3.12.83 \(\int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x (-4+\frac {e^x}{3}+\log (2))}{-4+\frac {e^x}{3}+\log (2)}} (-\frac {16 e^x}{3}+e^x (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))))}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx\)

Optimal. Leaf size=24 \[ e^{17+e^x+\frac {16}{-4+\frac {e^x}{3}+\log (2)}}+x \]

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Rubi [F]  time = 5.63, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+\exp \left (\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}\right ) \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(16 + E^(2*x)/9 - 8*Log[2] + Log[2]^2 + (E^x*(-8 + 2*Log[2]))/3 + E^((-52 + (17*E^x)/3 + 17*Log[2] + E^x*(
-4 + E^x/3 + Log[2]))/(-4 + E^x/3 + Log[2]))*((-16*E^x)/3 + E^x*(16 + E^(2*x)/9 - 8*Log[2] + Log[2]^2 + (E^x*(
-8 + 2*Log[2]))/3)))/(16 + E^(2*x)/9 - 8*Log[2] + Log[2]^2 + (E^x*(-8 + 2*Log[2]))/3),x]

[Out]

x + Defer[Subst][Defer[Int][2^(51/(-12 + x + Log[8]))*E^((-156 + x^2 + x*(5 + Log[8]))/(-12 + x + Log[8])), x]
, x, E^x] - (48 - 9*Log[2]^2 + 6*Log[2]*Log[8] - Log[8]^2)*Defer[Subst][Defer[Int][(2^(51/(-12 + x + Log[8]))*
E^((-156 + x^2 + x*(5 + Log[8]))/(-12 + x + Log[8])))/(-12 + x + Log[8])^2, x], x, E^x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {9 \left (\frac {x^2}{9}+\frac {2}{3} x (-4+\log (2))+16 \left (1+\frac {1}{16} (-8+\log (2)) \log (2)\right )+\frac {1}{9} 2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}} x \left (96+x^2+6 x (-4+\log (2))-72 \log (2)+9 \log ^2(2)\right )\right )}{x (12-x-\log (8))^2} \, dx,x,e^x\right )\\ &=9 \operatorname {Subst}\left (\int \frac {\frac {x^2}{9}+\frac {2}{3} x (-4+\log (2))+16 \left (1+\frac {1}{16} (-8+\log (2)) \log (2)\right )+\frac {1}{9} 2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}} x \left (96+x^2+6 x (-4+\log (2))-72 \log (2)+9 \log ^2(2)\right )}{x (12-x-\log (8))^2} \, dx,x,e^x\right )\\ &=9 \operatorname {Subst}\left (\int \left (\frac {1}{9 x}+\frac {2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}} \left (x^2-6 x (4-\log (2))+3 \left (32-24 \log (2)+3 \log ^2(2)\right )\right )}{9 (12-x-\log (8))^2}\right ) \, dx,x,e^x\right )\\ &=x+\operatorname {Subst}\left (\int \frac {2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}} \left (x^2-6 x (4-\log (2))+3 \left (32-24 \log (2)+3 \log ^2(2)\right )\right )}{(12-x-\log (8))^2} \, dx,x,e^x\right )\\ &=x+\operatorname {Subst}\left (\int \left (2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}}+\frac {2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}} \left (-48+9 \log ^2(2)-6 \log (2) \log (8)+\log ^2(8)\right )}{(-12+x+\log (8))^2}\right ) \, dx,x,e^x\right )\\ &=x+\left (-48+9 \log ^2(2)-6 \log (2) \log (8)+\log ^2(8)\right ) \operatorname {Subst}\left (\int \frac {2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}}}{(-12+x+\log (8))^2} \, dx,x,e^x\right )+\operatorname {Subst}\left (\int 2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}} \, dx,x,e^x\right )\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 5.20, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(16 + E^(2*x)/9 - 8*Log[2] + Log[2]^2 + (E^x*(-8 + 2*Log[2]))/3 + E^((-52 + (17*E^x)/3 + 17*Log[2] +
E^x*(-4 + E^x/3 + Log[2]))/(-4 + E^x/3 + Log[2]))*((-16*E^x)/3 + E^x*(16 + E^(2*x)/9 - 8*Log[2] + Log[2]^2 +
(E^x*(-8 + 2*Log[2]))/3)))/(16 + E^(2*x)/9 - 8*Log[2] + Log[2]^2 + (E^x*(-8 + 2*Log[2]))/3),x]

[Out]

Integrate[(16 + E^(2*x)/9 - 8*Log[2] + Log[2]^2 + (E^x*(-8 + 2*Log[2]))/3 + E^((-52 + (17*E^x)/3 + 17*Log[2] +
E^x*(-4 + E^x/3 + Log[2]))/(-4 + E^x/3 + Log[2]))*((-16*E^x)/3 + E^x*(16 + E^(2*x)/9 - 8*Log[2] + Log[2]^2 +
(E^x*(-8 + 2*Log[2]))/3)))/(16 + E^(2*x)/9 - 8*Log[2] + Log[2]^2 + (E^x*(-8 + 2*Log[2]))/3), x]

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fricas [A]  time = 0.73, size = 33, normalized size = 1.38 \begin {gather*} x + e^{\left (\frac {{\left (3 \, \log \relax (2) + 5\right )} e^{x} + e^{\left (2 \, x\right )} + 51 \, \log \relax (2) - 156}{e^{x} + 3 \, \log \relax (2) - 12}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(-log(3)+x)^2+(2*log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log(2)+16)*exp(x)-16*exp(-log(3)+x))*exp(
((exp(-log(3)+x)+log(2)-4)*exp(x)+17*exp(-log(3)+x)+17*log(2)-52)/(exp(-log(3)+x)+log(2)-4))+exp(-log(3)+x)^2+
(2*log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log(2)+16)/(exp(-log(3)+x)^2+(2*log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log
(2)+16),x, algorithm="fricas")

[Out]

x + e^(((3*log(2) + 5)*e^x + e^(2*x) + 51*log(2) - 156)/(e^x + 3*log(2) - 12))

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giac [B]  time = 3.41, size = 141, normalized size = 5.88 \begin {gather*} \frac {1}{2} \, {\left (2 \, x e^{x} + 2 \, e^{x} \log \left (e^{x} + 3 \, \log \relax (2) - 12\right ) - 2 \, e^{x} \log \left (-e^{x} - 3 \, \log \relax (2) + 12\right ) + e^{\left (\frac {x e^{x} \log \relax (2) + 3 \, x \log \relax (2)^{2} + 3 \, e^{x} \log \relax (2)^{2} - 4 \, x e^{x} - 24 \, x \log \relax (2) + e^{\left (2 \, x\right )} \log \relax (2) - 24 \, e^{x} \log \relax (2) + 48 \, x - 4 \, e^{\left (2 \, x\right )} + 32 \, e^{x}}{e^{x} \log \relax (2) + 3 \, \log \relax (2)^{2} - 4 \, e^{x} - 24 \, \log \relax (2) + 48} + \frac {\log \relax (2)^{2} + 13 \, \log \relax (2) - 52}{\log \relax (2) - 4}\right )}\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(-log(3)+x)^2+(2*log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log(2)+16)*exp(x)-16*exp(-log(3)+x))*exp(
((exp(-log(3)+x)+log(2)-4)*exp(x)+17*exp(-log(3)+x)+17*log(2)-52)/(exp(-log(3)+x)+log(2)-4))+exp(-log(3)+x)^2+
(2*log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log(2)+16)/(exp(-log(3)+x)^2+(2*log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log
(2)+16),x, algorithm="giac")

[Out]

1/2*(2*x*e^x + 2*e^x*log(e^x + 3*log(2) - 12) - 2*e^x*log(-e^x - 3*log(2) + 12) + e^((x*e^x*log(2) + 3*x*log(2
)^2 + 3*e^x*log(2)^2 - 4*x*e^x - 24*x*log(2) + e^(2*x)*log(2) - 24*e^x*log(2) + 48*x - 4*e^(2*x) + 32*e^x)/(e^
x*log(2) + 3*log(2)^2 - 4*e^x - 24*log(2) + 48) + (log(2)^2 + 13*log(2) - 52)/(log(2) - 4)))*e^(-x)

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maple [A]  time = 0.54, size = 35, normalized size = 1.46

method	result	size
risch	\(x +{\mathrm e}^{\frac {{\mathrm e}^{2 x}+3 \,{\mathrm e}^{x} \ln \relax (2)+5 \,{\mathrm e}^{x}+51 \ln \relax (2)-156}{{\mathrm e}^{x}+3 \ln \relax (2)-12}}\)	\(35\)
norman	\(\frac {\left (3 \ln \relax (2)-12\right ) x +\left (3 \ln \relax (2)-12\right ) {\mathrm e}^{\frac {\left (\frac {{\mathrm e}^{x}}{3}+\ln \relax (2)-4\right ) {\mathrm e}^{x}+\frac {17 \,{\mathrm e}^{x}}{3}+17 \ln \relax (2)-52}{\frac {{\mathrm e}^{x}}{3}+\ln \relax (2)-4}}+{\mathrm e}^{x} x +{\mathrm e}^{x} {\mathrm e}^{\frac {\left (\frac {{\mathrm e}^{x}}{3}+\ln \relax (2)-4\right ) {\mathrm e}^{x}+\frac {17 \,{\mathrm e}^{x}}{3}+17 \ln \relax (2)-52}{\frac {{\mathrm e}^{x}}{3}+\ln \relax (2)-4}}}{{\mathrm e}^{x}+3 \ln \relax (2)-12}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((exp(-ln(3)+x)^2+(2*ln(2)-8)*exp(-ln(3)+x)+ln(2)^2-8*ln(2)+16)*exp(x)-16*exp(-ln(3)+x))*exp(((exp(-ln(3)
+x)+ln(2)-4)*exp(x)+17*exp(-ln(3)+x)+17*ln(2)-52)/(exp(-ln(3)+x)+ln(2)-4))+exp(-ln(3)+x)^2+(2*ln(2)-8)*exp(-ln
(3)+x)+ln(2)^2-8*ln(2)+16)/(exp(-ln(3)+x)^2+(2*ln(2)-8)*exp(-ln(3)+x)+ln(2)^2-8*ln(2)+16),x,method=_RETURNVERB
OSE)

[Out]

x+exp((exp(2*x)+3*exp(x)*ln(2)+5*exp(x)+51*ln(2)-156)/(exp(x)+3*ln(2)-12))

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maxima [B]  time = 0.73, size = 261, normalized size = 10.88 \begin {gather*} {\left (\frac {x}{\log \relax (2)^{2} - 8 \, \log \relax (2) + 16} - \frac {\log \left (e^{x} + 3 \, \log \relax (2) - 12\right )}{\log \relax (2)^{2} - 8 \, \log \relax (2) + 16} + \frac {3}{{\left (\log \relax (2) - 4\right )} e^{x} + 3 \, \log \relax (2)^{2} - 24 \, \log \relax (2) + 48}\right )} \log \relax (2)^{2} - 8 \, {\left (\frac {x}{\log \relax (2)^{2} - 8 \, \log \relax (2) + 16} - \frac {\log \left (e^{x} + 3 \, \log \relax (2) - 12\right )}{\log \relax (2)^{2} - 8 \, \log \relax (2) + 16} + \frac {3}{{\left (\log \relax (2) - 4\right )} e^{x} + 3 \, \log \relax (2)^{2} - 24 \, \log \relax (2) + 48}\right )} \log \relax (2) + \frac {16 \, x}{\log \relax (2)^{2} - 8 \, \log \relax (2) + 16} + \frac {3 \, {\left (\log \relax (2) - 4\right )}}{e^{x} + 3 \, \log \relax (2) - 12} - \frac {6 \, \log \relax (2)}{e^{x} + 3 \, \log \relax (2) - 12} - \frac {16 \, \log \left (e^{x} + 3 \, \log \relax (2) - 12\right )}{\log \relax (2)^{2} - 8 \, \log \relax (2) + 16} + \frac {48}{{\left (\log \relax (2) - 4\right )} e^{x} + 3 \, \log \relax (2)^{2} - 24 \, \log \relax (2) + 48} + \frac {24}{e^{x} + 3 \, \log \relax (2) - 12} + e^{\left (\frac {48}{e^{x} + 3 \, \log \relax (2) - 12} + e^{x} + 17\right )} + \log \left (e^{x} + 3 \, \log \relax (2) - 12\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(-log(3)+x)^2+(2*log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log(2)+16)*exp(x)-16*exp(-log(3)+x))*exp(
((exp(-log(3)+x)+log(2)-4)*exp(x)+17*exp(-log(3)+x)+17*log(2)-52)/(exp(-log(3)+x)+log(2)-4))+exp(-log(3)+x)^2+
(2*log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log(2)+16)/(exp(-log(3)+x)^2+(2*log(2)-8)*exp(-log(3)+x)+log(2)^2-8*log
(2)+16),x, algorithm="maxima")

[Out]

(x/(log(2)^2 - 8*log(2) + 16) - log(e^x + 3*log(2) - 12)/(log(2)^2 - 8*log(2) + 16) + 3/((log(2) - 4)*e^x + 3*
log(2)^2 - 24*log(2) + 48))*log(2)^2 - 8*(x/(log(2)^2 - 8*log(2) + 16) - log(e^x + 3*log(2) - 12)/(log(2)^2 -
8*log(2) + 16) + 3/((log(2) - 4)*e^x + 3*log(2)^2 - 24*log(2) + 48))*log(2) + 16*x/(log(2)^2 - 8*log(2) + 16)
+ 3*(log(2) - 4)/(e^x + 3*log(2) - 12) - 6*log(2)/(e^x + 3*log(2) - 12) - 16*log(e^x + 3*log(2) - 12)/(log(2)^
2 - 8*log(2) + 16) + 48/((log(2) - 4)*e^x + 3*log(2)^2 - 24*log(2) + 48) + 24/(e^x + 3*log(2) - 12) + e^(48/(e
^x + 3*log(2) - 12) + e^x + 17) + log(e^x + 3*log(2) - 12)

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mupad [B]  time = 1.40, size = 75, normalized size = 3.12 \begin {gather*} x+2^{\frac {{\mathrm {e}}^x}{\ln \relax (2)+\frac {{\mathrm {e}}^x}{3}-4}}\,2^{\frac {17}{\ln \relax (2)+\frac {{\mathrm {e}}^x}{3}-4}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{2\,x}}{3\,\left (\ln \relax (2)+\frac {{\mathrm {e}}^x}{3}-4\right )}-\frac {52}{\ln \relax (2)+\frac {{\mathrm {e}}^x}{3}-4}+\frac {5\,{\mathrm {e}}^x}{3\,\left (\ln \relax (2)+\frac {{\mathrm {e}}^x}{3}-4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x - 2*log(3)) - 8*log(2) + exp(x - log(3))*(2*log(2) - 8) + log(2)^2 - exp((17*exp(x - log(3)) + 17
*log(2) + exp(x)*(exp(x - log(3)) + log(2) - 4) - 52)/(exp(x - log(3)) + log(2) - 4))*(16*exp(x - log(3)) - ex
p(x)*(exp(2*x - 2*log(3)) - 8*log(2) + exp(x - log(3))*(2*log(2) - 8) + log(2)^2 + 16)) + 16)/(exp(2*x - 2*log
(3)) - 8*log(2) + exp(x - log(3))*(2*log(2) - 8) + log(2)^2 + 16),x)

[Out]

x + 2^(exp(x)/(log(2) + exp(x)/3 - 4))*2^(17/(log(2) + exp(x)/3 - 4))*exp(exp(2*x)/(3*(log(2) + exp(x)/3 - 4))
- 52/(log(2) + exp(x)/3 - 4) + (5*exp(x))/(3*(log(2) + exp(x)/3 - 4)))

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sympy [B]  time = 0.44, size = 39, normalized size = 1.62 \begin {gather*} x + e^{\frac {\left (\frac {e^{x}}{3} - 4 + \log {\relax (2 )}\right ) e^{x} + \frac {17 e^{x}}{3} - 52 + 17 \log {\relax (2 )}}{\frac {e^{x}}{3} - 4 + \log {\relax (2 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(-ln(3)+x)**2+(2*ln(2)-8)*exp(-ln(3)+x)+ln(2)**2-8*ln(2)+16)*exp(x)-16*exp(-ln(3)+x))*exp(((ex
p(-ln(3)+x)+ln(2)-4)*exp(x)+17*exp(-ln(3)+x)+17*ln(2)-52)/(exp(-ln(3)+x)+ln(2)-4))+exp(-ln(3)+x)**2+(2*ln(2)-8
)*exp(-ln(3)+x)+ln(2)**2-8*ln(2)+16)/(exp(-ln(3)+x)**2+(2*ln(2)-8)*exp(-ln(3)+x)+ln(2)**2-8*ln(2)+16),x)

[Out]

x + exp(((exp(x)/3 - 4 + log(2))*exp(x) + 17*exp(x)/3 - 52 + 17*log(2))/(exp(x)/3 - 4 + log(2)))

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