3.97.17 \(\int \frac {(10+e^x (2-2 x)+e^{x^2} (2-4 x^2)+8 \log (2)) \log (\frac {4 x}{5+e^x+e^{x^2}-3 x+4 \log (2)})}{5 x+e^x x+e^{x^2} x-3 x^2+4 x \log (2)} \, dx\)

Optimal. Leaf size=28 \[ \log ^2\left (\frac {x}{-x+\frac {1}{4} \left (5+e^x+e^{x^2}+x\right )+\log (2)}\right ) \]

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Rubi [A]  time = 2.83, antiderivative size = 23, normalized size of antiderivative = 0.82, number of steps used = 2, number of rules used = 6, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {6, 6741, 12, 6742, 14, 6686} \begin {gather*} \log ^2\left (\frac {4 x}{e^{x^2}-3 x+e^x+5+\log (16)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((10 + E^x*(2 - 2*x) + E^x^2*(2 - 4*x^2) + 8*Log[2])*Log[(4*x)/(5 + E^x + E^x^2 - 3*x + 4*Log[2])])/(5*x +
E^x*x + E^x^2*x - 3*x^2 + 4*x*Log[2]),x]

[Out]

Log[(4*x)/(5 + E^x + E^x^2 - 3*x + Log[16])]^2

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6741

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

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 {\left (10+e^x (2-2 x)+e^{x^2} \left (2-4 x^2\right )+8 \log (2)\right ) \log \left (\frac {4 x}{5+e^x+e^{x^2}-3 x+4 \log (2)}\right )}{e^x x+e^{x^2} x-3 x^2+x (5+4 \log (2))} \, dx\\ &=\log ^2\left (\frac {4 x}{5+e^x+e^{x^2}-3 x+\log (16)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 30, normalized size = 1.07 \begin {gather*} \log ^2\left (\frac {4 x}{e^x+e^{x^2}-3 x+5 \left (1+\frac {4 \log (2)}{5}\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((10 + E^x*(2 - 2*x) + E^x^2*(2 - 4*x^2) + 8*Log[2])*Log[(4*x)/(5 + E^x + E^x^2 - 3*x + 4*Log[2])])/
(5*x + E^x*x + E^x^2*x - 3*x^2 + 4*x*Log[2]),x]

[Out]

Log[(4*x)/(E^x + E^x^2 - 3*x + 5*(1 + (4*Log[2])/5))]^2

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fricas [A]  time = 0.48, size = 27, normalized size = 0.96 \begin {gather*} \log \left (-\frac {4 \, x}{3 \, x - e^{\left (x^{2}\right )} - e^{x} - 4 \, \log \relax (2) - 5}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2+2)*exp(x^2)+(-2*x+2)*exp(x)+8*log(2)+10)*log(4*x/(exp(x^2)+exp(x)+4*log(2)-3*x+5))/(exp(x^2
)*x+exp(x)*x+4*x*log(2)-3*x^2+5*x),x, algorithm="fricas")

[Out]

log(-4*x/(3*x - e^(x^2) - e^x - 4*log(2) - 5))^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left ({\left (2 \, x^{2} - 1\right )} e^{\left (x^{2}\right )} + {\left (x - 1\right )} e^{x} - 4 \, \log \relax (2) - 5\right )} \log \left (-\frac {4 \, x}{3 \, x - e^{\left (x^{2}\right )} - e^{x} - 4 \, \log \relax (2) - 5}\right )}{3 \, x^{2} - x e^{\left (x^{2}\right )} - x e^{x} - 4 \, x \log \relax (2) - 5 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2+2)*exp(x^2)+(-2*x+2)*exp(x)+8*log(2)+10)*log(4*x/(exp(x^2)+exp(x)+4*log(2)-3*x+5))/(exp(x^2
)*x+exp(x)*x+4*x*log(2)-3*x^2+5*x),x, algorithm="giac")

[Out]

integrate(2*((2*x^2 - 1)*e^(x^2) + (x - 1)*e^x - 4*log(2) - 5)*log(-4*x/(3*x - e^(x^2) - e^x - 4*log(2) - 5))/
(3*x^2 - x*e^(x^2) - x*e^x - 4*x*log(2) - 5*x), x)

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maple [C]  time = 0.22, size = 468, normalized size = 16.71

method	result	size
risch	\(\ln \left (\ln \relax (2)-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}\right )^{2}-2 \ln \relax (x ) \ln \left (\ln \relax (2)-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}\right )-i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (2)-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (2)-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )+i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (2)-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )^{2}+i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i}{\ln \relax (2)-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (2)-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )^{2}-i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (2)-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )^{3}+i \pi \ln \left ({\mathrm e}^{x^{2}}+{\mathrm e}^{x}+4 \ln \relax (2)-3 x +5\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (2)-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (2)-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )-i \pi \ln \left ({\mathrm e}^{x^{2}}+{\mathrm e}^{x}+4 \ln \relax (2)-3 x +5\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (2)-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )^{2}-i \pi \ln \left ({\mathrm e}^{x^{2}}+{\mathrm e}^{x}+4 \ln \relax (2)-3 x +5\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (2)-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (2)-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )^{2}+i \pi \ln \left ({\mathrm e}^{x^{2}}+{\mathrm e}^{x}+4 \ln \relax (2)-3 x +5\right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (2)-\frac {3 x}{4}+\frac {5}{4}+\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{x^{2}}}{4}}\right )^{3}+\ln \relax (x )^{2}\)	\(468\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^2+2)*exp(x^2)+(-2*x+2)*exp(x)+8*ln(2)+10)*ln(4*x/(exp(x^2)+exp(x)+4*ln(2)-3*x+5))/(exp(x^2)*x+exp(x
)*x+4*x*ln(2)-3*x^2+5*x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(2)-3/4*x+5/4+1/4*exp(x)+1/4*exp(x^2))^2-2*ln(x)*ln(ln(2)-3/4*x+5/4+1/4*exp(x)+1/4*exp(x^2))-I*Pi*ln(x)*c
sgn(I*x)*csgn(I/(ln(2)-3/4*x+5/4+1/4*exp(x)+1/4*exp(x^2)))*csgn(I*x/(ln(2)-3/4*x+5/4+1/4*exp(x)+1/4*exp(x^2)))
+I*Pi*ln(x)*csgn(I*x)*csgn(I*x/(ln(2)-3/4*x+5/4+1/4*exp(x)+1/4*exp(x^2)))^2+I*Pi*ln(x)*csgn(I/(ln(2)-3/4*x+5/4
+1/4*exp(x)+1/4*exp(x^2)))*csgn(I*x/(ln(2)-3/4*x+5/4+1/4*exp(x)+1/4*exp(x^2)))^2-I*Pi*ln(x)*csgn(I*x/(ln(2)-3/
4*x+5/4+1/4*exp(x)+1/4*exp(x^2)))^3+I*Pi*ln(exp(x^2)+exp(x)+4*ln(2)-3*x+5)*csgn(I*x)*csgn(I/(ln(2)-3/4*x+5/4+1
/4*exp(x)+1/4*exp(x^2)))*csgn(I*x/(ln(2)-3/4*x+5/4+1/4*exp(x)+1/4*exp(x^2)))-I*Pi*ln(exp(x^2)+exp(x)+4*ln(2)-3
*x+5)*csgn(I*x)*csgn(I*x/(ln(2)-3/4*x+5/4+1/4*exp(x)+1/4*exp(x^2)))^2-I*Pi*ln(exp(x^2)+exp(x)+4*ln(2)-3*x+5)*c
sgn(I/(ln(2)-3/4*x+5/4+1/4*exp(x)+1/4*exp(x^2)))*csgn(I*x/(ln(2)-3/4*x+5/4+1/4*exp(x)+1/4*exp(x^2)))^2+I*Pi*ln
(exp(x^2)+exp(x)+4*ln(2)-3*x+5)*csgn(I*x/(ln(2)-3/4*x+5/4+1/4*exp(x)+1/4*exp(x^2)))^3+ln(x)^2

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maxima [B]  time = 0.49, size = 95, normalized size = 3.39 \begin {gather*} -\log \relax (x)^{2} + 2 \, \log \relax (x) \log \left (-3 \, x + e^{\left (x^{2}\right )} + e^{x} + 4 \, \log \relax (2) + 5\right ) - \log \left (-3 \, x + e^{\left (x^{2}\right )} + e^{x} + 4 \, \log \relax (2) + 5\right )^{2} + 2 \, {\left (\log \relax (x) - \log \left (-3 \, x + e^{\left (x^{2}\right )} + e^{x} + 4 \, \log \relax (2) + 5\right )\right )} \log \left (-\frac {4 \, x}{3 \, x - e^{\left (x^{2}\right )} - e^{x} - 4 \, \log \relax (2) - 5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2+2)*exp(x^2)+(-2*x+2)*exp(x)+8*log(2)+10)*log(4*x/(exp(x^2)+exp(x)+4*log(2)-3*x+5))/(exp(x^2
)*x+exp(x)*x+4*x*log(2)-3*x^2+5*x),x, algorithm="maxima")

[Out]

-log(x)^2 + 2*log(x)*log(-3*x + e^(x^2) + e^x + 4*log(2) + 5) - log(-3*x + e^(x^2) + e^x + 4*log(2) + 5)^2 + 2
*(log(x) - log(-3*x + e^(x^2) + e^x + 4*log(2) + 5))*log(-4*x/(3*x - e^(x^2) - e^x - 4*log(2) - 5))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\ln \left (\frac {4\,x}{{\mathrm {e}}^{x^2}-3\,x+4\,\ln \relax (2)+{\mathrm {e}}^x+5}\right )\,\left (8\,\ln \relax (2)-{\mathrm {e}}^{x^2}\,\left (4\,x^2-2\right )-{\mathrm {e}}^x\,\left (2\,x-2\right )+10\right )}{5\,x+x\,{\mathrm {e}}^{x^2}+4\,x\,\ln \relax (2)+x\,{\mathrm {e}}^x-3\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((4*x)/(exp(x^2) - 3*x + 4*log(2) + exp(x) + 5))*(8*log(2) - exp(x^2)*(4*x^2 - 2) - exp(x)*(2*x - 2) +
10))/(5*x + x*exp(x^2) + 4*x*log(2) + x*exp(x) - 3*x^2),x)

[Out]

int((log((4*x)/(exp(x^2) - 3*x + 4*log(2) + exp(x) + 5))*(8*log(2) - exp(x^2)*(4*x^2 - 2) - exp(x)*(2*x - 2) +
10))/(5*x + x*exp(x^2) + 4*x*log(2) + x*exp(x) - 3*x^2), x)

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sympy [A]  time = 4.97, size = 24, normalized size = 0.86 \begin {gather*} \log {\left (\frac {4 x}{- 3 x + e^{x} + e^{x^{2}} + 4 \log {\relax (2 )} + 5} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**2+2)*exp(x**2)+(-2*x+2)*exp(x)+8*ln(2)+10)*ln(4*x/(exp(x**2)+exp(x)+4*ln(2)-3*x+5))/(exp(x**
2)*x+exp(x)*x+4*x*ln(2)-3*x**2+5*x),x)

[Out]

log(4*x/(-3*x + exp(x) + exp(x**2) + 4*log(2) + 5))**2

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