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Maple [A]
time = 0.08, size = 45, normalized size = 0.79

method	result	size
default	\(-\frac {c \left (\frac {x}{2 c \,x^{2}+2 b}+\frac {3 \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}}\right )}{b^{2}}-\frac {1}{b^{2} x}\)	\(45\)
risch	\(\frac {-\frac {3 c \,x^{2}}{2 b^{2}}-\frac {1}{b}}{x \left (c \,x^{2}+b \right )}+\frac {3 \left (\munderset {\textit {\_R} =\RootOf \left (b^{5} \textit {\_Z}^{2}+c \right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} b^{5}+2 c \right ) x +b^{3} \textit {\_R} \right )\right )}{4}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.52, size = 49, normalized size = 0.86 \begin {gather*} -\frac {3 \, c x^{2} + 2 \, b}{2 \, {\left (b^{2} c x^{3} + b^{3} x\right )}} - \frac {3 \, c \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 136, normalized size = 2.39 \begin {gather*} \left [-\frac {6 \, c x^{2} - 3 \, {\left (c x^{3} + b x\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} - 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right ) + 4 \, b}{4 \, {\left (b^{2} c x^{3} + b^{3} x\right )}}, -\frac {3 \, c x^{2} + 3 \, {\left (c x^{3} + b x\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right ) + 2 \, b}{2 \, {\left (b^{2} c x^{3} + b^{3} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.15, size = 92, normalized size = 1.61 \begin {gather*} \frac {3 \sqrt {- \frac {c}{b^{5}}} \log {\left (- \frac {b^{3} \sqrt {- \frac {c}{b^{5}}}}{c} + x \right )}}{4} - \frac {3 \sqrt {- \frac {c}{b^{5}}} \log {\left (\frac {b^{3} \sqrt {- \frac {c}{b^{5}}}}{c} + x \right )}}{4} + \frac {- 2 b - 3 c x^{2}}{2 b^{3} x + 2 b^{2} c x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 6.70, size = 47, normalized size = 0.82 \begin {gather*} -\frac {3 \, c \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{2}} - \frac {3 \, c x^{2} + 2 \, b}{2 \, {\left (c x^{3} + b x\right )} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.06, size = 44, normalized size = 0.77 \begin {gather*} -\frac {\frac {1}{b}+\frac {3\,c\,x^2}{2\,b^2}}{c\,x^3+b\,x}-\frac {3\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{2\,b^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x^2 + c*x^4)^2,x)

[Out]

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

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