\(\int \log ^{-1+2251799813685248 e^{125+5 \log ^2(2)}}(x) (9007199254740992 e^{125+5 \log ^2(2)}+4 \log (x)) \, dx\) [901]

Optimal result

Integrand size = 35, antiderivative size = 18 \[ \int \log ^{-1+2251799813685248 e^{125+5 \log ^2(2)}}(x) \left (9007199254740992 e^{125+5 \log ^2(2)}+4 \log (x)\right ) \, dx=4 x \log ^{2 e^{5 (5+\log (2))^2}}(x) \] Output:

4*x*exp(exp(5*(ln(2)+5)^2)*ln(ln(x)))^2
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \log ^{-1+2251799813685248 e^{125+5 \log ^2(2)}}(x) \left (9007199254740992 e^{125+5 \log ^2(2)}+4 \log (x)\right ) \, dx=4 x \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \] Input:

Integrate[Log[x]^(-1 + 2251799813685248*E^(125 + 5*Log[2]^2))*(90071992547 
40992*E^(125 + 5*Log[2]^2) + 4*Log[x]),x]
 

Output:

4*x*Log[x]^(2251799813685248*E^(5*(25 + Log[2]^2)))
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.58 (sec) , antiderivative size = 141, normalized size of antiderivative = 7.83, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2808, 25, 2033, 3039, 7111}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \log ^{2251799813685248 e^{125+5 \log ^2(2)}-1}(x) \left (4 \log (x)+9007199254740992 e^{125+5 \log ^2(2)}\right ) \, dx\)

\(\Big \downarrow \) 2808

\(\displaystyle -4 \int -\frac {\Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right ) (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x)}{x}dx-4 (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \left (\log (x)+2251799813685248 e^{5 \left (25+\log ^2(2)\right )}\right ) \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 4 \int \frac {\Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right ) (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x)}{x}dx-4 (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \left (\log (x)+2251799813685248 e^{5 \left (25+\log ^2(2)\right )}\right ) \Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right )\)

\(\Big \downarrow \) 2033

\(\displaystyle 4 (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \int \frac {\Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right )}{x}dx-4 (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \left (\log (x)+2251799813685248 e^{5 \left (25+\log ^2(2)\right )}\right ) \Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right )\)

\(\Big \downarrow \) 3039

\(\displaystyle 4 (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \int \Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right )d\log (x)-4 (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \left (\log (x)+2251799813685248 e^{5 \left (25+\log ^2(2)\right )}\right ) \Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right )\)

\(\Big \downarrow \) 7111

\(\displaystyle 4 (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \left (\Gamma \left (1+2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right )+\log (x) \Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right )\right )-4 (-\log (x))^{-2251799813685248 e^{5 \left (25+\log ^2(2)\right )}} \log ^{2251799813685248 e^{5 \left (25+\log ^2(2)\right )}}(x) \left (\log (x)+2251799813685248 e^{5 \left (25+\log ^2(2)\right )}\right ) \Gamma \left (2251799813685248 e^{5 \left (25+\log ^2(2)\right )},-\log (x)\right )\)

Input:

Int[Log[x]^(-1 + 2251799813685248*E^(125 + 5*Log[2]^2))*(9007199254740992* 
E^(125 + 5*Log[2]^2) + 4*Log[x]),x]
 

Output:

(-4*Gamma[2251799813685248*E^(5*(25 + Log[2]^2)), -Log[x]]*Log[x]^(2251799 
813685248*E^(5*(25 + Log[2]^2)))*(2251799813685248*E^(5*(25 + Log[2]^2)) + 
 Log[x]))/(-Log[x])^(2251799813685248*E^(5*(25 + Log[2]^2))) + (4*Log[x]^( 
2251799813685248*E^(5*(25 + Log[2]^2)))*(Gamma[1 + 2251799813685248*E^(5*( 
25 + Log[2]^2)), -Log[x]] + Gamma[2251799813685248*E^(5*(25 + Log[2]^2)), 
-Log[x]]*Log[x]))/(-Log[x])^(2251799813685248*E^(5*(25 + Log[2]^2)))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2033

Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m + n) 
*((b*v)^n/(a*v)^n)   Int[v^(m + n)*Fx, x], x] /; FreeQ[{a, b, m, n}, x] && 
 !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[m + n]
 

rule 2808

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_ 
.)]*(e_.)), x_Symbol] :> With[{u = IntHide[(a + b*Log[c*x^n])^p, x]}, Simp[ 
(d + e*Log[f*x^r])   u, x] - Simp[e*r   Int[SimplifyIntegrand[u/x, x], x], 
x]] /; FreeQ[{a, b, c, d, e, f, n, p, r}, x]
 

rule 3039

Int[u_, x_Symbol] :> With[{lst = FunctionOfLog[Cancel[x*u], x]}, Simp[1/lst 
[[3]]   Subst[Int[lst[[1]], x], x, Log[lst[[2]]]], x] /;  !FalseQ[lst]] /; 
NonsumQ[u]
 

rule 7111

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

Maple [F(-1)]

Timed out.

\[\int \frac {\left (4 \ln \left (x \right )+8 \,{\mathrm e}^{5 \ln \left (2\right )^{2}+50 \ln \left (2\right )+125}\right ) {\mathrm e}^{2251799813685248 \,{\mathrm e}^{5 \ln \left (2\right )^{2}+125} \ln \left (\ln \left (x \right )\right )}}{\ln \left (x \right )}d x\]

Input:

int((4*ln(x)+8*exp(5*ln(2)^2+50*ln(2)+125))*exp(exp(5*ln(2)^2+50*ln(2)+125 
)*ln(ln(x)))^2/ln(x),x)
 

Output:

int((4*ln(x)+8*exp(5*ln(2)^2+50*ln(2)+125))*exp(exp(5*ln(2)^2+50*ln(2)+125 
)*ln(ln(x)))^2/ln(x),x)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \log ^{-1+2251799813685248 e^{125+5 \log ^2(2)}}(x) \left (9007199254740992 e^{125+5 \log ^2(2)}+4 \log (x)\right ) \, dx=4 \, x \log \left (x\right )^{2 \, e^{\left (5 \, \log \left (2\right )^{2} + 50 \, \log \left (2\right ) + 125\right )}} \] Input:

integrate((4*log(x)+8*exp(5*log(2)^2+50*log(2)+125))*exp(exp(5*log(2)^2+50 
*log(2)+125)*log(log(x)))^2/log(x),x, algorithm="fricas")
 

Output:

4*x*log(x)^(2*e^(5*log(2)^2 + 50*log(2) + 125))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (17) = 34\).

Time = 2.42 (sec) , antiderivative size = 128, normalized size of antiderivative = 7.11 \[ \int \log ^{-1+2251799813685248 e^{125+5 \log ^2(2)}}(x) \left (9007199254740992 e^{125+5 \log ^2(2)}+4 \log (x)\right ) \, dx=\frac {9007199254740992 e^{125} e^{5 \log {\left (2 \right )}^{2}} \log {\left (x \right )}^{-1 + 2251799813685248 e^{125} e^{5 \log {\left (2 \right )}^{2}}} \Gamma \left (2251799813685248 e^{125} e^{5 \log {\left (2 \right )}^{2}}, - \log {\left (x \right )}\right )}{\left (- \log {\left (x \right )}\right )^{-1 + 2251799813685248 e^{125} e^{5 \log {\left (2 \right )}^{2}}}} + \frac {4 \log {\left (x \right )}^{2251799813685248 e^{125} e^{5 \log {\left (2 \right )}^{2}}} \Gamma \left (1 + 2251799813685248 e^{125} e^{5 \log {\left (2 \right )}^{2}}, - \log {\left (x \right )}\right )}{\left (- \log {\left (x \right )}\right )^{2251799813685248 e^{125} e^{5 \log {\left (2 \right )}^{2}}}} \] Input:

integrate((4*ln(x)+8*exp(5*ln(2)**2+50*ln(2)+125))*exp(exp(5*ln(2)**2+50*l 
n(2)+125)*ln(ln(x)))**2/ln(x),x)
 

Output:

9007199254740992*(-log(x))**(-2251799813685248*exp(125)*exp(5*log(2)**2) + 
 1)*exp(125)*exp(5*log(2)**2)*log(x)**(-1 + 2251799813685248*exp(125)*exp( 
5*log(2)**2))*uppergamma(2251799813685248*exp(125)*exp(5*log(2)**2), -log( 
x)) + 4*log(x)**(2251799813685248*exp(125)*exp(5*log(2)**2))*uppergamma(1 
+ 2251799813685248*exp(125)*exp(5*log(2)**2), -log(x))/(-log(x))**(2251799 
813685248*exp(125)*exp(5*log(2)**2))
 

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 6.33 \[ \int \log ^{-1+2251799813685248 e^{125+5 \log ^2(2)}}(x) \left (9007199254740992 e^{125+5 \log ^2(2)}+4 \log (x)\right ) \, dx=-4 \, \left (-\log \left (x\right )\right )^{-2251799813685248 \, e^{\left (5 \, \log \left (2\right )^{2} + 125\right )} - 1} \log \left (x\right )^{2251799813685248 \, e^{\left (5 \, \log \left (2\right )^{2} + 125\right )} + 1} \Gamma \left (2251799813685248 \, e^{\left (5 \, \log \left (2\right )^{2} + 125\right )} + 1, -\log \left (x\right )\right ) - \frac {9007199254740992 \, \log \left (x\right )^{2251799813685248 \, e^{\left (5 \, \log \left (2\right )^{2} + 125\right )}} e^{\left (5 \, \log \left (2\right )^{2} + 125\right )} \Gamma \left (2251799813685248 \, e^{\left (5 \, \log \left (2\right )^{2} + 125\right )}, -\log \left (x\right )\right )}{\left (-\log \left (x\right )\right )^{2251799813685248 \, e^{\left (5 \, \log \left (2\right )^{2} + 125\right )}}} \] Input:

integrate((4*log(x)+8*exp(5*log(2)^2+50*log(2)+125))*exp(exp(5*log(2)^2+50 
*log(2)+125)*log(log(x)))^2/log(x),x, algorithm="maxima")
 

Output:

-4*(-log(x))^(-2251799813685248*e^(5*log(2)^2 + 125) - 1)*log(x)^(22517998 
13685248*e^(5*log(2)^2 + 125) + 1)*gamma(2251799813685248*e^(5*log(2)^2 + 
125) + 1, -log(x)) - 9007199254740992*log(x)^(2251799813685248*e^(5*log(2) 
^2 + 125))*e^(5*log(2)^2 + 125)*gamma(2251799813685248*e^(5*log(2)^2 + 125 
), -log(x))/(-log(x))^(2251799813685248*e^(5*log(2)^2 + 125))
 

Giac [F]

\[ \int \log ^{-1+2251799813685248 e^{125+5 \log ^2(2)}}(x) \left (9007199254740992 e^{125+5 \log ^2(2)}+4 \log (x)\right ) \, dx=\int { \frac {4 \, \log \left (x\right )^{2 \, e^{\left (5 \, \log \left (2\right )^{2} + 50 \, \log \left (2\right ) + 125\right )}} {\left (2 \, e^{\left (5 \, \log \left (2\right )^{2} + 50 \, \log \left (2\right ) + 125\right )} + \log \left (x\right )\right )}}{\log \left (x\right )} \,d x } \] Input:

integrate((4*log(x)+8*exp(5*log(2)^2+50*log(2)+125))*exp(exp(5*log(2)^2+50 
*log(2)+125)*log(log(x)))^2/log(x),x, algorithm="giac")
 

Output:

integrate(4*log(x)^(2*e^(5*log(2)^2 + 50*log(2) + 125))*(2*e^(5*log(2)^2 + 
 50*log(2) + 125) + log(x))/log(x), x)
 

Mupad [B] (verification not implemented)

Time = 0.68 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \log ^{-1+2251799813685248 e^{125+5 \log ^2(2)}}(x) \left (9007199254740992 e^{125+5 \log ^2(2)}+4 \log (x)\right ) \, dx=4\,x\,{\ln \left (x\right )}^{2251799813685248\,{\mathrm {e}}^{5\,{\ln \left (2\right )}^2+125}} \] Input:

int((log(x)^(2*exp(50*log(2) + 5*log(2)^2 + 125))*(8*exp(50*log(2) + 5*log 
(2)^2 + 125) + 4*log(x)))/log(x),x)
 

Output:

4*x*log(x)^(2251799813685248*exp(5*log(2)^2 + 125))
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \log ^{-1+2251799813685248 e^{125+5 \log ^2(2)}}(x) \left (9007199254740992 e^{125+5 \log ^2(2)}+4 \log (x)\right ) \, dx=4 e^{2251799813685248 e^{5 \mathrm {log}\left (2\right )^{2}} \mathrm {log}\left (\mathrm {log}\left (x \right )\right ) e^{125}} x \] Input:

int((4*log(x)+8*exp(5*log(2)^2+50*log(2)+125))*exp(exp(5*log(2)^2+50*log(2 
)+125)*log(log(x)))^2/log(x),x)
 

Output:

4*e**(2251799813685248*e**(5*log(2)**2)*log(log(x))*e**125)*x