\(\int \frac {4+9 x+(1+3 x) \log (x)+(3+\log (x)) \log (3+\log (x))}{3+\log (x)} \, dx\) [2502]

Optimal result

Integrand size = 30, antiderivative size = 14 \[ \int \frac {4+9 x+(1+3 x) \log (x)+(3+\log (x)) \log (3+\log (x))}{3+\log (x)} \, dx=x \left (1+\frac {3 x}{2}+\log (3+\log (x))\right ) \]

[Out]

x*(1+3/2*x+ln(3+ln(x)))

Rubi [F]

\[ \int \frac {4+9 x+(1+3 x) \log (x)+(3+\log (x)) \log (3+\log (x))}{3+\log (x)} \, dx=\int \frac {4+9 x+(1+3 x) \log (x)+(3+\log (x)) \log (3+\log (x))}{3+\log (x)} \, dx \]

[In]

Int[(4 + 9*x + (1 + 3*x)*Log[x] + (3 + Log[x])*Log[3 + Log[x]])/(3 + Log[x]),x]

[Out]

x + (3*x^2)/2 + ExpIntegralEi[3 + Log[x]]/E^3 + Defer[Int][Log[3 + Log[x]], x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4+9 x+\log (x)+3 x \log (x)}{3+\log (x)}+\log (3+\log (x))\right ) \, dx \\ & = \int \frac {4+9 x+\log (x)+3 x \log (x)}{3+\log (x)} \, dx+\int \log (3+\log (x)) \, dx \\ & = \int \left (1+3 x+\frac {1}{3+\log (x)}\right ) \, dx+\int \log (3+\log (x)) \, dx \\ & = x+\frac {3 x^2}{2}+\int \frac {1}{3+\log (x)} \, dx+\int \log (3+\log (x)) \, dx \\ & = x+\frac {3 x^2}{2}+\int \log (3+\log (x)) \, dx+\text {Subst}\left (\int \frac {e^x}{3+x} \, dx,x,\log (x)\right ) \\ & = x+\frac {3 x^2}{2}+\frac {\operatorname {ExpIntegralEi}(3+\log (x))}{e^3}+\int \log (3+\log (x)) \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {4+9 x+(1+3 x) \log (x)+(3+\log (x)) \log (3+\log (x))}{3+\log (x)} \, dx=x+\frac {3 x^2}{2}+x \log (3+\log (x)) \]

[In]

Integrate[(4 + 9*x + (1 + 3*x)*Log[x] + (3 + Log[x])*Log[3 + Log[x]])/(3 + Log[x]),x]

[Out]

x + (3*x^2)/2 + x*Log[3 + Log[x]]

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07

method	result	size
default	\(x +\frac {3 x^{2}}{2}+x \ln \left (3+\ln \left (x \right )\right )\)	\(15\)
norman	\(x +\frac {3 x^{2}}{2}+x \ln \left (3+\ln \left (x \right )\right )\)	\(15\)
risch	\(x +\frac {3 x^{2}}{2}+x \ln \left (3+\ln \left (x \right )\right )\)	\(15\)
parallelrisch	\(x +\frac {3 x^{2}}{2}+x \ln \left (3+\ln \left (x \right )\right )\)	\(15\)

[In]

int(((3+ln(x))*ln(3+ln(x))+(1+3*x)*ln(x)+9*x+4)/(3+ln(x)),x,method=_RETURNVERBOSE)

[Out]

x+3/2*x^2+x*ln(3+ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {4+9 x+(1+3 x) \log (x)+(3+\log (x)) \log (3+\log (x))}{3+\log (x)} \, dx=\frac {3}{2} \, x^{2} + x \log \left (\log \left (x\right ) + 3\right ) + x \]

[In]

integrate(((3+log(x))*log(3+log(x))+(1+3*x)*log(x)+9*x+4)/(3+log(x)),x, algorithm="fricas")

[Out]

3/2*x^2 + x*log(log(x) + 3) + x

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {4+9 x+(1+3 x) \log (x)+(3+\log (x)) \log (3+\log (x))}{3+\log (x)} \, dx=\frac {3 x^{2}}{2} + x \log {\left (\log {\left (x \right )} + 3 \right )} + x \]

[In]

integrate(((3+ln(x))*ln(3+ln(x))+(1+3*x)*ln(x)+9*x+4)/(3+ln(x)),x)

[Out]

3*x**2/2 + x*log(log(x) + 3) + x

Maxima [F]

\[ \int \frac {4+9 x+(1+3 x) \log (x)+(3+\log (x)) \log (3+\log (x))}{3+\log (x)} \, dx=\int { \frac {{\left (3 \, x + 1\right )} \log \left (x\right ) + {\left (\log \left (x\right ) + 3\right )} \log \left (\log \left (x\right ) + 3\right ) + 9 \, x + 4}{\log \left (x\right ) + 3} \,d x } \]

[In]

integrate(((3+log(x))*log(3+log(x))+(1+3*x)*log(x)+9*x+4)/(3+log(x)),x, algorithm="maxima")

[Out]

-e^(-3)*exp_integral_e(1, -log(x) - 3)*log(x) - 3*e^(-6)*exp_integral_e(1, -2*log(x) - 6)*log(x) + e^(-3)*exp_
integral_e(2, -log(x) - 3) + 3/2*e^(-6)*exp_integral_e(2, -2*log(x) - 6) - 4*e^(-3)*exp_integral_e(1, -log(x)
- 3) - 9*e^(-6)*exp_integral_e(1, -2*log(x) - 6) + x*log(log(x) + 3) - integrate(1/(log(x) + 3), x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {4+9 x+(1+3 x) \log (x)+(3+\log (x)) \log (3+\log (x))}{3+\log (x)} \, dx=\frac {3}{2} \, x^{2} + x \log \left (\log \left (x\right ) + 3\right ) + x \]

[In]

integrate(((3+log(x))*log(3+log(x))+(1+3*x)*log(x)+9*x+4)/(3+log(x)),x, algorithm="giac")

[Out]

3/2*x^2 + x*log(log(x) + 3) + x

Mupad [B] (verification not implemented)

Time = 9.73 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {4+9 x+(1+3 x) \log (x)+(3+\log (x)) \log (3+\log (x))}{3+\log (x)} \, dx=\frac {x\,\left (3\,x+2\,\ln \left (\ln \left (x\right )+3\right )+2\right )}{2} \]

[In]

int((9*x + log(log(x) + 3)*(log(x) + 3) + log(x)*(3*x + 1) + 4)/(log(x) + 3),x)

[Out]

(x*(3*x + 2*log(log(x) + 3) + 2))/2