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

Optimal. Leaf size=22 \[ \frac {4}{3} e^{\frac {e^{15}}{x}} x (3-2 \log (2))^2 \]

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Rubi [A]  time = 0.17, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12, 6688, 2288} \begin {gather*} \frac {4}{3} e^{\frac {e^{15}}{x}} x (3-\log (4))^2 \end {gather*}

Antiderivative was successfully verified.

[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 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} &=\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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.91 \begin {gather*} \frac {4}{3} e^{\frac {e^{15}}{x}} x (-3+\log (4))^2 \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.93, size = 25, normalized size = 1.14 \begin {gather*} \frac {4}{3} \, {\left (4 \, x \log \relax (2)^{2} - 12 \, x \log \relax (2) + 9 \, x\right )} e^{\left (\frac {e^{15}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.16, size = 45, normalized size = 2.05 \begin {gather*} \frac {4}{3} \, {\left (4 \, e^{\left (\frac {e^{15}}{x} + 45\right )} \log \relax (2)^{2} - 12 \, e^{\left (\frac {e^{15}}{x} + 45\right )} \log \relax (2) + 9 \, e^{\left (\frac {e^{15}}{x} + 45\right )}\right )} x e^{\left (-45\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.22, size = 22, normalized size = 1.00

method	result	size
norman	\(\left (\frac {16 \ln \relax (2)^{2}}{3}-16 \ln \relax (2)+12\right ) x \,{\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}\)	\(22\)
gosper	\(\frac {4 x \left (4 \ln \relax (2)^{2}-12 \ln \relax (2)+9\right ) {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}}{3}\)	\(23\)
risch	\(\frac {4 x \left (4 \ln \relax (2)^{2}-12 \ln \relax (2)+9\right ) {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}}{3}\)	\(23\)
derivativedivides	\(-12 \,{\mathrm e}^{15} \left (-x \,{\mathrm e}^{-15} {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}-\expIntegralEi \left (1, -\frac {{\mathrm e}^{15}}{x}\right )\right )-12 \,{\mathrm e}^{15} \expIntegralEi \left (1, -\frac {{\mathrm e}^{15}}{x}\right )-16 \,{\mathrm e}^{\frac {{\mathrm e}^{15}}{x}} x \ln \relax (2)+\frac {16 \,{\mathrm e}^{\frac {{\mathrm e}^{15}}{x}} x \ln \relax (2)^{2}}{3}\)	\(71\)
default	\(-12 \,{\mathrm e}^{15} \left (-x \,{\mathrm e}^{-15} {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}-\expIntegralEi \left (1, -\frac {{\mathrm e}^{15}}{x}\right )\right )-12 \,{\mathrm e}^{15} \expIntegralEi \left (1, -\frac {{\mathrm e}^{15}}{x}\right )-16 \,{\mathrm e}^{\frac {{\mathrm e}^{15}}{x}} x \ln \relax (2)+\frac {16 \,{\mathrm e}^{\frac {{\mathrm e}^{15}}{x}} x \ln \relax (2)^{2}}{3}\)	\(71\)
meijerg	\(\left (\frac {16 \ln \relax (2)^{2}}{3}-16 \ln \relax (2)+12\right ) {\mathrm e}^{15} \left (-\frac {x \,{\mathrm e}^{-15} \left (2+\frac {2 \,{\mathrm e}^{15}}{x}\right )}{2}+x \,{\mathrm e}^{-15+\frac {{\mathrm e}^{15}}{x}}+\ln \left (-\frac {{\mathrm e}^{15}}{x}\right )+\expIntegralEi \left (1, -\frac {{\mathrm e}^{15}}{x}\right )-14+\ln \relax (x )-i \pi +x \,{\mathrm e}^{-15}\right )+\frac {16 \ln \relax (2)^{2} {\mathrm e}^{15} \left (-\ln \left (-\frac {{\mathrm e}^{15}}{x}\right )-\expIntegralEi \left (1, -\frac {{\mathrm e}^{15}}{x}\right )-\ln \relax (x )+15+i \pi \right )}{3}-16 \ln \relax (2) {\mathrm e}^{15} \left (-\ln \left (-\frac {{\mathrm e}^{15}}{x}\right )-\expIntegralEi \left (1, -\frac {{\mathrm e}^{15}}{x}\right )-\ln \relax (x )+15+i \pi \right )+12 \,{\mathrm e}^{15} \left (-\ln \left (-\frac {{\mathrm e}^{15}}{x}\right )-\expIntegralEi \left (1, -\frac {{\mathrm e}^{15}}{x}\right )-\ln \relax (x )+15+i \pi \right )\)	\(182\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [C]  time = 0.41, size = 85, normalized size = 3.86 \begin {gather*} \frac {16}{3} \, {\rm Ei}\left (\frac {e^{15}}{x}\right ) e^{15} \log \relax (2)^{2} - \frac {16}{3} \, e^{15} \Gamma \left (-1, -\frac {e^{15}}{x}\right ) \log \relax (2)^{2} - 16 \, {\rm Ei}\left (\frac {e^{15}}{x}\right ) e^{15} \log \relax (2) + 16 \, e^{15} \Gamma \left (-1, -\frac {e^{15}}{x}\right ) \log \relax (2) + 12 \, {\rm Ei}\left (\frac {e^{15}}{x}\right ) e^{15} - 12 \, e^{15} \Gamma \left (-1, -\frac {e^{15}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.47, size = 16, normalized size = 0.73 \begin {gather*} \frac {4\,x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{15}}{x}}\,{\left (\ln \relax (4)-3\right )}^2}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.23, size = 26, normalized size = 1.18 \begin {gather*} \frac {\left (- 48 x \log {\relax (2 )} + 16 x \log {\relax (2 )}^{2} + 36 x\right ) e^{\frac {e^{15}}{x}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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