3.49.100 \(\int \frac {1-x+e x+2 x^2-3 x^3-x \log (5 x+x \log ^2(5))}{x} \, dx\)

Optimal. Leaf size=22 \[ (1-x) \left (-e+x^2+\log \left (x \left (5+\log ^2(5)\right )\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.45, number of steps used = 6, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6, 14, 2295} \begin {gather*} -x^3+x^2-(1-e) x+x-x \log \left (x \left (5+\log ^2(5)\right )\right )+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x - (1 - E)*x + x^2 - x^3 + Log[x] - x*Log[x*(5 + Log[5]^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 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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1+(-1+e) x+2 x^2-3 x^3-x \log \left (5 x+x \log ^2(5)\right )}{x} \, dx\\ &=\int \left (\frac {1-(1-e) x+2 x^2-3 x^3}{x}-\log \left (x \left (5+\log ^2(5)\right )\right )\right ) \, dx\\ &=\int \frac {1-(1-e) x+2 x^2-3 x^3}{x} \, dx-\int \log \left (x \left (5+\log ^2(5)\right )\right ) \, dx\\ &=x-x \log \left (x \left (5+\log ^2(5)\right )\right )+\int \left (-1+e+\frac {1}{x}+2 x-3 x^2\right ) \, dx\\ &=x-(1-e) x+x^2-x^3+\log (x)-x \log \left (x \left (5+\log ^2(5)\right )\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 26, normalized size = 1.18 \begin {gather*} e x+x^2-x^3+\log (x)-x \log \left (x \left (5+\log ^2(5)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.85, size = 29, normalized size = 1.32 \begin {gather*} -x^{3} + x^{2} + x e - {\left (x - 1\right )} \log \left (x \log \relax (5)^{2} + 5 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^3 + x^2 + x*e - (x - 1)*log(x*log(5)^2 + 5*x)

________________________________________________________________________________________

giac [A]  time = 0.13, size = 29, normalized size = 1.32 \begin {gather*} -x^{3} + x^{2} + x e - x \log \left (x \log \relax (5)^{2} + 5 \, x\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.09, size = 30, normalized size = 1.36

method	result	size
risch	\(-x \ln \left (x \ln \relax (5)^{2}+5 x \right )-x^{3}+x \,{\mathrm e}+x^{2}+\ln \relax (x )\)	\(30\)
norman	\(x^{2}+x \,{\mathrm e}+\ln \left (x \ln \relax (5)^{2}+5 x \right )-x^{3}-x \ln \left (x \ln \relax (5)^{2}+5 x \right )\)	\(39\)
derivativedivides	\(\frac {\ln \relax (5)^{6} \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {\ln \relax (5)^{4} {\mathrm e} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}-\frac {\ln \relax (5)^{4} \left (\left (\ln \relax (5)^{2}+5\right ) x \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )-\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}-\frac {\ln \relax (5)^{4} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}+\frac {15 \ln \relax (5)^{4} \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {10 \ln \relax (5)^{2} {\mathrm e} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}-\frac {10 \ln \relax (5)^{2} \left (\left (\ln \relax (5)^{2}+5\right ) x \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )-\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {\ln \relax (5)^{2} x^{2}}{\ln \relax (5)^{2}+5}-\frac {10 x \ln \relax (5)^{2}}{\left (\ln \relax (5)^{2}+5\right )^{2}}-x^{3}+\frac {75 \ln \relax (5)^{2} \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {25 \,{\mathrm e} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}-\frac {25 \left (\left (\ln \relax (5)^{2}+5\right ) x \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )-\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {5 x^{2}}{\ln \relax (5)^{2}+5}-\frac {25 x}{\left (\ln \relax (5)^{2}+5\right )^{2}}+\frac {125 \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}\)	\(329\)
default	\(\frac {\ln \relax (5)^{6} \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {\ln \relax (5)^{4} {\mathrm e} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}-\frac {\ln \relax (5)^{4} \left (\left (\ln \relax (5)^{2}+5\right ) x \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )-\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}-\frac {\ln \relax (5)^{4} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}+\frac {15 \ln \relax (5)^{4} \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {10 \ln \relax (5)^{2} {\mathrm e} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}-\frac {10 \ln \relax (5)^{2} \left (\left (\ln \relax (5)^{2}+5\right ) x \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )-\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {\ln \relax (5)^{2} x^{2}}{\ln \relax (5)^{2}+5}-\frac {10 x \ln \relax (5)^{2}}{\left (\ln \relax (5)^{2}+5\right )^{2}}-x^{3}+\frac {75 \ln \relax (5)^{2} \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {25 \,{\mathrm e} x}{\left (\ln \relax (5)^{2}+5\right )^{2}}-\frac {25 \left (\left (\ln \relax (5)^{2}+5\right ) x \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )-\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}+\frac {5 x^{2}}{\ln \relax (5)^{2}+5}-\frac {25 x}{\left (\ln \relax (5)^{2}+5\right )^{2}}+\frac {125 \ln \left (\left (\ln \relax (5)^{2}+5\right ) x \right )}{\left (\ln \relax (5)^{2}+5\right )^{3}}\)	\(329\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*ln(x*ln(5)^2+5*x)+x*exp(1)-3*x^3+2*x^2-x+1)/x,x,method=_RETURNVERBOSE)

[Out]

-x*ln(x*ln(5)^2+5*x)-x^3+x*exp(1)+x^2+ln(x)

________________________________________________________________________________________

maxima [B]  time = 0.37, size = 60, normalized size = 2.73 \begin {gather*} -x^{3} + x^{2} + x e - x + \frac {x \log \relax (5)^{2} - {\left (x \log \relax (5)^{2} + 5 \, x\right )} \log \left (x \log \relax (5)^{2} + 5 \, x\right ) + 5 \, x}{\log \relax (5)^{2} + 5} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.20, size = 29, normalized size = 1.32 \begin {gather*} \ln \relax (x)-x\,\ln \left (5\,x+x\,{\ln \relax (5)}^2\right )+x\,\mathrm {e}+x^2-x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - x*log(5*x + x*log(5)^2) + x*exp(1) + x^2 - x^3

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 27, normalized size = 1.23 \begin {gather*} - x^{3} + x^{2} - x \log {\left (x \log {\relax (5 )}^{2} + 5 x \right )} + e x + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*ln(x*ln(5)**2+5*x)+x*exp(1)-3*x**3+2*x**2-x+1)/x,x)

[Out]

-x**3 + x**2 - x*log(x*log(5)**2 + 5*x) + E*x + log(x)

________________________________________________________________________________________