3.46.57 \(\int \frac {-30 x+6 x^2+e^x (-45+54 x-9 x^2) \log (4)+(-90 x^2+18 x^3) \log (4)+(-10 x+2 x^2+e^x (-30+36 x-6 x^2) \log (4)+(-60 x^2+12 x^3) \log (4)) \log (5-x)+(e^x (-5+6 x-x^2) \log (4)+(-10 x^2+2 x^3) \log (4)) \log ^2(5-x)-2 x^2 \log (x)}{-45 x^2+9 x^3+(-30 x^2+6 x^3) \log (5-x)+(-5 x^2+x^3) \log ^2(5-x)} \, dx\)

Optimal. Leaf size=30 \[ -\left (\left (\frac {e^x}{x}-2 x\right ) \log (4)\right )+\frac {2 \log (x)}{3+\log (5-x)} \]

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Rubi [F]  time = 2.85, 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 {-30 x+6 x^2+e^x \left (-45+54 x-9 x^2\right ) \log (4)+\left (-90 x^2+18 x^3\right ) \log (4)+\left (-10 x+2 x^2+e^x \left (-30+36 x-6 x^2\right ) \log (4)+\left (-60 x^2+12 x^3\right ) \log (4)\right ) \log (5-x)+\left (e^x \left (-5+6 x-x^2\right ) \log (4)+\left (-10 x^2+2 x^3\right ) \log (4)\right ) \log ^2(5-x)-2 x^2 \log (x)}{-45 x^2+9 x^3+\left (-30 x^2+6 x^3\right ) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(5-x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-30*x + 6*x^2 + E^x*(-45 + 54*x - 9*x^2)*Log[4] + (-90*x^2 + 18*x^3)*Log[4] + (-10*x + 2*x^2 + E^x*(-30 +
36*x - 6*x^2)*Log[4] + (-60*x^2 + 12*x^3)*Log[4])*Log[5 - x] + (E^x*(-5 + 6*x - x^2)*Log[4] + (-10*x^2 + 2*x^
3)*Log[4])*Log[5 - x]^2 - 2*x^2*Log[x])/(-45*x^2 + 9*x^3 + (-30*x^2 + 6*x^3)*Log[5 - x] + (-5*x^2 + x^3)*Log[5
- x]^2),x]

[Out]

-((E^x*Log[4])/x) + 2*x*Log[4] + 6*Defer[Int][1/(x*(3 + Log[5 - x])^2), x] + 2*Defer[Int][Log[5 - x]/(x*(3 + L
og[5 - x])^2), x] - 2*Defer[Subst][Defer[Int][Log[5 - x]/(x*(3 + Log[x])^2), x], x, 5 - x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-3 (-5+x) \left (6 x^2 \log (4)+x \left (2-3 e^x \log (4)\right )+e^x \log (64)\right )-2 (-5+x) \left (x-3 e^x x \log (4)+6 x^2 \log (4)+e^x \log (64)\right ) \log (5-x)-(-5+x) \left (-e^x (-1+x)+2 x^2\right ) \log (4) \log ^2(5-x)+2 x^2 \log (x)}{(5-x) x^2 (3+\log (5-x))^2} \, dx\\ &=\int \left (\frac {6}{x (3+\log (5-x))^2}+\frac {18 \log (4)}{(3+\log (5-x))^2}+\frac {2 \log (5-x)}{x (3+\log (5-x))^2}+\frac {12 \log (4) \log (5-x)}{(3+\log (5-x))^2}+\frac {2 \log (4) \log ^2(5-x)}{(3+\log (5-x))^2}-\frac {e^x (-1+x) \left (\log (262144)+\log (4096) \log (5-x)+\log (4) \log ^2(5-x)\right )}{x^2 (3+\log (5-x))^2}-\frac {2 \log (x)}{(-5+x) (3+\log (5-x))^2}\right ) \, dx\\ &=2 \int \frac {\log (5-x)}{x (3+\log (5-x))^2} \, dx-2 \int \frac {\log (x)}{(-5+x) (3+\log (5-x))^2} \, dx+6 \int \frac {1}{x (3+\log (5-x))^2} \, dx+(2 \log (4)) \int \frac {\log ^2(5-x)}{(3+\log (5-x))^2} \, dx+(12 \log (4)) \int \frac {\log (5-x)}{(3+\log (5-x))^2} \, dx+(18 \log (4)) \int \frac {1}{(3+\log (5-x))^2} \, dx-\int \frac {e^x (-1+x) \left (\log (262144)+\log (4096) \log (5-x)+\log (4) \log ^2(5-x)\right )}{x^2 (3+\log (5-x))^2} \, dx\\ &=2 \int \frac {\log (5-x)}{x (3+\log (5-x))^2} \, dx-2 \operatorname {Subst}\left (\int \frac {\log (5-x)}{x (3+\log (x))^2} \, dx,x,5-x\right )+6 \int \frac {1}{x (3+\log (5-x))^2} \, dx-(2 \log (4)) \operatorname {Subst}\left (\int \frac {\log ^2(x)}{(3+\log (x))^2} \, dx,x,5-x\right )-(12 \log (4)) \operatorname {Subst}\left (\int \frac {\log (x)}{(3+\log (x))^2} \, dx,x,5-x\right )-(18 \log (4)) \operatorname {Subst}\left (\int \frac {1}{(3+\log (x))^2} \, dx,x,5-x\right )-\int \frac {e^x (-1+x) (\log (64)+\log (4) \log (5-x))}{x^2 (3+\log (5-x))} \, dx\\ &=\frac {18 (5-x) \log (4)}{3+\log (5-x)}+2 \int \frac {\log (5-x)}{x (3+\log (5-x))^2} \, dx-2 \operatorname {Subst}\left (\int \frac {\log (5-x)}{x (3+\log (x))^2} \, dx,x,5-x\right )+6 \int \frac {1}{x (3+\log (5-x))^2} \, dx-\log (4) \int \frac {e^x (-1+x)}{x^2} \, dx-(2 \log (4)) \operatorname {Subst}\left (\int \left (1+\frac {9}{(3+\log (x))^2}-\frac {6}{3+\log (x)}\right ) \, dx,x,5-x\right )-(12 \log (4)) \operatorname {Subst}\left (\int \left (-\frac {3}{(3+\log (x))^2}+\frac {1}{3+\log (x)}\right ) \, dx,x,5-x\right )-(18 \log (4)) \operatorname {Subst}\left (\int \frac {1}{3+\log (x)} \, dx,x,5-x\right )\\ &=-\frac {e^x \log (4)}{x}+2 x \log (4)+\frac {18 (5-x) \log (4)}{3+\log (5-x)}+2 \int \frac {\log (5-x)}{x (3+\log (5-x))^2} \, dx-2 \operatorname {Subst}\left (\int \frac {\log (5-x)}{x (3+\log (x))^2} \, dx,x,5-x\right )+6 \int \frac {1}{x (3+\log (5-x))^2} \, dx-(18 \log (4)) \operatorname {Subst}\left (\int \frac {e^x}{3+x} \, dx,x,\log (5-x)\right )-(18 \log (4)) \operatorname {Subst}\left (\int \frac {1}{(3+\log (x))^2} \, dx,x,5-x\right )+(36 \log (4)) \operatorname {Subst}\left (\int \frac {1}{(3+\log (x))^2} \, dx,x,5-x\right )\\ &=-\frac {e^x \log (4)}{x}+2 x \log (4)-\frac {18 \text {Ei}(3+\log (5-x)) \log (4)}{e^3}+2 \int \frac {\log (5-x)}{x (3+\log (5-x))^2} \, dx-2 \operatorname {Subst}\left (\int \frac {\log (5-x)}{x (3+\log (x))^2} \, dx,x,5-x\right )+6 \int \frac {1}{x (3+\log (5-x))^2} \, dx-(18 \log (4)) \operatorname {Subst}\left (\int \frac {1}{3+\log (x)} \, dx,x,5-x\right )+(36 \log (4)) \operatorname {Subst}\left (\int \frac {1}{3+\log (x)} \, dx,x,5-x\right )\\ &=-\frac {e^x \log (4)}{x}+2 x \log (4)-\frac {18 \text {Ei}(3+\log (5-x)) \log (4)}{e^3}+2 \int \frac {\log (5-x)}{x (3+\log (5-x))^2} \, dx-2 \operatorname {Subst}\left (\int \frac {\log (5-x)}{x (3+\log (x))^2} \, dx,x,5-x\right )+6 \int \frac {1}{x (3+\log (5-x))^2} \, dx-(18 \log (4)) \operatorname {Subst}\left (\int \frac {e^x}{3+x} \, dx,x,\log (5-x)\right )+(36 \log (4)) \operatorname {Subst}\left (\int \frac {e^x}{3+x} \, dx,x,\log (5-x)\right )\\ &=-\frac {e^x \log (4)}{x}+2 x \log (4)+2 \int \frac {\log (5-x)}{x (3+\log (5-x))^2} \, dx-2 \operatorname {Subst}\left (\int \frac {\log (5-x)}{x (3+\log (x))^2} \, dx,x,5-x\right )+6 \int \frac {1}{x (3+\log (5-x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 30, normalized size = 1.00 \begin {gather*} -\frac {e^x \log (4)}{x}+2 x \log (4)+\frac {2 \log (x)}{3+\log (5-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-30*x + 6*x^2 + E^x*(-45 + 54*x - 9*x^2)*Log[4] + (-90*x^2 + 18*x^3)*Log[4] + (-10*x + 2*x^2 + E^x*
(-30 + 36*x - 6*x^2)*Log[4] + (-60*x^2 + 12*x^3)*Log[4])*Log[5 - x] + (E^x*(-5 + 6*x - x^2)*Log[4] + (-10*x^2
+ 2*x^3)*Log[4])*Log[5 - x]^2 - 2*x^2*Log[x])/(-45*x^2 + 9*x^3 + (-30*x^2 + 6*x^3)*Log[5 - x] + (-5*x^2 + x^3)
*Log[5 - x]^2),x]

[Out]

-((E^x*Log[4])/x) + 2*x*Log[4] + (2*Log[x])/(3 + Log[5 - x])

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fricas [A]  time = 0.67, size = 55, normalized size = 1.83 \begin {gather*} \frac {2 \, {\left (6 \, x^{2} \log \relax (2) - 3 \, e^{x} \log \relax (2) + x \log \relax (x) + {\left (2 \, x^{2} \log \relax (2) - e^{x} \log \relax (2)\right )} \log \left (-x + 5\right )\right )}}{x \log \left (-x + 5\right ) + 3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^2*log(x)+(2*(-x^2+6*x-5)*log(2)*exp(x)+2*(2*x^3-10*x^2)*log(2))*log(5-x)^2+(2*(-6*x^2+36*x-30)
*log(2)*exp(x)+2*(12*x^3-60*x^2)*log(2)+2*x^2-10*x)*log(5-x)+2*(-9*x^2+54*x-45)*log(2)*exp(x)+2*(18*x^3-90*x^2
)*log(2)+6*x^2-30*x)/((x^3-5*x^2)*log(5-x)^2+(6*x^3-30*x^2)*log(5-x)+9*x^3-45*x^2),x, algorithm="fricas")

[Out]

2*(6*x^2*log(2) - 3*e^x*log(2) + x*log(x) + (2*x^2*log(2) - e^x*log(2))*log(-x + 5))/(x*log(-x + 5) + 3*x)

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giac [A]  time = 0.26, size = 59, normalized size = 1.97 \begin {gather*} \frac {2 \, {\left (2 \, x^{2} \log \relax (2) \log \left (-x + 5\right ) + 6 \, x^{2} \log \relax (2) - e^{x} \log \relax (2) \log \left (-x + 5\right ) - 3 \, e^{x} \log \relax (2) + x \log \relax (x)\right )}}{x \log \left (-x + 5\right ) + 3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^2*log(x)+(2*(-x^2+6*x-5)*log(2)*exp(x)+2*(2*x^3-10*x^2)*log(2))*log(5-x)^2+(2*(-6*x^2+36*x-30)
*log(2)*exp(x)+2*(12*x^3-60*x^2)*log(2)+2*x^2-10*x)*log(5-x)+2*(-9*x^2+54*x-45)*log(2)*exp(x)+2*(18*x^3-90*x^2
)*log(2)+6*x^2-30*x)/((x^3-5*x^2)*log(5-x)^2+(6*x^3-30*x^2)*log(5-x)+9*x^3-45*x^2),x, algorithm="giac")

[Out]

2*(2*x^2*log(2)*log(-x + 5) + 6*x^2*log(2) - e^x*log(2)*log(-x + 5) - 3*e^x*log(2) + x*log(x))/(x*log(-x + 5)
+ 3*x)

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maple [A]  time = 0.11, size = 33, normalized size = 1.10

method	result	size
risch	\(\frac {2 \ln \relax (2) \left (2 x^{2}-{\mathrm e}^{x}\right )}{x}+\frac {2 \ln \relax (x )}{\ln \left (5-x \right )+3}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x^2*ln(x)+(2*(-x^2+6*x-5)*ln(2)*exp(x)+2*(2*x^3-10*x^2)*ln(2))*ln(5-x)^2+(2*(-6*x^2+36*x-30)*ln(2)*exp
(x)+2*(12*x^3-60*x^2)*ln(2)+2*x^2-10*x)*ln(5-x)+2*(-9*x^2+54*x-45)*ln(2)*exp(x)+2*(18*x^3-90*x^2)*ln(2)+6*x^2-
30*x)/((x^3-5*x^2)*ln(5-x)^2+(6*x^3-30*x^2)*ln(5-x)+9*x^3-45*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.51, size = 55, normalized size = 1.83 \begin {gather*} \frac {2 \, {\left (6 \, x^{2} \log \relax (2) - 3 \, e^{x} \log \relax (2) + x \log \relax (x) + {\left (2 \, x^{2} \log \relax (2) - e^{x} \log \relax (2)\right )} \log \left (-x + 5\right )\right )}}{x \log \left (-x + 5\right ) + 3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^2*log(x)+(2*(-x^2+6*x-5)*log(2)*exp(x)+2*(2*x^3-10*x^2)*log(2))*log(5-x)^2+(2*(-6*x^2+36*x-30)
*log(2)*exp(x)+2*(12*x^3-60*x^2)*log(2)+2*x^2-10*x)*log(5-x)+2*(-9*x^2+54*x-45)*log(2)*exp(x)+2*(18*x^3-90*x^2
)*log(2)+6*x^2-30*x)/((x^3-5*x^2)*log(5-x)^2+(6*x^3-30*x^2)*log(5-x)+9*x^3-45*x^2),x, algorithm="maxima")

[Out]

2*(6*x^2*log(2) - 3*e^x*log(2) + x*log(x) + (2*x^2*log(2) - e^x*log(2))*log(-x + 5))/(x*log(-x + 5) + 3*x)

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mupad [B]  time = 3.31, size = 60, normalized size = 2.00 \begin {gather*} 4\,x\,\ln \relax (2)-\frac {10}{x}+\frac {\frac {2\,\left (x\,\ln \relax (x)-3\,x+15\right )}{x}-\frac {2\,\ln \left (5-x\right )\,\left (x-5\right )}{x}}{\ln \left (5-x\right )+3}-\frac {2\,{\mathrm {e}}^x\,\ln \relax (2)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((30*x + log(5 - x)^2*(2*log(2)*(10*x^2 - 2*x^3) + 2*exp(x)*log(2)*(x^2 - 6*x + 5)) + 2*x^2*log(x) + log(5
- x)*(10*x + 2*log(2)*(60*x^2 - 12*x^3) - 2*x^2 + 2*exp(x)*log(2)*(6*x^2 - 36*x + 30)) + 2*log(2)*(90*x^2 - 18
*x^3) - 6*x^2 + 2*exp(x)*log(2)*(9*x^2 - 54*x + 45))/(log(5 - x)^2*(5*x^2 - x^3) + log(5 - x)*(30*x^2 - 6*x^3)
+ 45*x^2 - 9*x^3),x)

[Out]

4*x*log(2) - 10/x + ((2*(x*log(x) - 3*x + 15))/x - (2*log(5 - x)*(x - 5))/x)/(log(5 - x) + 3) - (2*exp(x)*log(
2))/x

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sympy [A]  time = 0.57, size = 27, normalized size = 0.90 \begin {gather*} 4 x \log {\relax (2 )} + \frac {2 \log {\relax (x )}}{\log {\left (5 - x \right )} + 3} - \frac {2 e^{x} \log {\relax (2 )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x**2*ln(x)+(2*(-x**2+6*x-5)*ln(2)*exp(x)+2*(2*x**3-10*x**2)*ln(2))*ln(5-x)**2+(2*(-6*x**2+36*x-3
0)*ln(2)*exp(x)+2*(12*x**3-60*x**2)*ln(2)+2*x**2-10*x)*ln(5-x)+2*(-9*x**2+54*x-45)*ln(2)*exp(x)+2*(18*x**3-90*
x**2)*ln(2)+6*x**2-30*x)/((x**3-5*x**2)*ln(5-x)**2+(6*x**3-30*x**2)*ln(5-x)+9*x**3-45*x**2),x)

[Out]

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

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