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

Optimal. Leaf size=81 \[ -\frac {7 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{9/2}}-\frac {7 c^2}{2 b^4 x}+\frac {7 c}{6 b^3 x^3}-\frac {7}{10 b^2 x^5}+\frac {1}{2 b x^5 \left (b+c x^2\right )} \]

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Rubi [A]  time = 0.05, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1584, 290, 325, 205} \begin {gather*} -\frac {7 c^2}{2 b^4 x}-\frac {7 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{9/2}}+\frac {7 c}{6 b^3 x^3}-\frac {7}{10 b^2 x^5}+\frac {1}{2 b x^5 \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(b*x^2 + c*x^4)^2),x]

[Out]

-7/(10*b^2*x^5) + (7*c)/(6*b^3*x^3) - (7*c^2)/(2*b^4*x) + 1/(2*b*x^5*(b + c*x^2)) - (7*c^(5/2)*ArcTan[(Sqrt[c]
*x)/Sqrt[b]])/(2*b^(9/2))

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 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 {1}{x^2 \left (b x^2+c x^4\right )^2} \, dx &=\int \frac {1}{x^6 \left (b+c x^2\right )^2} \, dx\\ &=\frac {1}{2 b x^5 \left (b+c x^2\right )}+\frac {7 \int \frac {1}{x^6 \left (b+c x^2\right )} \, dx}{2 b}\\ &=-\frac {7}{10 b^2 x^5}+\frac {1}{2 b x^5 \left (b+c x^2\right )}-\frac {(7 c) \int \frac {1}{x^4 \left (b+c x^2\right )} \, dx}{2 b^2}\\ &=-\frac {7}{10 b^2 x^5}+\frac {7 c}{6 b^3 x^3}+\frac {1}{2 b x^5 \left (b+c x^2\right )}+\frac {\left (7 c^2\right ) \int \frac {1}{x^2 \left (b+c x^2\right )} \, dx}{2 b^3}\\ &=-\frac {7}{10 b^2 x^5}+\frac {7 c}{6 b^3 x^3}-\frac {7 c^2}{2 b^4 x}+\frac {1}{2 b x^5 \left (b+c x^2\right )}-\frac {\left (7 c^3\right ) \int \frac {1}{b+c x^2} \, dx}{2 b^4}\\ &=-\frac {7}{10 b^2 x^5}+\frac {7 c}{6 b^3 x^3}-\frac {7 c^2}{2 b^4 x}+\frac {1}{2 b x^5 \left (b+c x^2\right )}-\frac {7 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 80, normalized size = 0.99 \begin {gather*} -\frac {7 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{9/2}}-\frac {c^3 x}{2 b^4 \left (b+c x^2\right )}-\frac {3 c^2}{b^4 x}+\frac {2 c}{3 b^3 x^3}-\frac {1}{5 b^2 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(b*x^2 + c*x^4)^2),x]

[Out]

-1/5*1/(b^2*x^5) + (2*c)/(3*b^3*x^3) - (3*c^2)/(b^4*x) - (c^3*x)/(2*b^4*(b + c*x^2)) - (7*c^(5/2)*ArcTan[(Sqrt
[c]*x)/Sqrt[b]])/(2*b^(9/2))

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

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/(x^2*(b*x^2 + c*x^4)^2),x]

[Out]

IntegrateAlgebraic[1/(x^2*(b*x^2 + c*x^4)^2), x]

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fricas [A]  time = 0.66, size = 198, normalized size = 2.44 \begin {gather*} \left [-\frac {210 \, c^{3} x^{6} + 140 \, b c^{2} x^{4} - 28 \, b^{2} c x^{2} + 12 \, b^{3} - 105 \, {\left (c^{3} x^{7} + b c^{2} x^{5}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} - 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right )}{60 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}, -\frac {105 \, c^{3} x^{6} + 70 \, b c^{2} x^{4} - 14 \, b^{2} c x^{2} + 6 \, b^{3} + 105 \, {\left (c^{3} x^{7} + b c^{2} x^{5}\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right )}{30 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/60*(210*c^3*x^6 + 140*b*c^2*x^4 - 28*b^2*c*x^2 + 12*b^3 - 105*(c^3*x^7 + b*c^2*x^5)*sqrt(-c/b)*log((c*x^2
- 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)))/(b^4*c*x^7 + b^5*x^5), -1/30*(105*c^3*x^6 + 70*b*c^2*x^4 - 14*b^2*c*x^2
+ 6*b^3 + 105*(c^3*x^7 + b*c^2*x^5)*sqrt(c/b)*arctan(x*sqrt(c/b)))/(b^4*c*x^7 + b^5*x^5)]

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giac [A]  time = 0.18, size = 70, normalized size = 0.86 \begin {gather*} -\frac {7 \, c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{4}} - \frac {c^{3} x}{2 \, {\left (c x^{2} + b\right )} b^{4}} - \frac {45 \, c^{2} x^{4} - 10 \, b c x^{2} + 3 \, b^{2}}{15 \, b^{4} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/2*c^3*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^4) - 1/2*c^3*x/((c*x^2 + b)*b^4) - 1/15*(45*c^2*x^4 - 10*b*c*x^2 +
3*b^2)/(b^4*x^5)

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maple [A]  time = 0.01, size = 70, normalized size = 0.86 \begin {gather*} -\frac {c^{3} x}{2 \left (c \,x^{2}+b \right ) b^{4}}-\frac {7 c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}\, b^{4}}-\frac {3 c^{2}}{b^{4} x}+\frac {2 c}{3 b^{3} x^{3}}-\frac {1}{5 b^{2} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.03, size = 75, normalized size = 0.93 \begin {gather*} -\frac {105 \, c^{3} x^{6} + 70 \, b c^{2} x^{4} - 14 \, b^{2} c x^{2} + 6 \, b^{3}}{30 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}} - \frac {7 \, c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(105*c^3*x^6 + 70*b*c^2*x^4 - 14*b^2*c*x^2 + 6*b^3)/(b^4*c*x^7 + b^5*x^5) - 7/2*c^3*arctan(c*x/sqrt(b*c)
)/(sqrt(b*c)*b^4)

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mupad [B]  time = 4.28, size = 70, normalized size = 0.86 \begin {gather*} -\frac {\frac {1}{5\,b}-\frac {7\,c\,x^2}{15\,b^2}+\frac {7\,c^2\,x^4}{3\,b^3}+\frac {7\,c^3\,x^6}{2\,b^4}}{c\,x^7+b\,x^5}-\frac {7\,c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{2\,b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(b*x^2 + c*x^4)^2),x)

[Out]

- (1/(5*b) - (7*c*x^2)/(15*b^2) + (7*c^2*x^4)/(3*b^3) + (7*c^3*x^6)/(2*b^4))/(b*x^5 + c*x^7) - (7*c^(5/2)*atan
((c^(1/2)*x)/b^(1/2)))/(2*b^(9/2))

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sympy [A]  time = 0.43, size = 126, normalized size = 1.56 \begin {gather*} \frac {7 \sqrt {- \frac {c^{5}}{b^{9}}} \log {\left (- \frac {b^{5} \sqrt {- \frac {c^{5}}{b^{9}}}}{c^{3}} + x \right )}}{4} - \frac {7 \sqrt {- \frac {c^{5}}{b^{9}}} \log {\left (\frac {b^{5} \sqrt {- \frac {c^{5}}{b^{9}}}}{c^{3}} + x \right )}}{4} + \frac {- 6 b^{3} + 14 b^{2} c x^{2} - 70 b c^{2} x^{4} - 105 c^{3} x^{6}}{30 b^{5} x^{5} + 30 b^{4} c x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7*sqrt(-c**5/b**9)*log(-b**5*sqrt(-c**5/b**9)/c**3 + x)/4 - 7*sqrt(-c**5/b**9)*log(b**5*sqrt(-c**5/b**9)/c**3
+ x)/4 + (-6*b**3 + 14*b**2*c*x**2 - 70*b*c**2*x**4 - 105*c**3*x**6)/(30*b**5*x**5 + 30*b**4*c*x**7)

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