3.89.81 \(\int \frac {-1600 e^{4 x}+(-160 x-200 e^{4 x} x) \log (x)+(-160 x+e^{4 x} (-200 x+400 x^2)) \log ^2(x)}{16+40 e^{4 x}+25 e^{8 x}} \, dx\)

Optimal. Leaf size=24 \[ \frac {4 \left (4-x^2 \log ^2(x)\right )}{\frac {4}{5}+e^{4 x}} \]

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Rubi [F]  time = 1.25, 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 {-1600 e^{4 x}+\left (-160 x-200 e^{4 x} x\right ) \log (x)+\left (-160 x+e^{4 x} \left (-200 x+400 x^2\right )\right ) \log ^2(x)}{16+40 e^{4 x}+25 e^{8 x}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1600*E^(4*x) + (-160*x - 200*E^(4*x)*x)*Log[x] + (-160*x + E^(4*x)*(-200*x + 400*x^2))*Log[x]^2)/(16 + 4
0*E^(4*x) + 25*E^(8*x)),x]

[Out]

80/(4 + 5*E^(4*x)) - 40*Defer[Int][(x*Log[x])/(4 + 5*E^(4*x)), x] - 40*Defer[Int][(x*Log[x]^2)/(4 + 5*E^(4*x))
, x] - 320*Defer[Int][(x^2*Log[x]^2)/(4 + 5*E^(4*x))^2, x] + 80*Defer[Int][(x^2*Log[x]^2)/(4 + 5*E^(4*x)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1600 e^{4 x}+\left (-160 x-200 e^{4 x} x\right ) \log (x)+\left (-160 x+e^{4 x} \left (-200 x+400 x^2\right )\right ) \log ^2(x)}{\left (4+5 e^{4 x}\right )^2} \, dx\\ &=\int \left (-\frac {320 \left (-4+x^2 \log ^2(x)\right )}{\left (4+5 e^{4 x}\right )^2}+\frac {40 \left (-8-x \log (x)-x \log ^2(x)+2 x^2 \log ^2(x)\right )}{4+5 e^{4 x}}\right ) \, dx\\ &=40 \int \frac {-8-x \log (x)-x \log ^2(x)+2 x^2 \log ^2(x)}{4+5 e^{4 x}} \, dx-320 \int \frac {-4+x^2 \log ^2(x)}{\left (4+5 e^{4 x}\right )^2} \, dx\\ &=40 \int \left (-\frac {8}{4+5 e^{4 x}}-\frac {x \log (x)}{4+5 e^{4 x}}-\frac {x \log ^2(x)}{4+5 e^{4 x}}+\frac {2 x^2 \log ^2(x)}{4+5 e^{4 x}}\right ) \, dx-320 \int \left (-\frac {4}{\left (4+5 e^{4 x}\right )^2}+\frac {x^2 \log ^2(x)}{\left (4+5 e^{4 x}\right )^2}\right ) \, dx\\ &=-\left (40 \int \frac {x \log (x)}{4+5 e^{4 x}} \, dx\right )-40 \int \frac {x \log ^2(x)}{4+5 e^{4 x}} \, dx+80 \int \frac {x^2 \log ^2(x)}{4+5 e^{4 x}} \, dx-320 \int \frac {1}{4+5 e^{4 x}} \, dx-320 \int \frac {x^2 \log ^2(x)}{\left (4+5 e^{4 x}\right )^2} \, dx+1280 \int \frac {1}{\left (4+5 e^{4 x}\right )^2} \, dx\\ &=-\left (40 \int \frac {x \log (x)}{4+5 e^{4 x}} \, dx\right )-40 \int \frac {x \log ^2(x)}{4+5 e^{4 x}} \, dx+80 \int \frac {x^2 \log ^2(x)}{4+5 e^{4 x}} \, dx-80 \operatorname {Subst}\left (\int \frac {1}{x (4+5 x)} \, dx,x,e^{4 x}\right )-320 \int \frac {x^2 \log ^2(x)}{\left (4+5 e^{4 x}\right )^2} \, dx+320 \operatorname {Subst}\left (\int \frac {1}{x (4+5 x)^2} \, dx,x,e^{4 x}\right )\\ &=-\left (20 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{4 x}\right )\right )-40 \int \frac {x \log (x)}{4+5 e^{4 x}} \, dx-40 \int \frac {x \log ^2(x)}{4+5 e^{4 x}} \, dx+80 \int \frac {x^2 \log ^2(x)}{4+5 e^{4 x}} \, dx+100 \operatorname {Subst}\left (\int \frac {1}{4+5 x} \, dx,x,e^{4 x}\right )-320 \int \frac {x^2 \log ^2(x)}{\left (4+5 e^{4 x}\right )^2} \, dx+320 \operatorname {Subst}\left (\int \left (\frac {1}{16 x}-\frac {5}{4 (4+5 x)^2}-\frac {5}{16 (4+5 x)}\right ) \, dx,x,e^{4 x}\right )\\ &=\frac {80}{4+5 e^{4 x}}-40 \int \frac {x \log (x)}{4+5 e^{4 x}} \, dx-40 \int \frac {x \log ^2(x)}{4+5 e^{4 x}} \, dx+80 \int \frac {x^2 \log ^2(x)}{4+5 e^{4 x}} \, dx-320 \int \frac {x^2 \log ^2(x)}{\left (4+5 e^{4 x}\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 24, normalized size = 1.00 \begin {gather*} \frac {40 \left (4-x^2 \log ^2(x)\right )}{8+10 e^{4 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1600*E^(4*x) + (-160*x - 200*E^(4*x)*x)*Log[x] + (-160*x + E^(4*x)*(-200*x + 400*x^2))*Log[x]^2)/(
16 + 40*E^(4*x) + 25*E^(8*x)),x]

[Out]

(40*(4 - x^2*Log[x]^2))/(8 + 10*E^(4*x))

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fricas [A]  time = 0.60, size = 22, normalized size = 0.92 \begin {gather*} -\frac {20 \, {\left (x^{2} \log \relax (x)^{2} - 4\right )}}{5 \, e^{\left (4 \, x\right )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((400*x^2-200*x)*exp(2*x)^2-160*x)*log(x)^2+(-200*x*exp(2*x)^2-160*x)*log(x)-1600*exp(2*x)^2)/(25*e
xp(2*x)^4+40*exp(2*x)^2+16),x, algorithm="fricas")

[Out]

-20*(x^2*log(x)^2 - 4)/(5*e^(4*x) + 4)

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giac [A]  time = 0.93, size = 22, normalized size = 0.92 \begin {gather*} -\frac {20 \, {\left (x^{2} \log \relax (x)^{2} - 4\right )}}{5 \, e^{\left (4 \, x\right )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((400*x^2-200*x)*exp(2*x)^2-160*x)*log(x)^2+(-200*x*exp(2*x)^2-160*x)*log(x)-1600*exp(2*x)^2)/(25*e
xp(2*x)^4+40*exp(2*x)^2+16),x, algorithm="giac")

[Out]

-20*(x^2*log(x)^2 - 4)/(5*e^(4*x) + 4)

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maple [A]  time = 0.05, size = 33, normalized size = 1.38

method	result	size
risch	\(-\frac {20 x^{2} \ln \relax (x )^{2}}{5 \,{\mathrm e}^{4 x}+4}+\frac {80}{5 \,{\mathrm e}^{4 x}+4}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((400*x^2-200*x)*exp(2*x)^2-160*x)*ln(x)^2+(-200*x*exp(2*x)^2-160*x)*ln(x)-1600*exp(2*x)^2)/(25*exp(2*x)^
4+40*exp(2*x)^2+16),x,method=_RETURNVERBOSE)

[Out]

-20*x^2/(5*exp(4*x)+4)*ln(x)^2+80/(5*exp(4*x)+4)

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maxima [A]  time = 0.51, size = 32, normalized size = 1.33 \begin {gather*} -\frac {20 \, x^{2} \log \relax (x)^{2}}{5 \, e^{\left (4 \, x\right )} + 4} + \frac {80}{5 \, e^{\left (4 \, x\right )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((400*x^2-200*x)*exp(2*x)^2-160*x)*log(x)^2+(-200*x*exp(2*x)^2-160*x)*log(x)-1600*exp(2*x)^2)/(25*e
xp(2*x)^4+40*exp(2*x)^2+16),x, algorithm="maxima")

[Out]

-20*x^2*log(x)^2/(5*e^(4*x) + 4) + 80/(5*e^(4*x) + 4)

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mupad [B]  time = 5.24, size = 22, normalized size = 0.92 \begin {gather*} -\frac {20\,\left (x^2\,{\ln \relax (x)}^2-4\right )}{5\,{\mathrm {e}}^{4\,x}+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(1600*exp(4*x) + log(x)*(160*x + 200*x*exp(4*x)) + log(x)^2*(160*x + exp(4*x)*(200*x - 400*x^2)))/(40*exp
(4*x) + 25*exp(8*x) + 16),x)

[Out]

-(20*(x^2*log(x)^2 - 4))/(5*exp(4*x) + 4)

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sympy [A]  time = 0.26, size = 19, normalized size = 0.79 \begin {gather*} \frac {- 4 x^{2} \log {\relax (x )}^{2} + 16}{e^{4 x} + \frac {4}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((400*x**2-200*x)*exp(2*x)**2-160*x)*ln(x)**2+(-200*x*exp(2*x)**2-160*x)*ln(x)-1600*exp(2*x)**2)/(2
5*exp(2*x)**4+40*exp(2*x)**2+16),x)

[Out]

(-4*x**2*log(x)**2 + 16)/(exp(4*x) + 4/5)

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