3.43.37 \(\int \frac {e^{2 x}+(5+\frac {1}{16} e^{2 x} (88+16 x)) \log (i \pi +\log (2))}{\log (i \pi +\log (2))} \, dx\)

Optimal. Leaf size=26 \[ \left (5+\frac {e^{2 x}}{2}\right ) \left (5+x+\frac {1}{\log (i \pi +\log (2))}\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 47, normalized size of antiderivative = 1.81, number of steps used = 6, number of rules used = 3, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {12, 2194, 2176} \begin {gather*} 5 x-\frac {e^{2 x}}{4}+\frac {1}{4} e^{2 x} (2 x+11)+\frac {e^{2 x}}{2 \log (\log (2)+i \pi )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*x) + (5 + (E^(2*x)*(88 + 16*x))/16)*Log[I*Pi + Log[2]])/Log[I*Pi + Log[2]],x]

[Out]

-1/4*E^(2*x) + 5*x + (E^(2*x)*(11 + 2*x))/4 + E^(2*x)/(2*Log[I*Pi + Log[2]])

Rule 12

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (e^{2 x}+\left (5+\frac {1}{16} e^{2 x} (88+16 x)\right ) \log (i \pi +\log (2))\right ) \, dx}{\log (i \pi +\log (2))}\\ &=\frac {\int e^{2 x} \, dx}{\log (i \pi +\log (2))}+\int \left (5+\frac {1}{16} e^{2 x} (88+16 x)\right ) \, dx\\ &=5 x+\frac {e^{2 x}}{2 \log (i \pi +\log (2))}+\frac {1}{16} \int e^{2 x} (88+16 x) \, dx\\ &=5 x+\frac {1}{4} e^{2 x} (11+2 x)+\frac {e^{2 x}}{2 \log (i \pi +\log (2))}-\frac {1}{2} \int e^{2 x} \, dx\\ &=-\frac {e^{2 x}}{4}+5 x+\frac {1}{4} e^{2 x} (11+2 x)+\frac {e^{2 x}}{2 \log (i \pi +\log (2))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 27, normalized size = 1.04 \begin {gather*} 5 x+\frac {1}{2} e^{2 x} \left (5+x+\frac {1}{\log (i \pi +\log (2))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x) + (5 + (E^(2*x)*(88 + 16*x))/16)*Log[I*Pi + Log[2]])/Log[I*Pi + Log[2]],x]

[Out]

5*x + (E^(2*x)*(5 + x + Log[I*Pi + Log[2]]^(-1)))/2

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fricas [A]  time = 0.68, size = 48, normalized size = 1.85 \begin {gather*} \frac {{\left (8 \, {\left (x + 5\right )} e^{\left (2 \, x - 4 \, \log \relax (2)\right )} + 5 \, x\right )} \log \left (i \, \pi + \log \relax (2)\right ) + 8 \, e^{\left (2 \, x - 4 \, \log \relax (2)\right )}}{\log \left (i \, \pi + \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+88)*exp(x-2*log(2))^2+5)*log(log(2)+I*pi)+16*exp(x-2*log(2))^2)/log(log(2)+I*pi),x, algorith
m="fricas")

[Out]

((8*(x + 5)*e^(2*x - 4*log(2)) + 5*x)*log(I*pi + log(2)) + 8*e^(2*x - 4*log(2)))/log(I*pi + log(2))

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giac [A]  time = 0.23, size = 48, normalized size = 1.85 \begin {gather*} \frac {{\left (8 \, {\left (x + 5\right )} e^{\left (2 \, x - 4 \, \log \relax (2)\right )} + 5 \, x\right )} \log \left (i \, \pi + \log \relax (2)\right ) + 8 \, e^{\left (2 \, x - 4 \, \log \relax (2)\right )}}{\log \left (i \, \pi + \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+88)*exp(x-2*log(2))^2+5)*log(log(2)+I*pi)+16*exp(x-2*log(2))^2)/log(log(2)+I*pi),x, algorith
m="giac")

[Out]

((8*(x + 5)*e^(2*x - 4*log(2)) + 5*x)*log(I*pi + log(2)) + 8*e^(2*x - 4*log(2)))/log(I*pi + log(2))

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maple [A]  time = 0.09, size = 50, normalized size = 1.92

method	result	size
norman	\(5 x +\frac {x \,{\mathrm e}^{2 x}}{2}+\frac {\left (1+5 \ln \left (\ln \relax (2)+i \pi \right )\right ) {\mathrm e}^{2 x}}{2 \ln \left (\ln \relax (2)+i \pi \right )}\)	\(50\)
risch	\(5 x +\frac {\left (8 \ln \left (i \left (-i \ln \relax (2)+\pi \right )\right ) x +40 \ln \left (i \left (-i \ln \relax (2)+\pi \right )\right )+8\right ) {\mathrm e}^{2 x}}{16 \ln \left (i \left (-i \ln \relax (2)+\pi \right )\right )}\)	\(53\)
default	\(\frac {\frac {{\mathrm e}^{2 x} \ln \left (\ln \relax (2)+i \pi \right ) x}{2}+\frac {5 \,{\mathrm e}^{2 x} \ln \left (\ln \relax (2)+i \pi \right )}{2}+5 \ln \left (\ln \relax (2)+i \pi \right ) x +\frac {{\mathrm e}^{2 x}}{2}}{\ln \left (\ln \relax (2)+i \pi \right )}\)	\(74\)
derivativedivides	\(\frac {\ln \relax (2) \ln \left (\ln \relax (2)+i \pi \right ) {\mathrm e}^{2 x}+\frac {\ln \left (\ln \relax (2)+i \pi \right ) {\mathrm e}^{2 x} \left (x -2 \ln \relax (2)\right )}{2}+\frac {5 \,{\mathrm e}^{2 x} \ln \left (\ln \relax (2)+i \pi \right )}{2}+5 \ln \left (\ln \relax (2)+i \pi \right ) \left (x -2 \ln \relax (2)\right )+\frac {{\mathrm e}^{2 x}}{2}}{\ln \left (\ln \relax (2)+i \pi \right )}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x+88)*exp(x-2*ln(2))^2+5)*ln(ln(2)+I*Pi)+16*exp(x-2*ln(2))^2)/ln(ln(2)+I*Pi),x,method=_RETURNVERBOSE
)

[Out]

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

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maxima [A]  time = 0.34, size = 36, normalized size = 1.38 \begin {gather*} \frac {{\left ({\left (x + 5\right )} e^{\left (2 \, x\right )} + 10 \, x\right )} \log \left (i \, \pi + \log \relax (2)\right ) + e^{\left (2 \, x\right )}}{2 \, \log \left (i \, \pi + \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+88)*exp(x-2*log(2))^2+5)*log(log(2)+I*pi)+16*exp(x-2*log(2))^2)/log(log(2)+I*pi),x, algorith
m="maxima")

[Out]

1/2*(((x + 5)*e^(2*x) + 10*x)*log(I*pi + log(2)) + e^(2*x))/log(I*pi + log(2))

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mupad [B]  time = 0.21, size = 33, normalized size = 1.27 \begin {gather*} 5\,x+\frac {5\,{\mathrm {e}}^{2\,x}}{2}+\frac {x\,{\mathrm {e}}^{2\,x}}{2}+\frac {{\mathrm {e}}^{2\,x}}{2\,\ln \left (\ln \relax (2)+\Pi \,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*exp(2*x - 4*log(2)) + log(Pi*1i + log(2))*(exp(2*x - 4*log(2))*(16*x + 88) + 5))/log(Pi*1i + log(2)),x
)

[Out]

5*x + (5*exp(2*x))/2 + (x*exp(2*x))/2 + exp(2*x)/(2*log(Pi*1i + log(2)))

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sympy [A]  time = 0.20, size = 39, normalized size = 1.50 \begin {gather*} 5 x + \frac {\left (x \log {\left (\log {\relax (2 )} + i \pi \right )} + 1 + 5 \log {\left (\log {\relax (2 )} + i \pi \right )}\right ) e^{2 x}}{2 \log {\left (\log {\relax (2 )} + i \pi \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+88)*exp(x-2*ln(2))**2+5)*ln(ln(2)+I*pi)+16*exp(x-2*ln(2))**2)/ln(ln(2)+I*pi),x)

[Out]

5*x + (x*log(log(2) + I*pi) + 1 + 5*log(log(2) + I*pi))*exp(2*x)/(2*log(log(2) + I*pi))

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