3.62.28 \(\int \frac {-2 e^9+e^{9+\frac {1}{2} (4+x+\log (3))}+8 e^9 (i \pi +\log (3))}{8 (i \pi +\log (3))} \, dx\)

Optimal. Leaf size=39 \[ e^9 \left (x-\frac {2+\frac {1}{4} \left (-e^{\frac {1}{2} (4+x+\log (3))}+x\right )}{i \pi +\log (3)}\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 57, normalized size of antiderivative = 1.46, number of steps used = 3, number of rules used = 2, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {12, 2203} \begin {gather*} \frac {e^{\frac {1}{2} (x+4+\log (3))+9}}{4 (\log (3)+i \pi )}-\frac {e^9 x (1-4 i \pi -\log (81))}{4 (\log (3)+i \pi )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^9 + E^(9 + (4 + x + Log[3])/2) + 8*E^9*(I*Pi + Log[3]))/(8*(I*Pi + Log[3])),x]

[Out]

E^(9 + (4 + x + Log[3])/2)/(4*(I*Pi + Log[3])) - (E^9*x*(1 - (4*I)*Pi - Log[81]))/(4*(I*Pi + Log[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 2203

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (-2 e^9+e^{9+\frac {1}{2} (4+x+\log (3))}+8 e^9 (i \pi +\log (3))\right ) \, dx}{8 (i \pi +\log (3))}\\ &=-\frac {e^9 x (1-4 i \pi -\log (81))}{4 (i \pi +\log (3))}+\frac {\int e^{9+\frac {1}{2} (4+x+\log (3))} \, dx}{8 (i \pi +\log (3))}\\ &=\frac {e^{9+\frac {1}{2} (4+x+\log (3))}}{4 (i \pi +\log (3))}-\frac {e^9 x (1-4 i \pi -\log (81))}{4 (i \pi +\log (3))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 48, normalized size = 1.23 \begin {gather*} \frac {e^9 \left (2 \sqrt {3} e^{2+\frac {x}{2}}-2 x+8 i \pi x+8 x \log (3)\right )}{8 (i \pi +\log (3))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^9 + E^(9 + (4 + x + Log[3])/2) + 8*E^9*(I*Pi + Log[3]))/(8*(I*Pi + Log[3])),x]

[Out]

(E^9*(2*Sqrt[3]*E^(2 + x/2) - 2*x + (8*I)*Pi*x + 8*x*Log[3]))/(8*(I*Pi + Log[3]))

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fricas [A]  time = 0.92, size = 38, normalized size = 0.97 \begin {gather*} \frac {{\left (4 i \, \pi - 1\right )} x e^{9} + 4 \, x e^{9} \log \relax (3) + e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, \log \relax (3) + 11\right )}}{4 i \, \pi + 4 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(exp(9)*exp(1/2*log(3)+2+1/2*x)+8*exp(9)*(log(3)+I*pi)-2*exp(9))/(log(3)+I*pi),x, algorithm="fri
cas")

[Out]

((4*I*pi - 1)*x*e^9 + 4*x*e^9*log(3) + e^(1/2*x + 1/2*log(3) + 11))/(4*I*pi + 4*log(3))

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giac [A]  time = 0.14, size = 37, normalized size = 0.95 \begin {gather*} \frac {4 \, {\left (i \, \pi + \log \relax (3)\right )} x e^{9} - x e^{9} + e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, \log \relax (3) + 11\right )}}{4 \, {\left (i \, \pi + \log \relax (3)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(exp(9)*exp(1/2*log(3)+2+1/2*x)+8*exp(9)*(log(3)+I*pi)-2*exp(9))/(log(3)+I*pi),x, algorithm="gia
c")

[Out]

1/4*(4*(I*pi + log(3))*x*e^9 - x*e^9 + e^(1/2*x + 1/2*log(3) + 11))/(I*pi + log(3))

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maple [A]  time = 0.10, size = 46, normalized size = 1.18

method	result	size
default	\(\frac {2 \,{\mathrm e}^{9} {\mathrm e}^{\frac {\ln \relax (3)}{2}+2+\frac {x}{2}}+8 i \pi \,{\mathrm e}^{9} x +8 \ln \relax (3) {\mathrm e}^{9} x -2 x \,{\mathrm e}^{9}}{8 \ln \relax (3)+8 i \pi }\)	\(46\)
derivativedivides	\(\frac {{\mathrm e}^{9} \left ({\mathrm e}^{\frac {\ln \relax (3)}{2}+2+\frac {x}{2}}+\left (8 i \pi +8 \ln \relax (3)-2\right ) \ln \left ({\mathrm e}^{\frac {\ln \relax (3)}{2}+2+\frac {x}{2}}\right )\right )}{4 \ln \relax (3)+4 i \pi }\)	\(47\)
risch	\(\frac {i {\mathrm e}^{9} x \pi }{\ln \relax (3)+i \pi }+\frac {{\mathrm e}^{9} x \ln \relax (3)}{\ln \relax (3)+i \pi }-\frac {{\mathrm e}^{9} x}{4 \left (\ln \relax (3)+i \pi \right )}+\frac {\sqrt {3}\, {\mathrm e}^{11+\frac {x}{2}}}{4 \ln \relax (3)+4 i \pi }\)	\(67\)
norman	\(-\frac {{\mathrm e}^{9} \left (i \pi -\ln \relax (3)\right ) {\mathrm e}^{\frac {\ln \relax (3)}{2}+2+\frac {x}{2}}}{4 \left (\ln \relax (3)^{2}+\pi ^{2}\right )}+\frac {{\mathrm e}^{9} \left (4 \pi ^{2}+i \pi +4 \ln \relax (3)^{2}-\ln \relax (3)\right ) x}{4 \pi ^{2}+4 \ln \relax (3)^{2}}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*(exp(9)*exp(1/2*ln(3)+2+1/2*x)+8*exp(9)*(ln(3)+I*Pi)-2*exp(9))/(ln(3)+I*Pi),x,method=_RETURNVERBOSE)

[Out]

1/8/(ln(3)+I*Pi)*(2*exp(9)*exp(1/2*ln(3)+2+1/2*x)+8*I*Pi*exp(9)*x+8*ln(3)*exp(9)*x-2*x*exp(9))

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maxima [A]  time = 0.35, size = 37, normalized size = 0.95 \begin {gather*} \frac {4 \, {\left (i \, \pi + \log \relax (3)\right )} x e^{9} - x e^{9} + e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, \log \relax (3) + 11\right )}}{4 \, {\left (i \, \pi + \log \relax (3)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(exp(9)*exp(1/2*log(3)+2+1/2*x)+8*exp(9)*(log(3)+I*pi)-2*exp(9))/(log(3)+I*pi),x, algorithm="max
ima")

[Out]

1/4*(4*(I*pi + log(3))*x*e^9 - x*e^9 + e^(1/2*x + 1/2*log(3) + 11))/(I*pi + log(3))

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mupad [B]  time = 0.35, size = 41, normalized size = 1.05 \begin {gather*} \frac {{\mathrm {e}}^9\,\left (x\,1{}\mathrm {i}+4\,\Pi \,x-x\,\ln \relax (3)\,4{}\mathrm {i}-\sqrt {3}\,{\mathrm {e}}^{x/2}\,{\mathrm {e}}^2\,1{}\mathrm {i}\right )}{4\,\left (\Pi -\ln \relax (3)\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(9)*exp(x/2 + log(3)/2 + 2))/8 - exp(9)/4 + exp(9)*(Pi*1i + log(3)))/(Pi*1i + log(3)),x)

[Out]

(exp(9)*(x*1i + 4*Pi*x - x*log(3)*4i - 3^(1/2)*exp(x/2)*exp(2)*1i))/(4*(Pi - log(3)*1i))

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sympy [A]  time = 0.37, size = 54, normalized size = 1.38 \begin {gather*} \frac {x \left (- e^{9} + 4 e^{9} \log {\relax (3 )} + 4 i \pi e^{9}\right )}{4 \log {\relax (3 )} + 4 i \pi } + \frac {\sqrt {3} e^{11} e^{\frac {x}{2}}}{4 \log {\relax (3 )} + 4 i \pi } \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(exp(9)*exp(1/2*ln(3)+2+1/2*x)+8*exp(9)*(ln(3)+I*pi)-2*exp(9))/(ln(3)+I*pi),x)

[Out]

x*(-exp(9) + 4*exp(9)*log(3) + 4*I*pi*exp(9))/(4*log(3) + 4*I*pi) + sqrt(3)*exp(11)*exp(x/2)/(4*log(3) + 4*I*p
i)

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