3.1.84 \(\int \frac {1}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=68 \[ \frac {5 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{7/2}}+\frac {5 c}{2 b^3 x}-\frac {5}{6 b^2 x^3}+\frac {1}{2 b x^3 \left (b+c x^2\right )} \]

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Rubi [A]  time = 0.03, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1593, 290, 325, 205} \begin {gather*} \frac {5 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{7/2}}+\frac {5 c}{2 b^3 x}-\frac {5}{6 b^2 x^3}+\frac {1}{2 b x^3 \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(-2),x]

[Out]

-5/(6*b^2*x^3) + (5*c)/(2*b^3*x) + 1/(2*b*x^3*(b + c*x^2)) + (5*c^(3/2)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(7/2
))

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 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {1}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {1}{x^4 \left (b+c x^2\right )^2} \, dx\\ &=\frac {1}{2 b x^3 \left (b+c x^2\right )}+\frac {5 \int \frac {1}{x^4 \left (b+c x^2\right )} \, dx}{2 b}\\ &=-\frac {5}{6 b^2 x^3}+\frac {1}{2 b x^3 \left (b+c x^2\right )}-\frac {(5 c) \int \frac {1}{x^2 \left (b+c x^2\right )} \, dx}{2 b^2}\\ &=-\frac {5}{6 b^2 x^3}+\frac {5 c}{2 b^3 x}+\frac {1}{2 b x^3 \left (b+c x^2\right )}+\frac {\left (5 c^2\right ) \int \frac {1}{b+c x^2} \, dx}{2 b^3}\\ &=-\frac {5}{6 b^2 x^3}+\frac {5 c}{2 b^3 x}+\frac {1}{2 b x^3 \left (b+c x^2\right )}+\frac {5 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 67, normalized size = 0.99 \begin {gather*} \frac {5 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{7/2}}+\frac {c^2 x}{2 b^3 \left (b+c x^2\right )}+\frac {2 c}{b^3 x}-\frac {1}{3 b^2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(-2),x]

[Out]

-1/3*1/(b^2*x^3) + (2*c)/(b^3*x) + (c^2*x)/(2*b^3*(b + c*x^2)) + (5*c^(3/2)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^
(7/2))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b x^2+c x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.03, size = 172, normalized size = 2.53 \begin {gather*} \left [\frac {30 \, c^{2} x^{4} + 20 \, b c x^{2} + 15 \, {\left (c^{2} x^{5} + b c x^{3}\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^{2}}{12 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}, \frac {15 \, c^{2} x^{4} + 10 \, b c x^{2} + 15 \, {\left (c^{2} x^{5} + b c x^{3}\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right ) - 2 \, b^{2}}{6 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

[1/12*(30*c^2*x^4 + 20*b*c*x^2 + 15*(c^2*x^5 + b*c*x^3)*sqrt(-c/b)*log((c*x^2 + 2*b*x*sqrt(-c/b) - b)/(c*x^2 +
b)) - 4*b^2)/(b^3*c*x^5 + b^4*x^3), 1/6*(15*c^2*x^4 + 10*b*c*x^2 + 15*(c^2*x^5 + b*c*x^3)*sqrt(c/b)*arctan(x*
sqrt(c/b)) - 2*b^2)/(b^3*c*x^5 + b^4*x^3)]

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giac [A]  time = 0.15, size = 59, normalized size = 0.87 \begin {gather*} \frac {5 \, c^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{3}} + \frac {c^{2} x}{2 \, {\left (c x^{2} + b\right )} b^{3}} + \frac {6 \, c x^{2} - b}{3 \, b^{3} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 59, normalized size = 0.87 \begin {gather*} \frac {c^{2} x}{2 \left (c \,x^{2}+b \right ) b^{3}}+\frac {5 c^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}\, b^{3}}+\frac {2 c}{b^{3} x}-\frac {1}{3 b^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^4+b*x^2)^2,x)

[Out]

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

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maxima [A]  time = 2.96, size = 64, normalized size = 0.94 \begin {gather*} \frac {15 \, c^{2} x^{4} + 10 \, b c x^{2} - 2 \, b^{2}}{6 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}} + \frac {5 \, c^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

1/6*(15*c^2*x^4 + 10*b*c*x^2 - 2*b^2)/(b^3*c*x^5 + b^4*x^3) + 5/2*c^2*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^3)

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mupad [B]  time = 4.17, size = 58, normalized size = 0.85 \begin {gather*} \frac {\frac {5\,c\,x^2}{3\,b^2}-\frac {1}{3\,b}+\frac {5\,c^2\,x^4}{2\,b^3}}{c\,x^5+b\,x^3}+\frac {5\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{2\,b^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2 + c*x^4)^2,x)

[Out]

((5*c*x^2)/(3*b^2) - 1/(3*b) + (5*c^2*x^4)/(2*b^3))/(b*x^3 + c*x^5) + (5*c^(3/2)*atan((c^(1/2)*x)/b^(1/2)))/(2
*b^(7/2))

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sympy [A]  time = 0.39, size = 114, normalized size = 1.68 \begin {gather*} - \frac {5 \sqrt {- \frac {c^{3}}{b^{7}}} \log {\left (- \frac {b^{4} \sqrt {- \frac {c^{3}}{b^{7}}}}{c^{2}} + x \right )}}{4} + \frac {5 \sqrt {- \frac {c^{3}}{b^{7}}} \log {\left (\frac {b^{4} \sqrt {- \frac {c^{3}}{b^{7}}}}{c^{2}} + x \right )}}{4} + \frac {- 2 b^{2} + 10 b c x^{2} + 15 c^{2} x^{4}}{6 b^{4} x^{3} + 6 b^{3} c x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x**4+b*x**2)**2,x)

[Out]

-5*sqrt(-c**3/b**7)*log(-b**4*sqrt(-c**3/b**7)/c**2 + x)/4 + 5*sqrt(-c**3/b**7)*log(b**4*sqrt(-c**3/b**7)/c**2
+ x)/4 + (-2*b**2 + 10*b*c*x**2 + 15*c**2*x**4)/(6*b**4*x**3 + 6*b**3*c*x**5)

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