3.55.11 \(\int \frac {e^{\frac {e^3+3 x \log (\frac {-3 x+(11+x) \log (2)}{\log (2)})}{x}} (24 e^3 x-72 x^2+(e^3 (-88-8 x)+24 x^2) \log (2))}{-3 x^3+(11 x^2+x^3) \log (2)} \, dx\)

Optimal. Leaf size=29 \[ 8 \left (1+e^{\frac {e^3}{x}} \left (-4+x-3 \left (-5+\frac {x}{\log (2)}\right )\right )^3\right ) \]

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Rubi [A]  time = 0.50, antiderivative size = 30, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6688, 12, 2288} \begin {gather*} -\frac {8 e^{\frac {e^3}{x}} (x (3-\log (2))-\log (2048))^3}{\log ^3(2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((E^3 + 3*x*Log[(-3*x + (11 + x)*Log[2])/Log[2]])/x)*(24*E^3*x - 72*x^2 + (E^3*(-88 - 8*x) + 24*x^2)*Lo
g[2]))/(-3*x^3 + (11*x^2 + x^3)*Log[2]),x]

[Out]

(-8*E^(E^3/x)*(x*(3 - Log[2]) - Log[2048])^3)/Log[2]^3

Rule 12

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

Rule 2288

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]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8 e^{\frac {e^3}{x}} (x (-3+\log (2))+\log (2048))^2 \left (e^3 x (3-\log (2))-x^2 (9-\log (8))-e^3 \log (2048)\right )}{x^2 \log ^3(2)} \, dx\\ &=\frac {8 \int \frac {e^{\frac {e^3}{x}} (x (-3+\log (2))+\log (2048))^2 \left (e^3 x (3-\log (2))-x^2 (9-\log (8))-e^3 \log (2048)\right )}{x^2} \, dx}{\log ^3(2)}\\ &=-\frac {8 e^{\frac {e^3}{x}} (x (3-\log (2))-\log (2048))^3}{\log ^3(2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 26, normalized size = 0.90 \begin {gather*} \frac {8 e^{\frac {e^3}{x}} (x (-3+\log (2))+\log (2048))^3}{\log ^3(2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((E^3 + 3*x*Log[(-3*x + (11 + x)*Log[2])/Log[2]])/x)*(24*E^3*x - 72*x^2 + (E^3*(-88 - 8*x) + 24*x
^2)*Log[2]))/(-3*x^3 + (11*x^2 + x^3)*Log[2]),x]

[Out]

(8*E^(E^3/x)*(x*(-3 + Log[2]) + Log[2048])^3)/Log[2]^3

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fricas [A]  time = 0.63, size = 29, normalized size = 1.00 \begin {gather*} 8 \, e^{\left (\frac {3 \, x \log \left (\frac {{\left (x + 11\right )} \log \relax (2) - 3 \, x}{\log \relax (2)}\right ) + e^{3}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x-88)*exp(3)+24*x^2)*log(2)+24*x*exp(3)-72*x^2)*exp((3*x*log(((11+x)*log(2)-3*x)/log(2))+exp(3
))/x)/((x^3+11*x^2)*log(2)-3*x^3),x, algorithm="fricas")

[Out]

8*e^((3*x*log(((x + 11)*log(2) - 3*x)/log(2)) + e^3)/x)

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giac [B]  time = 0.44, size = 171, normalized size = 5.90 \begin {gather*} \frac {8 \, {\left (e^{\left (\frac {e^{3}}{x} + 12\right )} \log \relax (2)^{3} - 9 \, e^{\left (\frac {e^{3}}{x} + 12\right )} \log \relax (2)^{2} + \frac {33 \, e^{\left (\frac {e^{3}}{x} + 12\right )} \log \relax (2)^{3}}{x} + 27 \, e^{\left (\frac {e^{3}}{x} + 12\right )} \log \relax (2) - \frac {198 \, e^{\left (\frac {e^{3}}{x} + 12\right )} \log \relax (2)^{2}}{x} + \frac {363 \, e^{\left (\frac {e^{3}}{x} + 12\right )} \log \relax (2)^{3}}{x^{2}} + \frac {297 \, e^{\left (\frac {e^{3}}{x} + 12\right )} \log \relax (2)}{x} - \frac {1089 \, e^{\left (\frac {e^{3}}{x} + 12\right )} \log \relax (2)^{2}}{x^{2}} + \frac {1331 \, e^{\left (\frac {e^{3}}{x} + 12\right )} \log \relax (2)^{3}}{x^{3}} - 27 \, e^{\left (\frac {e^{3}}{x} + 12\right )}\right )} x^{3} e^{\left (-12\right )}}{\log \relax (2)^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x-88)*exp(3)+24*x^2)*log(2)+24*x*exp(3)-72*x^2)*exp((3*x*log(((11+x)*log(2)-3*x)/log(2))+exp(3
))/x)/((x^3+11*x^2)*log(2)-3*x^3),x, algorithm="giac")

[Out]

8*(e^(e^3/x + 12)*log(2)^3 - 9*e^(e^3/x + 12)*log(2)^2 + 33*e^(e^3/x + 12)*log(2)^3/x + 27*e^(e^3/x + 12)*log(
2) - 198*e^(e^3/x + 12)*log(2)^2/x + 363*e^(e^3/x + 12)*log(2)^3/x^2 + 297*e^(e^3/x + 12)*log(2)/x - 1089*e^(e
^3/x + 12)*log(2)^2/x^2 + 1331*e^(e^3/x + 12)*log(2)^3/x^3 - 27*e^(e^3/x + 12))*x^3*e^(-12)/log(2)^3

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maple [A]  time = 0.45, size = 30, normalized size = 1.03

method	result	size
norman	\(8 \,{\mathrm e}^{\frac {3 x \ln \left (\frac {\left (11+x \right ) \ln \relax (2)-3 x}{\ln \relax (2)}\right )+{\mathrm e}^{3}}{x}}\)	\(30\)
gosper	\(8 \,{\mathrm e}^{\frac {3 x \ln \left (\frac {x \ln \relax (2)+11 \ln \relax (2)-3 x}{\ln \relax (2)}\right )+{\mathrm e}^{3}}{x}}\)	\(32\)
risch	\(\frac {8 \left (x^{3} \ln \relax (2)^{3}+33 x^{2} \ln \relax (2)^{3}-9 x^{3} \ln \relax (2)^{2}+363 x \ln \relax (2)^{3}-198 x^{2} \ln \relax (2)^{2}+27 x^{3} \ln \relax (2)+1331 \ln \relax (2)^{3}-1089 x \ln \relax (2)^{2}+297 x^{2} \ln \relax (2)-27 x^{3}\right ) \left (\left (11+x \right ) \ln \relax (2)-3 x \right )^{3} {\mathrm e}^{\frac {{\mathrm e}^{3}}{x}}}{\left (x \ln \relax (2)+11 \ln \relax (2)-3 x \right )^{3} \ln \relax (2)^{3}}\)	\(115\)
default	\(\text {Expression too large to display}\)	\(3586\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-8*x-88)*exp(3)+24*x^2)*ln(2)+24*x*exp(3)-72*x^2)*exp((3*x*ln(((11+x)*ln(2)-3*x)/ln(2))+exp(3))/x)/((x^
3+11*x^2)*ln(2)-3*x^3),x,method=_RETURNVERBOSE)

[Out]

8*exp((3*x*ln(((11+x)*ln(2)-3*x)/ln(2))+exp(3))/x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {8 \, {\left ({\left (\log \relax (2)^{3} - 9 \, \log \relax (2)^{2} + 27 \, \log \relax (2) - 27\right )} x^{3} + 33 \, {\left (\log \relax (2)^{3} - 6 \, \log \relax (2)^{2} + 9 \, \log \relax (2)\right )} x^{2} + 363 \, {\left (\log \relax (2)^{3} - 3 \, \log \relax (2)^{2}\right )} x\right )} e^{\left (\frac {e^{3}}{x}\right )}}{\log \relax (2)^{3}} - 10648 \, \int \frac {e^{\left (\frac {e^{3}}{x} + 3\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x-88)*exp(3)+24*x^2)*log(2)+24*x*exp(3)-72*x^2)*exp((3*x*log(((11+x)*log(2)-3*x)/log(2))+exp(3
))/x)/((x^3+11*x^2)*log(2)-3*x^3),x, algorithm="maxima")

[Out]

8*((log(2)^3 - 9*log(2)^2 + 27*log(2) - 27)*x^3 + 33*(log(2)^3 - 6*log(2)^2 + 9*log(2))*x^2 + 363*(log(2)^3 -
3*log(2)^2)*x)*e^(e^3/x)/log(2)^3 - 10648*integrate(e^(e^3/x + 3)/x^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^3+3\,x\,\ln \left (-\frac {3\,x-\ln \relax (2)\,\left (x+11\right )}{\ln \relax (2)}\right )}{x}}\,\left (24\,x\,{\mathrm {e}}^3+\ln \relax (2)\,\left (24\,x^2-{\mathrm {e}}^3\,\left (8\,x+88\right )\right )-72\,x^2\right )}{\ln \relax (2)\,\left (x^3+11\,x^2\right )-3\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((exp(3) + 3*x*log(-(3*x - log(2)*(x + 11))/log(2)))/x)*(24*x*exp(3) + log(2)*(24*x^2 - exp(3)*(8*x +
88)) - 72*x^2))/(log(2)*(11*x^2 + x^3) - 3*x^3),x)

[Out]

int((exp((exp(3) + 3*x*log(-(3*x - log(2)*(x + 11))/log(2)))/x)*(24*x*exp(3) + log(2)*(24*x^2 - exp(3)*(8*x +
88)) - 72*x^2))/(log(2)*(11*x^2 + x^3) - 3*x^3), x)

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sympy [A]  time = 95.25, size = 26, normalized size = 0.90 \begin {gather*} 8 e^{\frac {3 x \log {\left (\frac {- 3 x + \left (x + 11\right ) \log {\relax (2 )}}{\log {\relax (2 )}} \right )} + e^{3}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x-88)*exp(3)+24*x**2)*ln(2)+24*x*exp(3)-72*x**2)*exp((3*x*ln(((11+x)*ln(2)-3*x)/ln(2))+exp(3))
/x)/((x**3+11*x**2)*ln(2)-3*x**3),x)

[Out]

8*exp((3*x*log((-3*x + (x + 11)*log(2))/log(2)) + exp(3))/x)

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