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

Optimal. Leaf size=23 \[ e^{2 x-(1+x) \log \left ((1+x)^2\right )}+\frac {5 x}{3} \]

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Rubi [F]  time = 0.18, 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 {1}{3} \left (5-3 e^{2 x+(-1-x) \log \left (1+2 x+x^2\right )} \log \left (1+2 x+x^2\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 27, normalized size = 1.17 \begin {gather*} \frac {5 x}{3}+e^{-2+2 (1+x)} \left ((1+x)^2\right )^{-1-x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.52, size = 23, normalized size = 1.00 \begin {gather*} \frac {5}{3} \, x + e^{\left (-{\left (x + 1\right )} \log \left (x^{2} + 2 \, x + 1\right ) + 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.31, size = 32, normalized size = 1.39 \begin {gather*} \frac {5}{3} \, x + e^{\left (-x \log \left (x^{2} + 2 \, x + 1\right ) + 2 \, x - \log \left (x^{2} + 2 \, x + 1\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 24, normalized size = 1.04

method	result	size
risch	\(\frac {5 x}{3}+\left (x^{2}+2 x +1\right )^{-x -1} {\mathrm e}^{2 x}\)	\(24\)
default	\(\frac {5 x}{3}+{\mathrm e}^{\left (-x -1\right ) \ln \left (x^{2}+2 x +1\right )+2 x}\)	\(25\)
norman	\(\frac {5 x}{3}+{\mathrm e}^{\left (-x -1\right ) \ln \left (x^{2}+2 x +1\right )+2 x}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.96, size = 27, normalized size = 1.17 \begin {gather*} \frac {5}{3} \, x + \frac {e^{\left (-2 \, x \log \left (x + 1\right ) + 2 \, x\right )}}{x^{2} + 2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.21, size = 26, normalized size = 1.13 \begin {gather*} \frac {5\,x}{3}+\frac {{\mathrm {e}}^{2\,x}}{{\left (x+1\right )}^2\,{\left (x^2+2\,x+1\right )}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.29, size = 24, normalized size = 1.04 \begin {gather*} \frac {5 x}{3} + e^{2 x + \left (- x - 1\right ) \log {\left (x^{2} + 2 x + 1 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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