3.40.67 \(\int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx\)

Optimal. Leaf size=25 \[ \frac {e^x}{x+\frac {2 x}{4-\log (\log (5) (4+\log (25)))}} \]

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Rubi [A]  time = 0.07, antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 7, number of rules used = 5, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {6, 12, 14, 2177, 2178} \begin {gather*} \frac {e^x (4-\log (\log (5) (4+\log (25))))}{x (6-\log (\log (5) (4+\log (25))))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(4 - 4*x) + E^x*(-1 + x)*Log[4*Log[5] + Log[5]*Log[25]])/(-6*x^2 + x^2*Log[4*Log[5] + Log[5]*Log[25]]
),x]

[Out]

(E^x*(4 - Log[Log[5]*(4 + Log[25])]))/(x*(6 - Log[Log[5]*(4 + Log[25])]))

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{x^2 (-6+\log (4 \log (5)+\log (5) \log (25)))} \, dx\\ &=\frac {\int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{x^2} \, dx}{-6+\log (4 \log (5)+\log (5) \log (25))}\\ &=\frac {\int \left (\frac {4 e^x \left (1-\frac {1}{4} \log (\log (5) (4+\log (25)))\right )}{x^2}-\frac {4 e^x \left (1-\frac {1}{4} \log (\log (5) (4+\log (25)))\right )}{x}\right ) \, dx}{-6+\log (4 \log (5)+\log (5) \log (25))}\\ &=-\frac {(4-\log (\log (5) (4+\log (25)))) \int \frac {e^x}{x^2} \, dx}{6-\log (\log (5) (4+\log (25)))}+\frac {(4-\log (\log (5) (4+\log (25)))) \int \frac {e^x}{x} \, dx}{6-\log (\log (5) (4+\log (25)))}\\ &=\frac {e^x (4-\log (\log (5) (4+\log (25))))}{x (6-\log (\log (5) (4+\log (25))))}+\frac {\text {Ei}(x) (4-\log (\log (5) (4+\log (25))))}{6-\log (\log (5) (4+\log (25)))}-\frac {(4-\log (\log (5) (4+\log (25)))) \int \frac {e^x}{x} \, dx}{6-\log (\log (5) (4+\log (25)))}\\ &=\frac {e^x (4-\log (\log (5) (4+\log (25))))}{x (6-\log (\log (5) (4+\log (25))))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 1.16 \begin {gather*} \frac {e^x (-4+\log (\log (5) (4+\log (25))))}{x (-6+\log (\log (5) (4+\log (25))))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(4 - 4*x) + E^x*(-1 + x)*Log[4*Log[5] + Log[5]*Log[25]])/(-6*x^2 + x^2*Log[4*Log[5] + Log[5]*Lo
g[25]]),x]

[Out]

(E^x*(-4 + Log[Log[5]*(4 + Log[25])]))/(x*(-6 + Log[Log[5]*(4 + Log[25])]))

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fricas [A]  time = 0.97, size = 41, normalized size = 1.64 \begin {gather*} \frac {e^{x} \log \left (2 \, \log \relax (5)^{2} + 4 \, \log \relax (5)\right ) - 4 \, e^{x}}{x \log \left (2 \, \log \relax (5)^{2} + 4 \, \log \relax (5)\right ) - 6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-1)*exp(x)*log(2*log(5)^2+4*log(5))+(-4*x+4)*exp(x))/(x^2*log(2*log(5)^2+4*log(5))-6*x^2),x, algo
rithm="fricas")

[Out]

(e^x*log(2*log(5)^2 + 4*log(5)) - 4*e^x)/(x*log(2*log(5)^2 + 4*log(5)) - 6*x)

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giac [A]  time = 0.20, size = 41, normalized size = 1.64 \begin {gather*} \frac {e^{x} \log \left (2 \, \log \relax (5)^{2} + 4 \, \log \relax (5)\right ) - 4 \, e^{x}}{x \log \left (2 \, \log \relax (5)^{2} + 4 \, \log \relax (5)\right ) - 6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-1)*exp(x)*log(2*log(5)^2+4*log(5))+(-4*x+4)*exp(x))/(x^2*log(2*log(5)^2+4*log(5))-6*x^2),x, algo
rithm="giac")

[Out]

(e^x*log(2*log(5)^2 + 4*log(5)) - 4*e^x)/(x*log(2*log(5)^2 + 4*log(5)) - 6*x)

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maple [A]  time = 0.14, size = 33, normalized size = 1.32

method	result	size
risch	\(\frac {\left (\ln \relax (2)+\ln \left (\ln \relax (5)\right )+\ln \left (2+\ln \relax (5)\right )-4\right ) {\mathrm e}^{x}}{x \left (\ln \relax (2)+\ln \left (\ln \relax (5)\right )+\ln \left (2+\ln \relax (5)\right )-6\right )}\)	\(33\)
gosper	\(\frac {{\mathrm e}^{x} \left (\ln \left (2 \ln \relax (5)^{2}+4 \ln \relax (5)\right )-4\right )}{x \left (\ln \left (2 \ln \relax (5)^{2}+4 \ln \relax (5)\right )-6\right )}\)	\(37\)
norman	\(\frac {\left (\ln \relax (2)+\ln \left (\ln \relax (5)^{2}+2 \ln \relax (5)\right )-4\right ) {\mathrm e}^{x}}{\left (\ln \relax (2)+\ln \left (\ln \relax (5)^{2}+2 \ln \relax (5)\right )-6\right ) x}\)	\(37\)
default	\(-\frac {4 \,{\mathrm e}^{x}}{x \left (\ln \relax (2)+\ln \left (\ln \relax (5)^{2}+2 \ln \relax (5)\right )-6\right )}+\frac {\ln \relax (2) {\mathrm e}^{x}}{\left (\ln \relax (2)+\ln \left (\ln \relax (5)^{2}+2 \ln \relax (5)\right )-6\right ) x}+\frac {\ln \left (\ln \relax (5)^{2}+2 \ln \relax (5)\right ) {\mathrm e}^{x}}{\left (\ln \relax (2)+\ln \left (\ln \relax (5)^{2}+2 \ln \relax (5)\right )-6\right ) x}\)	\(81\)
meijerg	\(\frac {\left (\ln \left (2 \ln \relax (5)^{2}+4 \ln \relax (5)\right )-4\right ) \left (\ln \relax (x )+i \pi -\ln \left (-x \right )-\expIntegralEi \left (1, -x \right )\right )}{\ln \left (2 \ln \relax (5)^{2}+4 \ln \relax (5)\right )-6}-\frac {\left (-\ln \left (2 \ln \relax (5)^{2}+4 \ln \relax (5)\right )+4\right ) \left (\frac {1}{x}+1-\ln \relax (x )-i \pi -\frac {2 x +2}{2 x}+\frac {{\mathrm e}^{x}}{x}+\ln \left (-x \right )+\expIntegralEi \left (1, -x \right )\right )}{\ln \left (2 \ln \relax (5)^{2}+4 \ln \relax (5)\right )-6}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x-1)*exp(x)*ln(2*ln(5)^2+4*ln(5))+(-4*x+4)*exp(x))/(x^2*ln(2*ln(5)^2+4*ln(5))-6*x^2),x,method=_RETURNVER
BOSE)

[Out]

(ln(2)+ln(ln(5))+ln(2+ln(5))-4)/x/(ln(2)+ln(ln(5))+ln(2+ln(5))-6)*exp(x)

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maxima [C]  time = 0.84, size = 98, normalized size = 3.92 \begin {gather*} \frac {{\rm Ei}\relax (x) \log \left (2 \, \log \relax (5)^{2} + 4 \, \log \relax (5)\right )}{\log \left (2 \, {\left (\log \relax (5) + 2\right )} \log \relax (5)\right ) - 6} - \frac {\Gamma \left (-1, -x\right ) \log \left (2 \, \log \relax (5)^{2} + 4 \, \log \relax (5)\right )}{\log \left (2 \, {\left (\log \relax (5) + 2\right )} \log \relax (5)\right ) - 6} - \frac {4 \, {\rm Ei}\relax (x)}{\log \left (2 \, {\left (\log \relax (5) + 2\right )} \log \relax (5)\right ) - 6} + \frac {4 \, \Gamma \left (-1, -x\right )}{\log \left (2 \, {\left (\log \relax (5) + 2\right )} \log \relax (5)\right ) - 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-1)*exp(x)*log(2*log(5)^2+4*log(5))+(-4*x+4)*exp(x))/(x^2*log(2*log(5)^2+4*log(5))-6*x^2),x, algo
rithm="maxima")

[Out]

Ei(x)*log(2*log(5)^2 + 4*log(5))/(log(2*(log(5) + 2)*log(5)) - 6) - gamma(-1, -x)*log(2*log(5)^2 + 4*log(5))/(
log(2*(log(5) + 2)*log(5)) - 6) - 4*Ei(x)/(log(2*(log(5) + 2)*log(5)) - 6) + 4*gamma(-1, -x)/(log(2*(log(5) +
2)*log(5)) - 6)

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mupad [B]  time = 0.10, size = 32, normalized size = 1.28 \begin {gather*} \frac {{\mathrm {e}}^x\,\left (\ln \left (\ln \left (625\right )+2\,{\ln \relax (5)}^2\right )-4\right )}{x\,\left (\ln \left (\ln \left (625\right )+2\,{\ln \relax (5)}^2\right )-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(4*x - 4) - exp(x)*log(4*log(5) + 2*log(5)^2)*(x - 1))/(6*x^2 - x^2*log(4*log(5) + 2*log(5)^2)),x)

[Out]

(exp(x)*(log(log(625) + 2*log(5)^2) - 4))/(x*(log(log(625) + 2*log(5)^2) - 6))

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sympy [B]  time = 0.17, size = 42, normalized size = 1.68 \begin {gather*} \frac {\left (-4 + \log {\left (\log {\relax (5 )} \right )} + \log {\relax (2 )} + \log {\left (\log {\relax (5 )} + 2 \right )}\right ) e^{x}}{- 6 x + x \log {\left (\log {\relax (5 )} \right )} + x \log {\relax (2 )} + x \log {\left (\log {\relax (5 )} + 2 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-1)*exp(x)*ln(2*ln(5)**2+4*ln(5))+(-4*x+4)*exp(x))/(x**2*ln(2*ln(5)**2+4*ln(5))-6*x**2),x)

[Out]

(-4 + log(log(5)) + log(2) + log(log(5) + 2))*exp(x)/(-6*x + x*log(log(5)) + x*log(2) + x*log(log(5) + 2))

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