3.35.45 \(\int \frac {2-i \pi -x^2-\log (-4+3 e^5)}{-2 x+10 x^2-x^3+x (i \pi +\log (-4+3 e^5))} \, dx\)

Optimal. Leaf size=25 \[ \log \left (10-x+\frac {-2+i \pi +\log \left (-4+3 e^5\right )}{x}\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 36, normalized size of antiderivative = 1.44, number of steps used = 5, number of rules used = 4, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {6, 1594, 1628, 628} \begin {gather*} -\log (x)+\log \left (i x^2-10 i x+\pi +i \left (2-\log \left (3 e^5-4\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - I*Pi - x^2 - Log[-4 + 3*E^5])/(-2*x + 10*x^2 - x^3 + x*(I*Pi + Log[-4 + 3*E^5])),x]

[Out]

-Log[x] + Log[Pi - (10*I)*x + I*x^2 + I*(2 - Log[-4 + 3*E^5])]

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 628

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

Rule 1594

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

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2-i \pi -x^2-\log \left (-4+3 e^5\right )}{10 x^2-x^3+x \left (-2+i \pi +\log \left (-4+3 e^5\right )\right )} \, dx\\ &=\int \frac {2-i \pi -x^2-\log \left (-4+3 e^5\right )}{x \left (-2+i \pi +10 x-x^2+\log \left (-4+3 e^5\right )\right )} \, dx\\ &=\int \left (-\frac {1}{x}+\frac {2 i (-5+x)}{\pi -10 i x+i x^2+i \left (2-\log \left (-4+3 e^5\right )\right )}\right ) \, dx\\ &=-\log (x)+2 i \int \frac {-5+x}{\pi -10 i x+i x^2+i \left (2-\log \left (-4+3 e^5\right )\right )} \, dx\\ &=-\log (x)+\log \left (\pi -10 i x+i x^2+i \left (2-\log \left (-4+3 e^5\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.02, size = 102, normalized size = 4.08 \begin {gather*} -i \tan ^{-1}\left (\frac {\pi }{2-10 x+x^2-\log \left (-4+3 e^5\right )}\right )-\log (x)+\frac {1}{2} \log \left (4+\pi ^2-40 x+104 x^2-20 x^3+x^4-4 \log \left (-4+3 e^5\right )+20 x \log \left (-4+3 e^5\right )-2 x^2 \log \left (-4+3 e^5\right )+\log ^2\left (-4+3 e^5\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - I*Pi - x^2 - Log[-4 + 3*E^5])/(-2*x + 10*x^2 - x^3 + x*(I*Pi + Log[-4 + 3*E^5])),x]

[Out]

(-I)*ArcTan[Pi/(2 - 10*x + x^2 - Log[-4 + 3*E^5])] - Log[x] + Log[4 + Pi^2 - 40*x + 104*x^2 - 20*x^3 + x^4 - 4
*Log[-4 + 3*E^5] + 20*x*Log[-4 + 3*E^5] - 2*x^2*Log[-4 + 3*E^5] + Log[-4 + 3*E^5]^2]/2

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fricas [A]  time = 0.47, size = 23, normalized size = 0.92 \begin {gather*} \log \left (x^{2} - 10 \, x - \log \left (-3 \, e^{5} + 4\right ) + 2\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(-3*exp(5)+4)-x^2+2)/(x*log(-3*exp(5)+4)-x^3+10*x^2-2*x),x, algorithm="fricas")

[Out]

log(x^2 - 10*x - log(-3*e^5 + 4) + 2) - log(x)

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giac [A]  time = 0.15, size = 24, normalized size = 0.96 \begin {gather*} \log \left (x^{2} - 10 \, x - \log \left (-3 \, e^{5} + 4\right ) + 2\right ) - \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(-3*exp(5)+4)-x^2+2)/(x*log(-3*exp(5)+4)-x^3+10*x^2-2*x),x, algorithm="giac")

[Out]

log(x^2 - 10*x - log(-3*e^5 + 4) + 2) - log(abs(x))

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maple [A]  time = 0.08, size = 24, normalized size = 0.96

method	result	size
default	\(-\ln \relax (x )+\ln \left (x^{2}-\ln \left (-3 \,{\mathrm e}^{5}+4\right )-10 x +2\right )\)	\(24\)
norman	\(-\ln \relax (x )+\ln \left (-x^{2}+\ln \left (-3 \,{\mathrm e}^{5}+4\right )+10 x -2\right )\)	\(24\)
risch	\(-\ln \left (-x \right )+\ln \left (x^{2}-\ln \left (-3 \,{\mathrm e}^{5}+4\right )-10 x +2\right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(-3*exp(5)+4)-x^2+2)/(x*ln(-3*exp(5)+4)-x^3+10*x^2-2*x),x,method=_RETURNVERBOSE)

[Out]

-ln(x)+ln(x^2-ln(-3*exp(5)+4)-10*x+2)

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maxima [A]  time = 0.40, size = 23, normalized size = 0.92 \begin {gather*} \log \left (x^{2} - 10 \, x - \log \left (-3 \, e^{5} + 4\right ) + 2\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(-3*exp(5)+4)-x^2+2)/(x*log(-3*exp(5)+4)-x^3+10*x^2-2*x),x, algorithm="maxima")

[Out]

log(x^2 - 10*x - log(-3*e^5 + 4) + 2) - log(x)

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mupad [B]  time = 0.20, size = 23, normalized size = 0.92 \begin {gather*} \ln \left (x^2-10\,x-\ln \left (4-3\,{\mathrm {e}}^5\right )+2\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(4 - 3*exp(5)) + x^2 - 2)/(2*x - x*log(4 - 3*exp(5)) - 10*x^2 + x^3),x)

[Out]

log(x^2 - log(4 - 3*exp(5)) - 10*x + 2) - log(x)

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sympy [A]  time = 1.16, size = 24, normalized size = 0.96 \begin {gather*} - \log {\relax (x )} + \log {\left (x^{2} - 10 x - \log {\left (-4 + 3 e^{5} \right )} + 2 - i \pi \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(-3*exp(5)+4)-x**2+2)/(x*ln(-3*exp(5)+4)-x**3+10*x**2-2*x),x)

[Out]

-log(x) + log(x**2 - 10*x - log(-4 + 3*exp(5)) + 2 - I*pi)

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