3.10.32 \(\int \frac {-32 x^2+e^x (18+9 x+(-12-6 x) \log (4)+(2+x) \log ^2(4))+(8 x^2+e^x (-9+6 \log (4)-\log ^2(4))) \log (\frac {1}{2} (-8 x^4+e^x (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4))))}{-8 x^4+e^x (9 x^2-6 x^2 \log (4)+x^2 \log ^2(4))} \, dx\)

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

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Rubi [A]  time = 3.09, antiderivative size = 31, normalized size of antiderivative = 1.11, number of steps used = 18, number of rules used = 5, integrand size = 124, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {6688, 6742, 14, 43, 2551} \begin {gather*} \frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (3-\log (4))^2\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-32*x^2 + E^x*(18 + 9*x + (-12 - 6*x)*Log[4] + (2 + x)*Log[4]^2) + (8*x^2 + E^x*(-9 + 6*Log[4] - Log[4]^2
))*Log[(-8*x^4 + E^x*(9*x^2 - 6*x^2*Log[4] + x^2*Log[4]^2))/2])/(-8*x^4 + E^x*(9*x^2 - 6*x^2*Log[4] + x^2*Log[
4]^2)),x]

[Out]

Log[-1/2*(x^2*(8*x^2 - E^x*(3 - Log[4])^2))]/x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32 x^2-e^x (2+x) (-3+\log (4))^2-\left (8 x^2-e^x (-3+\log (4))^2\right ) \log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )\right )}{x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )} \, dx\\ &=\int \left (\frac {8 (2-x)}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )}+\frac {2+x-\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )\right )}{x^2}\right ) \, dx\\ &=8 \int \frac {2-x}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx+\int \frac {2+x-\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )\right )}{x^2} \, dx\\ &=8 \int \left (\frac {2}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )}+\frac {x}{-8 x^2+9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )}\right ) \, dx+\int \left (\frac {2+x}{x^2}-\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )\right )}{x^2}\right ) \, dx\\ &=8 \int \frac {x}{-8 x^2+9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx+16 \int \frac {1}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx+\int \frac {2+x}{x^2} \, dx-\int \frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )\right )}{x^2} \, dx\\ &=\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (3-\log (4))^2\right )\right )}{x}+8 \int \frac {x}{-8 x^2+e^x (-3+\log (4))^2} \, dx+16 \int \frac {1}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx+\int \left (\frac {2}{x^2}+\frac {1}{x}\right ) \, dx-\int \frac {32 x^2-e^x (2+x) (-3+\log (4))^2}{x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )} \, dx\\ &=-\frac {2}{x}+\log (x)+\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (3-\log (4))^2\right )\right )}{x}+8 \int \frac {x}{-8 x^2+e^x (-3+\log (4))^2} \, dx+16 \int \frac {1}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx-\int \left (\frac {2+x}{x^2}+\frac {8 (2-x)}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )}\right ) \, dx\\ &=-\frac {2}{x}+\log (x)+\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (3-\log (4))^2\right )\right )}{x}+8 \int \frac {x}{-8 x^2+e^x (-3+\log (4))^2} \, dx-8 \int \frac {2-x}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx+16 \int \frac {1}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx-\int \frac {2+x}{x^2} \, dx\\ &=-\frac {2}{x}+\log (x)+\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (3-\log (4))^2\right )\right )}{x}+8 \int \frac {x}{-8 x^2+e^x (-3+\log (4))^2} \, dx-8 \int \left (\frac {2}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )}+\frac {x}{-8 x^2+9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )}\right ) \, dx+16 \int \frac {1}{8 x^2-9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx-\int \left (\frac {2}{x^2}+\frac {1}{x}\right ) \, dx\\ &=\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (3-\log (4))^2\right )\right )}{x}+8 \int \frac {x}{-8 x^2+e^x (-3+\log (4))^2} \, dx-8 \int \frac {x}{-8 x^2+9 e^x \left (1+\frac {2}{9} (-3+\log (2)) \log (4)\right )} \, dx\\ &=\frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (3-\log (4))^2\right )\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 29, normalized size = 1.04 \begin {gather*} \frac {\log \left (-\frac {1}{2} x^2 \left (8 x^2-e^x (-3+\log (4))^2\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32*x^2 + E^x*(18 + 9*x + (-12 - 6*x)*Log[4] + (2 + x)*Log[4]^2) + (8*x^2 + E^x*(-9 + 6*Log[4] - Lo
g[4]^2))*Log[(-8*x^4 + E^x*(9*x^2 - 6*x^2*Log[4] + x^2*Log[4]^2))/2])/(-8*x^4 + E^x*(9*x^2 - 6*x^2*Log[4] + x^
2*Log[4]^2)),x]

[Out]

Log[-1/2*(x^2*(8*x^2 - E^x*(-3 + Log[4])^2))]/x

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fricas [A]  time = 0.69, size = 37, normalized size = 1.32 \begin {gather*} \frac {\log \left (-4 \, x^{4} + \frac {1}{2} \, {\left (4 \, x^{2} \log \relax (2)^{2} - 12 \, x^{2} \log \relax (2) + 9 \, x^{2}\right )} e^{x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*log(2)^2+12*log(2)-9)*exp(x)+8*x^2)*log(1/2*(4*x^2*log(2)^2-12*x^2*log(2)+9*x^2)*exp(x)-4*x^4)
+(4*(2+x)*log(2)^2+2*(-6*x-12)*log(2)+9*x+18)*exp(x)-32*x^2)/((4*x^2*log(2)^2-12*x^2*log(2)+9*x^2)*exp(x)-8*x^
4),x, algorithm="fricas")

[Out]

log(-4*x^4 + 1/2*(4*x^2*log(2)^2 - 12*x^2*log(2) + 9*x^2)*e^x)/x

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giac [A]  time = 0.65, size = 44, normalized size = 1.57 \begin {gather*} -\frac {\log \relax (2) - \log \left (4 \, x^{2} e^{x} \log \relax (2)^{2} - 8 \, x^{4} - 12 \, x^{2} e^{x} \log \relax (2) + 9 \, x^{2} e^{x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*log(2)^2+12*log(2)-9)*exp(x)+8*x^2)*log(1/2*(4*x^2*log(2)^2-12*x^2*log(2)+9*x^2)*exp(x)-4*x^4)
+(4*(2+x)*log(2)^2+2*(-6*x-12)*log(2)+9*x+18)*exp(x)-32*x^2)/((4*x^2*log(2)^2-12*x^2*log(2)+9*x^2)*exp(x)-8*x^
4),x, algorithm="giac")

[Out]

-(log(2) - log(4*x^2*e^x*log(2)^2 - 8*x^4 - 12*x^2*e^x*log(2) + 9*x^2*e^x))/x

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maple [A]  time = 0.27, size = 38, normalized size = 1.36

method	result	size
norman	\(\frac {\ln \left (\frac {\left (4 x^{2} \ln \relax (2)^{2}-12 x^{2} \ln \relax (2)+9 x^{2}\right ) {\mathrm e}^{x}}{2}-4 x^{4}\right )}{x}\)	\(38\)
risch	\(\frac {\ln \left (\ln \relax (2)^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \relax (2)-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )}{x}-\frac {i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i \left (\ln \relax (2)^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \relax (2)-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right ) \mathrm {csgn}\left (i x^{2} \left (\ln \relax (2)^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \relax (2)-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right )-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} \left (\ln \relax (2)^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \relax (2)-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (\ln \relax (2)^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \relax (2)-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right ) \mathrm {csgn}\left (i x^{2} \left (\ln \relax (2)^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \relax (2)-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right )^{2}+i \pi \mathrm {csgn}\left (i x^{2} \left (\ln \relax (2)^{2} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} \ln \relax (2)-2 x^{2}+\frac {9 \,{\mathrm e}^{x}}{4}\right )\right )^{3}+2 \ln \relax (2)-4 \ln \relax (x )}{2 x}\)	\(303\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*ln(2)^2+12*ln(2)-9)*exp(x)+8*x^2)*ln(1/2*(4*x^2*ln(2)^2-12*x^2*ln(2)+9*x^2)*exp(x)-4*x^4)+(4*(2+x)*l
n(2)^2+2*(-6*x-12)*ln(2)+9*x+18)*exp(x)-32*x^2)/((4*x^2*ln(2)^2-12*x^2*ln(2)+9*x^2)*exp(x)-8*x^4),x,method=_RE
TURNVERBOSE)

[Out]

ln(1/2*(4*x^2*ln(2)^2-12*x^2*ln(2)+9*x^2)*exp(x)-4*x^4)/x

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maxima [A]  time = 0.61, size = 36, normalized size = 1.29 \begin {gather*} -\frac {\log \relax (2) - \log \left (-8 \, x^{2} + {\left (4 \, \log \relax (2)^{2} - 12 \, \log \relax (2) + 9\right )} e^{x}\right ) - 2 \, \log \relax (x)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*log(2)^2+12*log(2)-9)*exp(x)+8*x^2)*log(1/2*(4*x^2*log(2)^2-12*x^2*log(2)+9*x^2)*exp(x)-4*x^4)
+(4*(2+x)*log(2)^2+2*(-6*x-12)*log(2)+9*x+18)*exp(x)-32*x^2)/((4*x^2*log(2)^2-12*x^2*log(2)+9*x^2)*exp(x)-8*x^
4),x, algorithm="maxima")

[Out]

-(log(2) - log(-8*x^2 + (4*log(2)^2 - 12*log(2) + 9)*e^x) - 2*log(x))/x

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mupad [B]  time = 1.67, size = 37, normalized size = 1.32 \begin {gather*} \frac {\ln \left (\frac {{\mathrm {e}}^x\,\left (4\,x^2\,{\ln \relax (2)}^2-12\,x^2\,\ln \relax (2)+9\,x^2\right )}{2}-4\,x^4\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((exp(x)*(4*x^2*log(2)^2 - 12*x^2*log(2) + 9*x^2))/2 - 4*x^4)*(exp(x)*(4*log(2)^2 - 12*log(2) + 9) -
8*x^2) - exp(x)*(9*x - 2*log(2)*(6*x + 12) + 4*log(2)^2*(x + 2) + 18) + 32*x^2)/(exp(x)*(4*x^2*log(2)^2 - 12*x
^2*log(2) + 9*x^2) - 8*x^4),x)

[Out]

log((exp(x)*(4*x^2*log(2)^2 - 12*x^2*log(2) + 9*x^2))/2 - 4*x^4)/x

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sympy [A]  time = 0.66, size = 36, normalized size = 1.29 \begin {gather*} \frac {\log {\left (- 4 x^{4} + \left (- 6 x^{2} \log {\relax (2 )} + 2 x^{2} \log {\relax (2 )}^{2} + \frac {9 x^{2}}{2}\right ) e^{x} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*ln(2)**2+12*ln(2)-9)*exp(x)+8*x**2)*ln(1/2*(4*x**2*ln(2)**2-12*x**2*ln(2)+9*x**2)*exp(x)-4*x**
4)+(4*(2+x)*ln(2)**2+2*(-6*x-12)*ln(2)+9*x+18)*exp(x)-32*x**2)/((4*x**2*ln(2)**2-12*x**2*ln(2)+9*x**2)*exp(x)-
8*x**4),x)

[Out]

log(-4*x**4 + (-6*x**2*log(2) + 2*x**2*log(2)**2 + 9*x**2/2)*exp(x))/x

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