3.85.34 \(\int \frac {-5-i \pi -\log (4)}{x^2+e^{5+e^5} x^2} \, dx\)

Optimal. Leaf size=23 \[ \frac {5+i \pi +\log (4)}{x+e^{5+e^5} x} \]

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6, 12, 30} \begin {gather*} \frac {5+i \pi +\log (4)}{\left (1+e^{5+e^5}\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 - I*Pi - Log[4])/(x^2 + E^(5 + E^5)*x^2),x]

[Out]

(5 + I*Pi + Log[4])/((1 + E^(5 + E^5))*x)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5-i \pi -\log (4)}{\left (1+e^{5+e^5}\right ) x^2} \, dx\\ &=-\frac {(5+i \pi +\log (4)) \int \frac {1}{x^2} \, dx}{1+e^{5+e^5}}\\ &=\frac {5+i \pi +\log (4)}{\left (1+e^{5+e^5}\right ) x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 1.04 \begin {gather*} \frac {5+i \pi +\log (4)}{\left (1+e^{5+e^5}\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 - I*Pi - Log[4])/(x^2 + E^(5 + E^5)*x^2),x]

[Out]

(5 + I*Pi + Log[4])/((1 + E^(5 + E^5))*x)

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fricas [A]  time = 0.70, size = 21, normalized size = 0.91 \begin {gather*} \frac {i \, \pi + 2 \, \log \relax (2) + 5}{x e^{\left (e^{5} + 5\right )} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(2)-I*pi-5)/(x*exp(5)*exp(log(x)+exp(5))+x^2),x, algorithm="fricas")

[Out]

(I*pi + 2*log(2) + 5)/(x*e^(e^5 + 5) + x)

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giac [A]  time = 0.16, size = 22, normalized size = 0.96 \begin {gather*} \frac {i \, \pi + 2 \, \log \relax (2) + 5}{x {\left (e^{\left (e^{5} + 5\right )} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(2)-I*pi-5)/(x*exp(5)*exp(log(x)+exp(5))+x^2),x, algorithm="giac")

[Out]

(I*pi + 2*log(2) + 5)/(x*(e^(e^5 + 5) + 1))

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maple [A]  time = 0.05, size = 25, normalized size = 1.09

method	result	size
gosper	\(\frac {2 \ln \relax (2)+i \pi +5}{{\mathrm e}^{\ln \relax (x )+{\mathrm e}^{5}} {\mathrm e}^{5}+x}\)	\(25\)
default	\(-\frac {-2 \ln \relax (2)-i \pi -5}{\left ({\mathrm e}^{{\mathrm e}^{5}+5}+1\right ) x}\)	\(25\)
norman	\(\frac {2 \ln \relax (2)+i \pi +5}{\left ({\mathrm e}^{5} {\mathrm e}^{{\mathrm e}^{5}}+1\right ) x}\)	\(25\)
risch	\(\frac {i \pi }{\left ({\mathrm e}^{{\mathrm e}^{5}+5}+1\right ) x}+\frac {2 \ln \relax (2)}{\left ({\mathrm e}^{{\mathrm e}^{5}+5}+1\right ) x}+\frac {5}{\left ({\mathrm e}^{{\mathrm e}^{5}+5}+1\right ) x}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*ln(2)-I*Pi-5)/(x*exp(5)*exp(ln(x)+exp(5))+x^2),x,method=_RETURNVERBOSE)

[Out]

(2*ln(2)+I*Pi+5)/(exp(ln(x)+exp(5))*exp(5)+x)

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maxima [A]  time = 0.35, size = 23, normalized size = 1.00 \begin {gather*} -\frac {-i \, \pi - 2 \, \log \relax (2) - 5}{x {\left (e^{\left (e^{5} + 5\right )} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(2)-I*pi-5)/(x*exp(5)*exp(log(x)+exp(5))+x^2),x, algorithm="maxima")

[Out]

-(-I*pi - 2*log(2) - 5)/(x*(e^(e^5 + 5) + 1))

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mupad [B]  time = 0.15, size = 21, normalized size = 0.91 \begin {gather*} \frac {\ln \relax (4)+5+\Pi \,1{}\mathrm {i}}{x\,\left ({\mathrm {e}}^{{\mathrm {e}}^5+5}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(Pi*1i + 2*log(2) + 5)/(x^2 + x*exp(exp(5) + log(x))*exp(5)),x)

[Out]

(Pi*1i + log(4) + 5)/(x*(exp(exp(5) + 5) + 1))

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sympy [A]  time = 0.08, size = 24, normalized size = 1.04 \begin {gather*} - \frac {2 \log {\relax (2 )} + 5 + i \pi }{x \left (- e^{5} e^{e^{5}} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*ln(2)-I*pi-5)/(x*exp(5)*exp(ln(x)+exp(5))+x**2),x)

[Out]

-(2*log(2) + 5 + I*pi)/(x*(-exp(5)*exp(exp(5)) - 1))

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