3.52.37 \(\int \frac {(12+2 x+22 x^2-8 x^3) \log (\frac {4 e^x x^2}{1+8 x+16 x^2})+(-x-4 x^2) \log ^2(\frac {4 e^x x^2}{1+8 x+16 x^2})}{x+4 x^2} \, dx\)

Optimal. Leaf size=28 \[ (3-x) \log ^2\left (\frac {e^x}{\left (1+\frac {1}{4} \left (4+\frac {2}{x}\right )\right )^2}\right ) \]

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Rubi [F]  time = 0.80, 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 {\left (12+2 x+22 x^2-8 x^3\right ) \log \left (\frac {4 e^x x^2}{1+8 x+16 x^2}\right )+\left (-x-4 x^2\right ) \log ^2\left (\frac {4 e^x x^2}{1+8 x+16 x^2}\right )}{x+4 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((12 + 2*x + 22*x^2 - 8*x^3)*Log[(4*E^x*x^2)/(1 + 8*x + 16*x^2)] + (-x - 4*x^2)*Log[(4*E^x*x^2)/(1 + 8*x +
16*x^2)]^2)/(x + 4*x^2),x]

[Out]

x/2 - 3*x^2 + x^3/3 + 6*x*Log[(4*E^x*x^2)/(1 + 4*x)^2] - x^2*Log[(4*E^x*x^2)/(1 + 4*x)^2] - (25*Log[1 + 4*x])/
8 + 12*Defer[Int][Log[(4*E^x*x^2)/(1 + 4*x)^2]/x, x] - 52*Defer[Int][Log[(4*E^x*x^2)/(1 + 4*x)^2]/(1 + 4*x), x
] - Defer[Int][Log[(4*E^x*x^2)/(1 + 4*x)^2]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (12+2 x+22 x^2-8 x^3\right ) \log \left (\frac {4 e^x x^2}{1+8 x+16 x^2}\right )+\left (-x-4 x^2\right ) \log ^2\left (\frac {4 e^x x^2}{1+8 x+16 x^2}\right )}{x (1+4 x)} \, dx\\ &=\int \left (\frac {12+2 x+22 x^2-8 x^3}{x+4 x^2}-\log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )\right ) \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right ) \, dx\\ &=\int \left (-\frac {2 (-3+x) \left (2+x+4 x^2\right ) \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )}{x (1+4 x)}-\log ^2\left (\frac {4 e^x x^2}{(1+4 x)^2}\right )\right ) \, dx\\ &=-\left (2 \int \frac {(-3+x) \left (2+x+4 x^2\right ) \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )}{x (1+4 x)} \, dx\right )-\int \log ^2\left (\frac {4 e^x x^2}{(1+4 x)^2}\right ) \, dx\\ &=-\left (2 \int \left (-3 \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )-\frac {6 \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )}{x}+x \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )+\frac {26 \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )}{1+4 x}\right ) \, dx\right )-\int \log ^2\left (\frac {4 e^x x^2}{(1+4 x)^2}\right ) \, dx\\ &=-\left (2 \int x \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right ) \, dx\right )+6 \int \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right ) \, dx+12 \int \frac {\log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )}{x} \, dx-52 \int \frac {\log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )}{1+4 x} \, dx-\int \log ^2\left (\frac {4 e^x x^2}{(1+4 x)^2}\right ) \, dx\\ &=6 x \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )-x^2 \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )-6 \int \frac {2+x+4 x^2}{1+4 x} \, dx+12 \int \frac {\log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )}{x} \, dx-52 \int \frac {\log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )}{1+4 x} \, dx+\int \frac {x \left (2+x+4 x^2\right )}{1+4 x} \, dx-\int \log ^2\left (\frac {4 e^x x^2}{(1+4 x)^2}\right ) \, dx\\ &=6 x \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )-x^2 \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )-6 \int \left (x+\frac {2}{1+4 x}\right ) \, dx+12 \int \frac {\log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )}{x} \, dx-52 \int \frac {\log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )}{1+4 x} \, dx+\int \left (\frac {1}{2}+x^2-\frac {1}{2 (1+4 x)}\right ) \, dx-\int \log ^2\left (\frac {4 e^x x^2}{(1+4 x)^2}\right ) \, dx\\ &=\frac {x}{2}-3 x^2+\frac {x^3}{3}+6 x \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )-x^2 \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )-\frac {25}{8} \log (1+4 x)+12 \int \frac {\log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )}{x} \, dx-52 \int \frac {\log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )}{1+4 x} \, dx-\int \log ^2\left (\frac {4 e^x x^2}{(1+4 x)^2}\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.17, size = 151, normalized size = 5.39 \begin {gather*} -3 x^2+3 \log ^2\left (\frac {x^2}{(1+4 x)^2}\right )-12 \log (x) \left (x+\log \left (\frac {x^2}{(1+4 x)^2}\right )-\log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )\right )+6 x \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right )-x \log ^2\left (\frac {4 e^x x^2}{(1+4 x)^2}\right )+12 x \log (1+4 x)+12 \log \left (\frac {x^2}{(1+4 x)^2}\right ) \log (1+4 x)-12 \log \left (\frac {4 e^x x^2}{(1+4 x)^2}\right ) \log (1+4 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((12 + 2*x + 22*x^2 - 8*x^3)*Log[(4*E^x*x^2)/(1 + 8*x + 16*x^2)] + (-x - 4*x^2)*Log[(4*E^x*x^2)/(1 +
8*x + 16*x^2)]^2)/(x + 4*x^2),x]

[Out]

-3*x^2 + 3*Log[x^2/(1 + 4*x)^2]^2 - 12*Log[x]*(x + Log[x^2/(1 + 4*x)^2] - Log[(4*E^x*x^2)/(1 + 4*x)^2]) + 6*x*
Log[(4*E^x*x^2)/(1 + 4*x)^2] - x*Log[(4*E^x*x^2)/(1 + 4*x)^2]^2 + 12*x*Log[1 + 4*x] + 12*Log[x^2/(1 + 4*x)^2]*
Log[1 + 4*x] - 12*Log[(4*E^x*x^2)/(1 + 4*x)^2]*Log[1 + 4*x]

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fricas [A]  time = 0.73, size = 27, normalized size = 0.96 \begin {gather*} -{\left (x - 3\right )} \log \left (\frac {4 \, x^{2} e^{x}}{16 \, x^{2} + 8 \, x + 1}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2-x)*log(4*x^2*exp(x)/(16*x^2+8*x+1))^2+(-8*x^3+22*x^2+2*x+12)*log(4*x^2*exp(x)/(16*x^2+8*x+1
)))/(4*x^2+x),x, algorithm="fricas")

[Out]

-(x - 3)*log(4*x^2*e^x/(16*x^2 + 8*x + 1))^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (4 \, x^{2} + x\right )} \log \left (\frac {4 \, x^{2} e^{x}}{16 \, x^{2} + 8 \, x + 1}\right )^{2} + 2 \, {\left (4 \, x^{3} - 11 \, x^{2} - x - 6\right )} \log \left (\frac {4 \, x^{2} e^{x}}{16 \, x^{2} + 8 \, x + 1}\right )}{4 \, x^{2} + x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2-x)*log(4*x^2*exp(x)/(16*x^2+8*x+1))^2+(-8*x^3+22*x^2+2*x+12)*log(4*x^2*exp(x)/(16*x^2+8*x+1
)))/(4*x^2+x),x, algorithm="giac")

[Out]

integrate(-((4*x^2 + x)*log(4*x^2*e^x/(16*x^2 + 8*x + 1))^2 + 2*(4*x^3 - 11*x^2 - x - 6)*log(4*x^2*e^x/(16*x^2
+ 8*x + 1)))/(4*x^2 + x), x)

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maple [C]  time = 2.12, size = 6478, normalized size = 231.36

method	result	size
risch	\(\text {Expression too large to display}\)	\(6478\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^2-x)*ln(4*x^2*exp(x)/(16*x^2+8*x+1))^2+(-8*x^3+22*x^2+2*x+12)*ln(4*x^2*exp(x)/(16*x^2+8*x+1)))/(4*x
^2+x),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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maxima [B]  time = 0.54, size = 162, normalized size = 5.79 \begin {gather*} -x^{3} - x^{2} {\left (4 \, \log \relax (2) - 3\right )} - 4 \, x \log \left (4 \, x + 1\right )^{2} - 4 \, x \log \relax (x)^{2} - 4 \, {\left (\log \relax (2)^{2} - 3 \, \log \relax (2)\right )} x - 12 \, {\left (\log \left (4 \, x + 1\right ) - \log \relax (x)\right )} \log \left (\frac {4 \, x^{2} e^{x}}{16 \, x^{2} + 8 \, x + 1}\right ) + {\left (4 \, x^{2} + 4 \, x {\left (2 \, \log \relax (2) - 3\right )} + 8 \, x \log \relax (x) - 3\right )} \log \left (4 \, x + 1\right ) + 3 \, {\left (4 \, x + 8 \, \log \relax (x) + 1\right )} \log \left (4 \, x + 1\right ) - 12 \, \log \left (4 \, x + 1\right )^{2} - 4 \, {\left (x^{2} + x {\left (2 \, \log \relax (2) - 3\right )}\right )} \log \relax (x) - 12 \, x \log \relax (x) - 12 \, \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2-x)*log(4*x^2*exp(x)/(16*x^2+8*x+1))^2+(-8*x^3+22*x^2+2*x+12)*log(4*x^2*exp(x)/(16*x^2+8*x+1
)))/(4*x^2+x),x, algorithm="maxima")

[Out]

-x^3 - x^2*(4*log(2) - 3) - 4*x*log(4*x + 1)^2 - 4*x*log(x)^2 - 4*(log(2)^2 - 3*log(2))*x - 12*(log(4*x + 1) -
log(x))*log(4*x^2*e^x/(16*x^2 + 8*x + 1)) + (4*x^2 + 4*x*(2*log(2) - 3) + 8*x*log(x) - 3)*log(4*x + 1) + 3*(4
*x + 8*log(x) + 1)*log(4*x + 1) - 12*log(4*x + 1)^2 - 4*(x^2 + x*(2*log(2) - 3))*log(x) - 12*x*log(x) - 12*log
(x)^2

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mupad [B]  time = 3.60, size = 27, normalized size = 0.96 \begin {gather*} -{\ln \left (\frac {4\,x^2\,{\mathrm {e}}^x}{16\,x^2+8\,x+1}\right )}^2\,\left (x-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((4*x^2*exp(x))/(8*x + 16*x^2 + 1))^2*(x + 4*x^2) - log((4*x^2*exp(x))/(8*x + 16*x^2 + 1))*(2*x + 22*
x^2 - 8*x^3 + 12))/(x + 4*x^2),x)

[Out]

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

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sympy [A]  time = 0.39, size = 24, normalized size = 0.86 \begin {gather*} \left (3 - x\right ) \log {\left (\frac {4 x^{2} e^{x}}{16 x^{2} + 8 x + 1} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**2-x)*ln(4*x**2*exp(x)/(16*x**2+8*x+1))**2+(-8*x**3+22*x**2+2*x+12)*ln(4*x**2*exp(x)/(16*x**2
+8*x+1)))/(4*x**2+x),x)

[Out]

(3 - x)*log(4*x**2*exp(x)/(16*x**2 + 8*x + 1))**2

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