Integrand size = 106, antiderivative size = 22 \[ \int -\frac {256 e^{e^3+2 x+e \log (5)}}{-243+810 e^{e^3+2 x+e \log (5)}-1080 e^{2 e^3+4 x+2 e \log (5)}+720 e^{3 e^3+6 x+3 e \log (5)}-240 e^{4 e^3+8 x+4 e \log (5)}+32 e^{5 e^3+10 x+5 e \log (5)}} \, dx=\frac {16}{\left (-3+2 e^{2 x+e \left (e^2+\log (5)\right )}\right )^4} \] Output:
16/(2*exp((ln(5)+exp(2))*exp(1)+2*x)-3)^4
Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int -\frac {256 e^{e^3+2 x+e \log (5)}}{-243+810 e^{e^3+2 x+e \log (5)}-1080 e^{2 e^3+4 x+2 e \log (5)}+720 e^{3 e^3+6 x+3 e \log (5)}-240 e^{4 e^3+8 x+4 e \log (5)}+32 e^{5 e^3+10 x+5 e \log (5)}} \, dx=\frac {16}{\left (-3+2\ 5^e e^{e^3+2 x}\right )^4} \] Input:
Integrate[(-256*E^(E^3 + 2*x + E*Log[5]))/(-243 + 810*E^(E^3 + 2*x + E*Log [5]) - 1080*E^(2*E^3 + 4*x + 2*E*Log[5]) + 720*E^(3*E^3 + 6*x + 3*E*Log[5] ) - 240*E^(4*E^3 + 8*x + 4*E*Log[5]) + 32*E^(5*E^3 + 10*x + 5*E*Log[5])),x ]
Output:
16/(-3 + 2*5^E*E^(E^3 + 2*x))^4
Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {27, 25, 27, 2720, 17}
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.
\(\displaystyle \int -\frac {256 e^{2 x+e^3+e \log (5)}}{810 e^{2 x+e^3+e \log (5)}-1080 e^{4 x+2 e^3+2 e \log (5)}+720 e^{6 x+3 e^3+3 e \log (5)}-240 e^{8 x+4 e^3+4 e \log (5)}+32 e^{10 x+5 e^3+5 e \log (5)}-243} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -256 \int -\frac {5^e e^{2 x+e^3}}{243-162\ 5^{1+e} e^{2 x+e^3}+216\ 5^{1+2 e} e^{4 x+2 e^3}-144\ 5^{1+3 e} e^{6 x+3 e^3}+48\ 5^{1+4 e} e^{8 x+4 e^3}-32\ 5^{5 e} e^{10 x+5 e^3}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle 256 \int \frac {5^e e^{2 x+e^3}}{243-162\ 5^{1+e} e^{2 x+e^3}+216\ 5^{1+2 e} e^{4 x+2 e^3}-144\ 5^{1+3 e} e^{6 x+3 e^3}+48\ 5^{1+4 e} e^{8 x+4 e^3}-32\ 5^{5 e} e^{10 x+5 e^3}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 256\ 5^e \int \frac {e^{2 x+e^3}}{243-162\ 5^{1+e} e^{2 x+e^3}+216\ 5^{1+2 e} e^{4 x+2 e^3}-144\ 5^{1+3 e} e^{6 x+3 e^3}+48\ 5^{1+4 e} e^{8 x+4 e^3}-32\ 5^{5 e} e^{10 x+5 e^3}}dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle 128\ 5^e \int \frac {1}{\left (3-2\ 5^e e^{2 x+e^3}\right )^5}de^{2 x+e^3}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {16}{\left (3-2\ 5^e e^{2 x+e^3}\right )^4}\) |
Input:
Int[(-256*E^(E^3 + 2*x + E*Log[5]))/(-243 + 810*E^(E^3 + 2*x + E*Log[5]) - 1080*E^(2*E^3 + 4*x + 2*E*Log[5]) + 720*E^(3*E^3 + 6*x + 3*E*Log[5]) - 24 0*E^(4*E^3 + 8*x + 4*E*Log[5]) + 32*E^(5*E^3 + 10*x + 5*E*Log[5])),x]
Output:
16/(3 - 2*5^E*E^(E^3 + 2*x))^4
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]]]
Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\frac {16}{\left (2 \,5^{{\mathrm e}} {\mathrm e}^{2 x +{\mathrm e}^{3}}-3\right )^{4}}\) | \(20\) |
derivativedivides | \(\frac {16}{\left (2 \,{\mathrm e}^{{\mathrm e} \ln \left (5\right )+{\mathrm e} \,{\mathrm e}^{2}+2 x}-3\right )^{4}}\) | \(24\) |
default | \(\frac {16}{\left (2 \,{\mathrm e}^{{\mathrm e} \ln \left (5\right )+{\mathrm e} \,{\mathrm e}^{2}+2 x}-3\right )^{4}}\) | \(24\) |
norman | \(\frac {16}{\left (2 \,{\mathrm e}^{{\mathrm e} \ln \left (5\right )+{\mathrm e} \,{\mathrm e}^{2}+2 x}-3\right )^{4}}\) | \(24\) |
parallelrisch | \(\frac {16}{16 \,{\mathrm e}^{4 \,{\mathrm e} \ln \left (5\right )+4 \,{\mathrm e}^{3}+8 x}-96 \,{\mathrm e}^{3 \,{\mathrm e} \ln \left (5\right )+3 \,{\mathrm e}^{3}+6 x}+216 \,{\mathrm e}^{2 \,{\mathrm e} \ln \left (5\right )+2 \,{\mathrm e}^{3}+4 x}-216 \,{\mathrm e}^{{\mathrm e} \ln \left (5\right )+{\mathrm e} \,{\mathrm e}^{2}+2 x}+81}\) | \(81\) |
Input:
int(-256*exp(exp(1)*ln(5)+exp(1)*exp(2)+2*x)/(32*exp(exp(1)*ln(5)+exp(1)*e xp(2)+2*x)^5-240*exp(exp(1)*ln(5)+exp(1)*exp(2)+2*x)^4+720*exp(exp(1)*ln(5 )+exp(1)*exp(2)+2*x)^3-1080*exp(exp(1)*ln(5)+exp(1)*exp(2)+2*x)^2+810*exp( exp(1)*ln(5)+exp(1)*exp(2)+2*x)-243),x,method=_RETURNVERBOSE)
Output:
16/(2*5^exp(1)*exp(2*x+exp(3))-3)^4
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (21) = 42\).
Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 3.23 \[ \int -\frac {256 e^{e^3+2 x+e \log (5)}}{-243+810 e^{e^3+2 x+e \log (5)}-1080 e^{2 e^3+4 x+2 e \log (5)}+720 e^{3 e^3+6 x+3 e \log (5)}-240 e^{4 e^3+8 x+4 e \log (5)}+32 e^{5 e^3+10 x+5 e \log (5)}} \, dx=\frac {16}{16 \, e^{\left (4 \, e \log \left (5\right ) + 8 \, x + 4 \, e^{3}\right )} - 96 \, e^{\left (3 \, e \log \left (5\right ) + 6 \, x + 3 \, e^{3}\right )} + 216 \, e^{\left (2 \, e \log \left (5\right ) + 4 \, x + 2 \, e^{3}\right )} - 216 \, e^{\left (e \log \left (5\right ) + 2 \, x + e^{3}\right )} + 81} \] Input:
integrate(-256*exp(exp(1)*log(5)+exp(1)*exp(2)+2*x)/(32*exp(exp(1)*log(5)+ exp(1)*exp(2)+2*x)^5-240*exp(exp(1)*log(5)+exp(1)*exp(2)+2*x)^4+720*exp(ex p(1)*log(5)+exp(1)*exp(2)+2*x)^3-1080*exp(exp(1)*log(5)+exp(1)*exp(2)+2*x) ^2+810*exp(exp(1)*log(5)+exp(1)*exp(2)+2*x)-243),x, algorithm="fricas")
Output:
16/(16*e^(4*e*log(5) + 8*x + 4*e^3) - 96*e^(3*e*log(5) + 6*x + 3*e^3) + 21 6*e^(2*e*log(5) + 4*x + 2*e^3) - 216*e^(e*log(5) + 2*x + e^3) + 81)
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (20) = 40\).
Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.32 \[ \int -\frac {256 e^{e^3+2 x+e \log (5)}}{-243+810 e^{e^3+2 x+e \log (5)}-1080 e^{2 e^3+4 x+2 e \log (5)}+720 e^{3 e^3+6 x+3 e \log (5)}-240 e^{4 e^3+8 x+4 e \log (5)}+32 e^{5 e^3+10 x+5 e \log (5)}} \, dx=\frac {16}{- 216 \cdot 5^{e} e^{2 x + e^{3}} + 216 \cdot 5^{2 e} e^{4 x + 2 e^{3}} - 96 \cdot 5^{3 e} e^{6 x + 3 e^{3}} + 16 \cdot 5^{4 e} e^{8 x + 4 e^{3}} + 81} \] Input:
integrate(-256*exp(exp(1)*ln(5)+exp(1)*exp(2)+2*x)/(32*exp(exp(1)*ln(5)+ex p(1)*exp(2)+2*x)**5-240*exp(exp(1)*ln(5)+exp(1)*exp(2)+2*x)**4+720*exp(exp (1)*ln(5)+exp(1)*exp(2)+2*x)**3-1080*exp(exp(1)*ln(5)+exp(1)*exp(2)+2*x)** 2+810*exp(exp(1)*ln(5)+exp(1)*exp(2)+2*x)-243),x)
Output:
16/(-216*5**E*exp(2*x + exp(3)) + 216*5**(2*E)*exp(4*x + 2*exp(3)) - 96*5* *(3*E)*exp(6*x + 3*exp(3)) + 16*5**(4*E)*exp(8*x + 4*exp(3)) + 81)
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (21) = 42\).
Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 3.23 \[ \int -\frac {256 e^{e^3+2 x+e \log (5)}}{-243+810 e^{e^3+2 x+e \log (5)}-1080 e^{2 e^3+4 x+2 e \log (5)}+720 e^{3 e^3+6 x+3 e \log (5)}-240 e^{4 e^3+8 x+4 e \log (5)}+32 e^{5 e^3+10 x+5 e \log (5)}} \, dx=\frac {16}{16 \, e^{\left (4 \, e \log \left (5\right ) + 8 \, x + 4 \, e^{3}\right )} - 96 \, e^{\left (3 \, e \log \left (5\right ) + 6 \, x + 3 \, e^{3}\right )} + 216 \, e^{\left (2 \, e \log \left (5\right ) + 4 \, x + 2 \, e^{3}\right )} - 216 \, e^{\left (e \log \left (5\right ) + 2 \, x + e^{3}\right )} + 81} \] Input:
integrate(-256*exp(exp(1)*log(5)+exp(1)*exp(2)+2*x)/(32*exp(exp(1)*log(5)+ exp(1)*exp(2)+2*x)^5-240*exp(exp(1)*log(5)+exp(1)*exp(2)+2*x)^4+720*exp(ex p(1)*log(5)+exp(1)*exp(2)+2*x)^3-1080*exp(exp(1)*log(5)+exp(1)*exp(2)+2*x) ^2+810*exp(exp(1)*log(5)+exp(1)*exp(2)+2*x)-243),x, algorithm="maxima")
Output:
16/(16*e^(4*e*log(5) + 8*x + 4*e^3) - 96*e^(3*e*log(5) + 6*x + 3*e^3) + 21 6*e^(2*e*log(5) + 4*x + 2*e^3) - 216*e^(e*log(5) + 2*x + e^3) + 81)
Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int -\frac {256 e^{e^3+2 x+e \log (5)}}{-243+810 e^{e^3+2 x+e \log (5)}-1080 e^{2 e^3+4 x+2 e \log (5)}+720 e^{3 e^3+6 x+3 e \log (5)}-240 e^{4 e^3+8 x+4 e \log (5)}+32 e^{5 e^3+10 x+5 e \log (5)}} \, dx=\frac {16}{{\left (2 \, e^{\left (e \log \left (5\right ) + 2 \, x + e^{3}\right )} - 3\right )}^{4}} \] Input:
integrate(-256*exp(exp(1)*log(5)+exp(1)*exp(2)+2*x)/(32*exp(exp(1)*log(5)+ exp(1)*exp(2)+2*x)^5-240*exp(exp(1)*log(5)+exp(1)*exp(2)+2*x)^4+720*exp(ex p(1)*log(5)+exp(1)*exp(2)+2*x)^3-1080*exp(exp(1)*log(5)+exp(1)*exp(2)+2*x) ^2+810*exp(exp(1)*log(5)+exp(1)*exp(2)+2*x)-243),x, algorithm="giac")
Output:
16/(2*e^(e*log(5) + 2*x + e^3) - 3)^4
Time = 0.17 (sec) , antiderivative size = 86, normalized size of antiderivative = 3.91 \[ \int -\frac {256 e^{e^3+2 x+e \log (5)}}{-243+810 e^{e^3+2 x+e \log (5)}-1080 e^{2 e^3+4 x+2 e \log (5)}+720 e^{3 e^3+6 x+3 e \log (5)}-240 e^{4 e^3+8 x+4 e \log (5)}+32 e^{5 e^3+10 x+5 e \log (5)}} \, dx=-\frac {\frac {128\,5^{2\,\mathrm {e}}\,{\mathrm {e}}^{4\,x+2\,{\mathrm {e}}^3}}{3}-\frac {128\,5^{\mathrm {e}}\,{\mathrm {e}}^{2\,x+{\mathrm {e}}^3}\,\left (12\,5^{2\,\mathrm {e}}\,{\mathrm {e}}^{4\,x+2\,{\mathrm {e}}^3}-2\,5^{3\,\mathrm {e}}\,{\mathrm {e}}^{6\,x+3\,{\mathrm {e}}^3}+27\right )}{81}}{{\left (2\,5^{\mathrm {e}}\,{\mathrm {e}}^{2\,x+{\mathrm {e}}^3}-3\right )}^4} \] Input:
int(-(256*exp(2*x + exp(3) + exp(1)*log(5)))/(810*exp(2*x + exp(3) + exp(1 )*log(5)) - 1080*exp(4*x + 2*exp(3) + 2*exp(1)*log(5)) + 720*exp(6*x + 3*e xp(3) + 3*exp(1)*log(5)) - 240*exp(8*x + 4*exp(3) + 4*exp(1)*log(5)) + 32* exp(10*x + 5*exp(3) + 5*exp(1)*log(5)) - 243),x)
Output:
-((128*5^(2*exp(1))*exp(4*x + 2*exp(3)))/3 - (128*5^exp(1)*exp(2*x + exp(3 ))*(12*5^(2*exp(1))*exp(4*x + 2*exp(3)) - 2*5^(3*exp(1))*exp(6*x + 3*exp(3 )) + 27))/81)/(2*5^exp(1)*exp(2*x + exp(3)) - 3)^4
Time = 0.17 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.36 \[ \int -\frac {256 e^{e^3+2 x+e \log (5)}}{-243+810 e^{e^3+2 x+e \log (5)}-1080 e^{2 e^3+4 x+2 e \log (5)}+720 e^{3 e^3+6 x+3 e \log (5)}-240 e^{4 e^3+8 x+4 e \log (5)}+32 e^{5 e^3+10 x+5 e \log (5)}} \, dx=\frac {16}{16 e^{4 e^{3}+8 x} 5^{4 e}-96 e^{3 e^{3}+6 x} 5^{3 e}+216 e^{2 e^{3}+4 x} 5^{2 e}-216 e^{e^{3}+2 x} 5^{e}+81} \] Input:
int(-256*exp(exp(1)*log(5)+exp(1)*exp(2)+2*x)/(32*exp(exp(1)*log(5)+exp(1) *exp(2)+2*x)^5-240*exp(exp(1)*log(5)+exp(1)*exp(2)+2*x)^4+720*exp(exp(1)*l og(5)+exp(1)*exp(2)+2*x)^3-1080*exp(exp(1)*log(5)+exp(1)*exp(2)+2*x)^2+810 *exp(exp(1)*log(5)+exp(1)*exp(2)+2*x)-243),x)
Output:
16/(16*e**(4*e**3 + 8*x)*5**(4*e) - 96*e**(3*e**3 + 6*x)*5**(3*e) + 216*e* *(2*e**3 + 4*x)*5**(2*e) - 216*e**(e**3 + 2*x)*5**e + 81)