3.23.22 \(\int \frac {16 x^2+4 x^5+e^{1-x} (-12 x-4 x^2+2 x^5)+(8 x^2-4 x^5+e^{1-x} (-4 x+2 x^4)) \log (\frac {4 x-2 x^4+e^{1-x} (-2+x^3)}{x^3})}{(e^{1-x} (-2+x^3) \log ^2(2)+(4 x-2 x^4) \log ^2(2)) \log ^3(\frac {4 x-2 x^4+e^{1-x} (-2+x^3)}{x^3})} \, dx\)

Optimal. Leaf size=39 \[ \frac {x^2}{\log ^2(2) \log ^2\left (\frac {\left (-e^{1-x}+2 x\right ) \left (\frac {2}{x}-x^2\right )}{x^2}\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 5.14, 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 {16 x^2+4 x^5+e^{1-x} \left (-12 x-4 x^2+2 x^5\right )+\left (8 x^2-4 x^5+e^{1-x} \left (-4 x+2 x^4\right )\right ) \log \left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )}{\left (e^{1-x} \left (-2+x^3\right ) \log ^2(2)+\left (4 x-2 x^4\right ) \log ^2(2)\right ) \log ^3\left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(16*x^2 + 4*x^5 + E^(1 - x)*(-12*x - 4*x^2 + 2*x^5) + (8*x^2 - 4*x^5 + E^(1 - x)*(-4*x + 2*x^4))*Log[(4*x
- 2*x^4 + E^(1 - x)*(-2 + x^3))/x^3])/((E^(1 - x)*(-2 + x^3)*Log[2]^2 + (4*x - 2*x^4)*Log[2]^2)*Log[(4*x - 2*x
^4 + E^(1 - x)*(-2 + x^3))/x^3]^3),x]

[Out]

(2*2^(2/3)*Defer[Int][1/((2^(1/3) - x)*Log[((E - 2*E^x*x)*(-2 + x^3))/(E^x*x^3)]^3), x])/Log[2]^2 - (2*Defer[I
nt][x/Log[((E - 2*E^x*x)*(-2 + x^3))/(E^x*x^3)]^3, x])/Log[2]^2 + (2*(-2)^(2/3)*Defer[Int][1/((2^(1/3) + (-1)^
(1/3)*x)*Log[((E - 2*E^x*x)*(-2 + x^3))/(E^x*x^3)]^3), x])/Log[2]^2 - (2*(-1)^(1/3)*2^(2/3)*Defer[Int][1/((2^(
1/3) - (-1)^(2/3)*x)*Log[((E - 2*E^x*x)*(-2 + x^3))/(E^x*x^3)]^3), x])/Log[2]^2 - (2*E*Defer[Int][x/((-E + 2*E
^x*x)*Log[((E - 2*E^x*x)*(-2 + x^3))/(E^x*x^3)]^3), x])/Log[2]^2 - (2*E*Defer[Int][x^2/((-E + 2*E^x*x)*Log[((E
- 2*E^x*x)*(-2 + x^3))/(E^x*x^3)]^3), x])/Log[2]^2 + (2*Defer[Int][x/Log[((E - 2*E^x*x)*(-2 + x^3))/(E^x*x^3)
]^2, x])/Log[2]^2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 x \left (-2 e^x x \left (4+x^3\right )-e \left (-6-2 x+x^4\right )-\left (e-2 e^x x\right ) \left (-2+x^3\right ) \log \left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )\right )}{\left (e-2 e^x x\right ) \left (2-x^3\right ) \log ^2(2) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx\\ &=\frac {2 \int \frac {x \left (-2 e^x x \left (4+x^3\right )-e \left (-6-2 x+x^4\right )-\left (e-2 e^x x\right ) \left (-2+x^3\right ) \log \left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )\right )}{\left (e-2 e^x x\right ) \left (2-x^3\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}\\ &=\frac {2 \int \left (\frac {e x (1+x)}{\left (e-2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )}+\frac {x \left (-4-x^3-2 \log \left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )+x^3 \log \left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )\right )}{\left (-2+x^3\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )}\right ) \, dx}{\log ^2(2)}\\ &=\frac {2 \int \frac {x \left (-4-x^3-2 \log \left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )+x^3 \log \left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )\right )}{\left (-2+x^3\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}+\frac {(2 e) \int \frac {x (1+x)}{\left (e-2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}\\ &=\frac {2 \int \frac {x \left (4+x^3-\left (-2+x^3\right ) \log \left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )\right )}{\left (2-x^3\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}+\frac {(2 e) \int \left (-\frac {x}{\left (-e+2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )}-\frac {x^2}{\left (-e+2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )}\right ) \, dx}{\log ^2(2)}\\ &=\frac {2 \int \left (-\frac {x \left (4+x^3\right )}{\left (-2+x^3\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )}+\frac {x}{\log ^2\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )}\right ) \, dx}{\log ^2(2)}-\frac {(2 e) \int \frac {x}{\left (-e+2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}-\frac {(2 e) \int \frac {x^2}{\left (-e+2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}\\ &=-\frac {2 \int \frac {x \left (4+x^3\right )}{\left (-2+x^3\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}+\frac {2 \int \frac {x}{\log ^2\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}-\frac {(2 e) \int \frac {x}{\left (-e+2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}-\frac {(2 e) \int \frac {x^2}{\left (-e+2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}\\ &=-\frac {2 \int \left (\frac {x}{\log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )}+\frac {6 x}{\left (-2+x^3\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )}\right ) \, dx}{\log ^2(2)}+\frac {2 \int \frac {x}{\log ^2\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}-\frac {(2 e) \int \frac {x}{\left (-e+2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}-\frac {(2 e) \int \frac {x^2}{\left (-e+2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}\\ &=-\frac {2 \int \frac {x}{\log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}+\frac {2 \int \frac {x}{\log ^2\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}-\frac {12 \int \frac {x}{\left (-2+x^3\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}-\frac {(2 e) \int \frac {x}{\left (-e+2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}-\frac {(2 e) \int \frac {x^2}{\left (-e+2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}\\ &=-\frac {2 \int \frac {x}{\log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}+\frac {2 \int \frac {x}{\log ^2\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}-\frac {12 \int \left (-\frac {1}{3 \sqrt [3]{2} \left (\sqrt [3]{2}-x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )}-\frac {(-1)^{2/3}}{3 \sqrt [3]{2} \left (\sqrt [3]{2}+\sqrt [3]{-1} x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )}+\frac {\sqrt [3]{-\frac {1}{2}}}{3 \left (\sqrt [3]{2}-(-1)^{2/3} x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )}\right ) \, dx}{\log ^2(2)}-\frac {(2 e) \int \frac {x}{\left (-e+2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}-\frac {(2 e) \int \frac {x^2}{\left (-e+2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}\\ &=-\frac {2 \int \frac {x}{\log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}+\frac {2 \int \frac {x}{\log ^2\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}+\frac {\left (2 (-2)^{2/3}\right ) \int \frac {1}{\left (\sqrt [3]{2}+\sqrt [3]{-1} x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}+\frac {\left (2\ 2^{2/3}\right ) \int \frac {1}{\left (\sqrt [3]{2}-x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}-\frac {\left (2 \sqrt [3]{-1} 2^{2/3}\right ) \int \frac {1}{\left (\sqrt [3]{2}-(-1)^{2/3} x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}-\frac {(2 e) \int \frac {x}{\left (-e+2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}-\frac {(2 e) \int \frac {x^2}{\left (-e+2 e^x x\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \, dx}{\log ^2(2)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 33, normalized size = 0.85 \begin {gather*} \frac {x^2}{\log ^2(2) \log ^2\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*x^2 + 4*x^5 + E^(1 - x)*(-12*x - 4*x^2 + 2*x^5) + (8*x^2 - 4*x^5 + E^(1 - x)*(-4*x + 2*x^4))*Log
[(4*x - 2*x^4 + E^(1 - x)*(-2 + x^3))/x^3])/((E^(1 - x)*(-2 + x^3)*Log[2]^2 + (4*x - 2*x^4)*Log[2]^2)*Log[(4*x
- 2*x^4 + E^(1 - x)*(-2 + x^3))/x^3]^3),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.83, size = 38, normalized size = 0.97 \begin {gather*} \frac {x^{2}}{\log \relax (2)^{2} \log \left (-\frac {2 \, x^{4} - {\left (x^{3} - 2\right )} e^{\left (-x + 1\right )} - 4 \, x}{x^{3}}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^4-4*x)*exp(-x+1)-4*x^5+8*x^2)*log(((x^3-2)*exp(-x+1)-2*x^4+4*x)/x^3)+(2*x^5-4*x^2-12*x)*exp(-
x+1)+4*x^5+16*x^2)/((x^3-2)*log(2)^2*exp(-x+1)+(-2*x^4+4*x)*log(2)^2)/log(((x^3-2)*exp(-x+1)-2*x^4+4*x)/x^3)^3
,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 1.39, size = 91, normalized size = 2.33 \begin {gather*} \frac {x^{2}}{\log \relax (2)^{2} \log \left (-2 \, x^{4} + x^{3} e^{\left (-x + 1\right )} + 4 \, x - 2 \, e^{\left (-x + 1\right )}\right )^{2} - 2 \, \log \relax (2)^{2} \log \left (-2 \, x^{4} + x^{3} e^{\left (-x + 1\right )} + 4 \, x - 2 \, e^{\left (-x + 1\right )}\right ) \log \left (x^{3}\right ) + \log \relax (2)^{2} \log \left (x^{3}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^4-4*x)*exp(-x+1)-4*x^5+8*x^2)*log(((x^3-2)*exp(-x+1)-2*x^4+4*x)/x^3)+(2*x^5-4*x^2-12*x)*exp(-
x+1)+4*x^5+16*x^2)/((x^3-2)*log(2)^2*exp(-x+1)+(-2*x^4+4*x)*log(2)^2)/log(((x^3-2)*exp(-x+1)-2*x^4+4*x)/x^3)^3
,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.50, size = 411, normalized size = 10.54

method	result	size
risch	\(-\frac {4 x^{2}}{\left (-2 \pi \mathrm {csgn}\left (\frac {i \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )}{x^{3}}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{x^{3}}\right ) \mathrm {csgn}\left (i \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )}{x^{3}}\right )+\pi \,\mathrm {csgn}\left (\frac {i}{x^{3}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )}{x^{3}}\right )^{2}+\pi \,\mathrm {csgn}\left (i \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )}{x^{3}}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )}{x^{3}}\right )^{3}+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}-\pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\pi \mathrm {csgn}\left (i x^{3}\right )^{3}+2 \pi -2 i \ln \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )+6 i \ln \relax (x )-2 i \ln \relax (2)\right )^{2} \ln \relax (2)^{2}}\)	\(411\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^4-4*x)*exp(1-x)-4*x^5+8*x^2)*ln(((x^3-2)*exp(1-x)-2*x^4+4*x)/x^3)+(2*x^5-4*x^2-12*x)*exp(1-x)+4*x^5
+16*x^2)/((x^3-2)*ln(2)^2*exp(1-x)+(-2*x^4+4*x)*ln(2)^2)/ln(((x^3-2)*exp(1-x)-2*x^4+4*x)/x^3)^3,x,method=_RETU
RNVERBOSE)

[Out]

-4*x^2/(-2*Pi*csgn(I/x^3*(x^4-1/2*exp(1-x)*x^3-2*x+exp(1-x)))^2-Pi*csgn(I/x^3)*csgn(I*(x^4-1/2*exp(1-x)*x^3-2*
x+exp(1-x)))*csgn(I/x^3*(x^4-1/2*exp(1-x)*x^3-2*x+exp(1-x)))+Pi*csgn(I/x^3)*csgn(I/x^3*(x^4-1/2*exp(1-x)*x^3-2
*x+exp(1-x)))^2+Pi*csgn(I*(x^4-1/2*exp(1-x)*x^3-2*x+exp(1-x)))*csgn(I/x^3*(x^4-1/2*exp(1-x)*x^3-2*x+exp(1-x)))
^2+Pi*csgn(I/x^3*(x^4-1/2*exp(1-x)*x^3-2*x+exp(1-x)))^3+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^
2+Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)-Pi*csgn(I*x)*csgn(I*x^3)^2+Pi*csgn(I*x^2)^3-Pi*csgn(I*x^2)*csgn(I*x^3)^
2+Pi*csgn(I*x^3)^3+2*Pi-2*I*ln(x^4-1/2*exp(1-x)*x^3-2*x+exp(1-x))+6*I*ln(x)-2*I*ln(2))^2/ln(2)^2

________________________________________________________________________________________

maxima [B]  time = 1.08, size = 124, normalized size = 3.18 \begin {gather*} \frac {x^{2}}{x^{2} \log \relax (2)^{2} + \log \relax (2)^{2} \log \left (x^{3} - 2\right )^{2} + \log \relax (2)^{2} \log \left (-2 \, x e^{x} + e\right )^{2} + 6 \, x \log \relax (2)^{2} \log \relax (x) + 9 \, \log \relax (2)^{2} \log \relax (x)^{2} - 2 \, {\left (x \log \relax (2)^{2} + 3 \, \log \relax (2)^{2} \log \relax (x)\right )} \log \left (x^{3} - 2\right ) - 2 \, {\left (x \log \relax (2)^{2} - \log \relax (2)^{2} \log \left (x^{3} - 2\right ) + 3 \, \log \relax (2)^{2} \log \relax (x)\right )} \log \left (-2 \, x e^{x} + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^4-4*x)*exp(-x+1)-4*x^5+8*x^2)*log(((x^3-2)*exp(-x+1)-2*x^4+4*x)/x^3)+(2*x^5-4*x^2-12*x)*exp(-
x+1)+4*x^5+16*x^2)/((x^3-2)*log(2)^2*exp(-x+1)+(-2*x^4+4*x)*log(2)^2)/log(((x^3-2)*exp(-x+1)-2*x^4+4*x)/x^3)^3
,x, algorithm="maxima")

[Out]

x^2/(x^2*log(2)^2 + log(2)^2*log(x^3 - 2)^2 + log(2)^2*log(-2*x*e^x + e)^2 + 6*x*log(2)^2*log(x) + 9*log(2)^2*
log(x)^2 - 2*(x*log(2)^2 + 3*log(2)^2*log(x))*log(x^3 - 2) - 2*(x*log(2)^2 - log(2)^2*log(x^3 - 2) + 3*log(2)^
2*log(x))*log(-2*x*e^x + e))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {{\mathrm {e}}^{1-x}\,\left (-2\,x^5+4\,x^2+12\,x\right )-16\,x^2-4\,x^5+\ln \left (\frac {4\,x+{\mathrm {e}}^{1-x}\,\left (x^3-2\right )-2\,x^4}{x^3}\right )\,\left ({\mathrm {e}}^{1-x}\,\left (4\,x-2\,x^4\right )-8\,x^2+4\,x^5\right )}{{\ln \left (\frac {4\,x+{\mathrm {e}}^{1-x}\,\left (x^3-2\right )-2\,x^4}{x^3}\right )}^3\,\left ({\ln \relax (2)}^2\,\left (4\,x-2\,x^4\right )+{\mathrm {e}}^{1-x}\,{\ln \relax (2)}^2\,\left (x^3-2\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(1 - x)*(12*x + 4*x^2 - 2*x^5) - 16*x^2 - 4*x^5 + log((4*x + exp(1 - x)*(x^3 - 2) - 2*x^4)/x^3)*(exp(
1 - x)*(4*x - 2*x^4) - 8*x^2 + 4*x^5))/(log((4*x + exp(1 - x)*(x^3 - 2) - 2*x^4)/x^3)^3*(log(2)^2*(4*x - 2*x^4
) + exp(1 - x)*log(2)^2*(x^3 - 2))),x)

[Out]

-int((exp(1 - x)*(12*x + 4*x^2 - 2*x^5) - 16*x^2 - 4*x^5 + log((4*x + exp(1 - x)*(x^3 - 2) - 2*x^4)/x^3)*(exp(
1 - x)*(4*x - 2*x^4) - 8*x^2 + 4*x^5))/(log((4*x + exp(1 - x)*(x^3 - 2) - 2*x^4)/x^3)^3*(log(2)^2*(4*x - 2*x^4
) + exp(1 - x)*log(2)^2*(x^3 - 2))), x)

________________________________________________________________________________________

sympy [A]  time = 0.38, size = 32, normalized size = 0.82 \begin {gather*} \frac {x^{2}}{\log {\relax (2 )}^{2} \log {\left (\frac {- 2 x^{4} + 4 x + \left (x^{3} - 2\right ) e^{1 - x}}{x^{3}} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**4-4*x)*exp(-x+1)-4*x**5+8*x**2)*ln(((x**3-2)*exp(-x+1)-2*x**4+4*x)/x**3)+(2*x**5-4*x**2-12*x
)*exp(-x+1)+4*x**5+16*x**2)/((x**3-2)*ln(2)**2*exp(-x+1)+(-2*x**4+4*x)*ln(2)**2)/ln(((x**3-2)*exp(-x+1)-2*x**4
+4*x)/x**3)**3,x)

[Out]

x**2/(log(2)**2*log((-2*x**4 + 4*x + (x**3 - 2)*exp(1 - x))/x**3)**2)

________________________________________________________________________________________