3.8.98 \(\int \frac {40-10 x-10 x \log (\frac {i \pi +\log (-8+2 e^2)}{x})+5 x \log ^2(\frac {i \pi +\log (-8+2 e^2)}{x})}{x \log ^2(\frac {i \pi +\log (-8+2 e^2)}{x})} \, dx\)

Optimal. Leaf size=36 \[ 20+5 \left (x+\frac {2 (4-x)}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}\right ) \]

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Rubi [A]  time = 0.44, antiderivative size = 55, normalized size of antiderivative = 1.53, number of steps used = 13, number of rules used = 9, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6741, 12, 6742, 2353, 2297, 2299, 2178, 2302, 30} \begin {gather*} 5 x-\frac {10 x}{\log \left (\frac {\log \left (-2 \left (4-e^2\right )\right )+i \pi }{x}\right )}+\frac {40}{\log \left (\frac {\log \left (-2 \left (4-e^2\right )\right )+i \pi }{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(40 - 10*x - 10*x*Log[(I*Pi + Log[-8 + 2*E^2])/x] + 5*x*Log[(I*Pi + Log[-8 + 2*E^2])/x]^2)/(x*Log[(I*Pi +
Log[-8 + 2*E^2])/x]^2),x]

[Out]

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2299

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[1/(n*c^(1/n)), Subst[Int[E^(x/n)*(a + b*x)^p
, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[1/n]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 \left (8-2 x-2 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx\\ &=5 \int \frac {8-2 x-2 x \log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )+x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx\\ &=5 \int \left (1-\frac {2 (-4+x)}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}-\frac {2}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}\right ) \, dx\\ &=5 x-10 \int \frac {-4+x}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx-10 \int \frac {1}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx\\ &=5 x-10 \int \left (\frac {1}{\log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}-\frac {4}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}\right ) \, dx+\left (10 \left (i \pi +\log \left (-2 \left (4-e^2\right )\right )\right )\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )\right )\\ &=5 x+10 \text {Ei}\left (-\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )\right ) \left (i \pi +\log \left (-2 \left (4-e^2\right )\right )\right )-10 \int \frac {1}{\log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx+40 \int \frac {1}{x \log ^2\left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx\\ &=5 x+10 \text {Ei}\left (-\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )\right ) \left (i \pi +\log \left (-2 \left (4-e^2\right )\right )\right )-\frac {10 x}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}+10 \int \frac {1}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \, dx-40 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )\right )\\ &=5 x+10 \text {Ei}\left (-\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )\right ) \left (i \pi +\log \left (-2 \left (4-e^2\right )\right )\right )+\frac {40}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}-\frac {10 x}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}-\left (10 \left (i \pi +\log \left (-2 \left (4-e^2\right )\right )\right )\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )\right )\\ &=5 x+\frac {40}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}-\frac {10 x}{\log \left (\frac {i \pi +\log \left (-2 \left (4-e^2\right )\right )}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 51, normalized size = 1.42 \begin {gather*} 5 x+\frac {40}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )}-\frac {10 x}{\log \left (\frac {i \pi +\log \left (-8+2 e^2\right )}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(40 - 10*x - 10*x*Log[(I*Pi + Log[-8 + 2*E^2])/x] + 5*x*Log[(I*Pi + Log[-8 + 2*E^2])/x]^2)/(x*Log[(I
*Pi + Log[-8 + 2*E^2])/x]^2),x]

[Out]

5*x + 40/Log[(I*Pi + Log[-8 + 2*E^2])/x] - (10*x)/Log[(I*Pi + Log[-8 + 2*E^2])/x]

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fricas [A]  time = 0.60, size = 35, normalized size = 0.97 \begin {gather*} \frac {5 \, {\left (x \log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right ) - 2 \, x + 8\right )}}{\log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*log(log(-2*exp(2)+8)/x)^2-10*x*log(log(-2*exp(2)+8)/x)-10*x+40)/x/log(log(-2*exp(2)+8)/x)^2,x,
algorithm="fricas")

[Out]

5*(x*log(log(-2*e^2 + 8)/x) - 2*x + 8)/log(log(-2*e^2 + 8)/x)

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giac [B]  time = 0.69, size = 74, normalized size = 2.06 \begin {gather*} \frac {5 \, {\left (\log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right ) \log \left (-2 \, e^{2} + 8\right )^{3} - 2 \, \log \left (-2 \, e^{2} + 8\right )^{3} + \frac {8 \, \log \left (-2 \, e^{2} + 8\right )^{3}}{x}\right )} x}{\log \left (\frac {\log \left (-2 \, e^{2} + 8\right )}{x}\right ) \log \left (-2 \, e^{2} + 8\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*log(log(-2*exp(2)+8)/x)^2-10*x*log(log(-2*exp(2)+8)/x)-10*x+40)/x/log(log(-2*exp(2)+8)/x)^2,x,
algorithm="giac")

[Out]

5*(log(log(-2*e^2 + 8)/x)*log(-2*e^2 + 8)^3 - 2*log(-2*e^2 + 8)^3 + 8*log(-2*e^2 + 8)^3/x)*x/(log(log(-2*e^2 +
8)/x)*log(-2*e^2 + 8)^3)

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maple [A]  time = 0.18, size = 27, normalized size = 0.75

method	result	size
risch	\(5 x -\frac {10 \left (x -4\right )}{\ln \left (\frac {\ln \relax (2)+\ln \left (-{\mathrm e}^{2}+4\right )}{x}\right )}\)	\(27\)
norman	\(\frac {40-10 x +5 x \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}{\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}\)	\(36\)
derivativedivides	\(\frac {5 \ln \relax (2) x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right )}+\frac {5 \ln \left (-{\mathrm e}^{2}+4\right ) x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right )}-10 \ln \relax (2) \expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )-10 \ln \left (-{\mathrm e}^{2}+4\right ) \expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )+10 \ln \relax (2) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+10 \ln \left (-{\mathrm e}^{2}+4\right ) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+\frac {40}{\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}\)	\(187\)
default	\(\frac {5 \ln \relax (2) x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right )}+\frac {5 \ln \left (-{\mathrm e}^{2}+4\right ) x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right )}-10 \ln \relax (2) \expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )-10 \ln \left (-{\mathrm e}^{2}+4\right ) \expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )+10 \ln \relax (2) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+10 \ln \left (-{\mathrm e}^{2}+4\right ) \left (-\frac {x}{\ln \left (-2 \,{\mathrm e}^{2}+8\right ) \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}+\expIntegralEi \left (1, \ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )\right )\right )+\frac {40}{\ln \left (\frac {\ln \left (-2 \,{\mathrm e}^{2}+8\right )}{x}\right )}\)	\(187\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x*ln(ln(-2*exp(2)+8)/x)^2-10*x*ln(ln(-2*exp(2)+8)/x)-10*x+40)/x/ln(ln(-2*exp(2)+8)/x)^2,x,method=_RETUR
NVERBOSE)

[Out]

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

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maxima [C]  time = 0.61, size = 33, normalized size = 0.92 \begin {gather*} 5 \, x - \frac {10 \, {\left (x - 4\right )}}{\log \left (i \, \pi + \log \relax (2) + \log \left (e + 2\right ) + \log \left (e - 2\right )\right ) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*log(log(-2*exp(2)+8)/x)^2-10*x*log(log(-2*exp(2)+8)/x)-10*x+40)/x/log(log(-2*exp(2)+8)/x)^2,x,
algorithm="maxima")

[Out]

5*x - 10*(x - 4)/(log(I*pi + log(2) + log(e + 2) + log(e - 2)) - log(x))

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mupad [B]  time = 0.67, size = 25, normalized size = 0.69 \begin {gather*} 5\,x-\frac {10\,x-40}{\ln \left (\frac {\ln \left (8-2\,{\mathrm {e}}^2\right )}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*x + 10*x*log(log(8 - 2*exp(2))/x) - 5*x*log(log(8 - 2*exp(2))/x)^2 - 40)/(x*log(log(8 - 2*exp(2))/x)^
2),x)

[Out]

5*x - (10*x - 40)/log(log(8 - 2*exp(2))/x)

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sympy [A]  time = 0.12, size = 27, normalized size = 0.75 \begin {gather*} 5 x + \frac {40 - 10 x}{\log {\left (\frac {\log {\relax (2 )}}{x} + \frac {\log {\left (-4 + e^{2} \right )}}{x} + \frac {i \pi }{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*ln(ln(-2*exp(2)+8)/x)**2-10*x*ln(ln(-2*exp(2)+8)/x)-10*x+40)/x/ln(ln(-2*exp(2)+8)/x)**2,x)

[Out]

5*x + (40 - 10*x)/log(log(2)/x + log(-4 + exp(2))/x + I*pi/x)

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