3.82.41 \(\int \frac {-20000 x^2-2500 x^3+e (800+96 x)+e^{3 x} (-640 x^2-80 x^3)+e^{4 x} (-32 x^2-4 x^3)+e^{2 x} (-4800 x^2-600 x^3+e (32+68 x+8 x^2))+e^x (-16000 x^2-2000 x^3+e (320+360 x+40 x^2))+(e (-32-4 x)+1600 x^2+200 x^3+e^{2 x} (64 x^2+8 x^3)+e^x (640 x^2+80 x^3)) \log (8+x)+(-32 x^2-4 x^3) \log ^2(8+x)}{5000 x^4+625 x^5+e^2 (8+x)+e (400 x^2+50 x^3)+e^{4 x} (8 x^4+x^5)+e^{3 x} (160 x^4+20 x^5)+e^{2 x} (1200 x^4+150 x^5+e (16 x^2+2 x^3))+e^x (4000 x^4+500 x^5+e (160 x^2+20 x^3))+(-400 x^4-50 x^5+e (-16 x^2-2 x^3)+e^x (-160 x^4-20 x^5)+e^{2 x} (-16 x^4-2 x^5)) \log (8+x)+(8 x^4+x^5) \log ^2(8+x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {4 x}{x^2+\frac {e}{\left (5+e^x\right )^2-\log (8+x)}} \]

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Rubi [F]  time = 38.95, 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 {-20000 x^2-2500 x^3+e (800+96 x)+e^{3 x} \left (-640 x^2-80 x^3\right )+e^{4 x} \left (-32 x^2-4 x^3\right )+e^{2 x} \left (-4800 x^2-600 x^3+e \left (32+68 x+8 x^2\right )\right )+e^x \left (-16000 x^2-2000 x^3+e \left (320+360 x+40 x^2\right )\right )+\left (e (-32-4 x)+1600 x^2+200 x^3+e^{2 x} \left (64 x^2+8 x^3\right )+e^x \left (640 x^2+80 x^3\right )\right ) \log (8+x)+\left (-32 x^2-4 x^3\right ) \log ^2(8+x)}{5000 x^4+625 x^5+e^2 (8+x)+e \left (400 x^2+50 x^3\right )+e^{4 x} \left (8 x^4+x^5\right )+e^{3 x} \left (160 x^4+20 x^5\right )+e^{2 x} \left (1200 x^4+150 x^5+e \left (16 x^2+2 x^3\right )\right )+e^x \left (4000 x^4+500 x^5+e \left (160 x^2+20 x^3\right )\right )+\left (-400 x^4-50 x^5+e \left (-16 x^2-2 x^3\right )+e^x \left (-160 x^4-20 x^5\right )+e^{2 x} \left (-16 x^4-2 x^5\right )\right ) \log (8+x)+\left (8 x^4+x^5\right ) \log ^2(8+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-20000*x^2 - 2500*x^3 + E*(800 + 96*x) + E^(3*x)*(-640*x^2 - 80*x^3) + E^(4*x)*(-32*x^2 - 4*x^3) + E^(2*x
)*(-4800*x^2 - 600*x^3 + E*(32 + 68*x + 8*x^2)) + E^x*(-16000*x^2 - 2000*x^3 + E*(320 + 360*x + 40*x^2)) + (E*
(-32 - 4*x) + 1600*x^2 + 200*x^3 + E^(2*x)*(64*x^2 + 8*x^3) + E^x*(640*x^2 + 80*x^3))*Log[8 + x] + (-32*x^2 -
4*x^3)*Log[8 + x]^2)/(5000*x^4 + 625*x^5 + E^2*(8 + x) + E*(400*x^2 + 50*x^3) + E^(4*x)*(8*x^4 + x^5) + E^(3*x
)*(160*x^4 + 20*x^5) + E^(2*x)*(1200*x^4 + 150*x^5 + E*(16*x^2 + 2*x^3)) + E^x*(4000*x^4 + 500*x^5 + E*(160*x^
2 + 20*x^3)) + (-400*x^4 - 50*x^5 + E*(-16*x^2 - 2*x^3) + E^x*(-160*x^4 - 20*x^5) + E^(2*x)*(-16*x^4 - 2*x^5))
*Log[8 + x] + (8*x^4 + x^5)*Log[8 + x]^2),x]

[Out]

4/x - 4*E*Defer[Int][(E + 25*x^2 + 10*E^x*x^2 + E^(2*x)*x^2 - x^2*Log[8 + x])^(-2), x] - 8*E^2*Defer[Int][1/(x
^2*(E + 25*x^2 + 10*E^x*x^2 + E^(2*x)*x^2 - x^2*Log[8 + x])^2), x] - 8*E^2*Defer[Int][1/(x*(E + 25*x^2 + 10*E^
x*x^2 + E^(2*x)*x^2 - x^2*Log[8 + x])^2), x] - 200*E*Defer[Int][x/(E + 25*x^2 + 10*E^x*x^2 + E^(2*x)*x^2 - x^2
*Log[8 + x])^2, x] - 40*E*Defer[Int][(E^x*x)/(E + 25*x^2 + 10*E^x*x^2 + E^(2*x)*x^2 - x^2*Log[8 + x])^2, x] +
32*E*Defer[Int][1/((8 + x)*(E + 25*x^2 + 10*E^x*x^2 + E^(2*x)*x^2 - x^2*Log[8 + x])^2), x] + 12*E*Defer[Int][1
/(x^2*(E + 25*x^2 + 10*E^x*x^2 + E^(2*x)*x^2 - x^2*Log[8 + x])), x] + 8*E*Defer[Int][1/(x*(E + 25*x^2 + 10*E^x
*x^2 + E^(2*x)*x^2 - x^2*Log[8 + x])), x] + 8*E*Defer[Int][(x*Log[8 + x])/(-E - 25*x^2 - 10*E^x*x^2 - E^(2*x)*
x^2 + x^2*Log[8 + x])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \left (-625 x^2 (8+x)-500 e^x x^2 (8+x)-150 e^{2 x} x^2 (8+x)-20 e^{3 x} x^2 (8+x)-e^{4 x} x^2 (8+x)+8 e (25+3 x)+10 e^{1+x} \left (8+9 x+x^2\right )+e^{1+2 x} \left (8+17 x+2 x^2\right )-(8+x) \left (e-50 x^2-20 e^x x^2-2 e^{2 x} x^2\right ) \log (8+x)-x^2 (8+x) \log ^2(8+x)\right )}{(8+x) \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )^2} \, dx\\ &=4 \int \frac {-625 x^2 (8+x)-500 e^x x^2 (8+x)-150 e^{2 x} x^2 (8+x)-20 e^{3 x} x^2 (8+x)-e^{4 x} x^2 (8+x)+8 e (25+3 x)+10 e^{1+x} \left (8+9 x+x^2\right )+e^{1+2 x} \left (8+17 x+2 x^2\right )-(8+x) \left (e-50 x^2-20 e^x x^2-2 e^{2 x} x^2\right ) \log (8+x)-x^2 (8+x) \log ^2(8+x)}{(8+x) \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )^2} \, dx\\ &=4 \int \left (-\frac {1}{x^2}+\frac {e (3+2 x)}{x^2 \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )}-\frac {e \left (16 e+18 e x+2 e x^2+401 x^3+80 e^x x^3+50 x^4+10 e^x x^4-16 x^3 \log (8+x)-2 x^4 \log (8+x)\right )}{x^2 (8+x) \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )^2}\right ) \, dx\\ &=\frac {4}{x}+(4 e) \int \frac {3+2 x}{x^2 \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )} \, dx-(4 e) \int \frac {16 e+18 e x+2 e x^2+401 x^3+80 e^x x^3+50 x^4+10 e^x x^4-16 x^3 \log (8+x)-2 x^4 \log (8+x)}{x^2 (8+x) \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )^2} \, dx\\ &=\frac {4}{x}+(4 e) \int \left (\frac {3}{x^2 \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )}+\frac {2}{x \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )}\right ) \, dx-(4 e) \int \left (\frac {16 e+18 e x+2 e x^2+401 x^3+80 e^x x^3+50 x^4+10 e^x x^4-16 x^3 \log (8+x)-2 x^4 \log (8+x)}{8 x^2 \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )^2}-\frac {16 e+18 e x+2 e x^2+401 x^3+80 e^x x^3+50 x^4+10 e^x x^4-16 x^3 \log (8+x)-2 x^4 \log (8+x)}{64 x \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )^2}+\frac {16 e+18 e x+2 e x^2+401 x^3+80 e^x x^3+50 x^4+10 e^x x^4-16 x^3 \log (8+x)-2 x^4 \log (8+x)}{64 (8+x) \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )^2}\right ) \, dx\\ &=\frac {4}{x}+\frac {1}{16} e \int \frac {16 e+18 e x+2 e x^2+401 x^3+80 e^x x^3+50 x^4+10 e^x x^4-16 x^3 \log (8+x)-2 x^4 \log (8+x)}{x \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )^2} \, dx-\frac {1}{16} e \int \frac {16 e+18 e x+2 e x^2+401 x^3+80 e^x x^3+50 x^4+10 e^x x^4-16 x^3 \log (8+x)-2 x^4 \log (8+x)}{(8+x) \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )^2} \, dx-\frac {1}{2} e \int \frac {16 e+18 e x+2 e x^2+401 x^3+80 e^x x^3+50 x^4+10 e^x x^4-16 x^3 \log (8+x)-2 x^4 \log (8+x)}{x^2 \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )^2} \, dx+(8 e) \int \frac {1}{x \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )} \, dx+(12 e) \int \frac {1}{x^2 \left (e+25 x^2+10 e^x x^2+e^{2 x} x^2-x^2 \log (8+x)\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 51, normalized size = 1.89 \begin {gather*} -4 \left (-\frac {1}{x}-\frac {e}{x \left (-e-25 x^2-10 e^x x^2-e^{2 x} x^2+x^2 \log (8+x)\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20000*x^2 - 2500*x^3 + E*(800 + 96*x) + E^(3*x)*(-640*x^2 - 80*x^3) + E^(4*x)*(-32*x^2 - 4*x^3) +
E^(2*x)*(-4800*x^2 - 600*x^3 + E*(32 + 68*x + 8*x^2)) + E^x*(-16000*x^2 - 2000*x^3 + E*(320 + 360*x + 40*x^2))
+ (E*(-32 - 4*x) + 1600*x^2 + 200*x^3 + E^(2*x)*(64*x^2 + 8*x^3) + E^x*(640*x^2 + 80*x^3))*Log[8 + x] + (-32*
x^2 - 4*x^3)*Log[8 + x]^2)/(5000*x^4 + 625*x^5 + E^2*(8 + x) + E*(400*x^2 + 50*x^3) + E^(4*x)*(8*x^4 + x^5) +
E^(3*x)*(160*x^4 + 20*x^5) + E^(2*x)*(1200*x^4 + 150*x^5 + E*(16*x^2 + 2*x^3)) + E^x*(4000*x^4 + 500*x^5 + E*(
160*x^2 + 20*x^3)) + (-400*x^4 - 50*x^5 + E*(-16*x^2 - 2*x^3) + E^x*(-160*x^4 - 20*x^5) + E^(2*x)*(-16*x^4 - 2
*x^5))*Log[8 + x] + (8*x^4 + x^5)*Log[8 + x]^2),x]

[Out]

-4*(-x^(-1) - E/(x*(-E - 25*x^2 - 10*E^x*x^2 - E^(2*x)*x^2 + x^2*Log[8 + x])))

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fricas [B]  time = 0.62, size = 58, normalized size = 2.15 \begin {gather*} \frac {4 \, {\left (x e^{\left (2 \, x\right )} + 10 \, x e^{x} - x \log \left (x + 8\right ) + 25 \, x\right )}}{x^{2} e^{\left (2 \, x\right )} + 10 \, x^{2} e^{x} - x^{2} \log \left (x + 8\right ) + 25 \, x^{2} + e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3-32*x^2)*log(x+8)^2+((8*x^3+64*x^2)*exp(x)^2+(80*x^3+640*x^2)*exp(x)+(-4*x-32)*exp(1)+200*x^
3+1600*x^2)*log(x+8)+(-4*x^3-32*x^2)*exp(x)^4+(-80*x^3-640*x^2)*exp(x)^3+((8*x^2+68*x+32)*exp(1)-600*x^3-4800*
x^2)*exp(x)^2+((40*x^2+360*x+320)*exp(1)-2000*x^3-16000*x^2)*exp(x)+(96*x+800)*exp(1)-2500*x^3-20000*x^2)/((x^
5+8*x^4)*log(x+8)^2+((-2*x^5-16*x^4)*exp(x)^2+(-20*x^5-160*x^4)*exp(x)+(-2*x^3-16*x^2)*exp(1)-50*x^5-400*x^4)*
log(x+8)+(x^5+8*x^4)*exp(x)^4+(20*x^5+160*x^4)*exp(x)^3+((2*x^3+16*x^2)*exp(1)+150*x^5+1200*x^4)*exp(x)^2+((20
*x^3+160*x^2)*exp(1)+500*x^5+4000*x^4)*exp(x)+(x+8)*exp(1)^2+(50*x^3+400*x^2)*exp(1)+625*x^5+5000*x^4),x, algo
rithm="fricas")

[Out]

4*(x*e^(2*x) + 10*x*e^x - x*log(x + 8) + 25*x)/(x^2*e^(2*x) + 10*x^2*e^x - x^2*log(x + 8) + 25*x^2 + e)

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giac [B]  time = 0.51, size = 58, normalized size = 2.15 \begin {gather*} \frac {4 \, {\left (x e^{\left (2 \, x\right )} + 10 \, x e^{x} - x \log \left (x + 8\right ) + 25 \, x\right )}}{x^{2} e^{\left (2 \, x\right )} + 10 \, x^{2} e^{x} - x^{2} \log \left (x + 8\right ) + 25 \, x^{2} + e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3-32*x^2)*log(x+8)^2+((8*x^3+64*x^2)*exp(x)^2+(80*x^3+640*x^2)*exp(x)+(-4*x-32)*exp(1)+200*x^
3+1600*x^2)*log(x+8)+(-4*x^3-32*x^2)*exp(x)^4+(-80*x^3-640*x^2)*exp(x)^3+((8*x^2+68*x+32)*exp(1)-600*x^3-4800*
x^2)*exp(x)^2+((40*x^2+360*x+320)*exp(1)-2000*x^3-16000*x^2)*exp(x)+(96*x+800)*exp(1)-2500*x^3-20000*x^2)/((x^
5+8*x^4)*log(x+8)^2+((-2*x^5-16*x^4)*exp(x)^2+(-20*x^5-160*x^4)*exp(x)+(-2*x^3-16*x^2)*exp(1)-50*x^5-400*x^4)*
log(x+8)+(x^5+8*x^4)*exp(x)^4+(20*x^5+160*x^4)*exp(x)^3+((2*x^3+16*x^2)*exp(1)+150*x^5+1200*x^4)*exp(x)^2+((20
*x^3+160*x^2)*exp(1)+500*x^5+4000*x^4)*exp(x)+(x+8)*exp(1)^2+(50*x^3+400*x^2)*exp(1)+625*x^5+5000*x^4),x, algo
rithm="giac")

[Out]

4*(x*e^(2*x) + 10*x*e^x - x*log(x + 8) + 25*x)/(x^2*e^(2*x) + 10*x^2*e^x - x^2*log(x + 8) + 25*x^2 + e)

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maple [A]  time = 0.08, size = 48, normalized size = 1.78

method	result	size
risch	\(\frac {4}{x}-\frac {4 \,{\mathrm e}}{x \left ({\mathrm e}^{2 x} x^{2}+10 \,{\mathrm e}^{x} x^{2}-x^{2} \ln \left (x +8\right )+25 x^{2}+{\mathrm e}\right )}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^3-32*x^2)*ln(x+8)^2+((8*x^3+64*x^2)*exp(x)^2+(80*x^3+640*x^2)*exp(x)+(-4*x-32)*exp(1)+200*x^3+1600*
x^2)*ln(x+8)+(-4*x^3-32*x^2)*exp(x)^4+(-80*x^3-640*x^2)*exp(x)^3+((8*x^2+68*x+32)*exp(1)-600*x^3-4800*x^2)*exp
(x)^2+((40*x^2+360*x+320)*exp(1)-2000*x^3-16000*x^2)*exp(x)+(96*x+800)*exp(1)-2500*x^3-20000*x^2)/((x^5+8*x^4)
*ln(x+8)^2+((-2*x^5-16*x^4)*exp(x)^2+(-20*x^5-160*x^4)*exp(x)+(-2*x^3-16*x^2)*exp(1)-50*x^5-400*x^4)*ln(x+8)+(
x^5+8*x^4)*exp(x)^4+(20*x^5+160*x^4)*exp(x)^3+((2*x^3+16*x^2)*exp(1)+150*x^5+1200*x^4)*exp(x)^2+((20*x^3+160*x
^2)*exp(1)+500*x^5+4000*x^4)*exp(x)+(x+8)*exp(1)^2+(50*x^3+400*x^2)*exp(1)+625*x^5+5000*x^4),x,method=_RETURNV
ERBOSE)

[Out]

4/x-4/x*exp(1)/(exp(2*x)*x^2+10*exp(x)*x^2-x^2*ln(x+8)+25*x^2+exp(1))

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maxima [B]  time = 0.61, size = 58, normalized size = 2.15 \begin {gather*} \frac {4 \, {\left (x e^{\left (2 \, x\right )} + 10 \, x e^{x} - x \log \left (x + 8\right ) + 25 \, x\right )}}{x^{2} e^{\left (2 \, x\right )} + 10 \, x^{2} e^{x} - x^{2} \log \left (x + 8\right ) + 25 \, x^{2} + e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3-32*x^2)*log(x+8)^2+((8*x^3+64*x^2)*exp(x)^2+(80*x^3+640*x^2)*exp(x)+(-4*x-32)*exp(1)+200*x^
3+1600*x^2)*log(x+8)+(-4*x^3-32*x^2)*exp(x)^4+(-80*x^3-640*x^2)*exp(x)^3+((8*x^2+68*x+32)*exp(1)-600*x^3-4800*
x^2)*exp(x)^2+((40*x^2+360*x+320)*exp(1)-2000*x^3-16000*x^2)*exp(x)+(96*x+800)*exp(1)-2500*x^3-20000*x^2)/((x^
5+8*x^4)*log(x+8)^2+((-2*x^5-16*x^4)*exp(x)^2+(-20*x^5-160*x^4)*exp(x)+(-2*x^3-16*x^2)*exp(1)-50*x^5-400*x^4)*
log(x+8)+(x^5+8*x^4)*exp(x)^4+(20*x^5+160*x^4)*exp(x)^3+((2*x^3+16*x^2)*exp(1)+150*x^5+1200*x^4)*exp(x)^2+((20
*x^3+160*x^2)*exp(1)+500*x^5+4000*x^4)*exp(x)+(x+8)*exp(1)^2+(50*x^3+400*x^2)*exp(1)+625*x^5+5000*x^4),x, algo
rithm="maxima")

[Out]

4*(x*e^(2*x) + 10*x*e^x - x*log(x + 8) + 25*x)/(x^2*e^(2*x) + 10*x^2*e^x - x^2*log(x + 8) + 25*x^2 + e)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int \frac {{\mathrm {e}}^x\,\left (16000\,x^2-\mathrm {e}\,\left (40\,x^2+360\,x+320\right )+2000\,x^3\right )+{\mathrm {e}}^{4\,x}\,\left (4\,x^3+32\,x^2\right )+{\mathrm {e}}^{3\,x}\,\left (80\,x^3+640\,x^2\right )+{\mathrm {e}}^{2\,x}\,\left (4800\,x^2-\mathrm {e}\,\left (8\,x^2+68\,x+32\right )+600\,x^3\right )+{\ln \left (x+8\right )}^2\,\left (4\,x^3+32\,x^2\right )+20000\,x^2+2500\,x^3-\ln \left (x+8\right )\,\left ({\mathrm {e}}^x\,\left (80\,x^3+640\,x^2\right )+{\mathrm {e}}^{2\,x}\,\left (8\,x^3+64\,x^2\right )+1600\,x^2+200\,x^3-\mathrm {e}\,\left (4\,x+32\right )\right )-\mathrm {e}\,\left (96\,x+800\right )}{{\mathrm {e}}^{4\,x}\,\left (x^5+8\,x^4\right )+{\mathrm {e}}^{3\,x}\,\left (20\,x^5+160\,x^4\right )+{\ln \left (x+8\right )}^2\,\left (x^5+8\,x^4\right )-\ln \left (x+8\right )\,\left ({\mathrm {e}}^x\,\left (20\,x^5+160\,x^4\right )+{\mathrm {e}}^{2\,x}\,\left (2\,x^5+16\,x^4\right )+\mathrm {e}\,\left (2\,x^3+16\,x^2\right )+400\,x^4+50\,x^5\right )+\mathrm {e}\,\left (50\,x^3+400\,x^2\right )+{\mathrm {e}}^x\,\left (\mathrm {e}\,\left (20\,x^3+160\,x^2\right )+4000\,x^4+500\,x^5\right )+{\mathrm {e}}^2\,\left (x+8\right )+5000\,x^4+625\,x^5+{\mathrm {e}}^{2\,x}\,\left (\mathrm {e}\,\left (2\,x^3+16\,x^2\right )+1200\,x^4+150\,x^5\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(16000*x^2 - exp(1)*(360*x + 40*x^2 + 320) + 2000*x^3) + exp(4*x)*(32*x^2 + 4*x^3) + exp(3*x)*(64
0*x^2 + 80*x^3) + exp(2*x)*(4800*x^2 - exp(1)*(68*x + 8*x^2 + 32) + 600*x^3) + log(x + 8)^2*(32*x^2 + 4*x^3) +
20000*x^2 + 2500*x^3 - log(x + 8)*(exp(x)*(640*x^2 + 80*x^3) + exp(2*x)*(64*x^2 + 8*x^3) + 1600*x^2 + 200*x^3
- exp(1)*(4*x + 32)) - exp(1)*(96*x + 800))/(exp(4*x)*(8*x^4 + x^5) + exp(3*x)*(160*x^4 + 20*x^5) + log(x + 8
)^2*(8*x^4 + x^5) - log(x + 8)*(exp(x)*(160*x^4 + 20*x^5) + exp(2*x)*(16*x^4 + 2*x^5) + exp(1)*(16*x^2 + 2*x^3
) + 400*x^4 + 50*x^5) + exp(1)*(400*x^2 + 50*x^3) + exp(x)*(exp(1)*(160*x^2 + 20*x^3) + 4000*x^4 + 500*x^5) +
exp(2)*(x + 8) + 5000*x^4 + 625*x^5 + exp(2*x)*(exp(1)*(16*x^2 + 2*x^3) + 1200*x^4 + 150*x^5)),x)

[Out]

-int((exp(x)*(16000*x^2 - exp(1)*(360*x + 40*x^2 + 320) + 2000*x^3) + exp(4*x)*(32*x^2 + 4*x^3) + exp(3*x)*(64
0*x^2 + 80*x^3) + exp(2*x)*(4800*x^2 - exp(1)*(68*x + 8*x^2 + 32) + 600*x^3) + log(x + 8)^2*(32*x^2 + 4*x^3) +
20000*x^2 + 2500*x^3 - log(x + 8)*(exp(x)*(640*x^2 + 80*x^3) + exp(2*x)*(64*x^2 + 8*x^3) + 1600*x^2 + 200*x^3
- exp(1)*(4*x + 32)) - exp(1)*(96*x + 800))/(exp(4*x)*(8*x^4 + x^5) + exp(3*x)*(160*x^4 + 20*x^5) + log(x + 8
)^2*(8*x^4 + x^5) - log(x + 8)*(exp(x)*(160*x^4 + 20*x^5) + exp(2*x)*(16*x^4 + 2*x^5) + exp(1)*(16*x^2 + 2*x^3
) + 400*x^4 + 50*x^5) + exp(1)*(400*x^2 + 50*x^3) + exp(x)*(exp(1)*(160*x^2 + 20*x^3) + 4000*x^4 + 500*x^5) +
exp(2)*(x + 8) + 5000*x^4 + 625*x^5 + exp(2*x)*(exp(1)*(16*x^2 + 2*x^3) + 1200*x^4 + 150*x^5)), x)

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sympy [B]  time = 0.94, size = 42, normalized size = 1.56 \begin {gather*} - \frac {4 e}{x^{3} e^{2 x} + 10 x^{3} e^{x} - x^{3} \log {\left (x + 8 \right )} + 25 x^{3} + e x} + \frac {4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**3-32*x**2)*ln(x+8)**2+((8*x**3+64*x**2)*exp(x)**2+(80*x**3+640*x**2)*exp(x)+(-4*x-32)*exp(1)
+200*x**3+1600*x**2)*ln(x+8)+(-4*x**3-32*x**2)*exp(x)**4+(-80*x**3-640*x**2)*exp(x)**3+((8*x**2+68*x+32)*exp(1
)-600*x**3-4800*x**2)*exp(x)**2+((40*x**2+360*x+320)*exp(1)-2000*x**3-16000*x**2)*exp(x)+(96*x+800)*exp(1)-250
0*x**3-20000*x**2)/((x**5+8*x**4)*ln(x+8)**2+((-2*x**5-16*x**4)*exp(x)**2+(-20*x**5-160*x**4)*exp(x)+(-2*x**3-
16*x**2)*exp(1)-50*x**5-400*x**4)*ln(x+8)+(x**5+8*x**4)*exp(x)**4+(20*x**5+160*x**4)*exp(x)**3+((2*x**3+16*x**
2)*exp(1)+150*x**5+1200*x**4)*exp(x)**2+((20*x**3+160*x**2)*exp(1)+500*x**5+4000*x**4)*exp(x)+(x+8)*exp(1)**2+
(50*x**3+400*x**2)*exp(1)+625*x**5+5000*x**4),x)

[Out]

-4*E/(x**3*exp(2*x) + 10*x**3*exp(x) - x**3*log(x + 8) + 25*x**3 + E*x) + 4/x

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