3.17.44 $\int \frac{e^{3+x\log (\frac{e+40x+40\log (2)+(5x+5\log (2))\log (x)}{8+\log (x)})}(-e+320x+80x\log (x)+5x{\log }^2(x)+(8e+320x+320\log (2)+(e+80x+80\log (2))\log (x)+(5x+5\log (2)){\log }^2(x))\log (\frac{e+40x+40\log (2)+(5x+5\log (2))\log (x)}{8+\log (x)}))}{8e+320x+320\log (2)+(e+80x+80\log (2))\log (x)+(5x+5\log (2)){\log }^2(x)}dx$

Optimal. Leaf size=22 $$e^{3+x\log (5(x+\log (2))+\frac{e}{8+\log (x)})}$$

________________________________________________________________________________________

Rubi [F]  time = 3.78, 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 = {} $$\begin{array}{c}\int \frac{\exp (3+x\log \left(\frac{e+40x+40\log (2)+(5x+5\log (2))\log (x)}{8+\log (x)}\right))(-e+320x+80x\log (x)+5x{\log }^2(x)+(8e+320x+320\log (2)+(e+80x+80\log (2))\log (x)+(5x+5\log (2)){\log }^2(x))\log \left(\frac{e+40x+40\log (2)+(5x+5\log (2))\log (x)}{8+\log (x)}\right))}{8e+320x+320\log (2)+(e+80x+80\log (2))\log (x)+(5x+5\log (2)){\log }^2(x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(E^(3 + x*Log[(E + 40*x + 40*Log[2] + (5*x + 5*Log[2])*Log[x])/(8 + Log[x])])*(-E + 320*x + 80*x*Log[x] +
5*x*Log[x]^2 + (8*E + 320*x + 320*Log[2] + (E + 80*x + 80*Log[2])*Log[x] + (5*x + 5*Log[2])*Log[x]^2)*Log[(E +
40*x + 40*Log[2] + (5*x + 5*Log[2])*Log[x])/(8 + Log[x])]))/(8*E + 320*x + 320*Log[2] + (E + 80*x + 80*Log[2]
)*Log[x] + (5*x + 5*Log[2])*Log[x]^2),x]

[Out]

5*E^3*Defer[Int][x*((E + 40*(x + Log[2]) + 5*(x + Log[2])*Log[x])/(8 + Log[x]))^(-1 + x), x] - E^4*Defer[Int][
((E + 40*(x + Log[2]) + 5*(x + Log[2])*Log[x])/(8 + Log[x]))^(-1 + x)/(8 + Log[x])^2, x] + E^3*Defer[Int][((E
+ 40*(x + Log[2]) + 5*(x + Log[2])*Log[x])/(8 + Log[x]))^x*Log[(E + 40*(x + Log[2]) + 5*(x + Log[2])*Log[x])/(
8 + Log[x])], x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^3{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}(-e+320x+80x\log (x)+5x{\log }^2(x)+(8+\log (x))(e+40(x+\log (2))+5(x+\log (2))\log (x))\log \left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right))}{(8+\log (x){)}^2}dx\\ & =e^3\int \frac{{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}(-e+320x+80x\log (x)+5x{\log }^2(x)+(8+\log (x))(e+40(x+\log (2))+5(x+\log (2))\log (x))\log \left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right))}{(8+\log (x){)}^2}dx\\ & =e^3\int (-\frac{e{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{(8+\log (x){)}^2}+\frac{320x{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{(8+\log (x){)}^2}+\frac{80x\log (x){\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{(8+\log (x){)}^2}+\frac{5x{\log }^2(x){\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{(8+\log (x){)}^2}+\frac{(40x+e(1+\frac{40\log (2)}{e})+5x\log (x)+5\log (2)\log (x)){\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}\log \left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}{8+\log (x)})dx\\ & =e^3\int \frac{(40x+e(1+\frac{40\log (2)}{e})+5x\log (x)+5\log (2)\log (x)){\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}\log \left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}{8+\log (x)}dx+\left(5e^3\right)\int \frac{x{\log }^2(x){\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{(8+\log (x){)}^2}dx+\left(80e^3\right)\int \frac{x\log (x){\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{(8+\log (x){)}^2}dx+\left(320e^3\right)\int \frac{x{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{(8+\log (x){)}^2}dx-e^4\int \frac{{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{(8+\log (x){)}^2}dx\\ & =e^3\int {\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^x\log \left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)dx+\left(5e^3\right)\int (x{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}+\frac{64x{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{(8+\log (x){)}^2}-\frac{16x{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{8+\log (x)})dx+\left(80e^3\right)\int (-\frac{8x{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{(8+\log (x){)}^2}+\frac{x{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{8+\log (x)})dx+\left(320e^3\right)\int \frac{x{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{(8+\log (x){)}^2}dx-e^4\int \frac{{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{(8+\log (x){)}^2}dx\\ & =e^3\int {\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^x\log \left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)dx+\left(5e^3\right)\int x{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}dx+2(\left(320e^3\right)\int \frac{x{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{(8+\log (x){)}^2}dx)-\left(640e^3\right)\int \frac{x{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{(8+\log (x){)}^2}dx-e^4\int \frac{{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^{-1+x}}{(8+\log (x){)}^2}dx\end{array}\end{array}$$

________________________________________________________________________________________

Mathematica [A]  time = 0.14, size = 29, normalized size = 1.32 $$\begin{array}{c}{e}^3{\left(\frac{e+40(x+\log (2))+5(x+\log (2))\log (x)}{8+\log (x)}\right)}^x\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(3 + x*Log[(E + 40*x + 40*Log[2] + (5*x + 5*Log[2])*Log[x])/(8 + Log[x])])*(-E + 320*x + 80*x*Log
[x] + 5*x*Log[x]^2 + (8*E + 320*x + 320*Log[2] + (E + 80*x + 80*Log[2])*Log[x] + (5*x + 5*Log[2])*Log[x]^2)*Lo
g[(E + 40*x + 40*Log[2] + (5*x + 5*Log[2])*Log[x])/(8 + Log[x])]))/(8*E + 320*x + 320*Log[2] + (E + 80*x + 80*
Log[2])*Log[x] + (5*x + 5*Log[2])*Log[x]^2),x]

[Out]

E^3*((E + 40*(x + Log[2]) + 5*(x + Log[2])*Log[x])/(8 + Log[x]))^x

________________________________________________________________________________________

fricas [A]  time = 0.88, size = 31, normalized size = 1.41 $$\begin{array}{c}{e}^{(x\log \left(\frac{5(x+\log (2))\log (x)+40x+e+40\log (2)}{\log (x)+8}\right)+3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log(2)+8*exp(1)+320*x)*log(((5*log(2)+5
*x)*log(x)+40*log(2)+exp(1)+40*x)/(log(x)+8))+5*x*log(x)^2+80*x*log(x)-exp(1)+320*x)*exp(x*log(((5*log(2)+5*x)
*log(x)+40*log(2)+exp(1)+40*x)/(log(x)+8))+3)/((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log(
2)+8*exp(1)+320*x),x, algorithm="fricas")

[Out]

e^(x*log((5*(x + log(2))*log(x) + 40*x + e + 40*log(2))/(log(x) + 8)) + 3)

________________________________________________________________________________________

giac [B]  time = 0.88, size = 58, normalized size = 2.64 $$\begin{array}{c}{e}^{(x\log (\frac{5x\log (x)}{\log (x)+8}+\frac{5\log (2)\log (x)}{\log (x)+8}+\frac{40x}{\log (x)+8}+\frac{e}{\log (x)+8}+\frac{40\log (2)}{\log (x)+8})+3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log(2)+8*exp(1)+320*x)*log(((5*log(2)+5
*x)*log(x)+40*log(2)+exp(1)+40*x)/(log(x)+8))+5*x*log(x)^2+80*x*log(x)-exp(1)+320*x)*exp(x*log(((5*log(2)+5*x)
*log(x)+40*log(2)+exp(1)+40*x)/(log(x)+8))+3)/((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log(
2)+8*exp(1)+320*x),x, algorithm="giac")

[Out]

e^(x*log(5*x*log(x)/(log(x) + 8) + 5*log(2)*log(x)/(log(x) + 8) + 40*x/(log(x) + 8) + e/(log(x) + 8) + 40*log(
2)/(log(x) + 8)) + 3)

________________________________________________________________________________________

maple [C]  time = 0.65, size = 249, normalized size = 11.32

method	result	size
risch	${(\ln (x)+8)}^{-x}{(\mathrm{e}+(5\ln (x)+40)\ln (2)+(5\ln (x)+40)x)}^x{\mathrm{e}}^{3-\frac{ix\pi \mathrm{csgn}{\left(\frac{i(\mathrm{e}+(5\ln (x)+40)\ln (2)+(5\ln (x)+40)x)}{\ln (x)+8}\right)}^3}{2}+\frac{ix\pi \mathrm{csgn}{\left(\frac{i(\mathrm{e}+(5\ln (x)+40)\ln (2)+(5\ln (x)+40)x)}{\ln (x)+8}\right)}^2\mathrm{csgn}\left(\frac{i}{\ln (x)+8}\right)}{2}+\frac{ix\pi \mathrm{csgn}{\left(\frac{i(\mathrm{e}+(5\ln (x)+40)\ln (2)+(5\ln (x)+40)x)}{\ln (x)+8}\right)}^2\mathrm{csgn}\left(i(\mathrm{e}+(5\ln (x)+40)\ln (2)+(5\ln (x)+40)x)\right)}{2}-\frac{ix\pi \mathrm{csgn}\left(\frac{i(\mathrm{e}+(5\ln (x)+40)\ln (2)+(5\ln (x)+40)x)}{\ln (x)+8}\right)\mathrm{csgn}\left(\frac{i}{\ln (x)+8}\right)\mathrm{csgn}\left(i(\mathrm{e}+(5\ln (x)+40)\ln (2)+(5\ln (x)+40)x)\right)}{2}}$	$249$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((5*ln(2)+5*x)*ln(x)^2+(80*ln(2)+exp(1)+80*x)*ln(x)+320*ln(2)+8*exp(1)+320*x)*ln(((5*ln(2)+5*x)*ln(x)+40*
ln(2)+exp(1)+40*x)/(ln(x)+8))+5*x*ln(x)^2+80*x*ln(x)-exp(1)+320*x)*exp(x*ln(((5*ln(2)+5*x)*ln(x)+40*ln(2)+exp(
1)+40*x)/(ln(x)+8))+3)/((5*ln(2)+5*x)*ln(x)^2+(80*ln(2)+exp(1)+80*x)*ln(x)+320*ln(2)+8*exp(1)+320*x),x,method=
_RETURNVERBOSE)

[Out]

(ln(x)+8)^(-x)*(exp(1)+(5*ln(x)+40)*ln(2)+(5*ln(x)+40)*x)^x*exp(3-1/2*I*x*Pi*csgn(I/(ln(x)+8)*(exp(1)+(5*ln(x)
+40)*ln(2)+(5*ln(x)+40)*x))^3+1/2*I*x*Pi*csgn(I/(ln(x)+8)*(exp(1)+(5*ln(x)+40)*ln(2)+(5*ln(x)+40)*x))^2*csgn(I
/(ln(x)+8))+1/2*I*x*Pi*csgn(I/(ln(x)+8)*(exp(1)+(5*ln(x)+40)*ln(2)+(5*ln(x)+40)*x))^2*csgn(I*(exp(1)+(5*ln(x)+
40)*ln(2)+(5*ln(x)+40)*x))-1/2*I*x*Pi*csgn(I/(ln(x)+8)*(exp(1)+(5*ln(x)+40)*ln(2)+(5*ln(x)+40)*x))*csgn(I/(ln(
x)+8))*csgn(I*(exp(1)+(5*ln(x)+40)*ln(2)+(5*ln(x)+40)*x)))

________________________________________________________________________________________

maxima [A]  time = 0.74, size = 32, normalized size = 1.45 $$\begin{array}{c}{e}^{(x\log (5(x+\log (2))\log (x)+40x+e+40\log (2))-x\log (\log (x)+8)+3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log(2)+8*exp(1)+320*x)*log(((5*log(2)+5
*x)*log(x)+40*log(2)+exp(1)+40*x)/(log(x)+8))+5*x*log(x)^2+80*x*log(x)-exp(1)+320*x)*exp(x*log(((5*log(2)+5*x)
*log(x)+40*log(2)+exp(1)+40*x)/(log(x)+8))+3)/((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log(
2)+8*exp(1)+320*x),x, algorithm="maxima")

[Out]

e^(x*log(5*(x + log(2))*log(x) + 40*x + e + 40*log(2)) - x*log(log(x) + 8) + 3)

________________________________________________________________________________________

mupad [B]  time = 2.15, size = 33, normalized size = 1.50 $$\begin{array}{c}{\mathrm{e}}^3{\left(\frac{40x+\mathrm{e}+40\ln (2)+5\ln (2)\ln (x)+5x\ln (x)}{\ln (x)+8}\right)}^x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x*log((40*x + exp(1) + 40*log(2) + log(x)*(5*x + 5*log(2)))/(log(x) + 8)) + 3)*(320*x - exp(1) + 5*x*
log(x)^2 + log((40*x + exp(1) + 40*log(2) + log(x)*(5*x + 5*log(2)))/(log(x) + 8))*(320*x + 8*exp(1) + 320*log
(2) + log(x)^2*(5*x + 5*log(2)) + log(x)*(80*x + exp(1) + 80*log(2))) + 80*x*log(x)))/(320*x + 8*exp(1) + 320*
log(2) + log(x)^2*(5*x + 5*log(2)) + log(x)*(80*x + exp(1) + 80*log(2))),x)

[Out]

exp(3)*((40*x + exp(1) + 40*log(2) + 5*log(2)*log(x) + 5*x*log(x))/(log(x) + 8))^x

________________________________________________________________________________________

sympy [A]  time = 2.69, size = 34, normalized size = 1.55 $$\begin{array}{c}{e}^{x\log \left(\frac{40x+(5x+5\log (2))\log (x)+e+40\log (2)}{\log (x)+8}\right)+3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*ln(2)+5*x)*ln(x)**2+(80*ln(2)+exp(1)+80*x)*ln(x)+320*ln(2)+8*exp(1)+320*x)*ln(((5*ln(2)+5*x)*ln
(x)+40*ln(2)+exp(1)+40*x)/(ln(x)+8))+5*x*ln(x)**2+80*x*ln(x)-exp(1)+320*x)*exp(x*ln(((5*ln(2)+5*x)*ln(x)+40*ln
(2)+exp(1)+40*x)/(ln(x)+8))+3)/((5*ln(2)+5*x)*ln(x)**2+(80*ln(2)+exp(1)+80*x)*ln(x)+320*ln(2)+8*exp(1)+320*x),
x)

[Out]

exp(x*log((40*x + (5*x + 5*log(2))*log(x) + E + 40*log(2))/(log(x) + 8)) + 3)

________________________________________________________________________________________