\(\int \frac {e^{\frac {e^{15}}{x}} (-36 e^{15}+36 x+(48 e^{15}-48 x) \log (2)+(-16 e^{15}+16 x) \log ^2(2))}{3 x} \, dx\) [4811]

Optimal result

Integrand size = 51, antiderivative size = 22 \[ \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{3 x} \, dx=\frac {4}{3} e^{\frac {e^{15}}{x}} x (3-2 \log (2))^2 \]

[Out]

4/3*exp(exp(15)/x)*(3-2*ln(2))^2*x

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12, 6820, 2326} \[ \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{3 x} \, dx=\frac {4}{3} e^{\frac {e^{15}}{x}} x (3-\log (4))^2 \]

[In]

Int[(E^(E^15/x)*(-36*E^15 + 36*x + (48*E^15 - 48*x)*Log[2] + (-16*E^15 + 16*x)*Log[2]^2))/(3*x),x]

[Out]

(4*E^(E^15/x)*x*(3 - Log[4])^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 2326

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 6820

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{x} \, dx \\ & = \frac {1}{3} \int \frac {4 e^{\frac {e^{15}}{x}} \left (-e^{15}+x\right ) (3-\log (4))^2}{x} \, dx \\ & = \frac {1}{3} \left (4 (3-\log (4))^2\right ) \int \frac {e^{\frac {e^{15}}{x}} \left (-e^{15}+x\right )}{x} \, dx \\ & = \frac {4}{3} e^{\frac {e^{15}}{x}} x (3-\log (4))^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{3 x} \, dx=\frac {4}{3} e^{\frac {e^{15}}{x}} x (-3+\log (4))^2 \]

[In]

Integrate[(E^(E^15/x)*(-36*E^15 + 36*x + (48*E^15 - 48*x)*Log[2] + (-16*E^15 + 16*x)*Log[2]^2))/(3*x),x]

[Out]

(4*E^(E^15/x)*x*(-3 + Log[4])^2)/3

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00

method	result	size
norman	\(\left (\frac {16 \ln \left (2\right )^{2}}{3}-16 \ln \left (2\right )+12\right ) x \,{\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}\)	\(22\)
gosper	\(\frac {4 x \left (4 \ln \left (2\right )^{2}-12 \ln \left (2\right )+9\right ) {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}}{3}\)	\(23\)
risch	\(\frac {4 x \left (4 \ln \left (2\right )^{2}-12 \ln \left (2\right )+9\right ) {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}}{3}\)	\(23\)
parallelrisch	\(\frac {16 \,{\mathrm e}^{\frac {{\mathrm e}^{15}}{x}} \ln \left (2\right )^{2} x}{3}-16 \ln \left (2\right ) x \,{\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}+12 x \,{\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}\)	\(38\)
derivativedivides	\(-12 \,{\mathrm e}^{15} \left (-x \,{\mathrm e}^{-15} {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )-12 \,{\mathrm e}^{15} \operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )+16 \ln \left (2\right ) {\mathrm e}^{15} \left (-x \,{\mathrm e}^{-15} {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )-\frac {16 \ln \left (2\right )^{2} {\mathrm e}^{15} \left (-x \,{\mathrm e}^{-15} {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )}{3}+16 \ln \left (2\right ) {\mathrm e}^{15} \operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )-\frac {16 \ln \left (2\right )^{2} {\mathrm e}^{15} \operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )}{3}\)	\(143\)
default	\(-12 \,{\mathrm e}^{15} \left (-x \,{\mathrm e}^{-15} {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )-12 \,{\mathrm e}^{15} \operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )+16 \ln \left (2\right ) {\mathrm e}^{15} \left (-x \,{\mathrm e}^{-15} {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )-\frac {16 \ln \left (2\right )^{2} {\mathrm e}^{15} \left (-x \,{\mathrm e}^{-15} {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )}{3}+16 \ln \left (2\right ) {\mathrm e}^{15} \operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )-\frac {16 \ln \left (2\right )^{2} {\mathrm e}^{15} \operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )}{3}\)	\(143\)
meijerg	\(\left (\frac {16 \ln \left (2\right )^{2}}{3}-16 \ln \left (2\right )+12\right ) {\mathrm e}^{15} \left (x \,{\mathrm e}^{-15}-14+\ln \left (x \right )-i \pi -\frac {x \,{\mathrm e}^{-15} \left (2+\frac {2 \,{\mathrm e}^{15}}{x}\right )}{2}+x \,{\mathrm e}^{-15+\frac {{\mathrm e}^{15}}{x}}+\ln \left (-\frac {{\mathrm e}^{15}}{x}\right )+\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )+\frac {16 \ln \left (2\right )^{2} {\mathrm e}^{15} \left (-\ln \left (x \right )+15+i \pi -\ln \left (-\frac {{\mathrm e}^{15}}{x}\right )-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )}{3}-16 \ln \left (2\right ) {\mathrm e}^{15} \left (-\ln \left (x \right )+15+i \pi -\ln \left (-\frac {{\mathrm e}^{15}}{x}\right )-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )+12 \,{\mathrm e}^{15} \left (-\ln \left (x \right )+15+i \pi -\ln \left (-\frac {{\mathrm e}^{15}}{x}\right )-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )\)	\(182\)

[In]

int(1/3*((-16*exp(15)+16*x)*ln(2)^2+(48*exp(15)-48*x)*ln(2)-36*exp(15)+36*x)*exp(exp(15)/x)/x,x,method=_RETURN
VERBOSE)

[Out]

(16/3*ln(2)^2-16*ln(2)+12)*x*exp(exp(15)/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{3 x} \, dx=\frac {4}{3} \, {\left (4 \, x \log \left (2\right )^{2} - 12 \, x \log \left (2\right ) + 9 \, x\right )} e^{\left (\frac {e^{15}}{x}\right )} \]

[In]

integrate(1/3*((-16*exp(15)+16*x)*log(2)^2+(48*exp(15)-48*x)*log(2)-36*exp(15)+36*x)*exp(exp(15)/x)/x,x, algor
ithm="fricas")

[Out]

4/3*(4*x*log(2)^2 - 12*x*log(2) + 9*x)*e^(e^15/x)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{3 x} \, dx=\frac {\left (- 48 x \log {\left (2 \right )} + 16 x \log {\left (2 \right )}^{2} + 36 x\right ) e^{\frac {e^{15}}{x}}}{3} \]

[In]

integrate(1/3*((-16*exp(15)+16*x)*ln(2)**2+(48*exp(15)-48*x)*ln(2)-36*exp(15)+36*x)*exp(exp(15)/x)/x,x)

[Out]

(-48*x*log(2) + 16*x*log(2)**2 + 36*x)*exp(exp(15)/x)/3

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.23 (sec) , antiderivative size = 85, normalized size of antiderivative = 3.86 \[ \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{3 x} \, dx=\frac {16}{3} \, {\rm Ei}\left (\frac {e^{15}}{x}\right ) e^{15} \log \left (2\right )^{2} - \frac {16}{3} \, e^{15} \Gamma \left (-1, -\frac {e^{15}}{x}\right ) \log \left (2\right )^{2} - 16 \, {\rm Ei}\left (\frac {e^{15}}{x}\right ) e^{15} \log \left (2\right ) + 16 \, e^{15} \Gamma \left (-1, -\frac {e^{15}}{x}\right ) \log \left (2\right ) + 12 \, {\rm Ei}\left (\frac {e^{15}}{x}\right ) e^{15} - 12 \, e^{15} \Gamma \left (-1, -\frac {e^{15}}{x}\right ) \]

[In]

integrate(1/3*((-16*exp(15)+16*x)*log(2)^2+(48*exp(15)-48*x)*log(2)-36*exp(15)+36*x)*exp(exp(15)/x)/x,x, algor
ithm="maxima")

[Out]

16/3*Ei(e^15/x)*e^15*log(2)^2 - 16/3*e^15*gamma(-1, -e^15/x)*log(2)^2 - 16*Ei(e^15/x)*e^15*log(2) + 16*e^15*ga
mma(-1, -e^15/x)*log(2) + 12*Ei(e^15/x)*e^15 - 12*e^15*gamma(-1, -e^15/x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (18) = 36\).

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.05 \[ \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{3 x} \, dx=\frac {4}{3} \, {\left (4 \, e^{\left (\frac {e^{15}}{x} + 45\right )} \log \left (2\right )^{2} - 12 \, e^{\left (\frac {e^{15}}{x} + 45\right )} \log \left (2\right ) + 9 \, e^{\left (\frac {e^{15}}{x} + 45\right )}\right )} x e^{\left (-45\right )} \]

[In]

integrate(1/3*((-16*exp(15)+16*x)*log(2)^2+(48*exp(15)-48*x)*log(2)-36*exp(15)+36*x)*exp(exp(15)/x)/x,x, algor
ithm="giac")

[Out]

4/3*(4*e^(e^15/x + 45)*log(2)^2 - 12*e^(e^15/x + 45)*log(2) + 9*e^(e^15/x + 45))*x*e^(-45)

Mupad [B] (verification not implemented)

Time = 10.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{3 x} \, dx=\frac {4\,x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{15}}{x}}\,{\left (\ln \left (4\right )-3\right )}^2}{3} \]

[In]

int((exp(exp(15)/x)*(36*x - 36*exp(15) + log(2)^2*(16*x - 16*exp(15)) - log(2)*(48*x - 48*exp(15))))/(3*x),x)

[Out]

(4*x*exp(exp(15)/x)*(log(4) - 3)^2)/3