3.60.10 \(\int \frac {5 \log ^4(2)+5 e^{2 x} \log ^2(3)+e^x (-3 x \log ^2(2)-10 \log ^2(2) \log (3))}{5 x \log ^4(2)-10 e^x x \log ^2(2) \log (3)+5 e^{2 x} x \log ^2(3)} \, dx\)

Optimal. Leaf size=24 \[ -\frac {3}{5 \left (e^{-x} \log ^2(2)-\log (3)\right )}+\log (x) \]

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Rubi [A]  time = 0.92, antiderivative size = 29, normalized size of antiderivative = 1.21, number of steps used = 11, number of rules used = 8, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.110, Rules used = {6741, 12, 6742, 2282, 44, 36, 29, 31} \begin {gather*} \log (x)-\frac {3 \log ^2(2)}{5 \log (3) \left (\log ^2(2)-e^x \log (3)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*Log[2]^4 + 5*E^(2*x)*Log[3]^2 + E^x*(-3*x*Log[2]^2 - 10*Log[2]^2*Log[3]))/(5*x*Log[2]^4 - 10*E^x*x*Log[
2]^2*Log[3] + 5*E^(2*x)*x*Log[3]^2),x]

[Out]

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

Rule 12

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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]]]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 \log ^4(2)+5 e^{2 x} \log ^2(3)+e^x \left (-3 x \log ^2(2)-10 \log ^2(2) \log (3)\right )}{5 x \left (\log ^2(2)-e^x \log (3)\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {5 \log ^4(2)+5 e^{2 x} \log ^2(3)+e^x \left (-3 x \log ^2(2)-10 \log ^2(2) \log (3)\right )}{x \left (\log ^2(2)-e^x \log (3)\right )^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {5}{x}-\frac {3 \log ^4(2)}{\log (3) \left (\log ^2(2)-e^x \log (3)\right )^2}+\frac {3 \log ^2(2)}{\log (3) \left (\log ^2(2)-e^x \log (3)\right )}\right ) \, dx\\ &=\log (x)+\frac {\left (3 \log ^2(2)\right ) \int \frac {1}{\log ^2(2)-e^x \log (3)} \, dx}{5 \log (3)}-\frac {\left (3 \log ^4(2)\right ) \int \frac {1}{\left (\log ^2(2)-e^x \log (3)\right )^2} \, dx}{5 \log (3)}\\ &=\log (x)+\frac {\left (3 \log ^2(2)\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (\log ^2(2)-x \log (3)\right )} \, dx,x,e^x\right )}{5 \log (3)}-\frac {\left (3 \log ^4(2)\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (\log ^2(2)-x \log (3)\right )^2} \, dx,x,e^x\right )}{5 \log (3)}\\ &=\log (x)+\frac {3}{5} \operatorname {Subst}\left (\int \frac {1}{\log ^2(2)-x \log (3)} \, dx,x,e^x\right )+\frac {3 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )}{5 \log (3)}-\frac {\left (3 \log ^4(2)\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x \log ^4(2)}+\frac {\log (3)}{\log ^2(2) \left (\log ^2(2)-x \log (3)\right )^2}+\frac {\log (3)}{\log ^4(2) \left (\log ^2(2)-x \log (3)\right )}\right ) \, dx,x,e^x\right )}{5 \log (3)}\\ &=-\frac {3 \log ^2(2)}{5 \log (3) \left (\log ^2(2)-e^x \log (3)\right )}+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 33, normalized size = 1.38 \begin {gather*} \frac {1}{5} \left (-\frac {3 \log ^2(2)}{\log (3) \left (\log ^2(2)-e^x \log (3)\right )}+5 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*Log[2]^4 + 5*E^(2*x)*Log[3]^2 + E^x*(-3*x*Log[2]^2 - 10*Log[2]^2*Log[3]))/(5*x*Log[2]^4 - 10*E^x*
x*Log[2]^2*Log[3] + 5*E^(2*x)*x*Log[3]^2),x]

[Out]

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

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fricas [B]  time = 0.68, size = 47, normalized size = 1.96 \begin {gather*} \frac {3 \, \log \relax (2)^{2} + 5 \, {\left (e^{x} \log \relax (3)^{2} - \log \relax (3) \log \relax (2)^{2}\right )} \log \relax (x)}{5 \, {\left (e^{x} \log \relax (3)^{2} - \log \relax (3) \log \relax (2)^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(3)^2*exp(x)^2+(-10*log(2)^2*log(3)-3*x*log(2)^2)*exp(x)+5*log(2)^4)/(5*x*log(3)^2*exp(x)^2-10
*x*log(2)^2*log(3)*exp(x)+5*x*log(2)^4),x, algorithm="fricas")

[Out]

1/5*(3*log(2)^2 + 5*(e^x*log(3)^2 - log(3)*log(2)^2)*log(x))/(e^x*log(3)^2 - log(3)*log(2)^2)

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giac [B]  time = 0.17, size = 47, normalized size = 1.96 \begin {gather*} \frac {5 \, e^{x} \log \relax (3)^{2} \log \relax (x) - 5 \, \log \relax (3) \log \relax (2)^{2} \log \relax (x) + 3 \, \log \relax (2)^{2}}{5 \, {\left (e^{x} \log \relax (3)^{2} - \log \relax (3) \log \relax (2)^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(3)^2*exp(x)^2+(-10*log(2)^2*log(3)-3*x*log(2)^2)*exp(x)+5*log(2)^4)/(5*x*log(3)^2*exp(x)^2-10
*x*log(2)^2*log(3)*exp(x)+5*x*log(2)^4),x, algorithm="giac")

[Out]

1/5*(5*e^x*log(3)^2*log(x) - 5*log(3)*log(2)^2*log(x) + 3*log(2)^2)/(e^x*log(3)^2 - log(3)*log(2)^2)

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maple [A]  time = 0.19, size = 22, normalized size = 0.92

method	result	size
norman	\(\frac {3 \,{\mathrm e}^{x}}{5 \left (\ln \relax (3) {\mathrm e}^{x}-\ln \relax (2)^{2}\right )}+\ln \relax (x )\)	\(22\)
risch	\(\ln \relax (x )+\frac {3 \ln \relax (2)^{2}}{5 \ln \relax (3) \left (\ln \relax (3) {\mathrm e}^{x}-\ln \relax (2)^{2}\right )}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*ln(3)^2*exp(x)^2+(-10*ln(2)^2*ln(3)-3*x*ln(2)^2)*exp(x)+5*ln(2)^4)/(5*x*ln(3)^2*exp(x)^2-10*x*ln(2)^2*l
n(3)*exp(x)+5*x*ln(2)^4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.50, size = 27, normalized size = 1.12 \begin {gather*} \frac {3 \, \log \relax (2)^{2}}{5 \, {\left (e^{x} \log \relax (3)^{2} - \log \relax (3) \log \relax (2)^{2}\right )}} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(3)^2*exp(x)^2+(-10*log(2)^2*log(3)-3*x*log(2)^2)*exp(x)+5*log(2)^4)/(5*x*log(3)^2*exp(x)^2-10
*x*log(2)^2*log(3)*exp(x)+5*x*log(2)^4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.22, size = 22, normalized size = 0.92 \begin {gather*} \ln \relax (x)+\frac {3\,{\mathrm {e}}^x}{5\,{\mathrm {e}}^x\,\ln \relax (3)-5\,{\ln \relax (2)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*exp(2*x)*log(3)^2 + 5*log(2)^4 - exp(x)*(10*log(2)^2*log(3) + 3*x*log(2)^2))/(5*x*log(2)^4 + 5*x*exp(2*
x)*log(3)^2 - 10*x*exp(x)*log(2)^2*log(3)),x)

[Out]

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

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sympy [A]  time = 0.11, size = 29, normalized size = 1.21 \begin {gather*} \log {\relax (x )} + \frac {3 \log {\relax (2 )}^{2}}{5 e^{x} \log {\relax (3 )}^{2} - 5 \log {\relax (2 )}^{2} \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*ln(3)**2*exp(x)**2+(-10*ln(2)**2*ln(3)-3*x*ln(2)**2)*exp(x)+5*ln(2)**4)/(5*x*ln(3)**2*exp(x)**2-1
0*x*ln(2)**2*ln(3)*exp(x)+5*x*ln(2)**4),x)

[Out]

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

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