3.10.75 \(\int \frac {-6-3 e^x+e^{\frac {1}{3} (3 x+\log (2+e^x-\log (\frac {5}{2})))} (6+4 e^x-3 \log (\frac {5}{2}))+3 \log (\frac {5}{2})+(-6-3 e^x+3 \log (\frac {5}{2})) \log (x)}{6+3 e^x-3 \log (\frac {5}{2})} \, dx\)

Optimal. Leaf size=26 \[ 5+e^x \sqrt [3]{2+e^x-\log \left (\frac {5}{2}\right )}-x \log (x) \]

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Rubi [A]  time = 0.80, antiderivative size = 25, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 6, integrand size = 81, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {6741, 12, 6742, 2282, 34, 2295} \begin {gather*} e^x \sqrt [3]{e^x+2-\log \left (\frac {5}{2}\right )}-x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6 - 3*E^x + E^((3*x + Log[2 + E^x - Log[5/2]])/3)*(6 + 4*E^x - 3*Log[5/2]) + 3*Log[5/2] + (-6 - 3*E^x +
3*Log[5/2])*Log[x])/(6 + 3*E^x - 3*Log[5/2]),x]

[Out]

E^x*(2 + E^x - Log[5/2])^(1/3) - x*Log[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 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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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 {-3 e^x+e^{\frac {1}{3} \left (3 x+\log \left (2+e^x-\log \left (\frac {5}{2}\right )\right )\right )} \left (6+4 e^x-3 \log \left (\frac {5}{2}\right )\right )-6 \left (1-\frac {1}{2} \log \left (\frac {5}{2}\right )\right )+\left (-6-3 e^x+3 \log \left (\frac {5}{2}\right )\right ) \log (x)}{3 \left (e^x+2 \left (1-\frac {1}{2} \log \left (\frac {5}{2}\right )\right )\right )} \, dx\\ &=\frac {1}{3} \int \frac {-3 e^x+e^{\frac {1}{3} \left (3 x+\log \left (2+e^x-\log \left (\frac {5}{2}\right )\right )\right )} \left (6+4 e^x-3 \log \left (\frac {5}{2}\right )\right )-6 \left (1-\frac {1}{2} \log \left (\frac {5}{2}\right )\right )+\left (-6-3 e^x+3 \log \left (\frac {5}{2}\right )\right ) \log (x)}{e^x+2 \left (1-\frac {1}{2} \log \left (\frac {5}{2}\right )\right )} \, dx\\ &=\frac {1}{3} \int \left (\frac {e^x \left (4 e^x+6 \left (1-\frac {1}{2} \log \left (\frac {5}{2}\right )\right )\right )}{\left (e^x+2 \left (1-\frac {1}{2} \log \left (\frac {5}{2}\right )\right )\right )^{2/3}}-3 (1+\log (x))\right ) \, dx\\ &=\frac {1}{3} \int \frac {e^x \left (4 e^x+6 \left (1-\frac {1}{2} \log \left (\frac {5}{2}\right )\right )\right )}{\left (e^x+2 \left (1-\frac {1}{2} \log \left (\frac {5}{2}\right )\right )\right )^{2/3}} \, dx-\int (1+\log (x)) \, dx\\ &=-x+\frac {1}{3} \operatorname {Subst}\left (\int \frac {6+4 x-3 \log \left (\frac {5}{2}\right )}{\left (2+x-\log \left (\frac {5}{2}\right )\right )^{2/3}} \, dx,x,e^x\right )-\int \log (x) \, dx\\ &=e^x \sqrt [3]{2+e^x-\log \left (\frac {5}{2}\right )}-x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.43, size = 30, normalized size = 1.15 \begin {gather*} \frac {1}{3} \left (3 e^x \sqrt [3]{2+e^x-\log \left (\frac {5}{2}\right )}-3 x \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6 - 3*E^x + E^((3*x + Log[2 + E^x - Log[5/2]])/3)*(6 + 4*E^x - 3*Log[5/2]) + 3*Log[5/2] + (-6 - 3*
E^x + 3*Log[5/2])*Log[x])/(6 + 3*E^x - 3*Log[5/2]),x]

[Out]

(3*E^x*(2 + E^x - Log[5/2])^(1/3) - 3*x*Log[x])/3

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)+3*log(2/5)+6)*exp(1/3*log(exp(x)+log(2/5)+2)+x)+(-3*exp(x)-3*log(2/5)-6)*log(x)-3*exp(x)-
3*log(2/5)-6)/(3*exp(x)+3*log(2/5)+6),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (4 \, e^{x} + 3 \, \log \left (\frac {2}{5}\right ) + 6\right )} e^{\left (x + \frac {1}{3} \, \log \left (e^{x} + \log \left (\frac {2}{5}\right ) + 2\right )\right )} - 3 \, {\left (e^{x} + \log \left (\frac {2}{5}\right ) + 2\right )} \log \relax (x) - 3 \, e^{x} - 3 \, \log \left (\frac {2}{5}\right ) - 6}{3 \, {\left (e^{x} + \log \left (\frac {2}{5}\right ) + 2\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)+3*log(2/5)+6)*exp(1/3*log(exp(x)+log(2/5)+2)+x)+(-3*exp(x)-3*log(2/5)-6)*log(x)-3*exp(x)-
3*log(2/5)-6)/(3*exp(x)+3*log(2/5)+6),x, algorithm="giac")

[Out]

integrate(1/3*((4*e^x + 3*log(2/5) + 6)*e^(x + 1/3*log(e^x + log(2/5) + 2)) - 3*(e^x + log(2/5) + 2)*log(x) -
3*e^x - 3*log(2/5) - 6)/(e^x + log(2/5) + 2), x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (4 \,{\mathrm e}^{x}+3 \ln \left (\frac {2}{5}\right )+6\right ) {\mathrm e}^{\frac {\ln \left ({\mathrm e}^{x}+\ln \left (\frac {2}{5}\right )+2\right )}{3}+x}+\left (-3 \,{\mathrm e}^{x}-3 \ln \left (\frac {2}{5}\right )-6\right ) \ln \relax (x )-3 \,{\mathrm e}^{x}-3 \ln \left (\frac {2}{5}\right )-6}{3 \,{\mathrm e}^{x}+3 \ln \left (\frac {2}{5}\right )+6}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*exp(x)+3*ln(2/5)+6)*exp(1/3*ln(exp(x)+ln(2/5)+2)+x)+(-3*exp(x)-3*ln(2/5)-6)*ln(x)-3*exp(x)-3*ln(2/5)-6
)/(3*exp(x)+3*ln(2/5)+6),x)

[Out]

int(((4*exp(x)+3*ln(2/5)+6)*exp(1/3*ln(exp(x)+ln(2/5)+2)+x)+(-3*exp(x)-3*ln(2/5)-6)*ln(x)-3*exp(x)-3*ln(2/5)-6
)/(3*exp(x)+3*ln(2/5)+6),x)

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maxima [B]  time = 0.69, size = 123, normalized size = 4.73 \begin {gather*} -{\left (\frac {x}{\log \left (\frac {2}{5}\right ) + 2} - \frac {\log \left (e^{x} + \log \left (\frac {2}{5}\right ) + 2\right )}{\log \left (\frac {2}{5}\right ) + 2}\right )} \log \left (\frac {2}{5}\right ) - x \log \relax (x) + {\left (e^{x} + 3 \, \log \relax (5) - 3 \, \log \relax (2) - 6\right )} {\left (e^{x} - \log \relax (5) + \log \relax (2) + 2\right )}^{\frac {1}{3}} + 3 \, {\left (e^{x} - \log \relax (5) + \log \relax (2) + 2\right )}^{\frac {1}{3}} \log \left (\frac {2}{5}\right ) + x + 6 \, {\left (e^{x} - \log \relax (5) + \log \relax (2) + 2\right )}^{\frac {1}{3}} - \frac {2 \, x}{\log \left (\frac {2}{5}\right ) + 2} + \frac {2 \, \log \left (e^{x} + \log \left (\frac {2}{5}\right ) + 2\right )}{\log \left (\frac {2}{5}\right ) + 2} - \log \left (e^{x} + \log \left (\frac {2}{5}\right ) + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)+3*log(2/5)+6)*exp(1/3*log(exp(x)+log(2/5)+2)+x)+(-3*exp(x)-3*log(2/5)-6)*log(x)-3*exp(x)-
3*log(2/5)-6)/(3*exp(x)+3*log(2/5)+6),x, algorithm="maxima")

[Out]

-(x/(log(2/5) + 2) - log(e^x + log(2/5) + 2)/(log(2/5) + 2))*log(2/5) - x*log(x) + (e^x + 3*log(5) - 3*log(2)
- 6)*(e^x - log(5) + log(2) + 2)^(1/3) + 3*(e^x - log(5) + log(2) + 2)^(1/3)*log(2/5) + x + 6*(e^x - log(5) +
log(2) + 2)^(1/3) - 2*x/(log(2/5) + 2) + 2*log(e^x + log(2/5) + 2)/(log(2/5) + 2) - log(e^x + log(2/5) + 2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {3\,\ln \left (\frac {2}{5}\right )+3\,{\mathrm {e}}^x-{\mathrm {e}}^{x+\frac {\ln \left (\ln \left (\frac {2}{5}\right )+{\mathrm {e}}^x+2\right )}{3}}\,\left (3\,\ln \left (\frac {2}{5}\right )+4\,{\mathrm {e}}^x+6\right )+\ln \relax (x)\,\left (3\,\ln \left (\frac {2}{5}\right )+3\,{\mathrm {e}}^x+6\right )+6}{3\,\ln \left (\frac {2}{5}\right )+3\,{\mathrm {e}}^x+6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*log(2/5) + 3*exp(x) - exp(x + log(log(2/5) + exp(x) + 2)/3)*(3*log(2/5) + 4*exp(x) + 6) + log(x)*(3*lo
g(2/5) + 3*exp(x) + 6) + 6)/(3*log(2/5) + 3*exp(x) + 6),x)

[Out]

int(-(3*log(2/5) + 3*exp(x) - exp(x + log(log(2/5) + exp(x) + 2)/3)*(3*log(2/5) + 4*exp(x) + 6) + log(x)*(3*lo
g(2/5) + 3*exp(x) + 6) + 6)/(3*log(2/5) + 3*exp(x) + 6), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)+3*ln(2/5)+6)*exp(1/3*ln(exp(x)+ln(2/5)+2)+x)+(-3*exp(x)-3*ln(2/5)-6)*ln(x)-3*exp(x)-3*ln(
2/5)-6)/(3*exp(x)+3*ln(2/5)+6),x)

[Out]

Timed out

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