3.31.43 \(\int \frac {-15 x^3+10 x^4-x^5+e^x (-250+100 x-260 x^2+100 x^3-10 x^4)+e^{2 x} (50 x^3-20 x^4+2 x^5)+(-15 x+10 x^2-x^3+e^x (250-100 x+10 x^2)+e^{2 x} (50 x-20 x^2+2 x^3)) \log (x)+(1250-500 x+1300 x^2-500 x^3+50 x^4+e^x (-250 x^3+100 x^4-10 x^5)+(-1250+500 x-50 x^2+e^x (-250 x+100 x^2-10 x^3)) \log (x)) \log (\frac {x^2+\log (x)}{2 x})}{25 x^3-10 x^4+x^5+(25 x-10 x^2+x^3) \log (x)} \, dx\)

Optimal. Leaf size=38 \[ -2-x+\frac {2 x}{5-x}+\left (-e^x+5 \log \left (\frac {1}{2} \left (x+\frac {\log (x)}{x}\right )\right )\right )^2 \]

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Rubi [F]  time = 15.76, 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 {-15 x^3+10 x^4-x^5+e^x \left (-250+100 x-260 x^2+100 x^3-10 x^4\right )+e^{2 x} \left (50 x^3-20 x^4+2 x^5\right )+\left (-15 x+10 x^2-x^3+e^x \left (250-100 x+10 x^2\right )+e^{2 x} \left (50 x-20 x^2+2 x^3\right )\right ) \log (x)+\left (1250-500 x+1300 x^2-500 x^3+50 x^4+e^x \left (-250 x^3+100 x^4-10 x^5\right )+\left (-1250+500 x-50 x^2+e^x \left (-250 x+100 x^2-10 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x^2+\log (x)}{2 x}\right )}{25 x^3-10 x^4+x^5+\left (25 x-10 x^2+x^3\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-15*x^3 + 10*x^4 - x^5 + E^x*(-250 + 100*x - 260*x^2 + 100*x^3 - 10*x^4) + E^(2*x)*(50*x^3 - 20*x^4 + 2*x
^5) + (-15*x + 10*x^2 - x^3 + E^x*(250 - 100*x + 10*x^2) + E^(2*x)*(50*x - 20*x^2 + 2*x^3))*Log[x] + (1250 - 5
00*x + 1300*x^2 - 500*x^3 + 50*x^4 + E^x*(-250*x^3 + 100*x^4 - 10*x^5) + (-1250 + 500*x - 50*x^2 + E^x*(-250*x
+ 100*x^2 - 10*x^3))*Log[x])*Log[(x^2 + Log[x])/(2*x)])/(25*x^3 - 10*x^4 + x^5 + (25*x - 10*x^2 + x^3)*Log[x]
),x]

[Out]

E^(2*x) + 10/(5 - x) - x - (10*E^x*(x^3*Log[(x^2 + Log[x])/(2*x)] + x*Log[x]*Log[(x^2 + Log[x])/(2*x)]))/(x*(x
^2 + Log[x])) + 50*Defer[Int][Log[(x^2 + Log[x])/(2*x)]/(x*(x^2 + Log[x])), x] + 50*Defer[Int][(x*Log[(x^2 + L
og[x])/(2*x)])/(x^2 + Log[x]), x] - 50*Defer[Int][(Log[x]*Log[(x^2 + Log[x])/(2*x)])/(x*(x^2 + Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-15 x^3+10 x^4-x^5+e^x \left (-250+100 x-260 x^2+100 x^3-10 x^4\right )+e^{2 x} \left (50 x^3-20 x^4+2 x^5\right )+\left (-15 x+10 x^2-x^3+e^x \left (250-100 x+10 x^2\right )+e^{2 x} \left (50 x-20 x^2+2 x^3\right )\right ) \log (x)+\left (1250-500 x+1300 x^2-500 x^3+50 x^4+e^x \left (-250 x^3+100 x^4-10 x^5\right )+\left (-1250+500 x-50 x^2+e^x \left (-250 x+100 x^2-10 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x^2+\log (x)}{2 x}\right )}{(5-x)^2 x \left (x^2+\log (x)\right )} \, dx\\ &=\int \left (2 e^{2 x}-\frac {15 x^2}{(-5+x)^2 \left (x^2+\log (x)\right )}+\frac {10 x^3}{(-5+x)^2 \left (x^2+\log (x)\right )}-\frac {x^4}{(-5+x)^2 \left (x^2+\log (x)\right )}-\frac {15 \log (x)}{(-5+x)^2 \left (x^2+\log (x)\right )}+\frac {10 x \log (x)}{(-5+x)^2 \left (x^2+\log (x)\right )}-\frac {x^2 \log (x)}{(-5+x)^2 \left (x^2+\log (x)\right )}-\frac {500 \log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 \left (x^2+\log (x)\right )}+\frac {1250 \log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 x \left (x^2+\log (x)\right )}+\frac {1300 x \log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 \left (x^2+\log (x)\right )}-\frac {500 x^2 \log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 \left (x^2+\log (x)\right )}+\frac {50 x^3 \log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 \left (x^2+\log (x)\right )}+\frac {500 \log (x) \log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 \left (x^2+\log (x)\right )}-\frac {1250 \log (x) \log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 x \left (x^2+\log (x)\right )}-\frac {50 x \log (x) \log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 \left (x^2+\log (x)\right )}-\frac {10 e^x \left (1+x^2-\log (x)+x^3 \log \left (\frac {x^2+\log (x)}{2 x}\right )+x \log (x) \log \left (\frac {x^2+\log (x)}{2 x}\right )\right )}{x \left (x^2+\log (x)\right )}\right ) \, dx\\ &=2 \int e^{2 x} \, dx+10 \int \frac {x^3}{(-5+x)^2 \left (x^2+\log (x)\right )} \, dx+10 \int \frac {x \log (x)}{(-5+x)^2 \left (x^2+\log (x)\right )} \, dx-10 \int \frac {e^x \left (1+x^2-\log (x)+x^3 \log \left (\frac {x^2+\log (x)}{2 x}\right )+x \log (x) \log \left (\frac {x^2+\log (x)}{2 x}\right )\right )}{x \left (x^2+\log (x)\right )} \, dx-15 \int \frac {x^2}{(-5+x)^2 \left (x^2+\log (x)\right )} \, dx-15 \int \frac {\log (x)}{(-5+x)^2 \left (x^2+\log (x)\right )} \, dx+50 \int \frac {x^3 \log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 \left (x^2+\log (x)\right )} \, dx-50 \int \frac {x \log (x) \log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 \left (x^2+\log (x)\right )} \, dx-500 \int \frac {\log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 \left (x^2+\log (x)\right )} \, dx-500 \int \frac {x^2 \log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 \left (x^2+\log (x)\right )} \, dx+500 \int \frac {\log (x) \log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 \left (x^2+\log (x)\right )} \, dx+1250 \int \frac {\log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 x \left (x^2+\log (x)\right )} \, dx-1250 \int \frac {\log (x) \log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 x \left (x^2+\log (x)\right )} \, dx+1300 \int \frac {x \log \left (\frac {x^2+\log (x)}{2 x}\right )}{(-5+x)^2 \left (x^2+\log (x)\right )} \, dx-\int \frac {x^4}{(-5+x)^2 \left (x^2+\log (x)\right )} \, dx-\int \frac {x^2 \log (x)}{(-5+x)^2 \left (x^2+\log (x)\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 53, normalized size = 1.39 \begin {gather*} e^{2 x}-\frac {10}{-5+x}-x-10 e^x \log \left (\frac {x^2+\log (x)}{2 x}\right )+25 \log ^2\left (\frac {x^2+\log (x)}{2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*x^3 + 10*x^4 - x^5 + E^x*(-250 + 100*x - 260*x^2 + 100*x^3 - 10*x^4) + E^(2*x)*(50*x^3 - 20*x^4
+ 2*x^5) + (-15*x + 10*x^2 - x^3 + E^x*(250 - 100*x + 10*x^2) + E^(2*x)*(50*x - 20*x^2 + 2*x^3))*Log[x] + (12
50 - 500*x + 1300*x^2 - 500*x^3 + 50*x^4 + E^x*(-250*x^3 + 100*x^4 - 10*x^5) + (-1250 + 500*x - 50*x^2 + E^x*(
-250*x + 100*x^2 - 10*x^3))*Log[x])*Log[(x^2 + Log[x])/(2*x)])/(25*x^3 - 10*x^4 + x^5 + (25*x - 10*x^2 + x^3)*
Log[x]),x]

[Out]

E^(2*x) - 10/(-5 + x) - x - 10*E^x*Log[(x^2 + Log[x])/(2*x)] + 25*Log[(x^2 + Log[x])/(2*x)]^2

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fricas [A]  time = 0.65, size = 62, normalized size = 1.63 \begin {gather*} -\frac {10 \, {\left (x - 5\right )} e^{x} \log \left (\frac {x^{2} + \log \relax (x)}{2 \, x}\right ) - 25 \, {\left (x - 5\right )} \log \left (\frac {x^{2} + \log \relax (x)}{2 \, x}\right )^{2} + x^{2} - {\left (x - 5\right )} e^{\left (2 \, x\right )} - 5 \, x + 10}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-10*x^3+100*x^2-250*x)*exp(x)-50*x^2+500*x-1250)*log(x)+(-10*x^5+100*x^4-250*x^3)*exp(x)+50*x^4-
500*x^3+1300*x^2-500*x+1250)*log(1/2*(log(x)+x^2)/x)+((2*x^3-20*x^2+50*x)*exp(x)^2+(10*x^2-100*x+250)*exp(x)-x
^3+10*x^2-15*x)*log(x)+(2*x^5-20*x^4+50*x^3)*exp(x)^2+(-10*x^4+100*x^3-260*x^2+100*x-250)*exp(x)-x^5+10*x^4-15
*x^3)/((x^3-10*x^2+25*x)*log(x)+x^5-10*x^4+25*x^3),x, algorithm="fricas")

[Out]

-(10*(x - 5)*e^x*log(1/2*(x^2 + log(x))/x) - 25*(x - 5)*log(1/2*(x^2 + log(x))/x)^2 + x^2 - (x - 5)*e^(2*x) -
5*x + 10)/(x - 5)

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giac [B]  time = 0.40, size = 172, normalized size = 4.53 \begin {gather*} \frac {10 \, x e^{x} \log \relax (2) - 10 \, x e^{x} \log \left (x^{2} + \log \relax (x)\right ) - 50 \, x \log \relax (2) \log \left (x^{2} + \log \relax (x)\right ) + 25 \, x \log \left (x^{2} + \log \relax (x)\right )^{2} + 10 \, x e^{x} \log \relax (x) + 50 \, x \log \relax (2) \log \relax (x) - 50 \, x \log \left (x^{2} + \log \relax (x)\right ) \log \relax (x) + 25 \, x \log \relax (x)^{2} - x^{2} + x e^{\left (2 \, x\right )} - 50 \, e^{x} \log \relax (2) + 50 \, e^{x} \log \left (x^{2} + \log \relax (x)\right ) + 250 \, \log \relax (2) \log \left (x^{2} + \log \relax (x)\right ) - 125 \, \log \left (x^{2} + \log \relax (x)\right )^{2} - 50 \, e^{x} \log \relax (x) - 250 \, \log \relax (2) \log \relax (x) + 250 \, \log \left (x^{2} + \log \relax (x)\right ) \log \relax (x) - 125 \, \log \relax (x)^{2} + 5 \, x - 5 \, e^{\left (2 \, x\right )} - 10}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-10*x^3+100*x^2-250*x)*exp(x)-50*x^2+500*x-1250)*log(x)+(-10*x^5+100*x^4-250*x^3)*exp(x)+50*x^4-
500*x^3+1300*x^2-500*x+1250)*log(1/2*(log(x)+x^2)/x)+((2*x^3-20*x^2+50*x)*exp(x)^2+(10*x^2-100*x+250)*exp(x)-x
^3+10*x^2-15*x)*log(x)+(2*x^5-20*x^4+50*x^3)*exp(x)^2+(-10*x^4+100*x^3-260*x^2+100*x-250)*exp(x)-x^5+10*x^4-15
*x^3)/((x^3-10*x^2+25*x)*log(x)+x^5-10*x^4+25*x^3),x, algorithm="giac")

[Out]

(10*x*e^x*log(2) - 10*x*e^x*log(x^2 + log(x)) - 50*x*log(2)*log(x^2 + log(x)) + 25*x*log(x^2 + log(x))^2 + 10*
x*e^x*log(x) + 50*x*log(2)*log(x) - 50*x*log(x^2 + log(x))*log(x) + 25*x*log(x)^2 - x^2 + x*e^(2*x) - 50*e^x*l
og(2) + 50*e^x*log(x^2 + log(x)) + 250*log(2)*log(x^2 + log(x)) - 125*log(x^2 + log(x))^2 - 50*e^x*log(x) - 25
0*log(2)*log(x) + 250*log(x^2 + log(x))*log(x) - 125*log(x)^2 + 5*x - 5*e^(2*x) - 10)/(x - 5)

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maple [C]  time = 0.45, size = 1971, normalized size = 51.87

method	result	size
risch	\(\text {Expression too large to display}\)	\(1971\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-10*x^3+100*x^2-250*x)*exp(x)-50*x^2+500*x-1250)*ln(x)+(-10*x^5+100*x^4-250*x^3)*exp(x)+50*x^4-500*x^3
+1300*x^2-500*x+1250)*ln(1/2*(ln(x)+x^2)/x)+((2*x^3-20*x^2+50*x)*exp(x)^2+(10*x^2-100*x+250)*exp(x)-x^3+10*x^2
-15*x)*ln(x)+(2*x^5-20*x^4+50*x^3)*exp(x)^2+(-10*x^4+100*x^3-260*x^2+100*x-250)*exp(x)-x^5+10*x^4-15*x^3)/((x^
3-10*x^2+25*x)*ln(x)+x^5-10*x^4+25*x^3),x,method=_RETURNVERBOSE)

[Out]

25*ln(ln(x)+x^2)^2+(-50*ln(x)-10*exp(x))*ln(ln(x)+x^2)-(10-5*x-10*x*exp(x)*ln(x)+5*exp(2*x)+125*ln(x)^2+x^2+50
*exp(x)*ln(2)-x*exp(2*x)+50*exp(x)*ln(x)-25*x*ln(x)^2+125*I*Pi*ln((Pi*csgn(I/x*(ln(x)+x^2))^3-Pi*csgn(I/x)*csg
n(I/x*(ln(x)+x^2))^2-Pi*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2))^2-2*I*ln(2)+Pi*csgn(I/x)*csgn(I*(ln(x)+x^2))
*csgn(I/x*(ln(x)+x^2)))*x)*csgn(I/x*(ln(x)+x^2))^3-125*I*Pi*ln(ln(x)+x^2)*csgn(I/x*(ln(x)+x^2))^3+25*I*Pi*csgn
(I/x*(ln(x)+x^2))^3*exp(x)-10*x*ln(2)*exp(x)+50*ln(2)*ln(ln(x)+x^2)*x+250*ln(2)*ln((Pi*csgn(I/x*(ln(x)+x^2))^3
-Pi*csgn(I/x)*csgn(I/x*(ln(x)+x^2))^2-Pi*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2))^2-2*I*ln(2)+Pi*csgn(I/x)*cs
gn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2)))*x)-50*ln(2)*ln((Pi*csgn(I/x*(ln(x)+x^2))^3-Pi*csgn(I/x)*csgn(I/x*(ln(
x)+x^2))^2-Pi*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2))^2-2*I*ln(2)+Pi*csgn(I/x)*csgn(I*(ln(x)+x^2))*csgn(I/x*
(ln(x)+x^2)))*x)*x-250*ln(2)*ln(ln(x)+x^2)-25*I*Pi*ln((Pi*csgn(I/x*(ln(x)+x^2))^3-Pi*csgn(I/x)*csgn(I/x*(ln(x)
+x^2))^2-Pi*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2))^2-2*I*ln(2)+Pi*csgn(I/x)*csgn(I*(ln(x)+x^2))*csgn(I/x*(l
n(x)+x^2)))*x)*csgn(I/x)*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2))*x+25*I*Pi*ln(ln(x)+x^2)*csgn(I/x)*csgn(I*(l
n(x)+x^2))*csgn(I/x*(ln(x)+x^2))*x-5*I*Pi*x*csgn(I/x)*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2))*exp(x)-25*I*Pi
*ln((Pi*csgn(I/x*(ln(x)+x^2))^3-Pi*csgn(I/x)*csgn(I/x*(ln(x)+x^2))^2-Pi*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^
2))^2-2*I*ln(2)+Pi*csgn(I/x)*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2)))*x)*csgn(I/x*(ln(x)+x^2))^3*x-125*I*Pi*
ln((Pi*csgn(I/x*(ln(x)+x^2))^3-Pi*csgn(I/x)*csgn(I/x*(ln(x)+x^2))^2-Pi*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2
))^2-2*I*ln(2)+Pi*csgn(I/x)*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2)))*x)*csgn(I/x)*csgn(I/x*(ln(x)+x^2))^2-12
5*I*Pi*ln((Pi*csgn(I/x*(ln(x)+x^2))^3-Pi*csgn(I/x)*csgn(I/x*(ln(x)+x^2))^2-Pi*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln
(x)+x^2))^2-2*I*ln(2)+Pi*csgn(I/x)*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2)))*x)*csgn(I*(ln(x)+x^2))*csgn(I/x*
(ln(x)+x^2))^2-5*I*Pi*x*csgn(I/x*(ln(x)+x^2))^3*exp(x)-25*I*Pi*csgn(I/x)*csgn(I/x*(ln(x)+x^2))^2*exp(x)-25*I*P
i*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2))^2*exp(x)+125*I*Pi*ln(ln(x)+x^2)*csgn(I/x)*csgn(I/x*(ln(x)+x^2))^2+
125*I*Pi*ln(ln(x)+x^2)*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2))^2+25*I*Pi*ln(ln(x)+x^2)*csgn(I/x*(ln(x)+x^2))
^3*x-125*I*Pi*ln(ln(x)+x^2)*csgn(I/x)*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2))+5*I*Pi*x*csgn(I/x)*csgn(I/x*(l
n(x)+x^2))^2*exp(x)+5*I*Pi*x*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2))^2*exp(x)+25*I*Pi*csgn(I/x)*csgn(I*(ln(x
)+x^2))*csgn(I/x*(ln(x)+x^2))*exp(x)-25*I*Pi*ln(ln(x)+x^2)*csgn(I/x)*csgn(I/x*(ln(x)+x^2))^2*x-25*I*Pi*ln(ln(x
)+x^2)*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2))^2*x+25*I*Pi*ln((Pi*csgn(I/x*(ln(x)+x^2))^3-Pi*csgn(I/x)*csgn(
I/x*(ln(x)+x^2))^2-Pi*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2))^2-2*I*ln(2)+Pi*csgn(I/x)*csgn(I*(ln(x)+x^2))*c
sgn(I/x*(ln(x)+x^2)))*x)*csgn(I/x)*csgn(I/x*(ln(x)+x^2))^2*x+25*I*Pi*ln((Pi*csgn(I/x*(ln(x)+x^2))^3-Pi*csgn(I/
x)*csgn(I/x*(ln(x)+x^2))^2-Pi*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2))^2-2*I*ln(2)+Pi*csgn(I/x)*csgn(I*(ln(x)
+x^2))*csgn(I/x*(ln(x)+x^2)))*x)*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2))^2*x+125*I*Pi*ln((Pi*csgn(I/x*(ln(x)
+x^2))^3-Pi*csgn(I/x)*csgn(I/x*(ln(x)+x^2))^2-Pi*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2))^2-2*I*ln(2)+Pi*csgn
(I/x)*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2)))*x)*csgn(I/x)*csgn(I*(ln(x)+x^2))*csgn(I/x*(ln(x)+x^2)))/(x-5)

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maxima [B]  time = 0.82, size = 111, normalized size = 2.92 \begin {gather*} \frac {25 \, {\left (x - 5\right )} \log \left (x^{2} + \log \relax (x)\right )^{2} + 25 \, {\left (x - 5\right )} \log \relax (x)^{2} - x^{2} + {\left (x - 5\right )} e^{\left (2 \, x\right )} + 10 \, {\left (x \log \relax (2) + {\left (x - 5\right )} \log \relax (x) - 5 \, \log \relax (2)\right )} e^{x} - 10 \, {\left ({\left (x - 5\right )} e^{x} + 5 \, x \log \relax (2) + 5 \, {\left (x - 5\right )} \log \relax (x) - 25 \, \log \relax (2)\right )} \log \left (x^{2} + \log \relax (x)\right ) + 50 \, {\left (x \log \relax (2) - 5 \, \log \relax (2)\right )} \log \relax (x) + 5 \, x - 10}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-10*x^3+100*x^2-250*x)*exp(x)-50*x^2+500*x-1250)*log(x)+(-10*x^5+100*x^4-250*x^3)*exp(x)+50*x^4-
500*x^3+1300*x^2-500*x+1250)*log(1/2*(log(x)+x^2)/x)+((2*x^3-20*x^2+50*x)*exp(x)^2+(10*x^2-100*x+250)*exp(x)-x
^3+10*x^2-15*x)*log(x)+(2*x^5-20*x^4+50*x^3)*exp(x)^2+(-10*x^4+100*x^3-260*x^2+100*x-250)*exp(x)-x^5+10*x^4-15
*x^3)/((x^3-10*x^2+25*x)*log(x)+x^5-10*x^4+25*x^3),x, algorithm="maxima")

[Out]

(25*(x - 5)*log(x^2 + log(x))^2 + 25*(x - 5)*log(x)^2 - x^2 + (x - 5)*e^(2*x) + 10*(x*log(2) + (x - 5)*log(x)
- 5*log(2))*e^x - 10*((x - 5)*e^x + 5*x*log(2) + 5*(x - 5)*log(x) - 25*log(2))*log(x^2 + log(x)) + 50*(x*log(2
) - 5*log(2))*log(x) + 5*x - 10)/(x - 5)

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mupad [B]  time = 2.16, size = 53, normalized size = 1.39 \begin {gather*} {\mathrm {e}}^{2\,x}-x-\frac {10}{x-5}-10\,{\mathrm {e}}^x\,\ln \left (\frac {\frac {\ln \relax (x)}{2}+\frac {x^2}{2}}{x}\right )+25\,{\ln \left (\frac {\frac {\ln \relax (x)}{2}+\frac {x^2}{2}}{x}\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(260*x^2 - 100*x - 100*x^3 + 10*x^4 + 250) + log((log(x)/2 + x^2/2)/x)*(500*x + exp(x)*(250*x^3 -
100*x^4 + 10*x^5) + log(x)*(50*x^2 - 500*x + exp(x)*(250*x - 100*x^2 + 10*x^3) + 1250) - 1300*x^2 + 500*x^3 -
50*x^4 - 1250) - exp(2*x)*(50*x^3 - 20*x^4 + 2*x^5) - log(x)*(exp(2*x)*(50*x - 20*x^2 + 2*x^3) - 15*x + exp(x
)*(10*x^2 - 100*x + 250) + 10*x^2 - x^3) + 15*x^3 - 10*x^4 + x^5)/(log(x)*(25*x - 10*x^2 + x^3) + 25*x^3 - 10*
x^4 + x^5),x)

[Out]

exp(2*x) - x - 10/(x - 5) - 10*exp(x)*log((log(x)/2 + x^2/2)/x) + 25*log((log(x)/2 + x^2/2)/x)^2

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sympy [A]  time = 7.96, size = 46, normalized size = 1.21 \begin {gather*} - x + e^{2 x} - 10 e^{x} \log {\left (\frac {\frac {x^{2}}{2} + \frac {\log {\relax (x )}}{2}}{x} \right )} + 25 \log {\left (\frac {\frac {x^{2}}{2} + \frac {\log {\relax (x )}}{2}}{x} \right )}^{2} - \frac {10}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-10*x**3+100*x**2-250*x)*exp(x)-50*x**2+500*x-1250)*ln(x)+(-10*x**5+100*x**4-250*x**3)*exp(x)+50
*x**4-500*x**3+1300*x**2-500*x+1250)*ln(1/2*(ln(x)+x**2)/x)+((2*x**3-20*x**2+50*x)*exp(x)**2+(10*x**2-100*x+25
0)*exp(x)-x**3+10*x**2-15*x)*ln(x)+(2*x**5-20*x**4+50*x**3)*exp(x)**2+(-10*x**4+100*x**3-260*x**2+100*x-250)*e
xp(x)-x**5+10*x**4-15*x**3)/((x**3-10*x**2+25*x)*ln(x)+x**5-10*x**4+25*x**3),x)

[Out]

-x + exp(2*x) - 10*exp(x)*log((x**2/2 + log(x)/2)/x) + 25*log((x**2/2 + log(x)/2)/x)**2 - 10/(x - 5)

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