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

Optimal. Leaf size=25 \[ -5+e^{\left (\frac {1}{3}+x\right )^{5-x}}+12 x \left (x+(3+x)^2\right ) \]

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Rubi [F]  time = 7.27, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.86, size = 25, normalized size = 1.00 \begin {gather*} e^{\left (\frac {1}{3}+x\right )^{5-x}}+12 x \left (9+7 x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.62, size = 24, normalized size = 0.96 \begin {gather*} 12 \, x^{3} + 84 \, x^{2} + 108 \, x + e^{\left ({\left (x + \frac {1}{3}\right )}^{-x + 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)/(3*x+1),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)/(3*x+1),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.10, size = 25, normalized size = 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	\(108 x +{\mathrm e}^{{\mathrm e}^{\left (5-x \right ) \ln \left (x +\frac {1}{3}\right )}}+84 x^{2}+12 x^{3}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[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)/(3*
x+1),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.75, size = 130, normalized size = 5.20 \begin {gather*} 12 \, x^{3} + 84 \, x^{2} + 108 \, x + e^{\left (x^{5} e^{\left (x \log \relax (3) - x \log \left (3 \, x + 1\right )\right )} + \frac {5}{3} \, x^{4} e^{\left (x \log \relax (3) - x \log \left (3 \, x + 1\right )\right )} + \frac {10}{9} \, x^{3} e^{\left (x \log \relax (3) - x \log \left (3 \, x + 1\right )\right )} + \frac {10}{27} \, x^{2} e^{\left (x \log \relax (3) - x \log \left (3 \, x + 1\right )\right )} + \frac {5}{81} \, x e^{\left (x \log \relax (3) - x \log \left (3 \, x + 1\right )\right )} + \frac {1}{243} \, e^{\left (x \log \relax (3) - x \log \left (3 \, x + 1\right )\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)/(3*x+1),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)))

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mupad [B]  time = 4.46, size = 82, normalized size = 3.28 \begin {gather*} 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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 1.03, size = 26, normalized size = 1.04 \begin {gather*} 12 x^{3} + 84 x^{2} + 108 x + e^{e^{\left (5 - x\right ) \log {\left (x + \frac {1}{3} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)/(3*x+1),x)

[Out]

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

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