3.85.50 $\int \frac{4e^x+(e^x(-4x+x^2)+4e^x{x}^2(i\pi +\log (\log (4)))+e^x{x}^2{\log }^2(\log (4))(i\pi +\log (\log (4))))\log (\frac{-4+x+4x(i\pi +\log (\log (4)))+x{\log }^2(\log (4))(i\pi +\log (\log (4)))}{x(i\pi +\log (\log (4)))})}{-4x+x^2+4x^2(i\pi +\log (\log (4)))+x^2{\log }^2(\log (4))(i\pi +\log (\log (4)))}dx$

Optimal. Leaf size=30 $$e^x\log (4+{\log }^2(\log (4))+\frac{-4+x}{x(i\pi +\log (\log (4)))})$$

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Rubi [A]  time = 0.68, antiderivative size = 39, normalized size of antiderivative = 1.30, number of steps used = 5, number of rules used = 4, integrand size = 146, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.027, Rules used = {6, 1593, 6688, 2288} $$\begin{array}{c}{e}^x\log (-\frac{4}{x(\log (\log (4))+i\pi )}+4+{\log }^2(\log (4))+\frac{1}{\log (\log (4))+i\pi })\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{4e^x+(e^x(-4x+x^2)+4e^x{x}^2(i\pi +\log (\log (4)))+e^x{x}^2{\log }^2(\log (4))(i\pi +\log (\log (4))))\log \left(\frac{-4+x+4x(i\pi +\log (\log (4)))+x{\log }^2(\log (4))(i\pi +\log (\log (4)))}{x(i\pi +\log (\log (4)))}\right)}{-4x+x^2{\log }^2(\log (4))(i\pi +\log (\log (4)))+x^2(1+4(i\pi +\log (\log (4))))}dx\\ & =\int \frac{4e^x+(e^x(-4x+x^2)+4e^x{x}^2(i\pi +\log (\log (4)))+e^x{x}^2{\log }^2(\log (4))(i\pi +\log (\log (4))))\log \left(\frac{-4+x+4x(i\pi +\log (\log (4)))+x{\log }^2(\log (4))(i\pi +\log (\log (4)))}{x(i\pi +\log (\log (4)))}\right)}{-4x+x^2(1+4(i\pi +\log (\log (4)))+{\log }^2(\log (4))(i\pi +\log (\log (4))))}dx\\ & =\int \frac{4e^x+(e^x(-4x+x^2)+4e^x{x}^2(i\pi +\log (\log (4)))+e^x{x}^2{\log }^2(\log (4))(i\pi +\log (\log (4))))\log \left(\frac{-4+x+4x(i\pi +\log (\log (4)))+x{\log }^2(\log (4))(i\pi +\log (\log (4)))}{x(i\pi +\log (\log (4)))}\right)}{x(-4+x(1+4(i\pi +\log (\log (4)))+{\log }^2(\log (4))(i\pi +\log (\log (4)))))}dx\\ & =\int e^x(\frac{4}{x(-4+x(1+4\log (\log (4))+{\log }^3(\log (4))+i\pi (4+{\log }^2(\log (4)))))}+\log (4+{\log }^2(\log (4))+\frac{1}{i\pi +\log (\log (4))}-\frac{4}{x(i\pi +\log (\log (4)))}))dx\\ & =e^x\log (4+{\log }^2(\log (4))+\frac{1}{i\pi +\log (\log (4))}-\frac{4}{x(i\pi +\log (\log (4)))})\end{array}\end{array}$$

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Mathematica [A]  time = 0.09, size = 39, normalized size = 1.30 $$\begin{array}{c}{e}^x\log (4+{\log }^2(\log (4))+\frac{1}{i\pi +\log (\log (4))}-\frac{4}{x(i\pi +\log (\log (4)))})\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 1.01, size = 54, normalized size = 1.80 $$\begin{array}{c}{e}^x\log \left(\frac{2i\pi x\log {(-2\log (2))}^2+x\log {(-2\log (2))}^3-({\pi }^{2}x-4x)\log (-2\log (2))+x-4}{x\log (-2\log (2))}\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^x*log((2*I*pi*x*log(-2*log(2))^2 + x*log(-2*log(2))^3 - (pi^2*x - 4*x)*log(-2*log(2)) + x - 4)/(x*log(-2*log
(2))))

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giac [B]  time = 1.69, size = 443, normalized size = 14.77 $$\begin{array}{c}\frac{1}{2}{e}^x\log ({\pi }^2{x}^2\log (2{)}^4+x^2\log (2{)}^6+4{\pi }^2{x}^2\log (2{)}^3\log (\log (2))+6x^2\log (2{)}^5\log (\log (2))+6{\pi }^2{x}^2\log (2{)}^2\log {(\log (2))}^2+15x^2\log (2{)}^4\log {(\log (2))}^2+4{\pi }^2{x}^2\log (2)\log {(\log (2))}^3+20x^2\log (2{)}^3\log {(\log (2))}^3+{\pi }^2{x}^2\log {(\log (2))}^4+15x^2\log (2{)}^2\log {(\log (2))}^4+6x^2\log (2)\log {(\log (2))}^5+x^2\log {(\log (2))}^6+8{\pi }^2{x}^2\log (2{)}^2+8x^2\log (2{)}^4+16{\pi }^2{x}^2\log (2)\log (\log (2))+32x^2\log (2{)}^3\log (\log (2))+8{\pi }^2{x}^2\log {(\log (2))}^2+48x^2\log (2{)}^2\log {(\log (2))}^2+32x^2\log (2)\log {(\log (2))}^3+8x^2\log {(\log (2))}^4+2x^2\log (2{)}^3+6x^2\log (2{)}^2\log (\log (2))+6x^2\log (2)\log {(\log (2))}^2+2x^2\log {(\log (2))}^3+16{\pi }^2{x}^2+16x^2\log (2{)}^2-8x\log (2{)}^3+32x^2\log (2)\log (\log (2))-24x\log (2{)}^2\log (\log (2))+16x^2\log {(\log (2))}^2-24x\log (2)\log {(\log (2))}^2-8x\log {(\log (2))}^3+8x^2\log (2)+8x^2\log (\log (2))+x^2-32x\log (2)-32x\log (\log (2))-8x+16)-\frac{1}{2}{e}^x\log ({\pi }^2{x}^2+x^2\log (2{)}^2+2x^2\log (2)\log (\log (2))+x^2\log {(\log (2))}^2)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e^x*log(pi^2*x^2*log(2)^4 + x^2*log(2)^6 + 4*pi^2*x^2*log(2)^3*log(log(2)) + 6*x^2*log(2)^5*log(log(2)) +
6*pi^2*x^2*log(2)^2*log(log(2))^2 + 15*x^2*log(2)^4*log(log(2))^2 + 4*pi^2*x^2*log(2)*log(log(2))^3 + 20*x^2*l
og(2)^3*log(log(2))^3 + pi^2*x^2*log(log(2))^4 + 15*x^2*log(2)^2*log(log(2))^4 + 6*x^2*log(2)*log(log(2))^5 +
x^2*log(log(2))^6 + 8*pi^2*x^2*log(2)^2 + 8*x^2*log(2)^4 + 16*pi^2*x^2*log(2)*log(log(2)) + 32*x^2*log(2)^3*lo
g(log(2)) + 8*pi^2*x^2*log(log(2))^2 + 48*x^2*log(2)^2*log(log(2))^2 + 32*x^2*log(2)*log(log(2))^3 + 8*x^2*log
(log(2))^4 + 2*x^2*log(2)^3 + 6*x^2*log(2)^2*log(log(2)) + 6*x^2*log(2)*log(log(2))^2 + 2*x^2*log(log(2))^3 +
16*pi^2*x^2 + 16*x^2*log(2)^2 - 8*x*log(2)^3 + 32*x^2*log(2)*log(log(2)) - 24*x*log(2)^2*log(log(2)) + 16*x^2*
log(log(2))^2 - 24*x*log(2)*log(log(2))^2 - 8*x*log(log(2))^3 + 8*x^2*log(2) + 8*x^2*log(log(2)) + x^2 - 32*x*
log(2) - 32*x*log(log(2)) - 8*x + 16) - 1/2*e^x*log(pi^2*x^2 + x^2*log(2)^2 + 2*x^2*log(2)*log(log(2)) + x^2*l
og(log(2))^2)

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maple [A]  time = 1.23, size = 41, normalized size = 1.37

method	result	size
default	${\mathrm{e}}^x\ln \left(\frac{x\ln (-2\ln (2))\ln {(2\ln (2))}^2+4x\ln (-2\ln (2))+x-4}{x\ln (-2\ln (2))}\right)$	$41$
norman	${\mathrm{e}}^x\ln \left(\frac{x\ln (-2\ln (2))\ln {(2\ln (2))}^2+4x\ln (-2\ln (2))+x-4}{x\ln (-2\ln (2))}\right)$	$41$
risch	${\mathrm{e}}^x\ln ((x{(\ln (2)+\ln (\ln (2)))}^2+4x)(\ln (2)+\ln (\ln (2))+i\pi )+x-4)-{\mathrm{e}}^x\ln (x)-\frac{i\pi \mathrm{csgn}\left(\frac{i}{x}\right)\mathrm{csgn}\left(i((x\ln {(2\ln (2))}^2+4x)\ln (-2\ln (2))+x-4)\right)\mathrm{csgn}\left(\frac{i((x\ln {(2\ln (2))}^2+4x)\ln (-2\ln (2))+x-4)}{x}\right){\mathrm{e}}^x}{2}+\frac{i\pi \mathrm{csgn}\left(\frac{i}{x}\right)\mathrm{csgn}{\left(\frac{i((x\ln {(2\ln (2))}^2+4x)\ln (-2\ln (2))+x-4)}{x}\right)}^2{\mathrm{e}}^x}{2}+\frac{i\pi \mathrm{csgn}\left(i((x\ln {(2\ln (2))}^2+4x)\ln (-2\ln (2))+x-4)\right)\mathrm{csgn}{\left(\frac{i((x\ln {(2\ln (2))}^2+4x)\ln (-2\ln (2))+x-4)}{x}\right)}^2{\mathrm{e}}^x}{2}-\frac{i\pi \mathrm{csgn}\left(\frac{i((x\ln {(2\ln (2))}^2+4x)\ln (-2\ln (2))+x-4)}{x}\right)\mathrm{csgn}\left(\frac{i((x\ln {(2\ln (2))}^2+4x)\ln (-2\ln (2))+x-4)}{\ln (-2\ln (2))x}\right){\mathrm{e}}^x}{2}+\frac{i\pi \mathrm{csgn}{\left(\frac{i((x\ln {(2\ln (2))}^2+4x)\ln (-2\ln (2))+x-4)}{\ln (-2\ln (2))x}\right)}^2{\mathrm{e}}^x}{2}-\frac{i\pi \mathrm{csgn}{\left(\frac{i((x\ln {(2\ln (2))}^2+4x)\ln (-2\ln (2))+x-4)}{x}\right)}^3{\mathrm{e}}^x}{2}+\frac{i\pi \mathrm{csgn}\left(\frac{i((x\ln {(2\ln (2))}^2+4x)\ln (-2\ln (2))+x-4)}{x}\right)\mathrm{csgn}{\left(\frac{i((x\ln {(2\ln (2))}^2+4x)\ln (-2\ln (2))+x-4)}{\ln (-2\ln (2))x}\right)}^2{\mathrm{e}}^x}{2}-\frac{i\pi \mathrm{csgn}{\left(\frac{i((x\ln {(2\ln (2))}^2+4x)\ln (-2\ln (2))+x-4)}{\ln (-2\ln (2))x}\right)}^3{\mathrm{e}}^x}{2}-\ln (\ln (2)+\ln (\ln (2))+i\pi ){\mathrm{e}}^x$	$499$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)*ln((x*ln(-2*ln(2))*ln(2*ln(2))^2+4*x*ln(-2*ln(2))+x-4)/x/ln(-2*ln(2)))

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maxima [C]  time = 0.55, size = 82, normalized size = 2.73 $$\begin{array}{c}-(\log (i\pi +\log (2)+\log (\log (2)))+\log (x))e^x+e^x\log ((4i\pi +i\pi \log (2{)}^2+\log (2{)}^3+(i\pi +3\log (2))\log {(\log (2))}^2+\log {(\log (2))}^3+(2i\pi \log (2)+3\log (2{)}^2+4)\log (\log (2))+4\log (2)+1)x-4)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(I*pi + log(2) + log(log(2))) + log(x))*e^x + e^x*log((4*I*pi + I*pi*log(2)^2 + log(2)^3 + (I*pi + 3*log(
2))*log(log(2))^2 + log(log(2))^3 + (2*I*pi*log(2) + 3*log(2)^2 + 4)*log(log(2)) + 4*log(2) + 1)*x - 4)

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mupad [B]  time = 6.51, size = 38, normalized size = 1.27 $$\begin{array}{c}{\mathrm{e}}^x\ln \left(\frac{x+4x\ln (-\ln (4))+x\ln (-\ln (4)){\ln (\ln (4))}^2-4}{x\ln (-\ln (4))}\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 51.20, size = 313, normalized size = 10.43 $$\begin{array}{c}{e}^x\log (\frac{4x\log (\log (2))}{x\log (\log (2))+x\log (2)+i\pi x}+\frac{3x\log {(2)}^2\log (\log (2))}{x\log (\log (2))+x\log (2)+i\pi x}+\frac{x\log {(\log (2))}^3}{x\log (\log (2))+x\log (2)+i\pi x}+\frac{3x\log (2)\log {(\log (2))}^2}{x\log (\log (2))+x\log (2)+i\pi x}+\frac{x\log {(2)}^3}{x\log (\log (2))+x\log (2)+i\pi x}+\frac{x}{x\log (\log (2))+x\log (2)+i\pi x}+\frac{4x\log (2)}{x\log (\log (2))+x\log (2)+i\pi x}+\frac{2i\pi x\log (2)\log (\log (2))}{x\log (\log (2))+x\log (2)+i\pi x}+\frac{i\pi x\log {(\log (2))}^2}{x\log (\log (2))+x\log (2)+i\pi x}+\frac{i\pi x\log {(2)}^2}{x\log (\log (2))+x\log (2)+i\pi x}+\frac{4i\pi x}{x\log (\log (2))+x\log (2)+i\pi x}-\frac{4}{x\log (\log (2))+x\log (2)+i\pi x})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)*log(4*x*log(log(2))/(x*log(log(2)) + x*log(2) + I*pi*x) + 3*x*log(2)**2*log(log(2))/(x*log(log(2)) + x*
log(2) + I*pi*x) + x*log(log(2))**3/(x*log(log(2)) + x*log(2) + I*pi*x) + 3*x*log(2)*log(log(2))**2/(x*log(log
(2)) + x*log(2) + I*pi*x) + x*log(2)**3/(x*log(log(2)) + x*log(2) + I*pi*x) + x/(x*log(log(2)) + x*log(2) + I*
pi*x) + 4*x*log(2)/(x*log(log(2)) + x*log(2) + I*pi*x) + 2*I*pi*x*log(2)*log(log(2))/(x*log(log(2)) + x*log(2)
+ I*pi*x) + I*pi*x*log(log(2))**2/(x*log(log(2)) + x*log(2) + I*pi*x) + I*pi*x*log(2)**2/(x*log(log(2)) + x*l
og(2) + I*pi*x) + 4*I*pi*x/(x*log(log(2)) + x*log(2) + I*pi*x) - 4/(x*log(log(2)) + x*log(2) + I*pi*x))

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