\(\int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} (15-3 x+(-1-3 x) \log (\frac {1}{3} (1+3 x)))}{1+3 x} \, dx\) [7053]

Optimal result

Integrand size = 80, antiderivative size = 25 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=-5+e^{\left (\frac {1}{3}+x\right )^{5-x}}+12 x \left (x+(3+x)^2\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=\int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx \]

[In]

Int[(108 + 492*x + 540*x^2 + 108*x^3 + 3^(-5 + x)*E^(3^(-5 + x)*(1 + 3*x)^(5 - x))*(1 + 3*x)^(5 - x)*(15 - 3*x
+ (-1 - 3*x)*Log[(1 + 3*x)/3]))/(1 + 3*x),x]

[Out]

108*x + 84*x^2 + 12*x^3 + 5*Defer[Int][E^(1/3 + x)^(5 - x)*(1/3 + x)^(4 - x), x] - Defer[Int][E^(1/3 + x)^(5 -
x)*x*(1/3 + x)^(4 - x), x] - Log[1/3 + x]*Defer[Int][E^(1/3 + x)^(5 - x)*x*(1/3 + x)^(4 - x), x] - Log[1/3 +
x]*Defer[Int][3^(-5 + x)*E^(1/3 + x)^(5 - x)*(1 + 3*x)^(4 - x), x] + Defer[Int][Defer[Int][E^(1/3 + x)^(5 - x)
*x*(1/3 + x)^(4 - x), x]/(1/3 + x), x] + Defer[Int][Defer[Int][3^(-5 + x)*E^(1/3 + x)^(5 - x)*(1 + 3*x)^(4 - x
), x]/(1/3 + x), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (12 \left (9+14 x+3 x^2\right )-3^{-5+x} e^{\left (\frac {1}{3}+x\right )^{5-x}} (1+3 x)^{4-x} \left (-15+3 x+\log \left (\frac {1}{3}+x\right )+3 x \log \left (\frac {1}{3}+x\right )\right )\right ) \, dx \\ & = 12 \int \left (9+14 x+3 x^2\right ) \, dx-\int 3^{-5+x} e^{\left (\frac {1}{3}+x\right )^{5-x}} (1+3 x)^{4-x} \left (-15+3 x+\log \left (\frac {1}{3}+x\right )+3 x \log \left (\frac {1}{3}+x\right )\right ) \, dx \\ & = 108 x+84 x^2+12 x^3-\int \left (-5 3^{-4+x} e^{\left (\frac {1}{3}+x\right )^{5-x}} (1+3 x)^{4-x}+3^{-4+x} e^{\left (\frac {1}{3}+x\right )^{5-x}} x (1+3 x)^{4-x}+3^{-5+x} e^{\left (\frac {1}{3}+x\right )^{5-x}} (1+3 x)^{4-x} \log \left (\frac {1}{3}+x\right )+3^{-4+x} e^{\left (\frac {1}{3}+x\right )^{5-x}} x (1+3 x)^{4-x} \log \left (\frac {1}{3}+x\right )\right ) \, dx \\ & = 108 x+84 x^2+12 x^3+5 \int 3^{-4+x} e^{\left (\frac {1}{3}+x\right )^{5-x}} (1+3 x)^{4-x} \, dx-\int 3^{-4+x} e^{\left (\frac {1}{3}+x\right )^{5-x}} x (1+3 x)^{4-x} \, dx-\int 3^{-5+x} e^{\left (\frac {1}{3}+x\right )^{5-x}} (1+3 x)^{4-x} \log \left (\frac {1}{3}+x\right ) \, dx-\int 3^{-4+x} e^{\left (\frac {1}{3}+x\right )^{5-x}} x (1+3 x)^{4-x} \log \left (\frac {1}{3}+x\right ) \, dx \\ & = 108 x+84 x^2+12 x^3+5 \int e^{\left (\frac {1}{3}+x\right )^{5-x}} \left (\frac {1}{3}+x\right )^{4-x} \, dx-\log \left (\frac {1}{3}+x\right ) \int e^{\left (\frac {1}{3}+x\right )^{5-x}} x \left (\frac {1}{3}+x\right )^{4-x} \, dx-\log \left (\frac {1}{3}+x\right ) \int 3^{-5+x} e^{\left (\frac {1}{3}+x\right )^{5-x}} (1+3 x)^{4-x} \, dx-\int e^{\left (\frac {1}{3}+x\right )^{5-x}} x \left (\frac {1}{3}+x\right )^{4-x} \, dx+\int \frac {\int e^{\left (\frac {1}{3}+x\right )^{5-x}} x \left (\frac {1}{3}+x\right )^{4-x} \, dx}{\frac {1}{3}+x} \, dx+\int \frac {\int 3^{-5+x} e^{\left (\frac {1}{3}+x\right )^{5-x}} (1+3 x)^{4-x} \, dx}{\frac {1}{3}+x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 3.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=e^{\left (\frac {1}{3}+x\right )^{5-x}}+12 x \left (9+7 x+x^2\right ) \]

[In]

Integrate[(108 + 492*x + 540*x^2 + 108*x^3 + 3^(-5 + x)*E^(3^(-5 + x)*(1 + 3*x)^(5 - x))*(1 + 3*x)^(5 - x)*(15
- 3*x + (-1 - 3*x)*Log[(1 + 3*x)/3]))/(1 + 3*x),x]

[Out]

E^(1/3 + x)^(5 - x) + 12*x*(9 + 7*x + x^2)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00

method	result	size
risch	\(108 x +{\mathrm e}^{\left (x +\frac {1}{3}\right )^{5-x}}+84 x^{2}+12 x^{3}\)	\(25\)
default	\(108 x +{\mathrm e}^{{\mathrm e}^{\left (5-x \right ) \ln \left (x +\frac {1}{3}\right )}}+84 x^{2}+12 x^{3}\)	\(27\)
parts	\(108 x +{\mathrm e}^{{\mathrm e}^{\left (5-x \right ) \ln \left (x +\frac {1}{3}\right )}}+84 x^{2}+12 x^{3}\)	\(27\)
parallelrisch	\(12 x^{3}+84 x^{2}+108 x +{\mathrm e}^{{\mathrm e}^{\left (5-x \right ) \ln \left (x +\frac {1}{3}\right )}}-18\)	\(28\)

[In]

int((((-3*x-1)*ln(x+1/3)+15-3*x)*exp((5-x)*ln(x+1/3))*exp(exp((5-x)*ln(x+1/3)))+108*x^3+540*x^2+492*x+108)/(1+
3*x),x,method=_RETURNVERBOSE)

[Out]

108*x+exp((x+1/3)^(5-x))+84*x^2+12*x^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=12 \, x^{3} + 84 \, x^{2} + 108 \, x + e^{\left ({\left (x + \frac {1}{3}\right )}^{-x + 5}\right )} \]

[In]

integrate((((-3*x-1)*log(x+1/3)+15-3*x)*exp((5-x)*log(x+1/3))*exp(exp((5-x)*log(x+1/3)))+108*x^3+540*x^2+492*x
+108)/(1+3*x),x, algorithm="fricas")

[Out]

12*x^3 + 84*x^2 + 108*x + e^((x + 1/3)^(-x + 5))

Sympy [A] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=12 x^{3} + 84 x^{2} + 108 x + e^{e^{\left (5 - x\right ) \log {\left (x + \frac {1}{3} \right )}}} \]

[In]

integrate((((-3*x-1)*ln(x+1/3)+15-3*x)*exp((5-x)*ln(x+1/3))*exp(exp((5-x)*ln(x+1/3)))+108*x**3+540*x**2+492*x+
108)/(1+3*x),x)

[Out]

12*x**3 + 84*x**2 + 108*x + exp(exp((5 - x)*log(x + 1/3)))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (22) = 44\).

Time = 0.51 (sec) , antiderivative size = 130, normalized size of antiderivative = 5.20 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=12 \, x^{3} + 84 \, x^{2} + 108 \, x + e^{\left (x^{5} e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )} + \frac {5}{3} \, x^{4} e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )} + \frac {10}{9} \, x^{3} e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )} + \frac {10}{27} \, x^{2} e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )} + \frac {5}{81} \, x e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )} + \frac {1}{243} \, e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )}\right )} \]

[In]

integrate((((-3*x-1)*log(x+1/3)+15-3*x)*exp((5-x)*log(x+1/3))*exp(exp((5-x)*log(x+1/3)))+108*x^3+540*x^2+492*x
+108)/(1+3*x),x, algorithm="maxima")

[Out]

12*x^3 + 84*x^2 + 108*x + e^(x^5*e^(x*log(3) - x*log(3*x + 1)) + 5/3*x^4*e^(x*log(3) - x*log(3*x + 1)) + 10/9*
x^3*e^(x*log(3) - x*log(3*x + 1)) + 10/27*x^2*e^(x*log(3) - x*log(3*x + 1)) + 5/81*x*e^(x*log(3) - x*log(3*x +
1)) + 1/243*e^(x*log(3) - x*log(3*x + 1)))

Giac [F]

\[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=\int { \frac {108 \, x^{3} - {\left ({\left (3 \, x + 1\right )} \log \left (x + \frac {1}{3}\right ) + 3 \, x - 15\right )} {\left (x + \frac {1}{3}\right )}^{-x + 5} e^{\left ({\left (x + \frac {1}{3}\right )}^{-x + 5}\right )} + 540 \, x^{2} + 492 \, x + 108}{3 \, x + 1} \,d x } \]

[In]

integrate((((-3*x-1)*log(x+1/3)+15-3*x)*exp((5-x)*log(x+1/3))*exp(exp((5-x)*log(x+1/3)))+108*x^3+540*x^2+492*x
+108)/(1+3*x),x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 12.75 (sec) , antiderivative size = 82, normalized size of antiderivative = 3.28 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=108\,x+{\mathrm {e}}^{\frac {1}{243\,{\left (x+\frac {1}{3}\right )}^x}+\frac {10\,x^2}{27\,{\left (x+\frac {1}{3}\right )}^x}+\frac {10\,x^3}{9\,{\left (x+\frac {1}{3}\right )}^x}+\frac {5\,x^4}{3\,{\left (x+\frac {1}{3}\right )}^x}+\frac {x^5}{{\left (x+\frac {1}{3}\right )}^x}+\frac {5\,x}{81\,{\left (x+\frac {1}{3}\right )}^x}}+84\,x^2+12\,x^3 \]

[In]

int((492*x + 540*x^2 + 108*x^3 - exp(exp(-log(x + 1/3)*(x - 5)))*exp(-log(x + 1/3)*(x - 5))*(3*x + log(x + 1/3
)*(3*x + 1) - 15) + 108)/(3*x + 1),x)

[Out]

108*x + exp(1/(243*(x + 1/3)^x) + (10*x^2)/(27*(x + 1/3)^x) + (10*x^3)/(9*(x + 1/3)^x) + (5*x^4)/(3*(x + 1/3)^
x) + x^5/(x + 1/3)^x + (5*x)/(81*(x + 1/3)^x)) + 84*x^2 + 12*x^3