3.28.30 \(\int \frac {e^{\frac {1}{5} \log (1+x) \log (25-x+\log (5))} (-125-120 x+5 x^2+(-5-5 x) \log (5)+(-8-6 x+2 x^2+(4+3 x-x^2) \log (\frac {4-x}{2})) \log (1+x)+(200-58 x+2 x^2+(8-2 x) \log (5)+(-100+29 x-x^2+(-4+x) \log (5)) \log (\frac {4-x}{2})) \log (25-x+\log (5)))}{-2000-1420 x+560 x^2-20 x^3+(-80-60 x+20 x^2) \log (5)+(2000+1420 x-560 x^2+20 x^3+(80+60 x-20 x^2) \log (5)) \log (\frac {4-x}{2})+(-500-355 x+140 x^2-5 x^3+(-20-15 x+5 x^2) \log (5)) \log ^2(\frac {4-x}{2})} \, dx\)

Optimal. Leaf size=31 \[ \frac {e^{\frac {1}{5} \log (1+x) \log (25-x+\log (5))}}{-2+\log \left (2-\frac {x}{2}\right )} \]

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Rubi [B]  time = 0.65, antiderivative size = 255, normalized size of antiderivative = 8.23, number of steps used = 1, number of rules used = 1, integrand size = 237, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.004, Rules used = {2288} \begin {gather*} -\frac {e^{\frac {1}{5} \log (x+1) \log (-x+25+\log (5))} \left (\left (-2 x^2-\left (-x^2+3 x+4\right ) \log \left (\frac {4-x}{2}\right )+6 x+8\right ) \log (x+1)-\left (2 x^2-\left (x^2-29 x+(4-x) \log (5)+100\right ) \log \left (\frac {4-x}{2}\right )-58 x+2 (4-x) \log (5)+200\right ) \log (-x+25+\log (5))\right )}{\left (4 x^3-112 x^2+4 \left (-x^2+3 x+4\right ) \log (5)+\left (x^3-28 x^2+\left (-x^2+3 x+4\right ) \log (5)+71 x+100\right ) \log ^2\left (\frac {4-x}{2}\right )-4 \left (x^3-28 x^2+\left (-x^2+3 x+4\right ) \log (5)+71 x+100\right ) \log \left (\frac {4-x}{2}\right )+284 x+400\right ) \left (\frac {\log (x+1)}{-x+25+\log (5)}-\frac {\log (-x+25+\log (5))}{x+1}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((Log[1 + x]*Log[25 - x + Log[5]])/5)*(-125 - 120*x + 5*x^2 + (-5 - 5*x)*Log[5] + (-8 - 6*x + 2*x^2 + (
4 + 3*x - x^2)*Log[(4 - x)/2])*Log[1 + x] + (200 - 58*x + 2*x^2 + (8 - 2*x)*Log[5] + (-100 + 29*x - x^2 + (-4
+ x)*Log[5])*Log[(4 - x)/2])*Log[25 - x + Log[5]]))/(-2000 - 1420*x + 560*x^2 - 20*x^3 + (-80 - 60*x + 20*x^2)
*Log[5] + (2000 + 1420*x - 560*x^2 + 20*x^3 + (80 + 60*x - 20*x^2)*Log[5])*Log[(4 - x)/2] + (-500 - 355*x + 14
0*x^2 - 5*x^3 + (-20 - 15*x + 5*x^2)*Log[5])*Log[(4 - x)/2]^2),x]

[Out]

-((E^((Log[1 + x]*Log[25 - x + Log[5]])/5)*((8 + 6*x - 2*x^2 - (4 + 3*x - x^2)*Log[(4 - x)/2])*Log[1 + x] - (2
00 - 58*x + 2*x^2 + 2*(4 - x)*Log[5] - (100 - 29*x + x^2 + (4 - x)*Log[5])*Log[(4 - x)/2])*Log[25 - x + Log[5]
]))/((400 + 284*x - 112*x^2 + 4*x^3 + 4*(4 + 3*x - x^2)*Log[5] - 4*(100 + 71*x - 28*x^2 + x^3 + (4 + 3*x - x^2
)*Log[5])*Log[(4 - x)/2] + (100 + 71*x - 28*x^2 + x^3 + (4 + 3*x - x^2)*Log[5])*Log[(4 - x)/2]^2)*(Log[1 + x]/
(25 - x + Log[5]) - Log[25 - x + Log[5]]/(1 + x))))

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {e^{\frac {1}{5} \log (1+x) \log (25-x+\log (5))} \left (\left (8+6 x-2 x^2-\left (4+3 x-x^2\right ) \log \left (\frac {4-x}{2}\right )\right ) \log (1+x)-\left (200-58 x+2 x^2+2 (4-x) \log (5)-\left (100-29 x+x^2+(4-x) \log (5)\right ) \log \left (\frac {4-x}{2}\right )\right ) \log (25-x+\log (5))\right )}{\left (400+284 x-112 x^2+4 x^3+4 \left (4+3 x-x^2\right ) \log (5)-4 \left (100+71 x-28 x^2+x^3+\left (4+3 x-x^2\right ) \log (5)\right ) \log \left (\frac {4-x}{2}\right )+\left (100+71 x-28 x^2+x^3+\left (4+3 x-x^2\right ) \log (5)\right ) \log ^2\left (\frac {4-x}{2}\right )\right ) \left (\frac {\log (1+x)}{25-x+\log (5)}-\frac {\log (25-x+\log (5))}{1+x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.33, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {1}{5} \log (1+x) \log (25-x+\log (5))} \left (-125-120 x+5 x^2+(-5-5 x) \log (5)+\left (-8-6 x+2 x^2+\left (4+3 x-x^2\right ) \log \left (\frac {4-x}{2}\right )\right ) \log (1+x)+\left (200-58 x+2 x^2+(8-2 x) \log (5)+\left (-100+29 x-x^2+(-4+x) \log (5)\right ) \log \left (\frac {4-x}{2}\right )\right ) \log (25-x+\log (5))\right )}{-2000-1420 x+560 x^2-20 x^3+\left (-80-60 x+20 x^2\right ) \log (5)+\left (2000+1420 x-560 x^2+20 x^3+\left (80+60 x-20 x^2\right ) \log (5)\right ) \log \left (\frac {4-x}{2}\right )+\left (-500-355 x+140 x^2-5 x^3+\left (-20-15 x+5 x^2\right ) \log (5)\right ) \log ^2\left (\frac {4-x}{2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((Log[1 + x]*Log[25 - x + Log[5]])/5)*(-125 - 120*x + 5*x^2 + (-5 - 5*x)*Log[5] + (-8 - 6*x + 2*x
^2 + (4 + 3*x - x^2)*Log[(4 - x)/2])*Log[1 + x] + (200 - 58*x + 2*x^2 + (8 - 2*x)*Log[5] + (-100 + 29*x - x^2
+ (-4 + x)*Log[5])*Log[(4 - x)/2])*Log[25 - x + Log[5]]))/(-2000 - 1420*x + 560*x^2 - 20*x^3 + (-80 - 60*x + 2
0*x^2)*Log[5] + (2000 + 1420*x - 560*x^2 + 20*x^3 + (80 + 60*x - 20*x^2)*Log[5])*Log[(4 - x)/2] + (-500 - 355*
x + 140*x^2 - 5*x^3 + (-20 - 15*x + 5*x^2)*Log[5])*Log[(4 - x)/2]^2),x]

[Out]

Integrate[(E^((Log[1 + x]*Log[25 - x + Log[5]])/5)*(-125 - 120*x + 5*x^2 + (-5 - 5*x)*Log[5] + (-8 - 6*x + 2*x
^2 + (4 + 3*x - x^2)*Log[(4 - x)/2])*Log[1 + x] + (200 - 58*x + 2*x^2 + (8 - 2*x)*Log[5] + (-100 + 29*x - x^2
+ (-4 + x)*Log[5])*Log[(4 - x)/2])*Log[25 - x + Log[5]]))/(-2000 - 1420*x + 560*x^2 - 20*x^3 + (-80 - 60*x + 2
0*x^2)*Log[5] + (2000 + 1420*x - 560*x^2 + 20*x^3 + (80 + 60*x - 20*x^2)*Log[5])*Log[(4 - x)/2] + (-500 - 355*
x + 140*x^2 - 5*x^3 + (-20 - 15*x + 5*x^2)*Log[5])*Log[(4 - x)/2]^2), x]

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fricas [A]  time = 1.11, size = 26, normalized size = 0.84 \begin {gather*} \frac {e^{\left (\frac {1}{5} \, \log \left (x + 1\right ) \log \left (-x + \log \relax (5) + 25\right )\right )}}{\log \left (-\frac {1}{2} \, x + 2\right ) - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x-4)*log(5)-x^2+29*x-100)*log(2-1/2*x)+(-2*x+8)*log(5)+2*x^2-58*x+200)*log(log(5)-x+25)+((-x^2+3
*x+4)*log(2-1/2*x)+2*x^2-6*x-8)*log(x+1)+(-5*x-5)*log(5)+5*x^2-120*x-125)*exp(1/5*log(x+1)*log(log(5)-x+25))/(
((5*x^2-15*x-20)*log(5)-5*x^3+140*x^2-355*x-500)*log(2-1/2*x)^2+((-20*x^2+60*x+80)*log(5)+20*x^3-560*x^2+1420*
x+2000)*log(2-1/2*x)+(20*x^2-60*x-80)*log(5)-20*x^3+560*x^2-1420*x-2000),x, algorithm="fricas")

[Out]

e^(1/5*log(x + 1)*log(-x + log(5) + 25))/(log(-1/2*x + 2) - 2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (5 \, x^{2} - 5 \, {\left (x + 1\right )} \log \relax (5) + {\left (2 \, x^{2} - {\left (x^{2} - 3 \, x - 4\right )} \log \left (-\frac {1}{2} \, x + 2\right ) - 6 \, x - 8\right )} \log \left (x + 1\right ) + {\left (2 \, x^{2} - 2 \, {\left (x - 4\right )} \log \relax (5) - {\left (x^{2} - {\left (x - 4\right )} \log \relax (5) - 29 \, x + 100\right )} \log \left (-\frac {1}{2} \, x + 2\right ) - 58 \, x + 200\right )} \log \left (-x + \log \relax (5) + 25\right ) - 120 \, x - 125\right )} e^{\left (\frac {1}{5} \, \log \left (x + 1\right ) \log \left (-x + \log \relax (5) + 25\right )\right )}}{5 \, {\left (4 \, x^{3} + {\left (x^{3} - 28 \, x^{2} - {\left (x^{2} - 3 \, x - 4\right )} \log \relax (5) + 71 \, x + 100\right )} \log \left (-\frac {1}{2} \, x + 2\right )^{2} - 112 \, x^{2} - 4 \, {\left (x^{2} - 3 \, x - 4\right )} \log \relax (5) - 4 \, {\left (x^{3} - 28 \, x^{2} - {\left (x^{2} - 3 \, x - 4\right )} \log \relax (5) + 71 \, x + 100\right )} \log \left (-\frac {1}{2} \, x + 2\right ) + 284 \, x + 400\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x-4)*log(5)-x^2+29*x-100)*log(2-1/2*x)+(-2*x+8)*log(5)+2*x^2-58*x+200)*log(log(5)-x+25)+((-x^2+3
*x+4)*log(2-1/2*x)+2*x^2-6*x-8)*log(x+1)+(-5*x-5)*log(5)+5*x^2-120*x-125)*exp(1/5*log(x+1)*log(log(5)-x+25))/(
((5*x^2-15*x-20)*log(5)-5*x^3+140*x^2-355*x-500)*log(2-1/2*x)^2+((-20*x^2+60*x+80)*log(5)+20*x^3-560*x^2+1420*
x+2000)*log(2-1/2*x)+(20*x^2-60*x-80)*log(5)-20*x^3+560*x^2-1420*x-2000),x, algorithm="giac")

[Out]

integrate(-1/5*(5*x^2 - 5*(x + 1)*log(5) + (2*x^2 - (x^2 - 3*x - 4)*log(-1/2*x + 2) - 6*x - 8)*log(x + 1) + (2
*x^2 - 2*(x - 4)*log(5) - (x^2 - (x - 4)*log(5) - 29*x + 100)*log(-1/2*x + 2) - 58*x + 200)*log(-x + log(5) +
25) - 120*x - 125)*e^(1/5*log(x + 1)*log(-x + log(5) + 25))/(4*x^3 + (x^3 - 28*x^2 - (x^2 - 3*x - 4)*log(5) +
71*x + 100)*log(-1/2*x + 2)^2 - 112*x^2 - 4*(x^2 - 3*x - 4)*log(5) - 4*(x^3 - 28*x^2 - (x^2 - 3*x - 4)*log(5)
+ 71*x + 100)*log(-1/2*x + 2) + 284*x + 400), x)

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maple [A]  time = 0.08, size = 26, normalized size = 0.84

method	result	size
risch	\(\frac {\left (\ln \relax (5)-x +25\right )^{\frac {\ln \left (x +1\right )}{5}}}{\ln \left (2-\frac {x}{2}\right )-2}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((x-4)*ln(5)-x^2+29*x-100)*ln(2-1/2*x)+(-2*x+8)*ln(5)+2*x^2-58*x+200)*ln(ln(5)-x+25)+((-x^2+3*x+4)*ln(2-
1/2*x)+2*x^2-6*x-8)*ln(x+1)+(-5*x-5)*ln(5)+5*x^2-120*x-125)*exp(1/5*ln(x+1)*ln(ln(5)-x+25))/(((5*x^2-15*x-20)*
ln(5)-5*x^3+140*x^2-355*x-500)*ln(2-1/2*x)^2+((-20*x^2+60*x+80)*ln(5)+20*x^3-560*x^2+1420*x+2000)*ln(2-1/2*x)+
(20*x^2-60*x-80)*ln(5)-20*x^3+560*x^2-1420*x-2000),x,method=_RETURNVERBOSE)

[Out]

(ln(5)-x+25)^(1/5*ln(x+1))/(ln(2-1/2*x)-2)

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maxima [A]  time = 0.66, size = 31, normalized size = 1.00 \begin {gather*} -\frac {e^{\left (\frac {1}{5} \, \log \left (x + 1\right ) \log \left (-x + \log \relax (5) + 25\right )\right )}}{\log \relax (2) - \log \left (-x + 4\right ) + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x-4)*log(5)-x^2+29*x-100)*log(2-1/2*x)+(-2*x+8)*log(5)+2*x^2-58*x+200)*log(log(5)-x+25)+((-x^2+3
*x+4)*log(2-1/2*x)+2*x^2-6*x-8)*log(x+1)+(-5*x-5)*log(5)+5*x^2-120*x-125)*exp(1/5*log(x+1)*log(log(5)-x+25))/(
((5*x^2-15*x-20)*log(5)-5*x^3+140*x^2-355*x-500)*log(2-1/2*x)^2+((-20*x^2+60*x+80)*log(5)+20*x^3-560*x^2+1420*
x+2000)*log(2-1/2*x)+(20*x^2-60*x-80)*log(5)-20*x^3+560*x^2-1420*x-2000),x, algorithm="maxima")

[Out]

-e^(1/5*log(x + 1)*log(-x + log(5) + 25))/(log(2) - log(-x + 4) + 2)

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mupad [B]  time = 3.60, size = 25, normalized size = 0.81 \begin {gather*} \frac {{\left (x+1\right )}^{\frac {\ln \left (\ln \relax (5)-x+25\right )}{5}}}{\ln \left (2-\frac {x}{2}\right )-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((log(log(5) - x + 25)*log(x + 1))/5)*(120*x + log(5)*(5*x + 5) + log(x + 1)*(6*x - log(2 - x/2)*(3*x
- x^2 + 4) - 2*x^2 + 8) - 5*x^2 - log(log(5) - x + 25)*(log(2 - x/2)*(29*x + log(5)*(x - 4) - x^2 - 100) - log
(5)*(2*x - 8) - 58*x + 2*x^2 + 200) + 125))/(1420*x + log(5)*(60*x - 20*x^2 + 80) + log(2 - x/2)^2*(355*x + lo
g(5)*(15*x - 5*x^2 + 20) - 140*x^2 + 5*x^3 + 500) - 560*x^2 + 20*x^3 - log(2 - x/2)*(1420*x + log(5)*(60*x - 2
0*x^2 + 80) - 560*x^2 + 20*x^3 + 2000) + 2000),x)

[Out]

(x + 1)^(log(log(5) - x + 25)/5)/(log(2 - x/2) - 2)

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sympy [A]  time = 1.96, size = 24, normalized size = 0.77 \begin {gather*} \frac {e^{\frac {\log {\left (x + 1 \right )} \log {\left (- x + \log {\relax (5 )} + 25 \right )}}{5}}}{\log {\left (2 - \frac {x}{2} \right )} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x-4)*ln(5)-x**2+29*x-100)*ln(2-1/2*x)+(-2*x+8)*ln(5)+2*x**2-58*x+200)*ln(ln(5)-x+25)+((-x**2+3*x
+4)*ln(2-1/2*x)+2*x**2-6*x-8)*ln(x+1)+(-5*x-5)*ln(5)+5*x**2-120*x-125)*exp(1/5*ln(x+1)*ln(ln(5)-x+25))/(((5*x*
*2-15*x-20)*ln(5)-5*x**3+140*x**2-355*x-500)*ln(2-1/2*x)**2+((-20*x**2+60*x+80)*ln(5)+20*x**3-560*x**2+1420*x+
2000)*ln(2-1/2*x)+(20*x**2-60*x-80)*ln(5)-20*x**3+560*x**2-1420*x-2000),x)

[Out]

exp(log(x + 1)*log(-x + log(5) + 25)/5)/(log(2 - x/2) - 2)

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