3.53.82 \(\int \frac {-192-672 x-912 x^2-600 x^3-192 x^4-24 x^5+e^x (-12-30 x-18 x^2)+(96+144 x+72 x^2+12 x^3+e^x (6+3 x)) \log (\frac {e^{-2 x} (6400+25 e^{2 x}+12800 x+9600 x^2+3200 x^3+400 x^4+e^x (800+800 x+200 x^2))}{16 x^2+32 x^3+24 x^4+8 x^5+x^6})}{160+560 x+760 x^2+500 x^3+160 x^4+20 x^5+e^x (10+25 x+20 x^2+5 x^3)} \, dx\)

Optimal. Leaf size=36 \[ \frac {3 \log \left (\frac {25 \left (4 e^{-x}+\frac {4}{(4+2 x)^2}\right )^2}{x^2}\right )}{5+\frac {5}{x}} \]

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Rubi [F]  time = 4.02, 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 {-192-672 x-912 x^2-600 x^3-192 x^4-24 x^5+e^x \left (-12-30 x-18 x^2\right )+\left (96+144 x+72 x^2+12 x^3+e^x (6+3 x)\right ) \log \left (\frac {e^{-2 x} \left (6400+25 e^{2 x}+12800 x+9600 x^2+3200 x^3+400 x^4+e^x \left (800+800 x+200 x^2\right )\right )}{16 x^2+32 x^3+24 x^4+8 x^5+x^6}\right )}{160+560 x+760 x^2+500 x^3+160 x^4+20 x^5+e^x \left (10+25 x+20 x^2+5 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-192 - 672*x - 912*x^2 - 600*x^3 - 192*x^4 - 24*x^5 + E^x*(-12 - 30*x - 18*x^2) + (96 + 144*x + 72*x^2 +
12*x^3 + E^x*(6 + 3*x))*Log[(6400 + 25*E^(2*x) + 12800*x + 9600*x^2 + 3200*x^3 + 400*x^4 + E^x*(800 + 800*x +
200*x^2))/(E^(2*x)*(16*x^2 + 32*x^3 + 24*x^4 + 8*x^5 + x^6))])/(160 + 560*x + 760*x^2 + 500*x^3 + 160*x^4 + 20
*x^5 + E^x*(10 + 25*x + 20*x^2 + 5*x^3)),x]

[Out]

(-6*Log[x])/5 - (12*Log[2 + x])/5 - (3*Log[(25*(E^x + 4*(2 + x)^2)^2)/(E^(2*x)*x^2*(2 + x)^4)])/(5*(1 + x)) -
(48*Defer[Int][x/(16 + E^x + 16*x + 4*x^2), x])/5 - (24*Defer[Int][x^2/(16 + E^x + 16*x + 4*x^2), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 \left (-64-224 x-304 x^2-200 x^3-64 x^4-8 x^5-2 e^x \left (2+5 x+3 x^2\right )+(2+x) \left (e^x+4 (2+x)^2\right ) \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )\right )}{5 (1+x)^2 (2+x) \left (e^x+4 (2+x)^2\right )} \, dx\\ &=\frac {3}{5} \int \frac {-64-224 x-304 x^2-200 x^3-64 x^4-8 x^5-2 e^x \left (2+5 x+3 x^2\right )+(2+x) \left (e^x+4 (2+x)^2\right ) \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{(1+x)^2 (2+x) \left (e^x+4 (2+x)^2\right )} \, dx\\ &=\frac {3}{5} \int \left (-\frac {8 x^2 (2+x)}{(1+x) \left (16+e^x+16 x+4 x^2\right )}+\frac {-4-10 x-6 x^2+2 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )+x \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{(1+x)^2 (2+x)}\right ) \, dx\\ &=\frac {3}{5} \int \frac {-4-10 x-6 x^2+2 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )+x \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{(1+x)^2 (2+x)} \, dx-\frac {24}{5} \int \frac {x^2 (2+x)}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx\\ &=\frac {3}{5} \int \frac {-2 \left (2+5 x+3 x^2\right )+(2+x) \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{(1+x)^2 (2+x)} \, dx-\frac {24}{5} \int \left (-\frac {1}{16+e^x+16 x+4 x^2}+\frac {x}{16+e^x+16 x+4 x^2}+\frac {x^2}{16+e^x+16 x+4 x^2}+\frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )}\right ) \, dx\\ &=\frac {3}{5} \int \left (-\frac {2 (2+3 x)}{2+3 x+x^2}+\frac {\log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{(1+x)^2}\right ) \, dx+\frac {24}{5} \int \frac {1}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx\\ &=\frac {3}{5} \int \frac {\log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{(1+x)^2} \, dx-\frac {6}{5} \int \frac {2+3 x}{2+3 x+x^2} \, dx+\frac {24}{5} \int \frac {1}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx\\ &=-\frac {3 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{5 (1+x)}+\frac {3}{5} \int \frac {2 \left (-4 (1+x) (2+x)^3-e^x (2+3 x)\right )}{x (1+x) (2+x) \left (e^x+4 (2+x)^2\right )} \, dx+\frac {6}{5} \int \frac {1}{1+x} \, dx-\frac {24}{5} \int \frac {1}{2+x} \, dx+\frac {24}{5} \int \frac {1}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx\\ &=\frac {6}{5} \log (1+x)-\frac {24}{5} \log (2+x)-\frac {3 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{5 (1+x)}+\frac {6}{5} \int \frac {-4 (1+x) (2+x)^3-e^x (2+3 x)}{x (1+x) (2+x) \left (e^x+4 (2+x)^2\right )} \, dx+\frac {24}{5} \int \frac {1}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx\\ &=\frac {6}{5} \log (1+x)-\frac {24}{5} \log (2+x)-\frac {3 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{5 (1+x)}+\frac {6}{5} \int \left (\frac {-2-3 x}{x \left (2+3 x+x^2\right )}-\frac {4 x (2+x)}{(1+x) \left (16+e^x+16 x+4 x^2\right )}\right ) \, dx+\frac {24}{5} \int \frac {1}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx\\ &=\frac {6}{5} \log (1+x)-\frac {24}{5} \log (2+x)-\frac {3 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{5 (1+x)}+\frac {6}{5} \int \frac {-2-3 x}{x \left (2+3 x+x^2\right )} \, dx+\frac {24}{5} \int \frac {1}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx-\frac {24}{5} \int \frac {x (2+x)}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx\\ &=\frac {6}{5} \log (1+x)-\frac {24}{5} \log (2+x)-\frac {3 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{5 (1+x)}+\frac {6}{5} \int \left (\frac {1}{-1-x}-\frac {1}{x}+\frac {2}{2+x}\right ) \, dx+\frac {24}{5} \int \frac {1}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx-\frac {24}{5} \int \frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )} \, dx-\frac {24}{5} \int \left (\frac {1}{16+e^x+16 x+4 x^2}+\frac {x}{16+e^x+16 x+4 x^2}-\frac {1}{(1+x) \left (16+e^x+16 x+4 x^2\right )}\right ) \, dx\\ &=-\frac {6 \log (x)}{5}-\frac {12}{5} \log (2+x)-\frac {3 \log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )}{5 (1+x)}-2 \left (\frac {24}{5} \int \frac {x}{16+e^x+16 x+4 x^2} \, dx\right )-\frac {24}{5} \int \frac {x^2}{16+e^x+16 x+4 x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.17, size = 93, normalized size = 2.58 \begin {gather*} -\frac {3 \left (2+4 x+2 x^2+2 (1+x) \log (x)+4 (1+x) \log (2+x)-2 \log \left (e^x+4 (2+x)^2\right )-2 x \log \left (e^x+4 (2+x)^2\right )+\log \left (\frac {25 e^{-2 x} \left (e^x+4 (2+x)^2\right )^2}{x^2 (2+x)^4}\right )\right )}{5 (1+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-192 - 672*x - 912*x^2 - 600*x^3 - 192*x^4 - 24*x^5 + E^x*(-12 - 30*x - 18*x^2) + (96 + 144*x + 72*
x^2 + 12*x^3 + E^x*(6 + 3*x))*Log[(6400 + 25*E^(2*x) + 12800*x + 9600*x^2 + 3200*x^3 + 400*x^4 + E^x*(800 + 80
0*x + 200*x^2))/(E^(2*x)*(16*x^2 + 32*x^3 + 24*x^4 + 8*x^5 + x^6))])/(160 + 560*x + 760*x^2 + 500*x^3 + 160*x^
4 + 20*x^5 + E^x*(10 + 25*x + 20*x^2 + 5*x^3)),x]

[Out]

(-3*(2 + 4*x + 2*x^2 + 2*(1 + x)*Log[x] + 4*(1 + x)*Log[2 + x] - 2*Log[E^x + 4*(2 + x)^2] - 2*x*Log[E^x + 4*(2
+ x)^2] + Log[(25*(E^x + 4*(2 + x)^2)^2)/(E^(2*x)*x^2*(2 + x)^4)]))/(5*(1 + x))

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fricas [B]  time = 0.52, size = 77, normalized size = 2.14 \begin {gather*} \frac {3 \, x \log \left (\frac {25 \, {\left (16 \, x^{4} + 128 \, x^{3} + 384 \, x^{2} + 8 \, {\left (x^{2} + 4 \, x + 4\right )} e^{x} + 512 \, x + e^{\left (2 \, x\right )} + 256\right )} e^{\left (-2 \, x\right )}}{x^{6} + 8 \, x^{5} + 24 \, x^{4} + 32 \, x^{3} + 16 \, x^{2}}\right )}{5 \, {\left (x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*(6+3*x)+12*x^3+72*x^2+144*x+96)*log((25*exp(x)^2+(200*x^2+800*x+800)*exp(x)+400*x^4+3200*x^
3+9600*x^2+12800*x+6400)/(x^6+8*x^5+24*x^4+32*x^3+16*x^2)/exp(x)^2)+(-18*x^2-30*x-12)*exp(x)-24*x^5-192*x^4-60
0*x^3-912*x^2-672*x-192)/((5*x^3+20*x^2+25*x+10)*exp(x)+20*x^5+160*x^4+500*x^3+760*x^2+560*x+160),x, algorithm
="fricas")

[Out]

3/5*x*log(25*(16*x^4 + 128*x^3 + 384*x^2 + 8*(x^2 + 4*x + 4)*e^x + 512*x + e^(2*x) + 256)*e^(-2*x)/(x^6 + 8*x^
5 + 24*x^4 + 32*x^3 + 16*x^2))/(x + 1)

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giac [B]  time = 0.80, size = 163, normalized size = 4.53 \begin {gather*} -\frac {3 \, {\left (2 \, x^{2} - 2 \, x \log \left (-4 \, x^{2} - 16 \, x - e^{x} - 16\right ) + 4 \, x \log \left (x + 2\right ) + 2 \, x \log \relax (x) + 2 \, x - 2 \, \log \left (-4 \, x^{2} - 16 \, x - e^{x} - 16\right ) + 4 \, \log \left (x + 2\right ) + 2 \, \log \relax (x) + \log \left (\frac {25 \, {\left (16 \, x^{4} + 128 \, x^{3} + 8 \, x^{2} e^{x} + 384 \, x^{2} + 32 \, x e^{x} + 512 \, x + e^{\left (2 \, x\right )} + 32 \, e^{x} + 256\right )}}{x^{6} e^{\left (2 \, x\right )} + 8 \, x^{5} e^{\left (2 \, x\right )} + 24 \, x^{4} e^{\left (2 \, x\right )} + 32 \, x^{3} e^{\left (2 \, x\right )} + 16 \, x^{2} e^{\left (2 \, x\right )}}\right )\right )}}{5 \, {\left (x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*(6+3*x)+12*x^3+72*x^2+144*x+96)*log((25*exp(x)^2+(200*x^2+800*x+800)*exp(x)+400*x^4+3200*x^
3+9600*x^2+12800*x+6400)/(x^6+8*x^5+24*x^4+32*x^3+16*x^2)/exp(x)^2)+(-18*x^2-30*x-12)*exp(x)-24*x^5-192*x^4-60
0*x^3-912*x^2-672*x-192)/((5*x^3+20*x^2+25*x+10)*exp(x)+20*x^5+160*x^4+500*x^3+760*x^2+560*x+160),x, algorithm
="giac")

[Out]

-3/5*(2*x^2 - 2*x*log(-4*x^2 - 16*x - e^x - 16) + 4*x*log(x + 2) + 2*x*log(x) + 2*x - 2*log(-4*x^2 - 16*x - e^
x - 16) + 4*log(x + 2) + 2*log(x) + log(25*(16*x^4 + 128*x^3 + 8*x^2*e^x + 384*x^2 + 32*x*e^x + 512*x + e^(2*x
) + 32*e^x + 256)/(x^6*e^(2*x) + 8*x^5*e^(2*x) + 24*x^4*e^(2*x) + 32*x^3*e^(2*x) + 16*x^2*e^(2*x))))/(x + 1)

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maple [C]  time = 0.80, size = 1110, normalized size = 30.83

method	result	size
risch	\(\frac {6 \ln \left ({\mathrm e}^{x}\right )}{5 \left (x +1\right )}-\frac {3 \left (4 x +i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 \ln \relax (5)+8 \ln \relax (2)+4 x^{2}+4 x \ln \relax (x )+2 i \pi \,\mathrm {csgn}\left (i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )\right ) \mathrm {csgn}\left (i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}\right )^{2}-2 i \pi \,\mathrm {csgn}\left (i \left (2+x \right )\right ) \mathrm {csgn}\left (i \left (2+x \right )^{2}\right )^{2}+i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{-2 x} \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}}{\left (2+x \right )^{4}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2} {\mathrm e}^{-2 x}}{x^{2} \left (2+x \right )^{4}}\right )^{2}-2 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}+i \pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2} {\mathrm e}^{-2 x}}{x^{2} \left (2+x \right )^{4}}\right )^{2}+i \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )+i \pi \,\mathrm {csgn}\left (\frac {i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}}{\left (2+x \right )^{4}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-2 x} \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}}{\left (2+x \right )^{4}}\right )^{2}+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-2 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-2 x} \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}}{\left (2+x \right )^{4}}\right )^{2}+i \pi \,\mathrm {csgn}\left (\frac {i}{\left (2+x \right )^{4}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}}{\left (2+x \right )^{4}}\right )^{2}+i \pi \,\mathrm {csgn}\left (i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}}{\left (2+x \right )^{4}}\right )^{2}+i \pi \mathrm {csgn}\left (i \left (2+x \right )\right )^{2} \mathrm {csgn}\left (i \left (2+x \right )^{2}\right )-i \pi \,\mathrm {csgn}\left (i \left (2+x \right )\right ) \mathrm {csgn}\left (i \left (2+x \right )^{3}\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (2+x \right )^{2}\right ) \mathrm {csgn}\left (i \left (2+x \right )^{3}\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (2+x \right )\right ) \mathrm {csgn}\left (i \left (2+x \right )^{4}\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (2+x \right )^{3}\right ) \mathrm {csgn}\left (i \left (2+x \right )^{4}\right )^{2}-i \pi \mathrm {csgn}\left (i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )\right )^{2} \mathrm {csgn}\left (i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+8 x \ln \left (2+x \right )+i \pi \,\mathrm {csgn}\left (i \left (2+x \right )\right ) \mathrm {csgn}\left (i \left (2+x \right )^{3}\right ) \mathrm {csgn}\left (i \left (2+x \right )^{4}\right )-i \pi \,\mathrm {csgn}\left (\frac {i}{\left (2+x \right )^{4}}\right ) \mathrm {csgn}\left (i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}}{\left (2+x \right )^{4}}\right )+4 \ln \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )-i \pi \mathrm {csgn}\left (i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}\right )^{3}+i \pi \mathrm {csgn}\left (i \left (2+x \right )^{2}\right )^{3}+i \pi \mathrm {csgn}\left (i \left (2+x \right )^{4}\right )^{3}+i \pi \mathrm {csgn}\left (i \left (2+x \right )^{3}\right )^{3}-i \pi \mathrm {csgn}\left (\frac {i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}}{\left (2+x \right )^{4}}\right )^{3}+i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-4 \ln \left (4 x^{2}+{\mathrm e}^{x}+16 x +16\right )-i \pi \mathrm {csgn}\left (\frac {i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2} {\mathrm e}^{-2 x}}{x^{2} \left (2+x \right )^{4}}\right )^{3}-i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{-2 x} \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}}{\left (2+x \right )^{4}}\right )^{3}-4 \ln \left (4 x^{2}+{\mathrm e}^{x}+16 x +16\right ) x -i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-2 x}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}}{\left (2+x \right )^{4}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-2 x} \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}}{\left (2+x \right )^{4}}\right )-i \pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-2 x} \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2}}{\left (2+x \right )^{4}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+4 x +\frac {{\mathrm e}^{x}}{4}+4\right )^{2} {\mathrm e}^{-2 x}}{x^{2} \left (2+x \right )^{4}}\right )+i \pi \,\mathrm {csgn}\left (i \left (2+x \right )\right ) \mathrm {csgn}\left (i \left (2+x \right )^{2}\right ) \mathrm {csgn}\left (i \left (2+x \right )^{3}\right )\right )}{10 \left (x +1\right )}\)	\(1110\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)*(6+3*x)+12*x^3+72*x^2+144*x+96)*ln((25*exp(x)^2+(200*x^2+800*x+800)*exp(x)+400*x^4+3200*x^3+9600*
x^2+12800*x+6400)/(x^6+8*x^5+24*x^4+32*x^3+16*x^2)/exp(x)^2)+(-18*x^2-30*x-12)*exp(x)-24*x^5-192*x^4-600*x^3-9
12*x^2-672*x-192)/((5*x^3+20*x^2+25*x+10)*exp(x)+20*x^5+160*x^4+500*x^3+760*x^2+560*x+160),x,method=_RETURNVER
BOSE)

[Out]

6/5/(x+1)*ln(exp(x))-3/10*(4*x+4*ln(5)+8*ln(2)+4*x^2+4*x*ln(x)+I*Pi*csgn(I*exp(-2*x)*(x^2+4*x+1/4*exp(x)+4)^2/
(2+x)^4)*csgn(I/x^2*(x^2+4*x+1/4*exp(x)+4)^2/(2+x)^4*exp(-2*x))^2-2*I*Pi*csgn(I*exp(x))*csgn(I*exp(2*x))^2+I*P
i*csgn(I/x^2)*csgn(I/x^2*(x^2+4*x+1/4*exp(x)+4)^2/(2+x)^4*exp(-2*x))^2+I*Pi*csgn(I*exp(x))^2*csgn(I*exp(2*x))+
I*Pi*csgn(I/(2+x)^4*(x^2+4*x+1/4*exp(x)+4)^2)*csgn(I*exp(-2*x)*(x^2+4*x+1/4*exp(x)+4)^2/(2+x)^4)^2+I*Pi*csgn(I
*exp(-2*x))*csgn(I*exp(-2*x)*(x^2+4*x+1/4*exp(x)+4)^2/(2+x)^4)^2+I*Pi*csgn(I/(2+x)^4)*csgn(I/(2+x)^4*(x^2+4*x+
1/4*exp(x)+4)^2)^2+I*Pi*csgn(I*(x^2+4*x+1/4*exp(x)+4)^2)*csgn(I/(2+x)^4*(x^2+4*x+1/4*exp(x)+4)^2)^2+I*Pi*csgn(
I*(2+x))^2*csgn(I*(2+x)^2)-I*Pi*csgn(I*(2+x))*csgn(I*(2+x)^3)^2-I*Pi*csgn(I*(2+x)^2)*csgn(I*(2+x)^3)^2-I*Pi*cs
gn(I*(2+x))*csgn(I*(2+x)^4)^2-I*Pi*csgn(I*(2+x)^3)*csgn(I*(2+x)^4)^2-I*Pi*csgn(I*(x^2+4*x+1/4*exp(x)+4))^2*csg
n(I*(x^2+4*x+1/4*exp(x)+4)^2)+2*I*Pi*csgn(I*(x^2+4*x+1/4*exp(x)+4))*csgn(I*(x^2+4*x+1/4*exp(x)+4)^2)^2-2*I*Pi*
csgn(I*(2+x))*csgn(I*(2+x)^2)^2-2*I*Pi*csgn(I*x)*csgn(I*x^2)^2+I*Pi*csgn(I*x)^2*csgn(I*x^2)+8*x*ln(2+x)+4*ln(x
^2+4*x+1/4*exp(x)+4)-I*Pi*csgn(I/x^2*(x^2+4*x+1/4*exp(x)+4)^2/(2+x)^4*exp(-2*x))^3-4*ln(4*x^2+exp(x)+16*x+16)-
4*ln(4*x^2+exp(x)+16*x+16)*x-I*Pi*csgn(I*exp(-2*x)*(x^2+4*x+1/4*exp(x)+4)^2/(2+x)^4)^3+I*Pi*csgn(I*x^2)^3-I*Pi
*csgn(I*(x^2+4*x+1/4*exp(x)+4)^2)^3+I*Pi*csgn(I*(2+x)^2)^3+I*Pi*csgn(I*(2+x)^4)^3+I*Pi*csgn(I*(2+x)^3)^3-I*Pi*
csgn(I/(2+x)^4*(x^2+4*x+1/4*exp(x)+4)^2)^3-I*Pi*csgn(I*exp(-2*x))*csgn(I/(2+x)^4*(x^2+4*x+1/4*exp(x)+4)^2)*csg
n(I*exp(-2*x)*(x^2+4*x+1/4*exp(x)+4)^2/(2+x)^4)+I*Pi*csgn(I*exp(2*x))^3-I*Pi*csgn(I/x^2)*csgn(I*exp(-2*x)*(x^2
+4*x+1/4*exp(x)+4)^2/(2+x)^4)*csgn(I/x^2*(x^2+4*x+1/4*exp(x)+4)^2/(2+x)^4*exp(-2*x))+I*Pi*csgn(I*(2+x))*csgn(I
*(2+x)^2)*csgn(I*(2+x)^3)+I*Pi*csgn(I*(2+x))*csgn(I*(2+x)^3)*csgn(I*(2+x)^4)-I*Pi*csgn(I/(2+x)^4)*csgn(I*(x^2+
4*x+1/4*exp(x)+4)^2)*csgn(I/(2+x)^4*(x^2+4*x+1/4*exp(x)+4)^2))/(x+1)

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maxima [A]  time = 0.51, size = 42, normalized size = 1.17 \begin {gather*} -\frac {6 \, {\left (x^{2} - x \log \left (4 \, x^{2} + 16 \, x + e^{x} + 16\right ) + 2 \, x \log \left (x + 2\right ) + x \log \relax (x) + x + \log \relax (5) + 1\right )}}{5 \, {\left (x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*(6+3*x)+12*x^3+72*x^2+144*x+96)*log((25*exp(x)^2+(200*x^2+800*x+800)*exp(x)+400*x^4+3200*x^
3+9600*x^2+12800*x+6400)/(x^6+8*x^5+24*x^4+32*x^3+16*x^2)/exp(x)^2)+(-18*x^2-30*x-12)*exp(x)-24*x^5-192*x^4-60
0*x^3-912*x^2-672*x-192)/((5*x^3+20*x^2+25*x+10)*exp(x)+20*x^5+160*x^4+500*x^3+760*x^2+560*x+160),x, algorithm
="maxima")

[Out]

-6/5*(x^2 - x*log(4*x^2 + 16*x + e^x + 16) + 2*x*log(x + 2) + x*log(x) + x + log(5) + 1)/(x + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {672\,x-\ln \left (\frac {{\mathrm {e}}^{-2\,x}\,\left (12800\,x+25\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,\left (200\,x^2+800\,x+800\right )+9600\,x^2+3200\,x^3+400\,x^4+6400\right )}{x^6+8\,x^5+24\,x^4+32\,x^3+16\,x^2}\right )\,\left (144\,x+{\mathrm {e}}^x\,\left (3\,x+6\right )+72\,x^2+12\,x^3+96\right )+{\mathrm {e}}^x\,\left (18\,x^2+30\,x+12\right )+912\,x^2+600\,x^3+192\,x^4+24\,x^5+192}{560\,x+760\,x^2+500\,x^3+160\,x^4+20\,x^5+{\mathrm {e}}^x\,\left (5\,x^3+20\,x^2+25\,x+10\right )+160} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(672*x - log((exp(-2*x)*(12800*x + 25*exp(2*x) + exp(x)*(800*x + 200*x^2 + 800) + 9600*x^2 + 3200*x^3 + 4
00*x^4 + 6400))/(16*x^2 + 32*x^3 + 24*x^4 + 8*x^5 + x^6))*(144*x + exp(x)*(3*x + 6) + 72*x^2 + 12*x^3 + 96) +
exp(x)*(30*x + 18*x^2 + 12) + 912*x^2 + 600*x^3 + 192*x^4 + 24*x^5 + 192)/(560*x + 760*x^2 + 500*x^3 + 160*x^4
+ 20*x^5 + exp(x)*(25*x + 20*x^2 + 5*x^3 + 10) + 160),x)

[Out]

int(-(672*x - log((exp(-2*x)*(12800*x + 25*exp(2*x) + exp(x)*(800*x + 200*x^2 + 800) + 9600*x^2 + 3200*x^3 + 4
00*x^4 + 6400))/(16*x^2 + 32*x^3 + 24*x^4 + 8*x^5 + x^6))*(144*x + exp(x)*(3*x + 6) + 72*x^2 + 12*x^3 + 96) +
exp(x)*(30*x + 18*x^2 + 12) + 912*x^2 + 600*x^3 + 192*x^4 + 24*x^5 + 192)/(560*x + 760*x^2 + 500*x^3 + 160*x^4
+ 20*x^5 + exp(x)*(25*x + 20*x^2 + 5*x^3 + 10) + 160), x)

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sympy [B]  time = 1.16, size = 105, normalized size = 2.92 \begin {gather*} - \frac {6 \log {\relax (x )}}{5} + \frac {6 \log {\left (e^{- x} + \frac {1}{4 x^{2} + 16 x + 16} \right )}}{5} - \frac {3 \log {\left (\frac {\left (400 x^{4} + 3200 x^{3} + 9600 x^{2} + 12800 x + \left (200 x^{2} + 800 x + 800\right ) e^{x} + 25 e^{2 x} + 6400\right ) e^{- 2 x}}{x^{6} + 8 x^{5} + 24 x^{4} + 32 x^{3} + 16 x^{2}} \right )}}{5 x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*(6+3*x)+12*x**3+72*x**2+144*x+96)*ln((25*exp(x)**2+(200*x**2+800*x+800)*exp(x)+400*x**4+320
0*x**3+9600*x**2+12800*x+6400)/(x**6+8*x**5+24*x**4+32*x**3+16*x**2)/exp(x)**2)+(-18*x**2-30*x-12)*exp(x)-24*x
**5-192*x**4-600*x**3-912*x**2-672*x-192)/((5*x**3+20*x**2+25*x+10)*exp(x)+20*x**5+160*x**4+500*x**3+760*x**2+
560*x+160),x)

[Out]

-6*log(x)/5 + 6*log(exp(-x) + 1/(4*x**2 + 16*x + 16))/5 - 3*log((400*x**4 + 3200*x**3 + 9600*x**2 + 12800*x +
(200*x**2 + 800*x + 800)*exp(x) + 25*exp(2*x) + 6400)*exp(-2*x)/(x**6 + 8*x**5 + 24*x**4 + 32*x**3 + 16*x**2))
/(5*x + 5)

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