3.47.73 \(\int \frac {25-25 x+e^{2 e^x-2 x} (-5-5 x+12 x^2-2 x^3+e^x (10 x-12 x^2+2 x^3))+(25-55 x+10 x^2+e^{2 e^x-2 x} (-5+11 x-2 x^2)) \log (-\frac {x}{25+e^{2 e^x-2 x} (-5+x)-5 x})}{25+e^{2 e^x-2 x} (-5+x)-5 x} \, dx\)

Optimal. Leaf size=34 \[ -4+(1-x) x \log \left (\frac {x}{\left (5-e^{2 e^x-2 x}\right ) (-5+x)}\right ) \]

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Rubi [B]  time = 7.34, antiderivative size = 77, normalized size of antiderivative = 2.26, number of steps used = 27, number of rules used = 8, integrand size = 132, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {6742, 6688, 43, 2551, 2282, 6712, 31, 72} \begin {gather*} -\frac {1}{4} (1-2 x)^2 \log \left (-\frac {x}{-e^{2 e^x-2 x} (5-x)-5 x+25}\right )-\frac {1}{4} \log \left (5-e^{2 e^x-2 x}\right )-\frac {1}{4} \log (5-x)+\frac {\log (x)}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25 - 25*x + E^(2*E^x - 2*x)*(-5 - 5*x + 12*x^2 - 2*x^3 + E^x*(10*x - 12*x^2 + 2*x^3)) + (25 - 55*x + 10*x
^2 + E^(2*E^x - 2*x)*(-5 + 11*x - 2*x^2))*Log[-(x/(25 + E^(2*E^x - 2*x)*(-5 + x) - 5*x))])/(25 + E^(2*E^x - 2*
x)*(-5 + x) - 5*x),x]

[Out]

-1/4*Log[5 - E^(2*E^x - 2*x)] - Log[5 - x]/4 + Log[x]/4 - ((1 - 2*x)^2*Log[-(x/(25 - E^(2*E^x - 2*x)*(5 - x) -
5*x))])/4

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6712

Int[(u_)*(v_)^(r_.)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x
] - q*v*D[w, x])]}, -Dist[c*q, Subst[Int[(a + b*x^q)^m, x], x, v^(m*p + r + 1)*w], x] /; FreeQ[c, x]] /; FreeQ
[{a, b, m, p, q, r}, x] && EqQ[p + q*(m*p + r + 1), 0] && IntegerQ[q] && IntegerQ[m]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {2 e^{2 e^x} \left (-1+e^x\right ) (-1+x) x}{-e^{2 e^x}+5 e^{2 x}}+\frac {-5+5 x-5 \log \left (-\frac {x}{25+e^{2 e^x-2 x} (-5+x)-5 x}\right )+11 x \log \left (-\frac {x}{25+e^{2 e^x-2 x} (-5+x)-5 x}\right )-2 x^2 \log \left (-\frac {x}{25+e^{2 e^x-2 x} (-5+x)-5 x}\right )}{-5+x}\right ) \, dx\\ &=-\left (2 \int \frac {e^{2 e^x} \left (-1+e^x\right ) (-1+x) x}{-e^{2 e^x}+5 e^{2 x}} \, dx\right )+\int \frac {-5+5 x-5 \log \left (-\frac {x}{25+e^{2 e^x-2 x} (-5+x)-5 x}\right )+11 x \log \left (-\frac {x}{25+e^{2 e^x-2 x} (-5+x)-5 x}\right )-2 x^2 \log \left (-\frac {x}{25+e^{2 e^x-2 x} (-5+x)-5 x}\right )}{-5+x} \, dx\\ &=-\left (2 \int \left (-\frac {e^{2 e^x} \left (-1+e^x\right ) x}{-e^{2 e^x}+5 e^{2 x}}+\frac {e^{2 e^x} \left (-1+e^x\right ) x^2}{-e^{2 e^x}+5 e^{2 x}}\right ) \, dx\right )+\int \frac {-5 (-1+x)-\left (-5+11 x-2 x^2\right ) \log \left (-\frac {x}{25+e^{2 e^x-2 x} (-5+x)-5 x}\right )}{5-x} \, dx\\ &=2 \int \frac {e^{2 e^x} \left (-1+e^x\right ) x}{-e^{2 e^x}+5 e^{2 x}} \, dx-2 \int \frac {e^{2 e^x} \left (-1+e^x\right ) x^2}{-e^{2 e^x}+5 e^{2 x}} \, dx+\int \left (\frac {5 (-1+x)}{-5+x}-(-1+2 x) \log \left (-\frac {x}{25+e^{2 e^x-2 x} (-5+x)-5 x}\right )\right ) \, dx\\ &=2 \int \left (-\frac {e^{2 e^x} x}{-e^{2 e^x}+5 e^{2 x}}+\frac {e^{2 e^x+x} x}{-e^{2 e^x}+5 e^{2 x}}\right ) \, dx-2 \int \left (-\frac {e^{2 e^x} x^2}{-e^{2 e^x}+5 e^{2 x}}+\frac {e^{2 e^x+x} x^2}{-e^{2 e^x}+5 e^{2 x}}\right ) \, dx+5 \int \frac {-1+x}{-5+x} \, dx-\int (-1+2 x) \log \left (-\frac {x}{25+e^{2 e^x-2 x} (-5+x)-5 x}\right ) \, dx\\ &=-\frac {1}{4} (1-2 x)^2 \log \left (-\frac {x}{25-e^{2 e^x-2 x} (5-x)-5 x}\right )+\frac {1}{4} \int \frac {(1-2 x)^2 \left (-25 e^{2 x}+2 e^{2 e^x+x} (-5+x) x+e^{2 e^x} \left (5+10 x-2 x^2\right )\right )}{\left (e^{2 e^x}-5 e^{2 x}\right ) (5-x) x} \, dx-2 \int \frac {e^{2 e^x} x}{-e^{2 e^x}+5 e^{2 x}} \, dx+2 \int \frac {e^{2 e^x+x} x}{-e^{2 e^x}+5 e^{2 x}} \, dx+2 \int \frac {e^{2 e^x} x^2}{-e^{2 e^x}+5 e^{2 x}} \, dx-2 \int \frac {e^{2 e^x+x} x^2}{-e^{2 e^x}+5 e^{2 x}} \, dx+5 \int \left (1+\frac {4}{-5+x}\right ) \, dx\\ &=5 x+20 \log (5-x)-\frac {1}{4} (1-2 x)^2 \log \left (-\frac {x}{25-e^{2 e^x-2 x} (5-x)-5 x}\right )+\frac {1}{4} \int \left (\frac {2 e^{2 e^x} \left (-1+e^x\right ) (-1+2 x)^2}{-e^{2 e^x}+5 e^{2 x}}-\frac {5 (-1+2 x)^2}{(-5+x) x}\right ) \, dx-2 \int \frac {e^{2 e^x} x}{-e^{2 e^x}+5 e^{2 x}} \, dx+2 \int \frac {e^{2 e^x+x} x}{-e^{2 e^x}+5 e^{2 x}} \, dx+2 \int \frac {e^{2 e^x} x^2}{-e^{2 e^x}+5 e^{2 x}} \, dx-2 \int \frac {e^{2 e^x+x} x^2}{-e^{2 e^x}+5 e^{2 x}} \, dx\\ &=5 x+20 \log (5-x)-\frac {1}{4} (1-2 x)^2 \log \left (-\frac {x}{25-e^{2 e^x-2 x} (5-x)-5 x}\right )+\frac {1}{2} \int \frac {e^{2 e^x} \left (-1+e^x\right ) (-1+2 x)^2}{-e^{2 e^x}+5 e^{2 x}} \, dx-\frac {5}{4} \int \frac {(-1+2 x)^2}{(-5+x) x} \, dx-2 \int \frac {e^{2 e^x} x}{-e^{2 e^x}+5 e^{2 x}} \, dx+2 \int \frac {e^{2 e^x+x} x}{-e^{2 e^x}+5 e^{2 x}} \, dx+2 \int \frac {e^{2 e^x} x^2}{-e^{2 e^x}+5 e^{2 x}} \, dx-2 \int \frac {e^{2 e^x+x} x^2}{-e^{2 e^x}+5 e^{2 x}} \, dx\\ &=5 x+20 \log (5-x)-\frac {1}{4} (1-2 x)^2 \log \left (-\frac {x}{25-e^{2 e^x-2 x} (5-x)-5 x}\right )+\frac {1}{2} \int \left (\frac {e^{2 e^x} \left (-1+e^x\right )}{-e^{2 e^x}+5 e^{2 x}}-\frac {4 e^{2 e^x} \left (-1+e^x\right ) x}{-e^{2 e^x}+5 e^{2 x}}+\frac {4 e^{2 e^x} \left (-1+e^x\right ) x^2}{-e^{2 e^x}+5 e^{2 x}}\right ) \, dx-\frac {5}{4} \int \left (4+\frac {81}{5 (-5+x)}-\frac {1}{5 x}\right ) \, dx-2 \int \frac {e^{2 e^x} x}{-e^{2 e^x}+5 e^{2 x}} \, dx+2 \int \frac {e^{2 e^x+x} x}{-e^{2 e^x}+5 e^{2 x}} \, dx+2 \int \frac {e^{2 e^x} x^2}{-e^{2 e^x}+5 e^{2 x}} \, dx-2 \int \frac {e^{2 e^x+x} x^2}{-e^{2 e^x}+5 e^{2 x}} \, dx\\ &=-\frac {1}{4} \log (5-x)+\frac {\log (x)}{4}-\frac {1}{4} (1-2 x)^2 \log \left (-\frac {x}{25-e^{2 e^x-2 x} (5-x)-5 x}\right )+\frac {1}{2} \int \frac {e^{2 e^x} \left (-1+e^x\right )}{-e^{2 e^x}+5 e^{2 x}} \, dx-2 \int \frac {e^{2 e^x} x}{-e^{2 e^x}+5 e^{2 x}} \, dx+2 \int \frac {e^{2 e^x+x} x}{-e^{2 e^x}+5 e^{2 x}} \, dx-2 \int \frac {e^{2 e^x} \left (-1+e^x\right ) x}{-e^{2 e^x}+5 e^{2 x}} \, dx+2 \int \frac {e^{2 e^x} x^2}{-e^{2 e^x}+5 e^{2 x}} \, dx-2 \int \frac {e^{2 e^x+x} x^2}{-e^{2 e^x}+5 e^{2 x}} \, dx+2 \int \frac {e^{2 e^x} \left (-1+e^x\right ) x^2}{-e^{2 e^x}+5 e^{2 x}} \, dx\\ &=-\frac {1}{4} \log (5-x)+\frac {\log (x)}{4}-\frac {1}{4} (1-2 x)^2 \log \left (-\frac {x}{25-e^{2 e^x-2 x} (5-x)-5 x}\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {e^{2 x} (-1+x)}{x \left (-e^{2 x}+5 x^2\right )} \, dx,x,e^x\right )-2 \int \frac {e^{2 e^x} x}{-e^{2 e^x}+5 e^{2 x}} \, dx+2 \int \frac {e^{2 e^x+x} x}{-e^{2 e^x}+5 e^{2 x}} \, dx+2 \int \frac {e^{2 e^x} x^2}{-e^{2 e^x}+5 e^{2 x}} \, dx-2 \int \frac {e^{2 e^x+x} x^2}{-e^{2 e^x}+5 e^{2 x}} \, dx-2 \int \left (-\frac {e^{2 e^x} x}{-e^{2 e^x}+5 e^{2 x}}+\frac {e^{2 e^x+x} x}{-e^{2 e^x}+5 e^{2 x}}\right ) \, dx+2 \int \left (-\frac {e^{2 e^x} x^2}{-e^{2 e^x}+5 e^{2 x}}+\frac {e^{2 e^x+x} x^2}{-e^{2 e^x}+5 e^{2 x}}\right ) \, dx\\ &=-\frac {1}{4} \log (5-x)+\frac {\log (x)}{4}-\frac {1}{4} (1-2 x)^2 \log \left (-\frac {x}{25-e^{2 e^x-2 x} (5-x)-5 x}\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{5-x} \, dx,x,e^{2 e^x-2 x}\right )\\ &=-\frac {1}{4} \log \left (5-e^{2 e^x-2 x}\right )-\frac {1}{4} \log (5-x)+\frac {\log (x)}{4}-\frac {1}{4} (1-2 x)^2 \log \left (-\frac {x}{25-e^{2 e^x-2 x} (5-x)-5 x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 32, normalized size = 0.94 \begin {gather*} -\left ((-1+x) x \log \left (-\frac {x}{25+e^{2 e^x-2 x} (-5+x)-5 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 - 25*x + E^(2*E^x - 2*x)*(-5 - 5*x + 12*x^2 - 2*x^3 + E^x*(10*x - 12*x^2 + 2*x^3)) + (25 - 55*x
+ 10*x^2 + E^(2*E^x - 2*x)*(-5 + 11*x - 2*x^2))*Log[-(x/(25 + E^(2*E^x - 2*x)*(-5 + x) - 5*x))])/(25 + E^(2*E^
x - 2*x)*(-5 + x) - 5*x),x]

[Out]

-((-1 + x)*x*Log[-(x/(25 + E^(2*E^x - 2*x)*(-5 + x) - 5*x))])

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fricas [A]  time = 0.62, size = 33, normalized size = 0.97 \begin {gather*} -{\left (x^{2} - x\right )} \log \left (-\frac {x}{{\left (x - 5\right )} e^{\left (-2 \, x + 2 \, e^{x}\right )} - 5 \, x + 25}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+11*x-5)*exp(exp(x)-x)^2+10*x^2-55*x+25)*log(-x/((x-5)*exp(exp(x)-x)^2-5*x+25))+((2*x^3-12*
x^2+10*x)*exp(x)-2*x^3+12*x^2-5*x-5)*exp(exp(x)-x)^2-25*x+25)/((x-5)*exp(exp(x)-x)^2-5*x+25),x, algorithm="fri
cas")

[Out]

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

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giac [B]  time = 0.60, size = 88, normalized size = 2.59 \begin {gather*} -x^{2} \log \left (\frac {x e^{x}}{5 \, x e^{x} - x e^{\left (-x + 2 \, e^{x}\right )} - 25 \, e^{x} + 5 \, e^{\left (-x + 2 \, e^{x}\right )}}\right ) + x \log \left (\frac {x e^{x}}{5 \, x e^{x} - x e^{\left (-x + 2 \, e^{x}\right )} - 25 \, e^{x} + 5 \, e^{\left (-x + 2 \, e^{x}\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+11*x-5)*exp(exp(x)-x)^2+10*x^2-55*x+25)*log(-x/((x-5)*exp(exp(x)-x)^2-5*x+25))+((2*x^3-12*
x^2+10*x)*exp(x)-2*x^3+12*x^2-5*x-5)*exp(exp(x)-x)^2-25*x+25)/((x-5)*exp(exp(x)-x)^2-5*x+25),x, algorithm="gia
c")

[Out]

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

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maple [C]  time = 0.36, size = 814, normalized size = 23.94

method	result	size
risch	\(-x^{2} \ln \relax (x )+\left (x^{2}-x \right ) \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right )+x \ln \relax (x )+x^{2} \ln \left (x -5\right )-x \ln \left (x -5\right )+\frac {i \pi x \mathrm {csgn}\left (\frac {i x}{\left (x -5\right ) \left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right )}\right )^{3}}{2}-\frac {i \pi x \mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right ) \left (x -5\right )}\right )^{3}}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i x}{\left (x -5\right ) \left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right )}\right )^{3}}{2}+i \pi \,x^{2} \mathrm {csgn}\left (\frac {i x}{\left (x -5\right ) \left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right )}\right )^{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right ) \left (x -5\right )}\right )^{3}}{2}-i \pi x \mathrm {csgn}\left (\frac {i x}{\left (x -5\right ) \left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right )}\right )^{2}-i \pi \,x^{2}-\frac {i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right ) \left (x -5\right )}\right ) \mathrm {csgn}\left (\frac {i x}{\left (x -5\right ) \left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right )}\right )}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5}\right ) \mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right ) \left (x -5\right )}\right )}{2}-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5}\right ) \mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right ) \left (x -5\right )}\right )}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right ) \left (x -5\right )}\right ) \mathrm {csgn}\left (\frac {i x}{\left (x -5\right ) \left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right )}\right )}{2}+i x \pi -\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right ) \left (x -5\right )}\right ) \mathrm {csgn}\left (\frac {i x}{\left (x -5\right ) \left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right )}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right ) \left (x -5\right )}\right )^{2}}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5}\right ) \mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right ) \left (x -5\right )}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5}\right ) \mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right ) \left (x -5\right )}\right )^{2}}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right ) \left (x -5\right )}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right ) \left (x -5\right )}\right ) \mathrm {csgn}\left (\frac {i x}{\left (x -5\right ) \left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right )}\right )^{2}}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{\left (x -5\right ) \left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right )}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{\left (x -5\right ) \left ({\mathrm e}^{2 \,{\mathrm e}^{x}-2 x}-5\right )}\right )^{2}}{2}\)	\(814\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^2+11*x-5)*exp(exp(x)-x)^2+10*x^2-55*x+25)*ln(-x/((x-5)*exp(exp(x)-x)^2-5*x+25))+((2*x^3-12*x^2+10*
x)*exp(x)-2*x^3+12*x^2-5*x-5)*exp(exp(x)-x)^2-25*x+25)/((x-5)*exp(exp(x)-x)^2-5*x+25),x,method=_RETURNVERBOSE)

[Out]

-x^2*ln(x)+(x^2-x)*ln(exp(2*exp(x)-2*x)-5)+1/2*I*Pi*x*csgn(I/(exp(2*exp(x)-2*x)-5)/(x-5))*csgn(I*x/(x-5)/(exp(
2*exp(x)-2*x)-5))^2+x*ln(x)+x^2*ln(x-5)-x*ln(x-5)+1/2*I*Pi*x*csgn(I*x/(x-5)/(exp(2*exp(x)-2*x)-5))^3-1/2*I*Pi*
x*csgn(I/(exp(2*exp(x)-2*x)-5)/(x-5))^3-1/2*I*Pi*x^2*csgn(I*x/(x-5)/(exp(2*exp(x)-2*x)-5))^3+I*Pi*x^2*csgn(I*x
/(x-5)/(exp(2*exp(x)-2*x)-5))^2+1/2*I*Pi*x^2*csgn(I/(exp(2*exp(x)-2*x)-5)/(x-5))^3-I*Pi*x*csgn(I*x/(x-5)/(exp(
2*exp(x)-2*x)-5))^2-I*Pi*x^2-1/2*I*Pi*x^2*csgn(I*x)*csgn(I*x/(x-5)/(exp(2*exp(x)-2*x)-5))^2+1/2*I*Pi*x*csgn(I*
x)*csgn(I*x/(x-5)/(exp(2*exp(x)-2*x)-5))^2-1/2*I*Pi*x^2*csgn(I/(exp(2*exp(x)-2*x)-5)/(x-5))*csgn(I*x/(x-5)/(ex
p(2*exp(x)-2*x)-5))^2-1/2*I*Pi*x*csgn(I*x)*csgn(I/(exp(2*exp(x)-2*x)-5)/(x-5))*csgn(I*x/(x-5)/(exp(2*exp(x)-2*
x)-5))+1/2*I*Pi*x^2*csgn(I/(x-5))*csgn(I/(exp(2*exp(x)-2*x)-5))*csgn(I/(exp(2*exp(x)-2*x)-5)/(x-5))-1/2*I*Pi*x
*csgn(I/(x-5))*csgn(I/(exp(2*exp(x)-2*x)-5))*csgn(I/(exp(2*exp(x)-2*x)-5)/(x-5))+1/2*I*Pi*x^2*csgn(I*x)*csgn(I
/(exp(2*exp(x)-2*x)-5)/(x-5))*csgn(I*x/(x-5)/(exp(2*exp(x)-2*x)-5))+I*Pi*x+1/2*I*Pi*x*csgn(I/(x-5))*csgn(I/(ex
p(2*exp(x)-2*x)-5)/(x-5))^2-1/2*I*Pi*x^2*csgn(I/(exp(2*exp(x)-2*x)-5))*csgn(I/(exp(2*exp(x)-2*x)-5)/(x-5))^2+1
/2*I*Pi*x*csgn(I/(exp(2*exp(x)-2*x)-5))*csgn(I/(exp(2*exp(x)-2*x)-5)/(x-5))^2-1/2*I*Pi*x^2*csgn(I/(x-5))*csgn(
I/(exp(2*exp(x)-2*x)-5)/(x-5))^2

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maxima [A]  time = 0.42, size = 57, normalized size = 1.68 \begin {gather*} -2 \, x^{3} + 2 \, x^{2} + {\left (x^{2} - x\right )} \log \left (x - 5\right ) - {\left (x^{2} - x\right )} \log \relax (x) + {\left (x^{2} - x\right )} \log \left (5 \, e^{\left (2 \, x\right )} - e^{\left (2 \, e^{x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+11*x-5)*exp(exp(x)-x)^2+10*x^2-55*x+25)*log(-x/((x-5)*exp(exp(x)-x)^2-5*x+25))+((2*x^3-12*
x^2+10*x)*exp(x)-2*x^3+12*x^2-5*x-5)*exp(exp(x)-x)^2-25*x+25)/((x-5)*exp(exp(x)-x)^2-5*x+25),x, algorithm="max
ima")

[Out]

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

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mupad [B]  time = 3.49, size = 48, normalized size = 1.41 \begin {gather*} -\frac {\ln \left (-\frac {x}{{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (x-5\right )-5\,x+25}\right )\,\left (x^4-6\,x^3+5\,x^2\right )}{x\,\left (x-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(25*x + log(-x/(exp(2*exp(x) - 2*x)*(x - 5) - 5*x + 25))*(55*x + exp(2*exp(x) - 2*x)*(2*x^2 - 11*x + 5) -
10*x^2 - 25) + exp(2*exp(x) - 2*x)*(5*x - 12*x^2 + 2*x^3 - exp(x)*(10*x - 12*x^2 + 2*x^3) + 5) - 25)/(exp(2*e
xp(x) - 2*x)*(x - 5) - 5*x + 25),x)

[Out]

-(log(-x/(exp(-2*x)*exp(2*exp(x))*(x - 5) - 5*x + 25))*(5*x^2 - 6*x^3 + x^4))/(x*(x - 5))

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sympy [B]  time = 2.09, size = 63, normalized size = 1.85 \begin {gather*} \left (- x^{2} + x + \frac {10}{3}\right ) \log {\left (- \frac {x}{- 5 x + \left (x - 5\right ) e^{- 2 x + 2 e^{x}} + 25} \right )} - \frac {10 \log {\relax (x )}}{3} + \frac {10 \log {\left (x - 5 \right )}}{3} + \frac {10 \log {\left (e^{- 2 x + 2 e^{x}} - 5 \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**2+11*x-5)*exp(exp(x)-x)**2+10*x**2-55*x+25)*ln(-x/((x-5)*exp(exp(x)-x)**2-5*x+25))+((2*x**3
-12*x**2+10*x)*exp(x)-2*x**3+12*x**2-5*x-5)*exp(exp(x)-x)**2-25*x+25)/((x-5)*exp(exp(x)-x)**2-5*x+25),x)

[Out]

(-x**2 + x + 10/3)*log(-x/(-5*x + (x - 5)*exp(-2*x + 2*exp(x)) + 25)) - 10*log(x)/3 + 10*log(x - 5)/3 + 10*log
(exp(-2*x + 2*exp(x)) - 5)/3

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