3.41.77 \(\int \frac {1}{5} (9+6 x-3 x^2+e^x (-15+5 x)+(9+15 x-11 x^2+e^x (-15-5 x+5 x^2)) \log (x)+(-3 x+x^2+(-6 x+3 x^2) \log (x)) \log (x \log (x))) \, dx\)

Optimal. Leaf size=33 \[ (3-x) x \log (x) \left (-e^x+x+\frac {1}{5} (3-x-x \log (x \log (x)))\right ) \]

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Rubi [F]  time = 0.58, 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}{5} \left (9+6 x-3 x^2+e^x (-15+5 x)+\left (9+15 x-11 x^2+e^x \left (-15-5 x+5 x^2\right )\right ) \log (x)+\left (-3 x+x^2+\left (-6 x+3 x^2\right ) \log (x)\right ) \log (x \log (x))\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(9 + 6*x - 3*x^2 + E^x*(-15 + 5*x) + (9 + 15*x - 11*x^2 + E^x*(-15 - 5*x + 5*x^2))*Log[x] + (-3*x + x^2 +
(-6*x + 3*x^2)*Log[x])*Log[x*Log[x]])/5,x]

[Out]

x^3/45 + (9*x*Log[x])/5 - 3*E^x*x*Log[x] + (3*x^2*Log[x])/2 + E^x*x^2*Log[x] - (11*x^3*Log[x])/15 - (3*ExpInte
gralEi[2*Log[x]]*Log[x])/10 + (ExpIntegralEi[3*Log[x]]*Log[x])/15 + (3*ExpIntegralEi[2*Log[x]]*(1 + Log[x]))/1
0 - (ExpIntegralEi[3*Log[x]]*(1 + Log[x]))/15 - (3*x^2*Log[x*Log[x]])/10 + (x^3*Log[x*Log[x]])/15 - (6*Defer[I
nt][x*Log[x]*Log[x*Log[x]], x])/5 + (3*Defer[Int][x^2*Log[x]*Log[x*Log[x]], x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (9+6 x-3 x^2+e^x (-15+5 x)+\left (9+15 x-11 x^2+e^x \left (-15-5 x+5 x^2\right )\right ) \log (x)+\left (-3 x+x^2+\left (-6 x+3 x^2\right ) \log (x)\right ) \log (x \log (x))\right ) \, dx\\ &=\frac {9 x}{5}+\frac {3 x^2}{5}-\frac {x^3}{5}+\frac {1}{5} \int e^x (-15+5 x) \, dx+\frac {1}{5} \int \left (9+15 x-11 x^2+e^x \left (-15-5 x+5 x^2\right )\right ) \log (x) \, dx+\frac {1}{5} \int \left (-3 x+x^2+\left (-6 x+3 x^2\right ) \log (x)\right ) \log (x \log (x)) \, dx\\ &=-e^x (3-x)+\frac {9 x}{5}+\frac {3 x^2}{5}-\frac {x^3}{5}+\frac {9}{5} x \log (x)-3 e^x x \log (x)+\frac {3}{2} x^2 \log (x)+e^x x^2 \log (x)-\frac {11}{15} x^3 \log (x)-\frac {1}{5} \int \left (9+5 e^x (-3+x)+\frac {15 x}{2}-\frac {11 x^2}{3}\right ) \, dx+\frac {1}{5} \int x (-3+x+3 (-2+x) \log (x)) \log (x \log (x)) \, dx-\int e^x \, dx\\ &=-e^x-e^x (3-x)-\frac {3 x^2}{20}+\frac {2 x^3}{45}+\frac {9}{5} x \log (x)-3 e^x x \log (x)+\frac {3}{2} x^2 \log (x)+e^x x^2 \log (x)-\frac {11}{15} x^3 \log (x)+\frac {1}{5} \int \left (-3 x \log (x \log (x))+x^2 \log (x \log (x))-6 x \log (x) \log (x \log (x))+3 x^2 \log (x) \log (x \log (x))\right ) \, dx-\int e^x (-3+x) \, dx\\ &=-e^x-\frac {3 x^2}{20}+\frac {2 x^3}{45}+\frac {9}{5} x \log (x)-3 e^x x \log (x)+\frac {3}{2} x^2 \log (x)+e^x x^2 \log (x)-\frac {11}{15} x^3 \log (x)+\frac {1}{5} \int x^2 \log (x \log (x)) \, dx-\frac {3}{5} \int x \log (x \log (x)) \, dx+\frac {3}{5} \int x^2 \log (x) \log (x \log (x)) \, dx-\frac {6}{5} \int x \log (x) \log (x \log (x)) \, dx+\int e^x \, dx\\ &=-\frac {3 x^2}{20}+\frac {2 x^3}{45}+\frac {9}{5} x \log (x)-3 e^x x \log (x)+\frac {3}{2} x^2 \log (x)+e^x x^2 \log (x)-\frac {11}{15} x^3 \log (x)-\frac {3}{10} x^2 \log (x \log (x))+\frac {1}{15} x^3 \log (x \log (x))-\frac {1}{5} \int \frac {x^2 (1+\log (x))}{3 \log (x)} \, dx+\frac {3}{5} \int \frac {x (1+\log (x))}{2 \log (x)} \, dx+\frac {3}{5} \int x^2 \log (x) \log (x \log (x)) \, dx-\frac {6}{5} \int x \log (x) \log (x \log (x)) \, dx\\ &=-\frac {3 x^2}{20}+\frac {2 x^3}{45}+\frac {9}{5} x \log (x)-3 e^x x \log (x)+\frac {3}{2} x^2 \log (x)+e^x x^2 \log (x)-\frac {11}{15} x^3 \log (x)-\frac {3}{10} x^2 \log (x \log (x))+\frac {1}{15} x^3 \log (x \log (x))-\frac {1}{15} \int \frac {x^2 (1+\log (x))}{\log (x)} \, dx+\frac {3}{10} \int \frac {x (1+\log (x))}{\log (x)} \, dx+\frac {3}{5} \int x^2 \log (x) \log (x \log (x)) \, dx-\frac {6}{5} \int x \log (x) \log (x \log (x)) \, dx\\ &=-\frac {3 x^2}{20}+\frac {2 x^3}{45}+\frac {9}{5} x \log (x)-3 e^x x \log (x)+\frac {3}{2} x^2 \log (x)+e^x x^2 \log (x)-\frac {11}{15} x^3 \log (x)+\frac {3}{10} \text {Ei}(2 \log (x)) (1+\log (x))-\frac {1}{15} \text {Ei}(3 \log (x)) (1+\log (x))-\frac {3}{10} x^2 \log (x \log (x))+\frac {1}{15} x^3 \log (x \log (x))+\frac {1}{15} \int \frac {\text {Ei}(3 \log (x))}{x} \, dx-\frac {3}{10} \int \frac {\text {Ei}(2 \log (x))}{x} \, dx+\frac {3}{5} \int x^2 \log (x) \log (x \log (x)) \, dx-\frac {6}{5} \int x \log (x) \log (x \log (x)) \, dx\\ &=-\frac {3 x^2}{20}+\frac {2 x^3}{45}+\frac {9}{5} x \log (x)-3 e^x x \log (x)+\frac {3}{2} x^2 \log (x)+e^x x^2 \log (x)-\frac {11}{15} x^3 \log (x)+\frac {3}{10} \text {Ei}(2 \log (x)) (1+\log (x))-\frac {1}{15} \text {Ei}(3 \log (x)) (1+\log (x))-\frac {3}{10} x^2 \log (x \log (x))+\frac {1}{15} x^3 \log (x \log (x))+\frac {1}{15} \operatorname {Subst}(\int \text {Ei}(3 x) \, dx,x,\log (x))-\frac {3}{10} \operatorname {Subst}(\int \text {Ei}(2 x) \, dx,x,\log (x))+\frac {3}{5} \int x^2 \log (x) \log (x \log (x)) \, dx-\frac {6}{5} \int x \log (x) \log (x \log (x)) \, dx\\ &=\frac {x^3}{45}+\frac {9}{5} x \log (x)-3 e^x x \log (x)+\frac {3}{2} x^2 \log (x)+e^x x^2 \log (x)-\frac {11}{15} x^3 \log (x)-\frac {3}{10} \text {Ei}(2 \log (x)) \log (x)+\frac {1}{15} \text {Ei}(3 \log (x)) \log (x)+\frac {3}{10} \text {Ei}(2 \log (x)) (1+\log (x))-\frac {1}{15} \text {Ei}(3 \log (x)) (1+\log (x))-\frac {3}{10} x^2 \log (x \log (x))+\frac {1}{15} x^3 \log (x \log (x))+\frac {3}{5} \int x^2 \log (x) \log (x \log (x)) \, dx-\frac {6}{5} \int x \log (x) \log (x \log (x)) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 27, normalized size = 0.82 \begin {gather*} \frac {1}{5} (-3+x) x \log (x) \left (-3+5 e^x-4 x+x \log (x \log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9 + 6*x - 3*x^2 + E^x*(-15 + 5*x) + (9 + 15*x - 11*x^2 + E^x*(-15 - 5*x + 5*x^2))*Log[x] + (-3*x +
x^2 + (-6*x + 3*x^2)*Log[x])*Log[x*Log[x]])/5,x]

[Out]

((-3 + x)*x*Log[x]*(-3 + 5*E^x - 4*x + x*Log[x*Log[x]]))/5

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fricas [A]  time = 0.69, size = 48, normalized size = 1.45 \begin {gather*} \frac {1}{5} \, {\left (x^{3} - 3 \, x^{2}\right )} \log \left (x \log \relax (x)\right ) \log \relax (x) - \frac {1}{5} \, {\left (4 \, x^{3} - 9 \, x^{2} - 5 \, {\left (x^{2} - 3 \, x\right )} e^{x} - 9 \, x\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x^2-6*x)*log(x)+x^2-3*x)*log(x*log(x))+1/5*((5*x^2-5*x-15)*exp(x)-11*x^2+15*x+9)*log(x)+1/5*
(5*x-15)*exp(x)-3/5*x^2+6/5*x+9/5,x, algorithm="fricas")

[Out]

1/5*(x^3 - 3*x^2)*log(x*log(x))*log(x) - 1/5*(4*x^3 - 9*x^2 - 5*(x^2 - 3*x)*e^x - 9*x)*log(x)

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giac [B]  time = 0.20, size = 93, normalized size = 2.82 \begin {gather*} \frac {1}{5} \, {\left (x^{3} - 3 \, x^{2}\right )} \log \relax (x)^{2} + \frac {1}{5} \, {\left (x^{3} - 3 \, x^{2}\right )} \log \relax (x) \log \left (\log \relax (x)\right ) - {\left (x - 1\right )} e^{x} + {\left (x - 4\right )} e^{x} - \frac {1}{30} \, {\left (22 \, x^{3} - 45 \, x^{2} - 30 \, {\left (x^{2} - 3 \, x\right )} e^{x} - 54 \, x\right )} \log \relax (x) - \frac {1}{30} \, {\left (2 \, x^{3} - 9 \, x^{2}\right )} \log \relax (x) + 3 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x^2-6*x)*log(x)+x^2-3*x)*log(x*log(x))+1/5*((5*x^2-5*x-15)*exp(x)-11*x^2+15*x+9)*log(x)+1/5*
(5*x-15)*exp(x)-3/5*x^2+6/5*x+9/5,x, algorithm="giac")

[Out]

1/5*(x^3 - 3*x^2)*log(x)^2 + 1/5*(x^3 - 3*x^2)*log(x)*log(log(x)) - (x - 1)*e^x + (x - 4)*e^x - 1/30*(22*x^3 -
45*x^2 - 30*(x^2 - 3*x)*e^x - 54*x)*log(x) - 1/30*(2*x^3 - 9*x^2)*log(x) + 3*e^x

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maple [B]  time = 0.09, size = 60, normalized size = 1.82

method	result	size
default	\(\frac {9 x^{2} \ln \relax (x )}{5}-\frac {4 x^{3} \ln \relax (x )}{5}-\frac {3 \ln \relax (x ) \ln \left (x \ln \relax (x )\right ) x^{2}}{5}+\frac {\ln \relax (x ) x^{3} \ln \left (x \ln \relax (x )\right )}{5}+\frac {9 x \ln \relax (x )}{5}+x^{2} {\mathrm e}^{x} \ln \relax (x )-3 x \,{\mathrm e}^{x} \ln \relax (x )\)	\(60\)
risch	\(\frac {x^{3} \ln \relax (x ) \ln \left (\ln \relax (x )\right )}{5}-\frac {3 x^{2} \ln \relax (x ) \ln \left (\ln \relax (x )\right )}{5}+\frac {x^{3} \ln \relax (x )^{2}}{5}-\frac {3 x^{2} \ln \relax (x )^{2}}{5}-\frac {i \pi \,x^{3} \mathrm {csgn}\left (i x \ln \relax (x )\right )^{3} \ln \relax (x )}{10}+\frac {i \pi \,x^{3} \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2} \ln \relax (x )}{10}+\frac {i \pi \,x^{3} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2} \ln \relax (x )}{10}+\frac {3 i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i x \ln \relax (x )\right ) \ln \relax (x )}{10}-\frac {3 i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2} \ln \relax (x )}{10}-\frac {i \pi \,x^{3} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i x \ln \relax (x )\right ) \ln \relax (x )}{10}-\frac {3 i \pi \,x^{2} \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2} \ln \relax (x )}{10}+\frac {3 i \pi \,x^{2} \mathrm {csgn}\left (i x \ln \relax (x )\right )^{3} \ln \relax (x )}{10}-\frac {4 x^{3} \ln \relax (x )}{5}+\frac {9 x^{2} \ln \relax (x )}{5}+\frac {9 x \ln \relax (x )}{5}+x^{2} {\mathrm e}^{x} \ln \relax (x )-3 x \,{\mathrm e}^{x} \ln \relax (x )\)	\(258\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((3*x^2-6*x)*ln(x)+x^2-3*x)*ln(x*ln(x))+1/5*((5*x^2-5*x-15)*exp(x)-11*x^2+15*x+9)*ln(x)+1/5*(5*x-15)*e
xp(x)-3/5*x^2+6/5*x+9/5,x,method=_RETURNVERBOSE)

[Out]

9/5*x^2*ln(x)-4/5*x^3*ln(x)-3/5*ln(x)*ln(x*ln(x))*x^2+1/5*ln(x)*x^3*ln(x*ln(x))+9/5*x*ln(x)+x^2*exp(x)*ln(x)-3
*x*exp(x)*ln(x)

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maxima [B]  time = 0.39, size = 63, normalized size = 1.91 \begin {gather*} \frac {1}{5} \, {\left (x^{3} - 3 \, x^{2}\right )} \log \left (x \log \relax (x)\right ) \log \relax (x) - \frac {1}{30} \, {\left (22 \, x^{3} - 45 \, x^{2} - 30 \, {\left (x^{2} - 3 \, x\right )} e^{x} - 54 \, x\right )} \log \relax (x) - \frac {1}{30} \, {\left (2 \, x^{3} - 9 \, x^{2}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x^2-6*x)*log(x)+x^2-3*x)*log(x*log(x))+1/5*((5*x^2-5*x-15)*exp(x)-11*x^2+15*x+9)*log(x)+1/5*
(5*x-15)*exp(x)-3/5*x^2+6/5*x+9/5,x, algorithm="maxima")

[Out]

1/5*(x^3 - 3*x^2)*log(x*log(x))*log(x) - 1/30*(22*x^3 - 45*x^2 - 30*(x^2 - 3*x)*e^x - 54*x)*log(x) - 1/30*(2*x
^3 - 9*x^2)*log(x)

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mupad [B]  time = 3.33, size = 25, normalized size = 0.76 \begin {gather*} -\frac {x\,\ln \relax (x)\,\left (x-3\right )\,\left (4\,x-5\,{\mathrm {e}}^x-x\,\ln \left (x\,\ln \relax (x)\right )+3\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x)/5 + (log(x)*(15*x - exp(x)*(5*x - 5*x^2 + 15) - 11*x^2 + 9))/5 + (exp(x)*(5*x - 15))/5 - (log(x*log(
x))*(3*x + log(x)*(6*x - 3*x^2) - x^2))/5 - (3*x^2)/5 + 9/5,x)

[Out]

-(x*log(x)*(x - 3)*(4*x - 5*exp(x) - x*log(x*log(x)) + 3))/5

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sympy [B]  time = 0.76, size = 63, normalized size = 1.91 \begin {gather*} \left (x^{2} \log {\relax (x )} - 3 x \log {\relax (x )}\right ) e^{x} + \left (\frac {x^{3} \log {\relax (x )}}{5} - \frac {3 x^{2} \log {\relax (x )}}{5}\right ) \log {\left (x \log {\relax (x )} \right )} + \left (- \frac {4 x^{3}}{5} + \frac {9 x^{2}}{5} + \frac {9 x}{5}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((3*x**2-6*x)*ln(x)+x**2-3*x)*ln(x*ln(x))+1/5*((5*x**2-5*x-15)*exp(x)-11*x**2+15*x+9)*ln(x)+1/5*
(5*x-15)*exp(x)-3/5*x**2+6/5*x+9/5,x)

[Out]

(x**2*log(x) - 3*x*log(x))*exp(x) + (x**3*log(x)/5 - 3*x**2*log(x)/5)*log(x*log(x)) + (-4*x**3/5 + 9*x**2/5 +
9*x/5)*log(x)

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