Integrand size = 131, antiderivative size = 26 \begin {dmath*} \int \frac {e^{\frac {e^{8 x} \left (-3+\log (2)+e^x \log (2)\right )-\log (2) \log (5)}{-3+\log (2)+e^x \log (2)}} \left (e^{8 x} \left (72-48 \log (2)+8 \log ^2(2)+8 e^{2 x} \log ^2(2)+e^x \left (-48 \log (2)+16 \log ^2(2)\right )\right )+e^x \log ^2(2) \log (5)\right )}{9-6 \log (2)+\log ^2(2)+e^{2 x} \log ^2(2)+e^x \left (-6 \log (2)+2 \log ^2(2)\right )} \, dx=e^{e^{8 x}+\frac {\log (5)}{-1-e^x+\frac {3}{\log (2)}}} \end {dmath*}
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {e^{\frac {e^{8 x} \left (-3+\log (2)+e^x \log (2)\right )-\log (2) \log (5)}{-3+\log (2)+e^x \log (2)}} \left (e^{8 x} \left (72-48 \log (2)+8 \log ^2(2)+8 e^{2 x} \log ^2(2)+e^x \left (-48 \log (2)+16 \log ^2(2)\right )\right )+e^x \log ^2(2) \log (5)\right )}{9-6 \log (2)+\log ^2(2)+e^{2 x} \log ^2(2)+e^x \left (-6 \log (2)+2 \log ^2(2)\right )} \, dx=e^{e^{8 x}-\frac {\log (2) \log (5)}{-3+\log (2)+e^x \log (2)}} \end {dmath*}
Integrate[(E^((E^(8*x)*(-3 + Log[2] + E^x*Log[2]) - Log[2]*Log[5])/(-3 + L og[2] + E^x*Log[2]))*(E^(8*x)*(72 - 48*Log[2] + 8*Log[2]^2 + 8*E^(2*x)*Log [2]^2 + E^x*(-48*Log[2] + 16*Log[2]^2)) + E^x*Log[2]^2*Log[5]))/(9 - 6*Log [2] + Log[2]^2 + E^(2*x)*Log[2]^2 + E^x*(-6*Log[2] + 2*Log[2]^2)),x]
Time = 0.80 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {2720, 2726}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (e^{8 x} \left (8 e^{2 x} \log ^2(2)+e^x \left (16 \log ^2(2)-48 \log (2)\right )+72+8 \log ^2(2)-48 \log (2)\right )+e^x \log ^2(2) \log (5)\right ) \exp \left (\frac {e^{8 x} \left (e^x \log (2)-3+\log (2)\right )-\log (2) \log (5)}{e^x \log (2)-3+\log (2)}\right )}{e^{2 x} \log ^2(2)+e^x \left (2 \log ^2(2)-6 \log (2)\right )+9+\log ^2(2)-6 \log (2)} \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \int \frac {e^{e^{8 x}} 2^{\frac {\log (5)}{-e^x \log (2)+3-\log (2)}} \left (8 e^{9 x} \log ^2(2)+8 e^{7 x} (3-\log (2))^2-16 e^{8 x} (3-\log (2)) \log (2)+\log ^2(2) \log (5)\right )}{\left (-e^x \log (2)+3-\log (2)\right )^2}de^x\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle e^{e^{8 x}} 2^{\frac {\log (5)}{-e^x \log (2)+3-\log (2)}}\) |
Int[(E^((E^(8*x)*(-3 + Log[2] + E^x*Log[2]) - Log[2]*Log[5])/(-3 + Log[2] + E^x*Log[2]))*(E^(8*x)*(72 - 48*Log[2] + 8*Log[2]^2 + 8*E^(2*x)*Log[2]^2 + E^x*(-48*Log[2] + 16*Log[2]^2)) + E^x*Log[2]^2*Log[5]))/(9 - 6*Log[2] + Log[2]^2 + E^(2*x)*Log[2]^2 + E^x*(-6*Log[2] + 2*Log[2]^2)),x]
3.11.57.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 9.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left ({\mathrm e}^{x} \ln \left (2\right )+\ln \left (2\right )-3\right ) {\mathrm e}^{8 x}-\ln \left (2\right ) \ln \left (5\right )}{{\mathrm e}^{x} \ln \left (2\right )+\ln \left (2\right )-3}}\) | \(37\) |
risch | \({\mathrm e}^{-\frac {\ln \left (2\right ) \ln \left (5\right )-\ln \left (2\right ) {\mathrm e}^{9 x}-\ln \left (2\right ) {\mathrm e}^{8 x}+3 \,{\mathrm e}^{8 x}}{{\mathrm e}^{x} \ln \left (2\right )+\ln \left (2\right )-3}}\) | \(43\) |
int(((8*ln(2)^2*exp(x)^2+(16*ln(2)^2-48*ln(2))*exp(x)+8*ln(2)^2-48*ln(2)+7 2)*exp(4*x)^2+ln(2)^2*ln(5)*exp(x))*exp(((exp(x)*ln(2)+ln(2)-3)*exp(4*x)^2 -ln(2)*ln(5))/(exp(x)*ln(2)+ln(2)-3))/(ln(2)^2*exp(x)^2+(2*ln(2)^2-6*ln(2) )*exp(x)+ln(2)^2-6*ln(2)+9),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \begin {dmath*} \int \frac {e^{\frac {e^{8 x} \left (-3+\log (2)+e^x \log (2)\right )-\log (2) \log (5)}{-3+\log (2)+e^x \log (2)}} \left (e^{8 x} \left (72-48 \log (2)+8 \log ^2(2)+8 e^{2 x} \log ^2(2)+e^x \left (-48 \log (2)+16 \log ^2(2)\right )\right )+e^x \log ^2(2) \log (5)\right )}{9-6 \log (2)+\log ^2(2)+e^{2 x} \log ^2(2)+e^x \left (-6 \log (2)+2 \log ^2(2)\right )} \, dx=e^{\left (\frac {{\left (\log \left (2\right ) - 3\right )} e^{\left (8 \, x\right )} + e^{\left (9 \, x\right )} \log \left (2\right ) - \log \left (5\right ) \log \left (2\right )}{e^{x} \log \left (2\right ) + \log \left (2\right ) - 3}\right )} \end {dmath*}
integrate(((8*log(2)^2*exp(x)^2+(16*log(2)^2-48*log(2))*exp(x)+8*log(2)^2- 48*log(2)+72)*exp(4*x)^2+log(2)^2*log(5)*exp(x))*exp(((exp(x)*log(2)+log(2 )-3)*exp(4*x)^2-log(2)*log(5))/(exp(x)*log(2)+log(2)-3))/(log(2)^2*exp(x)^ 2+(2*log(2)^2-6*log(2))*exp(x)+log(2)^2-6*log(2)+9),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \begin {dmath*} \int \frac {e^{\frac {e^{8 x} \left (-3+\log (2)+e^x \log (2)\right )-\log (2) \log (5)}{-3+\log (2)+e^x \log (2)}} \left (e^{8 x} \left (72-48 \log (2)+8 \log ^2(2)+8 e^{2 x} \log ^2(2)+e^x \left (-48 \log (2)+16 \log ^2(2)\right )\right )+e^x \log ^2(2) \log (5)\right )}{9-6 \log (2)+\log ^2(2)+e^{2 x} \log ^2(2)+e^x \left (-6 \log (2)+2 \log ^2(2)\right )} \, dx=e^{\frac {\left (e^{x} \log {\left (2 \right )} - 3 + \log {\left (2 \right )}\right ) e^{8 x} - \log {\left (2 \right )} \log {\left (5 \right )}}{e^{x} \log {\left (2 \right )} - 3 + \log {\left (2 \right )}}} \end {dmath*}
integrate(((8*ln(2)**2*exp(x)**2+(16*ln(2)**2-48*ln(2))*exp(x)+8*ln(2)**2- 48*ln(2)+72)*exp(4*x)**2+ln(2)**2*ln(5)*exp(x))*exp(((exp(x)*ln(2)+ln(2)-3 )*exp(4*x)**2-ln(2)*ln(5))/(exp(x)*ln(2)+ln(2)-3))/(ln(2)**2*exp(x)**2+(2* ln(2)**2-6*ln(2))*exp(x)+ln(2)**2-6*ln(2)+9),x)
Time = 1.61 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \begin {dmath*} \int \frac {e^{\frac {e^{8 x} \left (-3+\log (2)+e^x \log (2)\right )-\log (2) \log (5)}{-3+\log (2)+e^x \log (2)}} \left (e^{8 x} \left (72-48 \log (2)+8 \log ^2(2)+8 e^{2 x} \log ^2(2)+e^x \left (-48 \log (2)+16 \log ^2(2)\right )\right )+e^x \log ^2(2) \log (5)\right )}{9-6 \log (2)+\log ^2(2)+e^{2 x} \log ^2(2)+e^x \left (-6 \log (2)+2 \log ^2(2)\right )} \, dx=e^{\left (-\frac {\log \left (5\right ) \log \left (2\right )}{e^{x} \log \left (2\right ) + \log \left (2\right ) - 3} + e^{\left (8 \, x\right )}\right )} \end {dmath*}
integrate(((8*log(2)^2*exp(x)^2+(16*log(2)^2-48*log(2))*exp(x)+8*log(2)^2- 48*log(2)+72)*exp(4*x)^2+log(2)^2*log(5)*exp(x))*exp(((exp(x)*log(2)+log(2 )-3)*exp(4*x)^2-log(2)*log(5))/(exp(x)*log(2)+log(2)-3))/(log(2)^2*exp(x)^ 2+(2*log(2)^2-6*log(2))*exp(x)+log(2)^2-6*log(2)+9),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (23) = 46\).
Time = 0.44 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.77 \begin {dmath*} \int \frac {e^{\frac {e^{8 x} \left (-3+\log (2)+e^x \log (2)\right )-\log (2) \log (5)}{-3+\log (2)+e^x \log (2)}} \left (e^{8 x} \left (72-48 \log (2)+8 \log ^2(2)+8 e^{2 x} \log ^2(2)+e^x \left (-48 \log (2)+16 \log ^2(2)\right )\right )+e^x \log ^2(2) \log (5)\right )}{9-6 \log (2)+\log ^2(2)+e^{2 x} \log ^2(2)+e^x \left (-6 \log (2)+2 \log ^2(2)\right )} \, dx=e^{\left (\frac {e^{\left (9 \, x\right )} \log \left (2\right )}{e^{x} \log \left (2\right ) + \log \left (2\right ) - 3} + \frac {e^{\left (8 \, x\right )} \log \left (2\right )}{e^{x} \log \left (2\right ) + \log \left (2\right ) - 3} - \frac {\log \left (5\right ) \log \left (2\right )}{e^{x} \log \left (2\right ) + \log \left (2\right ) - 3} - \frac {3 \, e^{\left (8 \, x\right )}}{e^{x} \log \left (2\right ) + \log \left (2\right ) - 3}\right )} \end {dmath*}
integrate(((8*log(2)^2*exp(x)^2+(16*log(2)^2-48*log(2))*exp(x)+8*log(2)^2- 48*log(2)+72)*exp(4*x)^2+log(2)^2*log(5)*exp(x))*exp(((exp(x)*log(2)+log(2 )-3)*exp(4*x)^2-log(2)*log(5))/(exp(x)*log(2)+log(2)-3))/(log(2)^2*exp(x)^ 2+(2*log(2)^2-6*log(2))*exp(x)+log(2)^2-6*log(2)+9),x, algorithm=\
e^(e^(9*x)*log(2)/(e^x*log(2) + log(2) - 3) + e^(8*x)*log(2)/(e^x*log(2) + log(2) - 3) - log(5)*log(2)/(e^x*log(2) + log(2) - 3) - 3*e^(8*x)/(e^x*lo g(2) + log(2) - 3))
Time = 16.71 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81 \begin {dmath*} \int \frac {e^{\frac {e^{8 x} \left (-3+\log (2)+e^x \log (2)\right )-\log (2) \log (5)}{-3+\log (2)+e^x \log (2)}} \left (e^{8 x} \left (72-48 \log (2)+8 \log ^2(2)+8 e^{2 x} \log ^2(2)+e^x \left (-48 \log (2)+16 \log ^2(2)\right )\right )+e^x \log ^2(2) \log (5)\right )}{9-6 \log (2)+\log ^2(2)+e^{2 x} \log ^2(2)+e^x \left (-6 \log (2)+2 \log ^2(2)\right )} \, dx={\mathrm {e}}^{\frac {\ln \left (2\right )\,\left ({\mathrm {e}}^{8\,x}+{\mathrm {e}}^{9\,x}-\ln \left (5\right )\right )}{\ln \left (2\right )+{\mathrm {e}}^x\,\ln \left (2\right )-3}}\,{\mathrm {e}}^{-\frac {3\,{\mathrm {e}}^{8\,x}}{\ln \left (2\right )+{\mathrm {e}}^x\,\ln \left (2\right )-3}} \end {dmath*}
int((exp((exp(8*x)*(log(2) + exp(x)*log(2) - 3) - log(2)*log(5))/(log(2) + exp(x)*log(2) - 3))*(exp(8*x)*(8*exp(2*x)*log(2)^2 - 48*log(2) - exp(x)*( 48*log(2) - 16*log(2)^2) + 8*log(2)^2 + 72) + exp(x)*log(2)^2*log(5)))/(ex p(2*x)*log(2)^2 - 6*log(2) - exp(x)*(6*log(2) - 2*log(2)^2) + log(2)^2 + 9 ),x)