3.4.15 \(\int \frac {(-7 x+\frac {1}{9} e^{5+2 x} x+e^5 (24 x-10 x^2+x^3)+\frac {1}{3} e^x (x+x^2+e^5 (11 x-2 x^2))) \log ^2(x)+e^{\frac {4}{\log (x)}} (-\frac {4 e^{5+x}}{3}+e^5 (-20+4 x)+(e^5 x-\frac {1}{3} e^{5+x} x) \log ^2(x))}{(\frac {1}{9} e^{5+2 x} x+\frac {1}{3} e^{5+x} (10 x-2 x^2)+e^5 (25 x-10 x^2+x^3)) \log ^2(x)} \, dx\)

Optimal. Leaf size=35 \[ x+\frac {-1+e^{\frac {4}{\log (x)}}-\frac {2+x}{e^5}}{5+\frac {e^x}{3}-x} \]

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Rubi [F]  time = 2.98, 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 {\left (-7 x+\frac {1}{9} e^{5+2 x} x+e^5 \left (24 x-10 x^2+x^3\right )+\frac {1}{3} e^x \left (x+x^2+e^5 \left (11 x-2 x^2\right )\right )\right ) \log ^2(x)+e^{\frac {4}{\log (x)}} \left (-\frac {4 e^{5+x}}{3}+e^5 (-20+4 x)+\left (e^5 x-\frac {1}{3} e^{5+x} x\right ) \log ^2(x)\right )}{\left (\frac {1}{9} e^{5+2 x} x+\frac {1}{3} e^{5+x} \left (10 x-2 x^2\right )+e^5 \left (25 x-10 x^2+x^3\right )\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-7*x + (E^(5 + 2*x)*x)/9 + E^5*(24*x - 10*x^2 + x^3) + (E^x*(x + x^2 + E^5*(11*x - 2*x^2)))/3)*Log[x]^2
+ E^(4/Log[x])*((-4*E^(5 + x))/3 + E^5*(-20 + 4*x) + (E^5*x - (E^(5 + x)*x)/3)*Log[x]^2))/(((E^(5 + 2*x)*x)/9
+ (E^(5 + x)*(10*x - 2*x^2))/3 + E^5*(25*x - 10*x^2 + x^3))*Log[x]^2),x]

[Out]

(3*E^(4/Log[x]))/(15 + E^x - 3*x) + 216*Defer[Int][(15 + E^x - 3*x)^(-2), x] - (63*Defer[Int][(15 + E^x - 3*x)
^(-2), x])/E^5 + 3*Defer[Int][E^(-5 + x)/(15 + E^x - 3*x)^2, x] + 33*Defer[Int][E^x/(15 + E^x - 3*x)^2, x] + D
efer[Int][E^(2*x)/(15 + E^x - 3*x)^2, x] - 90*Defer[Int][x/(15 + E^x - 3*x)^2, x] + 3*Defer[Int][(E^(-5 + x)*x
)/(15 + E^x - 3*x)^2, x] - 6*Defer[Int][(E^x*x)/(15 + E^x - 3*x)^2, x] + 9*Defer[Int][x^2/(15 + E^x - 3*x)^2,
x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-\frac {63}{e^5}+e^{2 x}-3 e^{x+\frac {4}{\log (x)}}+9 e^{\frac {4}{\log (x)}}+e^x (33-6 x)+3 e^{-5+x} (1+x)+9 \left (24-10 x+x^2\right )-\frac {12 e^{\frac {4}{\log (x)}} \left (15+e^x-3 x\right )}{x \log ^2(x)}}{\left (15+e^x-3 x\right )^2} \, dx\\ &=\int \left (-\frac {63}{e^5 \left (15+e^x-3 x\right )^2}+\frac {e^{2 x}}{\left (15+e^x-3 x\right )^2}+\frac {9 (-6+x) (-4+x)}{\left (15+e^x-3 x\right )^2}+\frac {3 e^{-5+x} (1+x)}{\left (15+e^x-3 x\right )^2}-\frac {3 e^x (-11+2 x)}{\left (15+e^x-3 x\right )^2}-\frac {3 e^{\frac {4}{\log (x)}} \left (60+4 e^x-12 x-3 x \log ^2(x)+e^x x \log ^2(x)\right )}{\left (15+e^x-3 x\right )^2 x \log ^2(x)}\right ) \, dx\\ &=3 \int \frac {e^{-5+x} (1+x)}{\left (15+e^x-3 x\right )^2} \, dx-3 \int \frac {e^x (-11+2 x)}{\left (15+e^x-3 x\right )^2} \, dx-3 \int \frac {e^{\frac {4}{\log (x)}} \left (60+4 e^x-12 x-3 x \log ^2(x)+e^x x \log ^2(x)\right )}{\left (15+e^x-3 x\right )^2 x \log ^2(x)} \, dx+9 \int \frac {(-6+x) (-4+x)}{\left (15+e^x-3 x\right )^2} \, dx-\frac {63 \int \frac {1}{\left (15+e^x-3 x\right )^2} \, dx}{e^5}+\int \frac {e^{2 x}}{\left (15+e^x-3 x\right )^2} \, dx\\ &=\frac {3 e^{\frac {4}{\log (x)}}}{15+e^x-3 x}+3 \int \left (\frac {e^{-5+x}}{\left (15+e^x-3 x\right )^2}+\frac {e^{-5+x} x}{\left (15+e^x-3 x\right )^2}\right ) \, dx-3 \int \left (-\frac {11 e^x}{\left (15+e^x-3 x\right )^2}+\frac {2 e^x x}{\left (15+e^x-3 x\right )^2}\right ) \, dx+9 \int \left (\frac {24}{\left (15+e^x-3 x\right )^2}-\frac {10 x}{\left (15+e^x-3 x\right )^2}+\frac {x^2}{\left (15+e^x-3 x\right )^2}\right ) \, dx-\frac {63 \int \frac {1}{\left (15+e^x-3 x\right )^2} \, dx}{e^5}+\int \frac {e^{2 x}}{\left (15+e^x-3 x\right )^2} \, dx\\ &=\frac {3 e^{\frac {4}{\log (x)}}}{15+e^x-3 x}+3 \int \frac {e^{-5+x}}{\left (15+e^x-3 x\right )^2} \, dx+3 \int \frac {e^{-5+x} x}{\left (15+e^x-3 x\right )^2} \, dx-6 \int \frac {e^x x}{\left (15+e^x-3 x\right )^2} \, dx+9 \int \frac {x^2}{\left (15+e^x-3 x\right )^2} \, dx+33 \int \frac {e^x}{\left (15+e^x-3 x\right )^2} \, dx-90 \int \frac {x}{\left (15+e^x-3 x\right )^2} \, dx+216 \int \frac {1}{\left (15+e^x-3 x\right )^2} \, dx-\frac {63 \int \frac {1}{\left (15+e^x-3 x\right )^2} \, dx}{e^5}+\int \frac {e^{2 x}}{\left (15+e^x-3 x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 43, normalized size = 1.23 \begin {gather*} \frac {3 e^{\frac {4}{\log (x)}}}{15+e^x-3 x}+x-\frac {3 \left (2+e^5+x\right )}{e^5 \left (15+e^x-3 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-7*x + (E^(5 + 2*x)*x)/9 + E^5*(24*x - 10*x^2 + x^3) + (E^x*(x + x^2 + E^5*(11*x - 2*x^2)))/3)*Log
[x]^2 + E^(4/Log[x])*((-4*E^(5 + x))/3 + E^5*(-20 + 4*x) + (E^5*x - (E^(5 + x)*x)/3)*Log[x]^2))/(((E^(5 + 2*x)
*x)/9 + (E^(5 + x)*(10*x - 2*x^2))/3 + E^5*(25*x - 10*x^2 + x^3))*Log[x]^2),x]

[Out]

(3*E^(4/Log[x]))/(15 + E^x - 3*x) + x - (3*(2 + E^5 + x))/(E^5*(15 + E^x - 3*x))

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fricas [A]  time = 0.62, size = 56, normalized size = 1.60 \begin {gather*} \frac {{\left (x^{2} - 5 \, x + 1\right )} e^{5} - x e^{\left (x - \log \relax (3) + 5\right )} + x - e^{\left (\frac {4}{\log \relax (x)} + 5\right )} + 2}{{\left (x - 5\right )} e^{5} - e^{\left (x - \log \relax (3) + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x*exp(5)*exp(-log(3)+x)+x*exp(5))*log(x)^2-4*exp(5)*exp(-log(3)+x)+(4*x-20)*exp(5))*exp(4/log(x)
)+(x*exp(5)*exp(-log(3)+x)^2+((-2*x^2+11*x)*exp(5)+x^2+x)*exp(-log(3)+x)+(x^3-10*x^2+24*x)*exp(5)-7*x)*log(x)^
2)/(x*exp(5)*exp(-log(3)+x)^2+(-2*x^2+10*x)*exp(5)*exp(-log(3)+x)+(x^3-10*x^2+25*x)*exp(5))/log(x)^2,x, algori
thm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x*exp(5)*exp(-log(3)+x)+x*exp(5))*log(x)^2-4*exp(5)*exp(-log(3)+x)+(4*x-20)*exp(5))*exp(4/log(x)
)+(x*exp(5)*exp(-log(3)+x)^2+((-2*x^2+11*x)*exp(5)+x^2+x)*exp(-log(3)+x)+(x^3-10*x^2+24*x)*exp(5)-7*x)*log(x)^
2)/(x*exp(5)*exp(-log(3)+x)^2+(-2*x^2+10*x)*exp(5)*exp(-log(3)+x)+(x^3-10*x^2+25*x)*exp(5))/log(x)^2,x, algori
thm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{12,[0,5,5,0]%%%}+%%%{-144,[0,5,4,0]%%%}+%%%{432,[0,5,3,0
]%%%}+%%%{4

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maple [A]  time = 0.62, size = 55, normalized size = 1.57

method	result	size
risch	\(\frac {\left (x^{2} {\mathrm e}^{5}-\frac {x \,{\mathrm e}^{5+x}}{3}-5 x \,{\mathrm e}^{5}+{\mathrm e}^{5}+x +2\right ) {\mathrm e}^{-5}}{x -\frac {{\mathrm e}^{x}}{3}-5}-\frac {{\mathrm e}^{\frac {4}{\ln \relax (x )}}}{x -\frac {{\mathrm e}^{x}}{3}-5}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x*exp(5)*exp(-ln(3)+x)+x*exp(5))*ln(x)^2-4*exp(5)*exp(-ln(3)+x)+(4*x-20)*exp(5))*exp(4/ln(x))+(x*exp(5
)*exp(-ln(3)+x)^2+((-2*x^2+11*x)*exp(5)+x^2+x)*exp(-ln(3)+x)+(x^3-10*x^2+24*x)*exp(5)-7*x)*ln(x)^2)/(x*exp(5)*
exp(-ln(3)+x)^2+(-2*x^2+10*x)*exp(5)*exp(-ln(3)+x)+(x^3-10*x^2+25*x)*exp(5))/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.45, size = 59, normalized size = 1.69 \begin {gather*} \frac {3 \, x^{2} e^{5} - 3 \, x {\left (5 \, e^{5} - 1\right )} - x e^{\left (x + 5\right )} + 3 \, e^{5} - 3 \, e^{\left (\frac {4}{\log \relax (x)} + 5\right )} + 6}{3 \, x e^{5} - 15 \, e^{5} - e^{\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x*exp(5)*exp(-log(3)+x)+x*exp(5))*log(x)^2-4*exp(5)*exp(-log(3)+x)+(4*x-20)*exp(5))*exp(4/log(x)
)+(x*exp(5)*exp(-log(3)+x)^2+((-2*x^2+11*x)*exp(5)+x^2+x)*exp(-log(3)+x)+(x^3-10*x^2+24*x)*exp(5)-7*x)*log(x)^
2)/(x*exp(5)*exp(-log(3)+x)^2+(-2*x^2+10*x)*exp(5)*exp(-log(3)+x)+(x^3-10*x^2+25*x)*exp(5))/log(x)^2,x, algori
thm="maxima")

[Out]

(3*x^2*e^5 - 3*x*(5*e^5 - 1) - x*e^(x + 5) + 3*e^5 - 3*e^(4/log(x) + 5) + 6)/(3*x*e^5 - 15*e^5 - e^(x + 5))

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mupad [B]  time = 0.75, size = 52, normalized size = 1.49 \begin {gather*} -\frac {{\mathrm {e}}^{-5}\,\left (3\,x+3\,{\mathrm {e}}^5-15\,x\,{\mathrm {e}}^5+3\,x^2\,{\mathrm {e}}^5-3\,{\mathrm {e}}^{\frac {4}{\ln \relax (x)}}\,{\mathrm {e}}^5-x\,{\mathrm {e}}^5\,{\mathrm {e}}^x+6\right )}{{\mathrm {e}}^x-3\,x+15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4/log(x))*(log(x)^2*(x*exp(5) - x*exp(x - log(3))*exp(5)) - 4*exp(x - log(3))*exp(5) + exp(5)*(4*x -
20)) + log(x)^2*(exp(5)*(24*x - 10*x^2 + x^3) - 7*x + exp(x - log(3))*(x + exp(5)*(11*x - 2*x^2) + x^2) + x*ex
p(5)*exp(2*x - 2*log(3))))/(log(x)^2*(exp(5)*(25*x - 10*x^2 + x^3) + exp(x - log(3))*exp(5)*(10*x - 2*x^2) + x
*exp(5)*exp(2*x - 2*log(3)))),x)

[Out]

-(exp(-5)*(3*x + 3*exp(5) - 15*x*exp(5) + 3*x^2*exp(5) - 3*exp(4/log(x))*exp(5) - x*exp(5)*exp(x) + 6))/(exp(x
) - 3*x + 15)

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sympy [A]  time = 0.39, size = 41, normalized size = 1.17 \begin {gather*} x + \frac {- 3 x + 3 e^{5} e^{\frac {4}{\log {\relax (x )}}} - 3 e^{5} - 6}{- 3 x e^{5} + e^{5} e^{x} + 15 e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x*exp(5)*exp(-ln(3)+x)+x*exp(5))*ln(x)**2-4*exp(5)*exp(-ln(3)+x)+(4*x-20)*exp(5))*exp(4/ln(x))+(
x*exp(5)*exp(-ln(3)+x)**2+((-2*x**2+11*x)*exp(5)+x**2+x)*exp(-ln(3)+x)+(x**3-10*x**2+24*x)*exp(5)-7*x)*ln(x)**
2)/(x*exp(5)*exp(-ln(3)+x)**2+(-2*x**2+10*x)*exp(5)*exp(-ln(3)+x)+(x**3-10*x**2+25*x)*exp(5))/ln(x)**2,x)

[Out]

x + (-3*x + 3*exp(5)*exp(4/log(x)) - 3*exp(5) - 6)/(-3*x*exp(5) + exp(5)*exp(x) + 15*exp(5))

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