3.92.54 \(\int \frac {-40+70 x+(10-20 x) \log (2 x)+e^{2 x^2+2 x \log (\frac {4}{x})} (-36+72 x+36 \log (\frac {4}{x})+(24-48 x-24 \log (\frac {4}{x})) \log (2 x)+(-4+8 x+4 \log (\frac {4}{x})) \log ^2(2 x))}{9-6 \log (2 x)+\log ^2(2 x)} \, dx\)

Optimal. Leaf size=35 \[ 2 \left (e^{2 x \left (x+\log \left (\frac {4}{x}\right )\right )}+\frac {5 \left (-x+x^2\right )}{3-\log (2 x)}\right ) \]

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Rubi [F]  time = 1.33, 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 {-40+70 x+(10-20 x) \log (2 x)+e^{2 x^2+2 x \log \left (\frac {4}{x}\right )} \left (-36+72 x+36 \log \left (\frac {4}{x}\right )+\left (24-48 x-24 \log \left (\frac {4}{x}\right )\right ) \log (2 x)+\left (-4+8 x+4 \log \left (\frac {4}{x}\right )\right ) \log ^2(2 x)\right )}{9-6 \log (2 x)+\log ^2(2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-40 + 70*x + (10 - 20*x)*Log[2*x] + E^(2*x^2 + 2*x*Log[4/x])*(-36 + 72*x + 36*Log[4/x] + (24 - 48*x - 24*
Log[4/x])*Log[2*x] + (-4 + 8*x + 4*Log[4/x])*Log[2*x]^2))/(9 - 6*Log[2*x] + Log[2*x]^2),x]

[Out]

(-10*(1 - x)*x)/(3 - Log[2*x]) - 4*Defer[Int][E^(2*x^2 + 2*x*Log[4])*(x^(-1))^(2*x), x] + 4*Log[4/x]*Defer[Int
][E^(2*x^2 + 2*x*Log[4])*(x^(-1))^(2*x), x] + 8*Defer[Int][E^(2*x^2 + 2*x*Log[4])*(x^(-1))^(-1 + 2*x), x] + 4*
Defer[Int][Defer[Int][16^x*E^(2*x^2)*(x^(-1))^(2*x), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-40+70 x+4^{1+2 x} e^{2 x^2} \left (\frac {1}{x}\right )^{2 x} \left (-1+2 x+\log \left (\frac {4}{x}\right )\right ) (-3+\log (2 x))^2+(10-20 x) \log (2 x)}{(3-\log (2 x))^2} \, dx\\ &=\int \left (4^{1+2 x} e^{2 x^2} \left (\frac {1}{x}\right )^{2 x} \left (-1+2 x+\log \left (\frac {4}{x}\right )\right )-\frac {10 (4-7 x-\log (2 x)+2 x \log (2 x))}{(-3+\log (2 x))^2}\right ) \, dx\\ &=-\left (10 \int \frac {4-7 x-\log (2 x)+2 x \log (2 x)}{(-3+\log (2 x))^2} \, dx\right )+\int 4^{1+2 x} e^{2 x^2} \left (\frac {1}{x}\right )^{2 x} \left (-1+2 x+\log \left (\frac {4}{x}\right )\right ) \, dx\\ &=-\left (10 \int \left (\frac {1-x}{(-3+\log (2 x))^2}+\frac {-1+2 x}{-3+\log (2 x)}\right ) \, dx\right )+\int 4 e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \left (-1+2 x+\log \left (\frac {4}{x}\right )\right ) \, dx\\ &=4 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \left (-1+2 x+\log \left (\frac {4}{x}\right )\right ) \, dx-10 \int \frac {1-x}{(-3+\log (2 x))^2} \, dx-10 \int \frac {-1+2 x}{-3+\log (2 x)} \, dx\\ &=-\frac {10 (1-x) x}{3-\log (2 x)}+4 \int \left (-e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x}+2 e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{-1+2 x}+e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \log \left (\frac {4}{x}\right )\right ) \, dx-10 \int \left (-\frac {1}{-3+\log (2 x)}+\frac {2 x}{-3+\log (2 x)}\right ) \, dx+10 \int \frac {1}{-3+\log (2 x)} \, dx-20 \int \frac {1-x}{-3+\log (2 x)} \, dx\\ &=-\frac {10 (1-x) x}{3-\log (2 x)}-4 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \, dx+4 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \log \left (\frac {4}{x}\right ) \, dx+5 \operatorname {Subst}\left (\int \frac {e^x}{-3+x} \, dx,x,\log (2 x)\right )+8 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{-1+2 x} \, dx+10 \int \frac {1}{-3+\log (2 x)} \, dx-20 \int \left (\frac {1}{-3+\log (2 x)}-\frac {x}{-3+\log (2 x)}\right ) \, dx-20 \int \frac {x}{-3+\log (2 x)} \, dx\\ &=5 e^3 \text {Ei}(-3+\log (2 x))-\frac {10 (1-x) x}{3-\log (2 x)}-4 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \, dx+4 \int \frac {\int 16^x e^{2 x^2} \left (\frac {1}{x}\right )^{2 x} \, dx}{x} \, dx+5 \operatorname {Subst}\left (\int \frac {e^x}{-3+x} \, dx,x,\log (2 x)\right )-5 \operatorname {Subst}\left (\int \frac {e^{2 x}}{-3+x} \, dx,x,\log (2 x)\right )+8 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{-1+2 x} \, dx-20 \int \frac {1}{-3+\log (2 x)} \, dx+20 \int \frac {x}{-3+\log (2 x)} \, dx+\left (4 \log \left (\frac {4}{x}\right )\right ) \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \, dx\\ &=-5 e^6 \text {Ei}(-2 (3-\log (2 x)))+10 e^3 \text {Ei}(-3+\log (2 x))-\frac {10 (1-x) x}{3-\log (2 x)}-4 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \, dx+4 \int \frac {\int e^{2 x^2+x \log (16)} \left (\frac {1}{x}\right )^{2 x} \, dx}{x} \, dx+5 \operatorname {Subst}\left (\int \frac {e^{2 x}}{-3+x} \, dx,x,\log (2 x)\right )+8 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{-1+2 x} \, dx-10 \operatorname {Subst}\left (\int \frac {e^x}{-3+x} \, dx,x,\log (2 x)\right )+\left (4 \log \left (\frac {4}{x}\right )\right ) \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \, dx\\ &=-\frac {10 (1-x) x}{3-\log (2 x)}-4 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \, dx+4 \int \frac {\int e^{2 x^2+x \log (16)} \left (\frac {1}{x}\right )^{2 x} \, dx}{x} \, dx+8 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{-1+2 x} \, dx+\left (4 \log \left (\frac {4}{x}\right )\right ) \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.76, size = 35, normalized size = 1.00 \begin {gather*} 2 \left (16^x e^{2 x^2} \left (\frac {1}{x}\right )^{2 x}-\frac {5 (-1+x) x}{-3+\log (2 x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-40 + 70*x + (10 - 20*x)*Log[2*x] + E^(2*x^2 + 2*x*Log[4/x])*(-36 + 72*x + 36*Log[4/x] + (24 - 48*x
- 24*Log[4/x])*Log[2*x] + (-4 + 8*x + 4*Log[4/x])*Log[2*x]^2))/(9 - 6*Log[2*x] + Log[2*x]^2),x]

[Out]

2*(16^x*E^(2*x^2)*(x^(-1))^(2*x) - (5*(-1 + x)*x)/(-3 + Log[2*x]))

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fricas [A]  time = 0.90, size = 59, normalized size = 1.69 \begin {gather*} -\frac {2 \, {\left (5 \, x^{2} - {\left (3 \, \log \relax (2) - \log \left (\frac {4}{x}\right ) - 3\right )} e^{\left (2 \, x^{2} + 2 \, x \log \left (\frac {4}{x}\right )\right )} - 5 \, x\right )}}{3 \, \log \relax (2) - \log \left (\frac {4}{x}\right ) - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*log(4/x)+8*x-4)*log(2*x)^2+(-24*log(4/x)-48*x+24)*log(2*x)+36*log(4/x)+72*x-36)*exp(x*log(4/x)+
x^2)^2+(-20*x+10)*log(2*x)+70*x-40)/(log(2*x)^2-6*log(2*x)+9),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.53, size = 61, normalized size = 1.74 \begin {gather*} -\frac {2 \, {\left (5 \, x^{2} - e^{\left (2 \, x^{2} + 4 \, x \log \relax (2) - 2 \, x \log \relax (x)\right )} \log \left (2 \, x\right ) - 5 \, x + 3 \, e^{\left (2 \, x^{2} + 4 \, x \log \relax (2) - 2 \, x \log \relax (x)\right )}\right )}}{\log \left (2 \, x\right ) - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*log(4/x)+8*x-4)*log(2*x)^2+(-24*log(4/x)-48*x+24)*log(2*x)+36*log(4/x)+72*x-36)*exp(x*log(4/x)+
x^2)^2+(-20*x+10)*log(2*x)+70*x-40)/(log(2*x)^2-6*log(2*x)+9),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.81, size = 44, normalized size = 1.26

method	result	size
risch	\(-\frac {20 i x \left (x -1\right )}{2 i \ln \relax (2)+2 i \ln \relax (x )-6 i}+2 \,4^{2 x} x^{-2 x} {\mathrm e}^{2 x^{2}}\)	\(44\)
default	\(\frac {-10 x^{2}+10 x}{\ln \relax (2)+\ln \relax (x )-3}+\frac {\left (2 \ln \relax (2)-6\right ) {\mathrm e}^{2 \left (\ln \left (\frac {4}{x}\right )+x \right ) x}+2 \ln \relax (x ) {\mathrm e}^{2 \left (\ln \left (\frac {4}{x}\right )+x \right ) x}}{\ln \relax (2)+\ln \relax (x )-3}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*ln(4/x)+8*x-4)*ln(2*x)^2+(-24*ln(4/x)-48*x+24)*ln(2*x)+36*ln(4/x)+72*x-36)*exp(x*ln(4/x)+x^2)^2+(-20*
x+10)*ln(2*x)+70*x-40)/(ln(2*x)^2-6*ln(2*x)+9),x,method=_RETURNVERBOSE)

[Out]

-20*I*x*(x-1)/(2*I*ln(2)+2*I*ln(x)-6*I)+2*(4^x)^2*(x^(-x))^2*exp(2*x^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {20 \, e^{3} E_{2}\left (-\log \left (2 \, x\right ) + 3\right )}{\log \left (2 \, x\right ) - 3} - \frac {35 \, e^{6} E_{2}\left (-2 \, \log \left (2 \, x\right ) + 6\right )}{2 \, {\left (\log \left (2 \, x\right ) - 3\right )}} + \frac {2 \, {\left (30 \, x^{2} + {\left (\log \relax (2) + \log \relax (x) - 3\right )} e^{\left (2 \, x^{2} + 4 \, x \log \relax (2) - 2 \, x \log \relax (x)\right )} - 15 \, x\right )}}{\log \relax (2) + \log \relax (x) - 3} - 2 \, \int \frac {10 \, {\left (7 \, x - 2\right )}}{\log \relax (2) + \log \relax (x) - 3}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*log(4/x)+8*x-4)*log(2*x)^2+(-24*log(4/x)-48*x+24)*log(2*x)+36*log(4/x)+72*x-36)*exp(x*log(4/x)+
x^2)^2+(-20*x+10)*log(2*x)+70*x-40)/(log(2*x)^2-6*log(2*x)+9),x, algorithm="maxima")

[Out]

20*e^3*exp_integral_e(2, -log(2*x) + 3)/(log(2*x) - 3) - 35/2*e^6*exp_integral_e(2, -2*log(2*x) + 6)/(log(2*x)
- 3) + 2*(30*x^2 + (log(2) + log(x) - 3)*e^(2*x^2 + 4*x*log(2) - 2*x*log(x)) - 15*x)/(log(2) + log(x) - 3) -
2*integrate(10*(7*x - 2)/(log(2) + log(x) - 3), x)

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mupad [B]  time = 7.67, size = 60, normalized size = 1.71 \begin {gather*} 10\,x-\frac {10\,x\,\left (7\,x-4\right )-10\,x\,\ln \left (2\,x\right )\,\left (2\,x-1\right )}{\ln \left (2\,x\right )-3}-20\,x^2+2\,2^{4\,x}\,{\mathrm {e}}^{2\,x^2}\,{\left (\frac {1}{x}\right )}^{2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((70*x + exp(2*x*log(4/x) + 2*x^2)*(72*x + 36*log(4/x) + log(2*x)^2*(8*x + 4*log(4/x) - 4) - log(2*x)*(48*x
+ 24*log(4/x) - 24) - 36) - log(2*x)*(20*x - 10) - 40)/(log(2*x)^2 - 6*log(2*x) + 9),x)

[Out]

10*x - (10*x*(7*x - 4) - 10*x*log(2*x)*(2*x - 1))/(log(2*x) - 3) - 20*x^2 + 2*2^(4*x)*exp(2*x^2)*(1/x)^(2*x)

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sympy [A]  time = 0.64, size = 34, normalized size = 0.97 \begin {gather*} \frac {- 10 x^{2} + 10 x}{\log {\left (2 x \right )} - 3} + 2 e^{2 x^{2} + 2 x \left (- \log {\left (2 x \right )} + \log {\relax (8 )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*ln(4/x)+8*x-4)*ln(2*x)**2+(-24*ln(4/x)-48*x+24)*ln(2*x)+36*ln(4/x)+72*x-36)*exp(x*ln(4/x)+x**2)
**2+(-20*x+10)*ln(2*x)+70*x-40)/(ln(2*x)**2-6*ln(2*x)+9),x)

[Out]

(-10*x**2 + 10*x)/(log(2*x) - 3) + 2*exp(2*x**2 + 2*x*(-log(2*x) + log(8)))

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