3.94.28 \(\int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx\)

Optimal. Leaf size=39 \[ -e^5+e^{e^{4 (4+x)}}+\frac {x (3-x+\log (4))}{i \pi +\log (5-\log (3))} \]

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Rubi [A]  time = 0.04, antiderivative size = 51, normalized size of antiderivative = 1.31, number of steps used = 4, number of rules used = 3, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {12, 2282, 2194} \begin {gather*} -\frac {x^2}{\log (5-\log (3))+i \pi }+e^{e^{4 x+16}}+\frac {x (3+\log (4))}{\log (5-\log (3))+i \pi } \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 - 2*x + Log[4] + 4*E^(16 + E^(16 + 4*x) + 4*x)*(I*Pi + Log[5 - Log[3]]))/(I*Pi + Log[5 - Log[3]]),x]

[Out]

E^E^(16 + 4*x) - x^2/(I*Pi + Log[5 - Log[3]]) + (x*(3 + Log[4]))/(I*Pi + Log[5 - 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 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]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))\right ) \, dx}{i \pi +\log (5-\log (3))}\\ &=-\frac {x^2}{i \pi +\log (5-\log (3))}+\frac {x (3+\log (4))}{i \pi +\log (5-\log (3))}+4 \int e^{16+e^{16+4 x}+4 x} \, dx\\ &=-\frac {x^2}{i \pi +\log (5-\log (3))}+\frac {x (3+\log (4))}{i \pi +\log (5-\log (3))}+\operatorname {Subst}\left (\int e^{16+e^{16} x} \, dx,x,e^{4 x}\right )\\ &=e^{e^{16+4 x}}-\frac {x^2}{i \pi +\log (5-\log (3))}+\frac {x (3+\log (4))}{i \pi +\log (5-\log (3))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 52, normalized size = 1.33 \begin {gather*} \frac {3 x-x^2+x \log (4)+e^{e^{4 (4+x)}} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 - 2*x + Log[4] + 4*E^(16 + E^(16 + 4*x) + 4*x)*(I*Pi + Log[5 - Log[3]]))/(I*Pi + Log[5 - Log[3]])
,x]

[Out]

(3*x - x^2 + x*Log[4] + E^E^(4*(4 + x))*(I*Pi + Log[5 - Log[3]]))/(I*Pi + Log[5 - Log[3]])

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fricas [A]  time = 0.59, size = 54, normalized size = 1.38 \begin {gather*} -\frac {{\left ({\left (x^{2} - 2 \, x \log \relax (2) - 3 \, x\right )} e^{\left (4 \, x + 16\right )} - e^{\left (4 \, x + e^{\left (4 \, x + 16\right )} + 16\right )} \log \left (\log \relax (3) - 5\right )\right )} e^{\left (-4 \, x - 16\right )}}{\log \left (\log \relax (3) - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(4*x+16)*log(log(3)-5)*exp(exp(4*x+16))+2*log(2)+3-2*x)/log(log(3)-5),x, algorithm="fricas")

[Out]

-((x^2 - 2*x*log(2) - 3*x)*e^(4*x + 16) - e^(4*x + e^(4*x + 16) + 16)*log(log(3) - 5))*e^(-4*x - 16)/log(log(3
) - 5)

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giac [A]  time = 0.20, size = 35, normalized size = 0.90 \begin {gather*} -\frac {x^{2} - 2 \, x \log \relax (2) - e^{\left (e^{\left (4 \, x + 16\right )}\right )} \log \left (\log \relax (3) - 5\right ) - 3 \, x}{\log \left (\log \relax (3) - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(4*x+16)*log(log(3)-5)*exp(exp(4*x+16))+2*log(2)+3-2*x)/log(log(3)-5),x, algorithm="giac")

[Out]

-(x^2 - 2*x*log(2) - e^(e^(4*x + 16))*log(log(3) - 5) - 3*x)/log(log(3) - 5)

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maple [A]  time = 0.06, size = 36, normalized size = 0.92

method	result	size
default	\(\frac {-x^{2}+3 x +\ln \left (\ln \relax (3)-5\right ) {\mathrm e}^{{\mathrm e}^{4 x +16}}+2 x \ln \relax (2)}{\ln \left (\ln \relax (3)-5\right )}\)	\(36\)
norman	\(\frac {\left (2 \ln \relax (2)+3\right ) x}{\ln \left (\ln \relax (3)-5\right )}-\frac {x^{2}}{\ln \left (\ln \relax (3)-5\right )}+{\mathrm e}^{{\mathrm e}^{4 x +16}}\)	\(36\)
risch	\(-\frac {x^{2}}{\ln \left (\ln \relax (3)-5\right )}+\frac {3 x}{\ln \left (\ln \relax (3)-5\right )}+{\mathrm e}^{{\mathrm e}^{4 x +16}}+\frac {2 x \ln \relax (2)}{\ln \left (\ln \relax (3)-5\right )}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(4*x+16)*ln(ln(3)-5)*exp(exp(4*x+16))+2*ln(2)+3-2*x)/ln(ln(3)-5),x,method=_RETURNVERBOSE)

[Out]

1/ln(ln(3)-5)*(-x^2+3*x+ln(ln(3)-5)*exp(exp(4*x+16))+2*x*ln(2))

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maxima [A]  time = 0.36, size = 35, normalized size = 0.90 \begin {gather*} -\frac {x^{2} - 2 \, x \log \relax (2) - e^{\left (e^{\left (4 \, x + 16\right )}\right )} \log \left (\log \relax (3) - 5\right ) - 3 \, x}{\log \left (\log \relax (3) - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(4*x+16)*log(log(3)-5)*exp(exp(4*x+16))+2*log(2)+3-2*x)/log(log(3)-5),x, algorithm="maxima")

[Out]

-(x^2 - 2*x*log(2) - e^(e^(4*x + 16))*log(log(3) - 5) - 3*x)/log(log(3) - 5)

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mupad [B]  time = 0.19, size = 35, normalized size = 0.90 \begin {gather*} \frac {3\,x+x\,\ln \relax (4)-x^2+\ln \left (\ln \relax (3)-5\right )\,{\mathrm {e}}^{{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{16}}}{\ln \left (\ln \relax (3)-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*log(2) - 2*x + 4*log(log(3) - 5)*exp(exp(4*x + 16))*exp(4*x + 16) + 3)/log(log(3) - 5),x)

[Out]

(3*x + x*log(4) - x^2 + log(log(3) - 5)*exp(exp(4*x)*exp(16)))/log(log(3) - 5)

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sympy [A]  time = 0.31, size = 42, normalized size = 1.08 \begin {gather*} \frac {x^{2}}{- \log {\left (5 - \log {\relax (3 )} \right )} - i \pi } + \frac {x \left (2 \log {\relax (2 )} + 3\right )}{\log {\left (5 - \log {\relax (3 )} \right )} + i \pi } + e^{e^{16} e^{4 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(4*x+16)*ln(ln(3)-5)*exp(exp(4*x+16))+2*ln(2)+3-2*x)/ln(ln(3)-5),x)

[Out]

x**2/(-log(5 - log(3)) - I*pi) + x*(2*log(2) + 3)/(log(5 - log(3)) + I*pi) + exp(exp(16)*exp(4*x))

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