3.21.32 \(\int \frac {1+4 x-16 x^2-16 x^3+(-8 x-12 x^2) \log (x)+(-x+4 x^2+4 x^3+(2 x+3 x^2) \log (x)) \log (2 x)+(-4+\log (2 x)) \log (-4+\log (2 x))}{-4+\log (2 x)} \, dx\)

Optimal. Leaf size=22 \[ x (-x+x (1+x) (x+\log (x))+\log (-4+\log (2 x))) \]

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Rubi [F]  time = 0.49, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1+4 x-16 x^2-16 x^3+\left (-8 x-12 x^2\right ) \log (x)+\left (-x+4 x^2+4 x^3+\left (2 x+3 x^2\right ) \log (x)\right ) \log (2 x)+(-4+\log (2 x)) \log (-4+\log (2 x))}{-4+\log (2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(1 + 4*x - 16*x^2 - 16*x^3 + (-8*x - 12*x^2)*Log[x] + (-x + 4*x^2 + 4*x^3 + (2*x + 3*x^2)*Log[x])*Log[2*x]
+ (-4 + Log[2*x])*Log[-4 + Log[2*x]])/(-4 + Log[2*x]),x]

[Out]

-x^2 + x^3 + x^4 + (E^4*ExpIntegralEi[-4 + Log[2*x]])/2 + (x^2 + x^3)*Log[x] + Defer[Subst][Defer[Int][Log[-4
+ Log[x]], x], x, 2*x]/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1+4 x-16 x^2-16 x^3-8 x \log (x)-12 x^2 \log (x)-x \log (2 x)+4 x^2 \log (2 x)+4 x^3 \log (2 x)+2 x \log (x) \log (2 x)+3 x^2 \log (x) \log (2 x)}{-4+\log (2 x)}+\log (-4+\log (2 x))\right ) \, dx\\ &=\int \frac {1+4 x-16 x^2-16 x^3-8 x \log (x)-12 x^2 \log (x)-x \log (2 x)+4 x^2 \log (2 x)+4 x^3 \log (2 x)+2 x \log (x) \log (2 x)+3 x^2 \log (x) \log (2 x)}{-4+\log (2 x)} \, dx+\int \log (-4+\log (2 x)) \, dx\\ &=\frac {1}{2} \operatorname {Subst}(\int \log (-4+\log (x)) \, dx,x,2 x)+\int \frac {-1-4 x+16 x^2+16 x^3-x (2+3 x) \log (x) (-4+\log (2 x))-x \left (-1+4 x+4 x^2\right ) \log (2 x)}{4-\log (2 x)} \, dx\\ &=\frac {1}{2} \operatorname {Subst}(\int \log (-4+\log (x)) \, dx,x,2 x)+\int \left (x \left (-1+4 x+4 x^2+2 \log (x)+3 x \log (x)\right )+\frac {1}{-4+\log (2 x)}\right ) \, dx\\ &=\frac {1}{2} \operatorname {Subst}(\int \log (-4+\log (x)) \, dx,x,2 x)+\int x \left (-1+4 x+4 x^2+2 \log (x)+3 x \log (x)\right ) \, dx+\int \frac {1}{-4+\log (2 x)} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {e^x}{-4+x} \, dx,x,\log (2 x)\right )+\frac {1}{2} \operatorname {Subst}(\int \log (-4+\log (x)) \, dx,x,2 x)+\int \left (x \left (-1+4 x+4 x^2\right )+x (2+3 x) \log (x)\right ) \, dx\\ &=\frac {1}{2} e^4 \text {Ei}(-4+\log (2 x))+\frac {1}{2} \operatorname {Subst}(\int \log (-4+\log (x)) \, dx,x,2 x)+\int x \left (-1+4 x+4 x^2\right ) \, dx+\int x (2+3 x) \log (x) \, dx\\ &=\frac {1}{2} e^4 \text {Ei}(-4+\log (2 x))+\left (x^2+x^3\right ) \log (x)+\frac {1}{2} \operatorname {Subst}(\int \log (-4+\log (x)) \, dx,x,2 x)-\int x (1+x) \, dx+\int \left (-x+4 x^2+4 x^3\right ) \, dx\\ &=-\frac {x^2}{2}+\frac {4 x^3}{3}+x^4+\frac {1}{2} e^4 \text {Ei}(-4+\log (2 x))+\left (x^2+x^3\right ) \log (x)+\frac {1}{2} \operatorname {Subst}(\int \log (-4+\log (x)) \, dx,x,2 x)-\int \left (x+x^2\right ) \, dx\\ &=-x^2+x^3+x^4+\frac {1}{2} e^4 \text {Ei}(-4+\log (2 x))+\left (x^2+x^3\right ) \log (x)+\frac {1}{2} \operatorname {Subst}(\int \log (-4+\log (x)) \, dx,x,2 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 25, normalized size = 1.14 \begin {gather*} x \left (x \left (-1+x+x^2\right )+x (1+x) \log (x)+\log (-4+\log (2 x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 4*x - 16*x^2 - 16*x^3 + (-8*x - 12*x^2)*Log[x] + (-x + 4*x^2 + 4*x^3 + (2*x + 3*x^2)*Log[x])*Lo
g[2*x] + (-4 + Log[2*x])*Log[-4 + Log[2*x]])/(-4 + Log[2*x]),x]

[Out]

x*(x*(-1 + x + x^2) + x*(1 + x)*Log[x] + Log[-4 + Log[2*x]])

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fricas [A]  time = 0.53, size = 31, normalized size = 1.41 \begin {gather*} x^{4} + x^{3} - x^{2} + {\left (x^{3} + x^{2}\right )} \log \relax (x) + x \log \left (\log \relax (2) + \log \relax (x) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(2*x)-4)*log(log(2*x)-4)+((3*x^2+2*x)*log(x)+4*x^3+4*x^2-x)*log(2*x)+(-12*x^2-8*x)*log(x)-16*x^
3-16*x^2+4*x+1)/(log(2*x)-4),x, algorithm="fricas")

[Out]

x^4 + x^3 - x^2 + (x^3 + x^2)*log(x) + x*log(log(2) + log(x) - 4)

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giac [A]  time = 0.18, size = 33, normalized size = 1.50 \begin {gather*} x^{4} + x^{3} \log \relax (x) + x^{3} + x^{2} \log \relax (x) - x^{2} + x \log \left (\log \relax (2) + \log \relax (x) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(2*x)-4)*log(log(2*x)-4)+((3*x^2+2*x)*log(x)+4*x^3+4*x^2-x)*log(2*x)+(-12*x^2-8*x)*log(x)-16*x^
3-16*x^2+4*x+1)/(log(2*x)-4),x, algorithm="giac")

[Out]

x^4 + x^3*log(x) + x^3 + x^2*log(x) - x^2 + x*log(log(2) + log(x) - 4)

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maple [A]  time = 0.10, size = 34, normalized size = 1.55

method	result	size
risch	\(x^{4}+x^{3} \ln \relax (x )+x^{3}+x^{2} \ln \relax (x )-x^{2}+x \ln \left (\ln \relax (2)+\ln \relax (x )-4\right )\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((ln(2*x)-4)*ln(ln(2*x)-4)+((3*x^2+2*x)*ln(x)+4*x^3+4*x^2-x)*ln(2*x)+(-12*x^2-8*x)*ln(x)-16*x^3-16*x^2+4*x
+1)/(ln(2*x)-4),x,method=_RETURNVERBOSE)

[Out]

x^4+x^3*ln(x)+x^3+x^2*ln(x)-x^2+x*ln(ln(2)+ln(x)-4)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {11}{3} \, x^{3} + \frac {1}{4} \, e^{8} E_{1}\left (-2 \, \log \left (2 \, x\right ) + 8\right ) \log \left (2 \, x\right ) - \frac {1}{2} \, e^{12} E_{1}\left (-3 \, \log \left (2 \, x\right ) + 12\right ) \log \left (2 \, x\right ) - \frac {1}{4} \, e^{16} E_{1}\left (-4 \, \log \left (2 \, x\right ) + 16\right ) \log \left (2 \, x\right ) + 2 \, e^{8} E_{1}\left (-2 \, \log \left (2 \, x\right ) + 8\right ) \log \relax (x) + \frac {3}{2} \, e^{12} E_{1}\left (-3 \, \log \left (2 \, x\right ) + 12\right ) \log \relax (x) + \frac {7}{2} \, x^{2} - \frac {9}{8} \, e^{8} E_{2}\left (-2 \, \log \left (2 \, x\right ) + 8\right ) - \frac {1}{3} \, e^{12} E_{2}\left (-3 \, \log \left (2 \, x\right ) + 12\right ) + \frac {1}{16} \, e^{16} E_{2}\left (-4 \, \log \left (2 \, x\right ) + 16\right ) - \frac {1}{2} \, e^{4} E_{1}\left (-\log \left (2 \, x\right ) + 4\right ) - e^{8} E_{1}\left (-2 \, \log \left (2 \, x\right ) + 8\right ) + 2 \, e^{12} E_{1}\left (-3 \, \log \left (2 \, x\right ) + 12\right ) + e^{16} E_{1}\left (-4 \, \log \left (2 \, x\right ) + 16\right ) + {\left (x^{3} + x^{2}\right )} \log \relax (x) + x \log \left (\log \relax (2) + \log \relax (x) - 4\right ) - \int \frac {12 \, x^{2} {\left (\log \relax (2) - 4\right )} + 8 \, x {\left (\log \relax (2) - 4\right )} + 1}{\log \relax (2) + \log \relax (x) - 4}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(2*x)-4)*log(log(2*x)-4)+((3*x^2+2*x)*log(x)+4*x^3+4*x^2-x)*log(2*x)+(-12*x^2-8*x)*log(x)-16*x^
3-16*x^2+4*x+1)/(log(2*x)-4),x, algorithm="maxima")

[Out]

11/3*x^3 + 1/4*e^8*exp_integral_e(1, -2*log(2*x) + 8)*log(2*x) - 1/2*e^12*exp_integral_e(1, -3*log(2*x) + 12)*
log(2*x) - 1/4*e^16*exp_integral_e(1, -4*log(2*x) + 16)*log(2*x) + 2*e^8*exp_integral_e(1, -2*log(2*x) + 8)*lo
g(x) + 3/2*e^12*exp_integral_e(1, -3*log(2*x) + 12)*log(x) + 7/2*x^2 - 9/8*e^8*exp_integral_e(2, -2*log(2*x) +
8) - 1/3*e^12*exp_integral_e(2, -3*log(2*x) + 12) + 1/16*e^16*exp_integral_e(2, -4*log(2*x) + 16) - 1/2*e^4*e
xp_integral_e(1, -log(2*x) + 4) - e^8*exp_integral_e(1, -2*log(2*x) + 8) + 2*e^12*exp_integral_e(1, -3*log(2*x
) + 12) + e^16*exp_integral_e(1, -4*log(2*x) + 16) + (x^3 + x^2)*log(x) + x*log(log(2) + log(x) - 4) - integra
te((12*x^2*(log(2) - 4) + 8*x*(log(2) - 4) + 1)/(log(2) + log(x) - 4), x)

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mupad [B]  time = 1.43, size = 31, normalized size = 1.41 \begin {gather*} \ln \relax (x)\,\left (x^3+x^2\right )-x^2+x^3+x^4+x\,\ln \left (\ln \left (2\,x\right )-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x + log(log(2*x) - 4)*(log(2*x) - 4) - log(x)*(8*x + 12*x^2) - 16*x^2 - 16*x^3 + log(2*x)*(log(x)*(2*x
+ 3*x^2) - x + 4*x^2 + 4*x^3) + 1)/(log(2*x) - 4),x)

[Out]

log(x)*(x^2 + x^3) - x^2 + x^3 + x^4 + x*log(log(2*x) - 4)

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sympy [A]  time = 0.42, size = 31, normalized size = 1.41 \begin {gather*} x^{4} + x^{3} - x^{2} + x \log {\left (\log {\relax (x )} - 4 + \log {\relax (2 )} \right )} + \left (x^{3} + x^{2}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((ln(2*x)-4)*ln(ln(2*x)-4)+((3*x**2+2*x)*ln(x)+4*x**3+4*x**2-x)*ln(2*x)+(-12*x**2-8*x)*ln(x)-16*x**3
-16*x**2+4*x+1)/(ln(2*x)-4),x)

[Out]

x**4 + x**3 - x**2 + x*log(log(x) - 4 + log(2)) + (x**3 + x**2)*log(x)

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