\(\int \frac {20 e^{4 x} \log (2)+e^{\frac {4 (x+(2+x) \log (2))}{\log (2)}} (20+20 \log (2))+e^{3 x+\frac {x+(2+x) \log (2)}{\log (2)}} (20+80 \log (2))+e^{x+\frac {3 (x+(2+x) \log (2))}{\log (2)}} (60+80 \log (2))+e^{2 x+\frac {2 (x+(2+x) \log (2))}{\log (2)}} (60+120 \log (2))}{\log (2)} \, dx\) [2148]

Optimal result

Integrand size = 116, antiderivative size = 19 \[ \int \frac {20 e^{4 x} \log (2)+e^{\frac {4 (x+(2+x) \log (2))}{\log (2)}} (20+20 \log (2))+e^{3 x+\frac {x+(2+x) \log (2)}{\log (2)}} (20+80 \log (2))+e^{x+\frac {3 (x+(2+x) \log (2))}{\log (2)}} (60+80 \log (2))+e^{2 x+\frac {2 (x+(2+x) \log (2))}{\log (2)}} (60+120 \log (2))}{\log (2)} \, dx=5 \left (e^x+e^{2+x+\frac {x}{\log (2)}}\right )^4 \]

[Out]

5*(exp(x)+exp(2+x/ln(2)+x))^4

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(75\) vs. \(2(19)=38\).

Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 3.95, number of steps used = 11, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {12, 2225, 2259} \[ \int \frac {20 e^{4 x} \log (2)+e^{\frac {4 (x+(2+x) \log (2))}{\log (2)}} (20+20 \log (2))+e^{3 x+\frac {x+(2+x) \log (2)}{\log (2)}} (20+80 \log (2))+e^{x+\frac {3 (x+(2+x) \log (2))}{\log (2)}} (60+80 \log (2))+e^{2 x+\frac {2 (x+(2+x) \log (2))}{\log (2)}} (60+120 \log (2))}{\log (2)} \, dx=5 e^{4 x}+5 e^{4 \left (\frac {x (1+\log (2))}{\log (2)}+2\right )}+30 e^{\frac {2 x (1+\log (4))}{\log (2)}+4}+20 e^{\frac {x (1+\log (16))}{\log (2)}+2}+20 e^{\frac {x (3+\log (16))}{\log (2)}+6} \]

[In]

Int[(20*E^(4*x)*Log[2] + E^((4*(x + (2 + x)*Log[2]))/Log[2])*(20 + 20*Log[2]) + E^(3*x + (x + (2 + x)*Log[2])/
Log[2])*(20 + 80*Log[2]) + E^(x + (3*(x + (2 + x)*Log[2]))/Log[2])*(60 + 80*Log[2]) + E^(2*x + (2*(x + (2 + x)
*Log[2]))/Log[2])*(60 + 120*Log[2]))/Log[2],x]

[Out]

5*E^(4*x) + 5*E^(4*(2 + (x*(1 + Log[2]))/Log[2])) + 30*E^(4 + (2*x*(1 + Log[4]))/Log[2]) + 20*E^(2 + (x*(1 + L
og[16]))/Log[2]) + 20*E^(6 + (x*(3 + Log[16]))/Log[2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2259

Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] &&  !PowerOfLinearMatchQ[v, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (20 e^{4 x} \log (2)+e^{\frac {4 (x+(2+x) \log (2))}{\log (2)}} (20+20 \log (2))+e^{3 x+\frac {x+(2+x) \log (2)}{\log (2)}} (20+80 \log (2))+e^{x+\frac {3 (x+(2+x) \log (2))}{\log (2)}} (60+80 \log (2))+e^{2 x+\frac {2 (x+(2+x) \log (2))}{\log (2)}} (60+120 \log (2))\right ) \, dx}{\log (2)} \\ & = 20 \int e^{4 x} \, dx+\frac {(20 (1+\log (2))) \int e^{\frac {4 (x+(2+x) \log (2))}{\log (2)}} \, dx}{\log (2)}+\frac {(60 (1+\log (4))) \int e^{2 x+\frac {2 (x+(2+x) \log (2))}{\log (2)}} \, dx}{\log (2)}+\frac {(20 (1+\log (16))) \int e^{3 x+\frac {x+(2+x) \log (2)}{\log (2)}} \, dx}{\log (2)}+\frac {(20 (3+\log (16))) \int e^{x+\frac {3 (x+(2+x) \log (2))}{\log (2)}} \, dx}{\log (2)} \\ & = 5 e^{4 x}+\frac {(20 (1+\log (2))) \int e^{4 \left (2+\frac {x (1+\log (2))}{\log (2)}\right )} \, dx}{\log (2)}+\frac {(60 (1+\log (4))) \int e^{4+\frac {2 x (1+\log (4))}{\log (2)}} \, dx}{\log (2)}+\frac {(20 (1+\log (16))) \int e^{2+\frac {x (1+\log (16))}{\log (2)}} \, dx}{\log (2)}+\frac {(20 (3+\log (16))) \int e^{6+\frac {x (3+\log (16))}{\log (2)}} \, dx}{\log (2)} \\ & = 5 e^{4 x}+5 e^{4 \left (2+\frac {x (1+\log (2))}{\log (2)}\right )}+30 e^{4+\frac {2 x (1+\log (4))}{\log (2)}}+20 e^{2+\frac {x (1+\log (16))}{\log (2)}}+20 e^{6+\frac {x (3+\log (16))}{\log (2)}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(69\) vs. \(2(19)=38\).

Time = 0.31 (sec) , antiderivative size = 69, normalized size of antiderivative = 3.63 \[ \int \frac {20 e^{4 x} \log (2)+e^{\frac {4 (x+(2+x) \log (2))}{\log (2)}} (20+20 \log (2))+e^{3 x+\frac {x+(2+x) \log (2)}{\log (2)}} (20+80 \log (2))+e^{x+\frac {3 (x+(2+x) \log (2))}{\log (2)}} (60+80 \log (2))+e^{2 x+\frac {2 (x+(2+x) \log (2))}{\log (2)}} (60+120 \log (2))}{\log (2)} \, dx=5 \left (e^{4 x}+6 e^{4+x \left (4+\frac {2}{\log (2)}\right )}+e^{4 \left (2+x+\frac {x}{\log (2)}\right )}+4 e^{2+\frac {x (1+\log (16))}{\log (2)}}+4 e^{6+\frac {x (3+\log (16))}{\log (2)}}\right ) \]

[In]

Integrate[(20*E^(4*x)*Log[2] + E^((4*(x + (2 + x)*Log[2]))/Log[2])*(20 + 20*Log[2]) + E^(3*x + (x + (2 + x)*Lo
g[2])/Log[2])*(20 + 80*Log[2]) + E^(x + (3*(x + (2 + x)*Log[2]))/Log[2])*(60 + 80*Log[2]) + E^(2*x + (2*(x + (
2 + x)*Log[2]))/Log[2])*(60 + 120*Log[2]))/Log[2],x]

[Out]

5*(E^(4*x) + 6*E^(4 + x*(4 + 2/Log[2])) + E^(4*(2 + x + x/Log[2])) + 4*E^(2 + (x*(1 + Log[16]))/Log[2]) + 4*E^
(6 + (x*(3 + Log[16]))/Log[2]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(86\) vs. \(2(17)=34\).

Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 4.58

method	result	size
risch	\(5 \,{\mathrm e}^{4 x}+20 \,{\mathrm e}^{\frac {4 x \ln \left (2\right )+2 \ln \left (2\right )+x}{\ln \left (2\right )}}+30 \,{\mathrm e}^{\frac {4 x \ln \left (2\right )+4 \ln \left (2\right )+2 x}{\ln \left (2\right )}}+20 \,{\mathrm e}^{\frac {4 x \ln \left (2\right )+6 \ln \left (2\right )+3 x}{\ln \left (2\right )}}+5 \,{\mathrm e}^{\frac {4 x \ln \left (2\right )+8 \ln \left (2\right )+4 x}{\ln \left (2\right )}}\)	\(87\)
parallelrisch	\(\frac {5 \ln \left (2\right ) {\mathrm e}^{4 x}+20 \,{\mathrm e}^{3 x} {\mathrm e}^{\frac {\left (2+x \right ) \ln \left (2\right )+x}{\ln \left (2\right )}} \ln \left (2\right )+30 \,{\mathrm e}^{2 x} {\mathrm e}^{\frac {2 \left (2+x \right ) \ln \left (2\right )+2 x}{\ln \left (2\right )}} \ln \left (2\right )+20 \,{\mathrm e}^{x} {\mathrm e}^{\frac {3 \left (2+x \right ) \ln \left (2\right )+3 x}{\ln \left (2\right )}} \ln \left (2\right )+5 \ln \left (2\right ) {\mathrm e}^{\frac {4 \left (2+x \right ) \ln \left (2\right )+4 x}{\ln \left (2\right )}}}{\ln \left (2\right )}\)	\(103\)
parts	\(5 \,{\mathrm e}^{4 x}+5 \,{\mathrm e}^{\frac {4 \left (2+x \right ) \ln \left (2\right )+4 x}{\ln \left (2\right )}}+\frac {60 \left (1+2 \ln \left (2\right )\right ) {\mathrm e}^{4+\left (2+\frac {2+2 \ln \left (2\right )}{\ln \left (2\right )}\right ) x}}{\ln \left (2\right ) \left (2+\frac {2+2 \ln \left (2\right )}{\ln \left (2\right )}\right )}+\frac {20 \left (1+4 \ln \left (2\right )\right ) {\mathrm e}^{2+\left (3+\frac {1+\ln \left (2\right )}{\ln \left (2\right )}\right ) x}}{\ln \left (2\right ) \left (3+\frac {1+\ln \left (2\right )}{\ln \left (2\right )}\right )}+\frac {20 \left (4 \ln \left (2\right )+3\right ) {\mathrm e}^{6+\left (1+\frac {3 \ln \left (2\right )+3}{\ln \left (2\right )}\right ) x}}{\ln \left (2\right ) \left (1+\frac {3 \ln \left (2\right )+3}{\ln \left (2\right )}\right )}\)	\(153\)
default	\(\frac {\frac {\left (20 \ln \left (2\right )+20\right ) {\mathrm e}^{\frac {4 \left (2+x \right ) \ln \left (2\right )+4 x}{\ln \left (2\right )}} \ln \left (2\right )}{4 \ln \left (2\right )+4}+\frac {\left (80 \ln \left (2\right )+20\right ) {\mathrm e}^{2+\left (3+\frac {1+\ln \left (2\right )}{\ln \left (2\right )}\right ) x}}{3+\frac {1+\ln \left (2\right )}{\ln \left (2\right )}}+\frac {\left (80 \ln \left (2\right )+60\right ) {\mathrm e}^{6+\left (1+\frac {3 \ln \left (2\right )+3}{\ln \left (2\right )}\right ) x}}{1+\frac {3 \ln \left (2\right )+3}{\ln \left (2\right )}}+\frac {\left (120 \ln \left (2\right )+60\right ) {\mathrm e}^{4+\left (2+\frac {2+2 \ln \left (2\right )}{\ln \left (2\right )}\right ) x}}{2+\frac {2+2 \ln \left (2\right )}{\ln \left (2\right )}}+5 \ln \left (2\right ) {\mathrm e}^{4 x}}{\ln \left (2\right )}\)	\(159\)

[In]

int(((20*ln(2)+20)*exp(((2+x)*ln(2)+x)/ln(2))^4+(80*ln(2)+60)*exp(x)*exp(((2+x)*ln(2)+x)/ln(2))^3+(120*ln(2)+6
0)*exp(x)^2*exp(((2+x)*ln(2)+x)/ln(2))^2+(80*ln(2)+20)*exp(x)^3*exp(((2+x)*ln(2)+x)/ln(2))+20*ln(2)*exp(x)^4)/
ln(2),x,method=_RETURNVERBOSE)

[Out]

5*exp(4*x)+20*exp((4*x*ln(2)+2*ln(2)+x)/ln(2))+30*exp(2*(2*x*ln(2)+2*ln(2)+x)/ln(2))+20*exp((4*x*ln(2)+6*ln(2)
+3*x)/ln(2))+5*exp(4*(x*ln(2)+2*ln(2)+x)/ln(2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (17) = 34\).

Time = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 8.58 \[ \int \frac {20 e^{4 x} \log (2)+e^{\frac {4 (x+(2+x) \log (2))}{\log (2)}} (20+20 \log (2))+e^{3 x+\frac {x+(2+x) \log (2)}{\log (2)}} (20+80 \log (2))+e^{x+\frac {3 (x+(2+x) \log (2))}{\log (2)}} (60+80 \log (2))+e^{2 x+\frac {2 (x+(2+x) \log (2))}{\log (2)}} (60+120 \log (2))}{\log (2)} \, dx=5 \, {\left (e^{\left (\frac {4 \, {\left (2 \, {\left (2 \, x + 3\right )} \log \left (2\right ) + 3 \, x\right )}}{\log \left (2\right )}\right )} + 4 \, e^{\left (\frac {3 \, {\left (2 \, {\left (2 \, x + 3\right )} \log \left (2\right ) + 3 \, x\right )}}{\log \left (2\right )} + \frac {4 \, {\left ({\left (x + 2\right )} \log \left (2\right ) + x\right )}}{\log \left (2\right )}\right )} + 6 \, e^{\left (\frac {2 \, {\left (2 \, {\left (2 \, x + 3\right )} \log \left (2\right ) + 3 \, x\right )}}{\log \left (2\right )} + \frac {8 \, {\left ({\left (x + 2\right )} \log \left (2\right ) + x\right )}}{\log \left (2\right )}\right )} + 4 \, e^{\left (\frac {2 \, {\left (2 \, x + 3\right )} \log \left (2\right ) + 3 \, x}{\log \left (2\right )} + \frac {12 \, {\left ({\left (x + 2\right )} \log \left (2\right ) + x\right )}}{\log \left (2\right )}\right )} + e^{\left (\frac {16 \, {\left ({\left (x + 2\right )} \log \left (2\right ) + x\right )}}{\log \left (2\right )}\right )}\right )} e^{\left (-\frac {12 \, {\left ({\left (x + 2\right )} \log \left (2\right ) + x\right )}}{\log \left (2\right )}\right )} \]

[In]

integrate(((20*log(2)+20)*exp(((2+x)*log(2)+x)/log(2))^4+(80*log(2)+60)*exp(x)*exp(((2+x)*log(2)+x)/log(2))^3+
(120*log(2)+60)*exp(x)^2*exp(((2+x)*log(2)+x)/log(2))^2+(80*log(2)+20)*exp(x)^3*exp(((2+x)*log(2)+x)/log(2))+2
0*log(2)*exp(x)^4)/log(2),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (15) = 30\).

Time = 1.79 (sec) , antiderivative size = 99, normalized size of antiderivative = 5.21 \[ \int \frac {20 e^{4 x} \log (2)+e^{\frac {4 (x+(2+x) \log (2))}{\log (2)}} (20+20 \log (2))+e^{3 x+\frac {x+(2+x) \log (2)}{\log (2)}} (20+80 \log (2))+e^{x+\frac {3 (x+(2+x) \log (2))}{\log (2)}} (60+80 \log (2))+e^{2 x+\frac {2 (x+(2+x) \log (2))}{\log (2)}} (60+120 \log (2))}{\log (2)} \, dx=\frac {5 e^{8} e^{4 x} e^{\frac {4 x}{\log {\left (2 \right )}}} \log {\left (2 \right )} + 20 e^{6} e^{4 x} e^{\frac {3 x}{\log {\left (2 \right )}}} \log {\left (2 \right )} + 30 e^{4} e^{4 x} e^{\frac {2 x}{\log {\left (2 \right )}}} \log {\left (2 \right )} + 20 e^{2} e^{4 x} e^{\frac {x}{\log {\left (2 \right )}}} \log {\left (2 \right )} + 5 e^{4 x} \log {\left (2 \right )}}{\log {\left (2 \right )}} \]

[In]

integrate(((20*ln(2)+20)*exp(((2+x)*ln(2)+x)/ln(2))**4+(80*ln(2)+60)*exp(x)*exp(((2+x)*ln(2)+x)/ln(2))**3+(120
*ln(2)+60)*exp(x)**2*exp(((2+x)*ln(2)+x)/ln(2))**2+(80*ln(2)+20)*exp(x)**3*exp(((2+x)*ln(2)+x)/ln(2))+20*ln(2)
*exp(x)**4)/ln(2),x)

[Out]

(5*exp(8)*exp(4*x)*exp(4*x/log(2))*log(2) + 20*exp(6)*exp(4*x)*exp(3*x/log(2))*log(2) + 30*exp(4)*exp(4*x)*exp
(2*x/log(2))*log(2) + 20*exp(2)*exp(4*x)*exp(x/log(2))*log(2) + 5*exp(4*x)*log(2))/log(2)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (17) = 34\).

Time = 0.19 (sec) , antiderivative size = 150, normalized size of antiderivative = 7.89 \[ \int \frac {20 e^{4 x} \log (2)+e^{\frac {4 (x+(2+x) \log (2))}{\log (2)}} (20+20 \log (2))+e^{3 x+\frac {x+(2+x) \log (2)}{\log (2)}} (20+80 \log (2))+e^{x+\frac {3 (x+(2+x) \log (2))}{\log (2)}} (60+80 \log (2))+e^{2 x+\frac {2 (x+(2+x) \log (2))}{\log (2)}} (60+120 \log (2))}{\log (2)} \, dx=\frac {5 \, {\left (e^{\left (4 \, x\right )} \log \left (2\right ) + e^{\left (\frac {4 \, {\left ({\left (x + 2\right )} \log \left (2\right ) + x\right )}}{\log \left (2\right )}\right )} \log \left (2\right ) + \frac {4 \, {\left (4 \, \log \left (2\right ) + 1\right )} e^{\left (3 \, x + \frac {{\left (x + 2\right )} \log \left (2\right ) + x}{\log \left (2\right )}\right )}}{\frac {\log \left (2\right ) + 1}{\log \left (2\right )} + 3} + \frac {6 \, {\left (2 \, \log \left (2\right ) + 1\right )} e^{\left (2 \, x + \frac {2 \, {\left ({\left (x + 2\right )} \log \left (2\right ) + x\right )}}{\log \left (2\right )}\right )}}{\frac {\log \left (2\right ) + 1}{\log \left (2\right )} + 1} + \frac {4 \, {\left (4 \, \log \left (2\right ) + 3\right )} e^{\left (x + \frac {3 \, {\left ({\left (x + 2\right )} \log \left (2\right ) + x\right )}}{\log \left (2\right )}\right )}}{\frac {3 \, {\left (\log \left (2\right ) + 1\right )}}{\log \left (2\right )} + 1}\right )}}{\log \left (2\right )} \]

[In]

integrate(((20*log(2)+20)*exp(((2+x)*log(2)+x)/log(2))^4+(80*log(2)+60)*exp(x)*exp(((2+x)*log(2)+x)/log(2))^3+
(120*log(2)+60)*exp(x)^2*exp(((2+x)*log(2)+x)/log(2))^2+(80*log(2)+20)*exp(x)^3*exp(((2+x)*log(2)+x)/log(2))+2
0*log(2)*exp(x)^4)/log(2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (17) = 34\).

Time = 0.26 (sec) , antiderivative size = 150, normalized size of antiderivative = 7.89 \[ \int \frac {20 e^{4 x} \log (2)+e^{\frac {4 (x+(2+x) \log (2))}{\log (2)}} (20+20 \log (2))+e^{3 x+\frac {x+(2+x) \log (2)}{\log (2)}} (20+80 \log (2))+e^{x+\frac {3 (x+(2+x) \log (2))}{\log (2)}} (60+80 \log (2))+e^{2 x+\frac {2 (x+(2+x) \log (2))}{\log (2)}} (60+120 \log (2))}{\log (2)} \, dx=\frac {5 \, {\left (e^{\left (4 \, x\right )} \log \left (2\right ) + e^{\left (\frac {4 \, {\left ({\left (x + 2\right )} \log \left (2\right ) + x\right )}}{\log \left (2\right )}\right )} \log \left (2\right ) + \frac {4 \, {\left (4 \, \log \left (2\right ) + 1\right )} e^{\left (3 \, x + \frac {{\left (x + 2\right )} \log \left (2\right ) + x}{\log \left (2\right )}\right )}}{\frac {\log \left (2\right ) + 1}{\log \left (2\right )} + 3} + \frac {6 \, {\left (2 \, \log \left (2\right ) + 1\right )} e^{\left (2 \, x + \frac {2 \, {\left ({\left (x + 2\right )} \log \left (2\right ) + x\right )}}{\log \left (2\right )}\right )}}{\frac {\log \left (2\right ) + 1}{\log \left (2\right )} + 1} + \frac {4 \, {\left (4 \, \log \left (2\right ) + 3\right )} e^{\left (x + \frac {3 \, {\left ({\left (x + 2\right )} \log \left (2\right ) + x\right )}}{\log \left (2\right )}\right )}}{\frac {3 \, {\left (\log \left (2\right ) + 1\right )}}{\log \left (2\right )} + 1}\right )}}{\log \left (2\right )} \]

[In]

integrate(((20*log(2)+20)*exp(((2+x)*log(2)+x)/log(2))^4+(80*log(2)+60)*exp(x)*exp(((2+x)*log(2)+x)/log(2))^3+
(120*log(2)+60)*exp(x)^2*exp(((2+x)*log(2)+x)/log(2))^2+(80*log(2)+20)*exp(x)^3*exp(((2+x)*log(2)+x)/log(2))+2
0*log(2)*exp(x)^4)/log(2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 3.47 \[ \int \frac {20 e^{4 x} \log (2)+e^{\frac {4 (x+(2+x) \log (2))}{\log (2)}} (20+20 \log (2))+e^{3 x+\frac {x+(2+x) \log (2)}{\log (2)}} (20+80 \log (2))+e^{x+\frac {3 (x+(2+x) \log (2))}{\log (2)}} (60+80 \log (2))+e^{2 x+\frac {2 (x+(2+x) \log (2))}{\log (2)}} (60+120 \log (2))}{\log (2)} \, dx=5\,{\mathrm {e}}^{4\,x}+20\,{\mathrm {e}}^{4\,x+\frac {x}{\ln \left (2\right )}+2}+30\,{\mathrm {e}}^{4\,x+\frac {2\,x}{\ln \left (2\right )}+4}+20\,{\mathrm {e}}^{4\,x+\frac {3\,x}{\ln \left (2\right )}+6}+5\,{\mathrm {e}}^{4\,x+\frac {4\,x}{\ln \left (2\right )}+8} \]

[In]

int((20*exp(4*x)*log(2) + exp((4*(x + log(2)*(x + 2)))/log(2))*(20*log(2) + 20) + exp((3*(x + log(2)*(x + 2)))
/log(2))*exp(x)*(80*log(2) + 60) + exp((x + log(2)*(x + 2))/log(2))*exp(3*x)*(80*log(2) + 20) + exp((2*(x + lo
g(2)*(x + 2)))/log(2))*exp(2*x)*(120*log(2) + 60))/log(2),x)

[Out]

5*exp(4*x) + 20*exp(4*x + x/log(2) + 2) + 30*exp(4*x + (2*x)/log(2) + 4) + 20*exp(4*x + (3*x)/log(2) + 6) + 5*
exp(4*x + (4*x)/log(2) + 8)