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

Optimal. Leaf size=58 \[ -\frac {c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{7/2}}-\frac {c^2}{b^3 x}+\frac {c}{3 b^2 x^3}-\frac {1}{5 b x^5} \]

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Rubi [A]  time = 0.04, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1584, 325, 205} \begin {gather*} -\frac {c^2}{b^3 x}-\frac {c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{7/2}}+\frac {c}{3 b^2 x^3}-\frac {1}{5 b x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(b*x^2 + c*x^4)),x]

[Out]

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

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Mathematica [A]  time = 0.03, size = 58, normalized size = 1.00 \begin {gather*} -\frac {c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{7/2}}-\frac {c^2}{b^3 x}+\frac {c}{3 b^2 x^3}-\frac {1}{5 b x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(b*x^2 + c*x^4)),x]

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 132, normalized size = 2.28 \begin {gather*} \left [\frac {15 \, c^{2} x^{5} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} - 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right ) - 30 \, c^{2} x^{4} + 10 \, b c x^{2} - 6 \, b^{2}}{30 \, b^{3} x^{5}}, -\frac {15 \, c^{2} x^{5} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right ) + 15 \, c^{2} x^{4} - 5 \, b c x^{2} + 3 \, b^{2}}{15 \, b^{3} x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 52, normalized size = 0.90 \begin {gather*} -\frac {c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} b^{3}} - \frac {15 \, c^{2} x^{4} - 5 \, b c x^{2} + 3 \, b^{2}}{15 \, b^{3} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 52, normalized size = 0.90 \begin {gather*} -\frac {c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}\, b^{3}}-\frac {c^{2}}{b^{3} x}+\frac {c}{3 b^{2} x^{3}}-\frac {1}{5 b \,x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.89, size = 52, normalized size = 0.90 \begin {gather*} -\frac {c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} b^{3}} - \frac {15 \, c^{2} x^{4} - 5 \, b c x^{2} + 3 \, b^{2}}{15 \, b^{3} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 48, normalized size = 0.83 \begin {gather*} -\frac {\frac {1}{5\,b}-\frac {c\,x^2}{3\,b^2}+\frac {c^2\,x^4}{b^3}}{x^5}-\frac {c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{b^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(b*x^2 + c*x^4)),x)

[Out]

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

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sympy [B]  time = 0.28, size = 100, normalized size = 1.72 \begin {gather*} \frac {\sqrt {- \frac {c^{5}}{b^{7}}} \log {\left (- \frac {b^{4} \sqrt {- \frac {c^{5}}{b^{7}}}}{c^{3}} + x \right )}}{2} - \frac {\sqrt {- \frac {c^{5}}{b^{7}}} \log {\left (\frac {b^{4} \sqrt {- \frac {c^{5}}{b^{7}}}}{c^{3}} + x \right )}}{2} + \frac {- 3 b^{2} + 5 b c x^{2} - 15 c^{2} x^{4}}{15 b^{3} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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