3.65.68 \(\int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x (-10-5 x+4 x^4)+(5 x+8 x^4+2 x^5+e^{2 x} (8 x^2+2 x^3)+e^x (5+16 x^3+4 x^4)) \log (\frac {5+8 x^3+2 x^4+e^x (8 x^2+2 x^3)}{2 e^x x^2+2 x^3})}{25 x+40 x^4+10 x^5+e^{2 x} (40 x^2+10 x^3)+e^x (25+80 x^3+20 x^4)} \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{5} x \log \left (4+x+\frac {5}{2 x^2 \left (e^x+x\right )}\right ) \]

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Rubi [A]  time = 3.03, antiderivative size = 43, normalized size of antiderivative = 1.87, number of steps used = 19, number of rules used = 4, integrand size = 176, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6742, 6688, 43, 2548} \begin {gather*} \frac {1}{5} x \log \left (\frac {2 x^4+8 x^3+2 e^x (x+4) x^2+5}{2 x^2 \left (x+e^x\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15*x + 2*E^(2*x)*x^3 + 2*x^5 + E^x*(-10 - 5*x + 4*x^4) + (5*x + 8*x^4 + 2*x^5 + E^(2*x)*(8*x^2 + 2*x^3)
+ E^x*(5 + 16*x^3 + 4*x^4))*Log[(5 + 8*x^3 + 2*x^4 + E^x*(8*x^2 + 2*x^3))/(2*E^x*x^2 + 2*x^3)])/(25*x + 40*x^4
+ 10*x^5 + E^(2*x)*(40*x^2 + 10*x^3) + E^x*(25 + 80*x^3 + 20*x^4)),x]

[Out]

(x*Log[(5 + 8*x^3 + 2*x^4 + 2*E^x*x^2*(4 + x))/(2*x^2*(E^x + x))])/5

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 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 6688

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

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 {(-1+x) x}{5 \left (e^x+x\right )}-\frac {40+35 x+5 x^2-32 x^3+16 x^4+14 x^5+2 x^6}{5 (4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )}+\frac {x+4 \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )+x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )}{5 (4+x)}\right ) \, dx\\ &=\frac {1}{5} \int \frac {(-1+x) x}{e^x+x} \, dx-\frac {1}{5} \int \frac {40+35 x+5 x^2-32 x^3+16 x^4+14 x^5+2 x^6}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )} \, dx+\frac {1}{5} \int \frac {x+4 \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )+x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )}{4+x} \, dx\\ &=\frac {1}{5} \int \left (-\frac {x}{e^x+x}+\frac {x^2}{e^x+x}\right ) \, dx-\frac {1}{5} \int \left (\frac {15}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}+\frac {5 x}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}-\frac {8 x^3}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}+\frac {6 x^4}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}+\frac {2 x^5}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}-\frac {20}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )}\right ) \, dx+\frac {1}{5} \int \left (\frac {x}{4+x}+\log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )\right ) \, dx\\ &=\frac {1}{5} \int \frac {x}{4+x} \, dx-\frac {1}{5} \int \frac {x}{e^x+x} \, dx+\frac {1}{5} \int \frac {x^2}{e^x+x} \, dx+\frac {1}{5} \int \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right ) \, dx-\frac {2}{5} \int \frac {x^5}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-\frac {6}{5} \int \frac {x^4}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+\frac {8}{5} \int \frac {x^3}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-3 \int \frac {1}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+4 \int \frac {1}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )} \, dx-\int \frac {x}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx\\ &=\frac {1}{5} x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )-\frac {1}{5} \int \frac {x}{e^x+x} \, dx+\frac {1}{5} \int \frac {x^2}{e^x+x} \, dx+\frac {1}{5} \int \left (1-\frac {4}{4+x}\right ) \, dx-\frac {1}{5} \int \frac {2 e^{2 x} x^3+x \left (-15+2 x^4\right )+e^x \left (-10-5 x+4 x^4\right )}{\left (e^x+x\right ) \left (5+8 x^3+2 x^4+2 e^x x^2 (4+x)\right )} \, dx-\frac {2}{5} \int \frac {x^5}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-\frac {6}{5} \int \frac {x^4}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+\frac {8}{5} \int \frac {x^3}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-3 \int \frac {1}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+4 \int \frac {1}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )} \, dx-\int \frac {x}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx\\ &=\frac {x}{5}-\frac {4}{5} \log (4+x)+\frac {1}{5} x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )-\frac {1}{5} \int \frac {x}{e^x+x} \, dx+\frac {1}{5} \int \frac {x^2}{e^x+x} \, dx-\frac {1}{5} \int \left (\frac {x}{4+x}+\frac {(-1+x) x}{e^x+x}-\frac {40+35 x+5 x^2-32 x^3+16 x^4+14 x^5+2 x^6}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )}\right ) \, dx-\frac {2}{5} \int \frac {x^5}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-\frac {6}{5} \int \frac {x^4}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+\frac {8}{5} \int \frac {x^3}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-3 \int \frac {1}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+4 \int \frac {1}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )} \, dx-\int \frac {x}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx\\ &=\frac {x}{5}-\frac {4}{5} \log (4+x)+\frac {1}{5} x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )-\frac {1}{5} \int \frac {x}{4+x} \, dx-\frac {1}{5} \int \frac {x}{e^x+x} \, dx-\frac {1}{5} \int \frac {(-1+x) x}{e^x+x} \, dx+\frac {1}{5} \int \frac {x^2}{e^x+x} \, dx+\frac {1}{5} \int \frac {40+35 x+5 x^2-32 x^3+16 x^4+14 x^5+2 x^6}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )} \, dx-\frac {2}{5} \int \frac {x^5}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-\frac {6}{5} \int \frac {x^4}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+\frac {8}{5} \int \frac {x^3}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-3 \int \frac {1}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+4 \int \frac {1}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )} \, dx-\int \frac {x}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx\\ &=\frac {x}{5}-\frac {4}{5} \log (4+x)+\frac {1}{5} x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )-\frac {1}{5} \int \frac {x}{e^x+x} \, dx+\frac {1}{5} \int \frac {x^2}{e^x+x} \, dx-\frac {1}{5} \int \left (1-\frac {4}{4+x}\right ) \, dx-\frac {1}{5} \int \left (-\frac {x}{e^x+x}+\frac {x^2}{e^x+x}\right ) \, dx+\frac {1}{5} \int \left (\frac {15}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}+\frac {5 x}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}-\frac {8 x^3}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}+\frac {6 x^4}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}+\frac {2 x^5}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}-\frac {20}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )}\right ) \, dx-\frac {2}{5} \int \frac {x^5}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-\frac {6}{5} \int \frac {x^4}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+\frac {8}{5} \int \frac {x^3}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-3 \int \frac {1}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+4 \int \frac {1}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )} \, dx-\int \frac {x}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx\\ &=\frac {1}{5} x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.85, size = 43, normalized size = 1.87 \begin {gather*} \frac {1}{5} x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*x + 2*E^(2*x)*x^3 + 2*x^5 + E^x*(-10 - 5*x + 4*x^4) + (5*x + 8*x^4 + 2*x^5 + E^(2*x)*(8*x^2 + 2
*x^3) + E^x*(5 + 16*x^3 + 4*x^4))*Log[(5 + 8*x^3 + 2*x^4 + E^x*(8*x^2 + 2*x^3))/(2*E^x*x^2 + 2*x^3)])/(25*x +
40*x^4 + 10*x^5 + E^(2*x)*(40*x^2 + 10*x^3) + E^x*(25 + 80*x^3 + 20*x^4)),x]

[Out]

(x*Log[(5 + 8*x^3 + 2*x^4 + 2*E^x*x^2*(4 + x))/(2*x^2*(E^x + x))])/5

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fricas [B]  time = 0.56, size = 43, normalized size = 1.87 \begin {gather*} \frac {1}{5} \, x \log \left (\frac {2 \, x^{4} + 8 \, x^{3} + 2 \, {\left (x^{3} + 4 \, x^{2}\right )} e^{x} + 5}{2 \, {\left (x^{3} + x^{2} e^{x}\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3+8*x^2)*exp(x)^2+(4*x^4+16*x^3+5)*exp(x)+2*x^5+8*x^4+5*x)*log(((2*x^3+8*x^2)*exp(x)+2*x^4+8*
x^3+5)/(2*exp(x)*x^2+2*x^3))+2*exp(x)^2*x^3+(4*x^4-5*x-10)*exp(x)+2*x^5-15*x)/((10*x^3+40*x^2)*exp(x)^2+(20*x^
4+80*x^3+25)*exp(x)+10*x^5+40*x^4+25*x),x, algorithm="fricas")

[Out]

1/5*x*log(1/2*(2*x^4 + 8*x^3 + 2*(x^3 + 4*x^2)*e^x + 5)/(x^3 + x^2*e^x))

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giac [B]  time = 1.49, size = 44, normalized size = 1.91 \begin {gather*} \frac {1}{5} \, x \log \left (\frac {2 \, x^{4} + 2 \, x^{3} e^{x} + 8 \, x^{3} + 8 \, x^{2} e^{x} + 5}{2 \, {\left (x^{3} + x^{2} e^{x}\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3+8*x^2)*exp(x)^2+(4*x^4+16*x^3+5)*exp(x)+2*x^5+8*x^4+5*x)*log(((2*x^3+8*x^2)*exp(x)+2*x^4+8*
x^3+5)/(2*exp(x)*x^2+2*x^3))+2*exp(x)^2*x^3+(4*x^4-5*x-10)*exp(x)+2*x^5-15*x)/((10*x^3+40*x^2)*exp(x)^2+(20*x^
4+80*x^3+25)*exp(x)+10*x^5+40*x^4+25*x),x, algorithm="giac")

[Out]

1/5*x*log(1/2*(2*x^4 + 2*x^3*e^x + 8*x^3 + 8*x^2*e^x + 5)/(x^3 + x^2*e^x))

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maple [C]  time = 0.26, size = 537, normalized size = 23.35

method	result	size
risch	\(\frac {x \ln \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{5}-\frac {x \ln \left ({\mathrm e}^{x}+x \right )}{5}-\frac {2 x \ln \relax (x )}{5}+\frac {i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{10}-\frac {i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}}{5}+\frac {i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}}{10}-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}+x}\right ) \mathrm {csgn}\left (i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right )}{10}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}+x}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right )^{2}}{10}+\frac {i \pi x \,\mathrm {csgn}\left (i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right )^{2}}{10}-\frac {i \pi x \mathrm {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right )^{3}}{10}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{x^{2} \left ({\mathrm e}^{x}+x \right )}\right )^{2}}{10}-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{x^{2} \left ({\mathrm e}^{x}+x \right )}\right ) \mathrm {csgn}\left (\frac {i}{x^{2}}\right )}{10}-\frac {i \pi x \mathrm {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{x^{2} \left ({\mathrm e}^{x}+x \right )}\right )^{3}}{10}+\frac {i \pi x \mathrm {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{x^{2} \left ({\mathrm e}^{x}+x \right )}\right )^{2} \mathrm {csgn}\left (\frac {i}{x^{2}}\right )}{10}\)	\(537\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^3+8*x^2)*exp(x)^2+(4*x^4+16*x^3+5)*exp(x)+2*x^5+8*x^4+5*x)*ln(((2*x^3+8*x^2)*exp(x)+2*x^4+8*x^3+5)/
(2*exp(x)*x^2+2*x^3))+2*exp(x)^2*x^3+(4*x^4-5*x-10)*exp(x)+2*x^5-15*x)/((10*x^3+40*x^2)*exp(x)^2+(20*x^4+80*x^
3+25)*exp(x)+10*x^5+40*x^4+25*x),x,method=_RETURNVERBOSE)

[Out]

1/5*x*ln(5/2+x^4+(exp(x)+4)*x^3+4*exp(x)*x^2)-1/5*x*ln(exp(x)+x)-2/5*x*ln(x)+1/10*I*Pi*x*csgn(I*x)^2*csgn(I*x^
2)-1/5*I*Pi*x*csgn(I*x)*csgn(I*x^2)^2+1/10*I*Pi*x*csgn(I*x^2)^3-1/10*I*Pi*x*csgn(I/(exp(x)+x))*csgn(I*(5/2+x^4
+(exp(x)+4)*x^3+4*exp(x)*x^2))*csgn(I/(exp(x)+x)*(5/2+x^4+(exp(x)+4)*x^3+4*exp(x)*x^2))+1/10*I*Pi*x*csgn(I/(ex
p(x)+x))*csgn(I/(exp(x)+x)*(5/2+x^4+(exp(x)+4)*x^3+4*exp(x)*x^2))^2+1/10*I*Pi*x*csgn(I*(5/2+x^4+(exp(x)+4)*x^3
+4*exp(x)*x^2))*csgn(I/(exp(x)+x)*(5/2+x^4+(exp(x)+4)*x^3+4*exp(x)*x^2))^2-1/10*I*Pi*x*csgn(I/(exp(x)+x)*(5/2+
x^4+(exp(x)+4)*x^3+4*exp(x)*x^2))^3+1/10*I*Pi*x*csgn(I/(exp(x)+x)*(5/2+x^4+(exp(x)+4)*x^3+4*exp(x)*x^2))*csgn(
I/x^2*(5/2+x^4+(exp(x)+4)*x^3+4*exp(x)*x^2)/(exp(x)+x))^2-1/10*I*Pi*x*csgn(I/(exp(x)+x)*(5/2+x^4+(exp(x)+4)*x^
3+4*exp(x)*x^2))*csgn(I/x^2*(5/2+x^4+(exp(x)+4)*x^3+4*exp(x)*x^2)/(exp(x)+x))*csgn(I/x^2)-1/10*I*Pi*x*csgn(I/x
^2*(5/2+x^4+(exp(x)+4)*x^3+4*exp(x)*x^2)/(exp(x)+x))^3+1/10*I*Pi*x*csgn(I/x^2*(5/2+x^4+(exp(x)+4)*x^3+4*exp(x)
*x^2)/(exp(x)+x))^2*csgn(I/x^2)

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maxima [B]  time = 0.54, size = 48, normalized size = 2.09 \begin {gather*} -\frac {1}{5} \, x \log \relax (2) + \frac {1}{5} \, x \log \left (2 \, x^{4} + 8 \, x^{3} + 2 \, {\left (x^{3} + 4 \, x^{2}\right )} e^{x} + 5\right ) - \frac {1}{5} \, x \log \left (x + e^{x}\right ) - \frac {2}{5} \, x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3+8*x^2)*exp(x)^2+(4*x^4+16*x^3+5)*exp(x)+2*x^5+8*x^4+5*x)*log(((2*x^3+8*x^2)*exp(x)+2*x^4+8*
x^3+5)/(2*exp(x)*x^2+2*x^3))+2*exp(x)^2*x^3+(4*x^4-5*x-10)*exp(x)+2*x^5-15*x)/((10*x^3+40*x^2)*exp(x)^2+(20*x^
4+80*x^3+25)*exp(x)+10*x^5+40*x^4+25*x),x, algorithm="maxima")

[Out]

-1/5*x*log(2) + 1/5*x*log(2*x^4 + 8*x^3 + 2*(x^3 + 4*x^2)*e^x + 5) - 1/5*x*log(x + e^x) - 2/5*x*log(x)

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mupad [B]  time = 4.52, size = 46, normalized size = 2.00 \begin {gather*} \frac {x\,\ln \left (\frac {{\mathrm {e}}^x\,\left (2\,x^3+8\,x^2\right )+8\,x^3+2\,x^4+5}{2\,x^2\,{\mathrm {e}}^x+2\,x^3}\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((exp(x)*(8*x^2 + 2*x^3) + 8*x^3 + 2*x^4 + 5)/(2*x^2*exp(x) + 2*x^3))*(5*x + exp(x)*(16*x^3 + 4*x^4 +
5) + exp(2*x)*(8*x^2 + 2*x^3) + 8*x^4 + 2*x^5) - 15*x + 2*x^3*exp(2*x) - exp(x)*(5*x - 4*x^4 + 10) + 2*x^5)/(2
5*x + exp(x)*(80*x^3 + 20*x^4 + 25) + exp(2*x)*(40*x^2 + 10*x^3) + 40*x^4 + 10*x^5),x)

[Out]

(x*log((exp(x)*(8*x^2 + 2*x^3) + 8*x^3 + 2*x^4 + 5)/(2*x^2*exp(x) + 2*x^3)))/5

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sympy [B]  time = 0.77, size = 42, normalized size = 1.83 \begin {gather*} \frac {x \log {\left (\frac {2 x^{4} + 8 x^{3} + \left (2 x^{3} + 8 x^{2}\right ) e^{x} + 5}{2 x^{3} + 2 x^{2} e^{x}} \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**3+8*x**2)*exp(x)**2+(4*x**4+16*x**3+5)*exp(x)+2*x**5+8*x**4+5*x)*ln(((2*x**3+8*x**2)*exp(x)+
2*x**4+8*x**3+5)/(2*exp(x)*x**2+2*x**3))+2*exp(x)**2*x**3+(4*x**4-5*x-10)*exp(x)+2*x**5-15*x)/((10*x**3+40*x**
2)*exp(x)**2+(20*x**4+80*x**3+25)*exp(x)+10*x**5+40*x**4+25*x),x)

[Out]

x*log((2*x**4 + 8*x**3 + (2*x**3 + 8*x**2)*exp(x) + 5)/(2*x**3 + 2*x**2*exp(x)))/5

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