3.11.63 \(\int \frac {-30+x^2+\frac {x^2 (-60-2 x^2)}{e^2 (30-x^2)^2}}{-30 x+x^3} \, dx\)

Optimal. Leaf size=21 \[ e^{2 \left (-1+\log (x)-\log \left (30-x^2\right )\right )}+\log (x) \]

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Rubi [A]  time = 0.24, antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1593, 6688, 444, 34} \begin {gather*} \frac {x^2}{e^2 \left (30-x^2\right )^2}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-30 + x^2 + (x^2*(-60 - 2*x^2))/(E^2*(30 - x^2)^2))/(-30*x + x^3),x]

[Out]

x^2/(E^2*(30 - x^2)^2) + Log[x]

Rule 34

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_)), x_Symbol] :> Simp[(d*x*(a + b*x)^(m + 1))/(b*(m + 2)), x] /
; FreeQ[{a, b, c, d, m}, x] && EqQ[a*d - b*c*(m + 2), 0]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

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]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-30+x^2+\frac {x^2 \left (-60-2 x^2\right )}{e^2 \left (30-x^2\right )^2}}{x \left (-30+x^2\right )} \, dx\\ &=\int \left (\frac {1}{x}-\frac {2 x \left (30+x^2\right )}{e^2 \left (-30+x^2\right )^3}\right ) \, dx\\ &=\log (x)-\frac {2 \int \frac {x \left (30+x^2\right )}{\left (-30+x^2\right )^3} \, dx}{e^2}\\ &=\log (x)-\frac {\operatorname {Subst}\left (\int \frac {30+x}{(-30+x)^3} \, dx,x,x^2\right )}{e^2}\\ &=\frac {x^2}{e^2 \left (30-x^2\right )^2}+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 26, normalized size = 1.24 \begin {gather*} \frac {30}{e^2 \left (-30+x^2\right )^2}+\frac {1}{e^2 \left (-30+x^2\right )}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-30 + x^2 + (x^2*(-60 - 2*x^2))/(E^2*(30 - x^2)^2))/(-30*x + x^3),x]

[Out]

30/(E^2*(-30 + x^2)^2) + 1/(E^2*(-30 + x^2)) + Log[x]

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fricas [A]  time = 0.60, size = 34, normalized size = 1.62 \begin {gather*} \frac {{\left ({\left (x^{4} - 60 \, x^{2} + 900\right )} e^{2} \log \relax (x) + x^{2}\right )} e^{\left (-2\right )}}{x^{4} - 60 \, x^{2} + 900} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-60)*exp(2*log(x)-2*log(-x^2+30)-2)+x^2-30)/(x^3-30*x),x, algorithm="fricas")

[Out]

((x^4 - 60*x^2 + 900)*e^2*log(x) + x^2)*e^(-2)/(x^4 - 60*x^2 + 900)

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giac [A]  time = 0.34, size = 20, normalized size = 0.95 \begin {gather*} \frac {x^{2} e^{\left (-2\right )}}{{\left (x^{2} - 30\right )}^{2}} + \frac {1}{2} \, \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-60)*exp(2*log(x)-2*log(-x^2+30)-2)+x^2-30)/(x^3-30*x),x, algorithm="giac")

[Out]

x^2*e^(-2)/(x^2 - 30)^2 + 1/2*log(x^2)

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maple [A]  time = 0.20, size = 17, normalized size = 0.81

method	result	size
risch	\(\frac {{\mathrm e}^{-2} x^{2}}{\left (x^{2}-30\right )^{2}}+\ln \relax (x )\)	\(17\)
norman	\(\frac {{\mathrm e}^{-2} x^{2}}{\left (x^{2}-30\right )^{2}}+\ln \relax (x )\)	\(19\)
default	\(\ln \relax (x )-2 \,{\mathrm e}^{-2} \left (-\frac {15}{\left (x^{2}-30\right )^{2}}-\frac {1}{2 \left (x^{2}-30\right )}\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2-60)*exp(2*ln(x)-2*ln(-x^2+30)-2)+x^2-30)/(x^3-30*x),x,method=_RETURNVERBOSE)

[Out]

exp(-2)*x^2/(x^2-30)^2+ln(x)

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maxima [A]  time = 0.48, size = 27, normalized size = 1.29 \begin {gather*} \frac {x^{2}}{x^{4} e^{2} - 60 \, x^{2} e^{2} + 900 \, e^{2}} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-60)*exp(2*log(x)-2*log(-x^2+30)-2)+x^2-30)/(x^3-30*x),x, algorithm="maxima")

[Out]

x^2/(x^4*e^2 - 60*x^2*e^2 + 900*e^2) + log(x)

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mupad [B]  time = 0.14, size = 16, normalized size = 0.76 \begin {gather*} \ln \relax (x)+\frac {x^2\,{\mathrm {e}}^{-2}}{{\left (x^2-30\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*log(x) - 2*log(30 - x^2) - 2)*(2*x^2 + 60) - x^2 + 30)/(30*x - x^3),x)

[Out]

log(x) + (x^2*exp(-2))/(x^2 - 30)^2

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sympy [A]  time = 0.49, size = 26, normalized size = 1.24 \begin {gather*} \frac {x^{2}}{x^{4} e^{2} - 60 x^{2} e^{2} + 900 e^{2}} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2-60)*exp(2*ln(x)-2*ln(-x**2+30)-2)+x**2-30)/(x**3-30*x),x)

[Out]

x**2/(x**4*exp(2) - 60*x**2*exp(2) + 900*exp(2)) + log(x)

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