3.75.16 \(\int \frac {e^{4+\frac {e^4-\log (4 e^{5+x})}{\log (4 e^{5+x})}}-36 \log ^2(4 e^{5+x})}{\log ^2(4 e^{5+x})} \, dx\)

Optimal. Leaf size=24 \[ -e^{-1+\frac {e^4}{\log \left (4 e^{5+x}\right )}}-36 x \]

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Rubi [A]  time = 0.12, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {2282, 14, 2209} \begin {gather*} -36 x-e^{\frac {e^4}{\log \left (4 e^{x+5}\right )}-1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(4 + (E^4 - Log[4*E^(5 + x)])/Log[4*E^(5 + x)]) - 36*Log[4*E^(5 + x)]^2)/Log[4*E^(5 + x)]^2,x]

[Out]

-E^(-1 + E^4/Log[4*E^(5 + x)]) - 36*x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

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} &=\operatorname {Subst}\left (\int \frac {-36+\frac {e^{3+\frac {e^4}{\log (4 x)}}}{\log ^2(4 x)}}{x} \, dx,x,e^{5+x}\right )\\ &=\operatorname {Subst}\left (\int \left (-\frac {36}{x}+\frac {e^{3+\frac {e^4}{\log (4 x)}}}{x \log ^2(4 x)}\right ) \, dx,x,e^{5+x}\right )\\ &=-36 x+\operatorname {Subst}\left (\int \frac {e^{3+\frac {e^4}{\log (4 x)}}}{x \log ^2(4 x)} \, dx,x,e^{5+x}\right )\\ &=-36 x+\operatorname {Subst}\left (\int \frac {e^{3+\frac {e^4}{x}}}{x^2} \, dx,x,\log \left (4 e^{5+x}\right )\right )\\ &=-e^{-1+\frac {e^4}{\log \left (4 e^{5+x}\right )}}-36 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 24, normalized size = 1.00 \begin {gather*} -e^{-1+\frac {e^4}{\log \left (4 e^{5+x}\right )}}-36 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4 + (E^4 - Log[4*E^(5 + x)])/Log[4*E^(5 + x)]) - 36*Log[4*E^(5 + x)]^2)/Log[4*E^(5 + x)]^2,x]

[Out]

-E^(-1 + E^4/Log[4*E^(5 + x)]) - 36*x

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fricas [A]  time = 0.76, size = 32, normalized size = 1.33 \begin {gather*} -{\left (36 \, x e^{4} + e^{\left (\frac {3 \, x + e^{4} + 6 \, \log \relax (2) + 15}{x + 2 \, \log \relax (2) + 5}\right )}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(4)*exp((-log(4*exp(5)*exp(x))+exp(4))/log(4*exp(5)*exp(x)))-36*log(4*exp(5)*exp(x))^2)/log(4*ex
p(5)*exp(x))^2,x, algorithm="fricas")

[Out]

-(36*x*e^4 + e^((3*x + e^4 + 6*log(2) + 15)/(x + 2*log(2) + 5)))*e^(-4)

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giac [A]  time = 0.33, size = 21, normalized size = 0.88 \begin {gather*} -36 \, x - e^{\left (\frac {e^{4}}{\log \left (4 \, e^{\left (x + 5\right )}\right )} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(4)*exp((-log(4*exp(5)*exp(x))+exp(4))/log(4*exp(5)*exp(x)))-36*log(4*exp(5)*exp(x))^2)/log(4*ex
p(5)*exp(x))^2,x, algorithm="giac")

[Out]

-36*x - e^(e^4/log(4*e^(x + 5)) - 1)

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maple [A]  time = 0.52, size = 28, normalized size = 1.17

method	result	size
derivativedivides	\(-36 \ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )-{\mathrm e}^{\frac {{\mathrm e}^{4}}{\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )}-1}\)	\(28\)
default	\(-36 \ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )-{\mathrm e}^{\frac {{\mathrm e}^{4}}{\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )}-1}\)	\(28\)
risch	\(-36 x -{\mathrm e}^{\frac {-2 \ln \relax (2)-5-\ln \left ({\mathrm e}^{x}\right )+{\mathrm e}^{4}}{2 \ln \relax (2)+5+\ln \left ({\mathrm e}^{x}\right )}}\)	\(33\)
norman	\(\frac {-36 \ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )^{2}-\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right ) {\mathrm e}^{\frac {-\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )+{\mathrm e}^{4}}{\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )}}}{\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4)*exp((-ln(4*exp(5)*exp(x))+exp(4))/ln(4*exp(5)*exp(x)))-36*ln(4*exp(5)*exp(x))^2)/ln(4*exp(5)*exp(x
))^2,x,method=_RETURNVERBOSE)

[Out]

-36*ln(4*exp(5)*exp(x))-exp(exp(4)/ln(4*exp(5)*exp(x))-1)

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maxima [A]  time = 0.37, size = 27, normalized size = 1.12 \begin {gather*} -e^{\left (\frac {e^{4}}{\log \left (4 \, e^{\left (x + 5\right )}\right )} - 1\right )} - 36 \, \log \left (4 \, e^{\left (x + 5\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(4)*exp((-log(4*exp(5)*exp(x))+exp(4))/log(4*exp(5)*exp(x)))-36*log(4*exp(5)*exp(x))^2)/log(4*ex
p(5)*exp(x))^2,x, algorithm="maxima")

[Out]

-e^(e^4/log(4*e^(x + 5)) - 1) - 36*log(4*e^(x + 5))

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mupad [B]  time = 5.36, size = 51, normalized size = 2.12 \begin {gather*} -36\,x-\frac {{\mathrm {e}}^{-\frac {5}{x+\ln \relax (4)+5}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^4}{x+\ln \relax (4)+5}}\,{\mathrm {e}}^{-\frac {x}{x+\ln \relax (4)+5}}}{2^{\frac {2}{x+\ln \relax (4)+5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4)*exp(-(log(4*exp(5)*exp(x)) - exp(4))/log(4*exp(5)*exp(x))) - 36*log(4*exp(5)*exp(x))^2)/log(4*exp(
5)*exp(x))^2,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(4)*exp((-ln(4*exp(5)*exp(x))+exp(4))/ln(4*exp(5)*exp(x)))-36*ln(4*exp(5)*exp(x))**2)/ln(4*exp(5
)*exp(x))**2,x)

[Out]

Timed out

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