3.27.59 \(\int \frac {e^x (2+4 x)+(1+2 x)^x (2 x+(1+2 x) \log (1+2 x))}{1+2 x} \, dx\) [2659]

Optimal. Leaf size=14 \[ 3+2 e^x+(1+2 x)^x \]

[Out]

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

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Rubi [F]
time = 180.01, 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*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(2 + 4*x) + (1 + 2*x)^x*(2*x + (1 + 2*x)*Log[1 + 2*x]))/(1 + 2*x),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [A]
time = 0.06, size = 13, normalized size = 0.93 \begin {gather*} 2 e^x+(1+2 x)^x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(2 + 4*x) + (1 + 2*x)^x*(2*x + (1 + 2*x)*Log[1 + 2*x]))/(1 + 2*x),x]

[Out]

2*E^x + (1 + 2*x)^x

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Maple [A]
time = 0.22, size = 15, normalized size = 1.07

method	result	size
risch	\(\left (2 x +1\right )^{x}+2 \,{\mathrm e}^{x}\)	\(13\)
default	\({\mathrm e}^{x \ln \left (2 x +1\right )}+2 \,{\mathrm e}^{x}\)	\(15\)
norman	\({\mathrm e}^{x \ln \left (2 x +1\right )}+2 \,{\mathrm e}^{x}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x+1)*ln(2*x+1)+2*x)*exp(x*ln(2*x+1))+(4*x+2)*exp(x))/(2*x+1),x,method=_RETURNVERBOSE)

[Out]

exp(x*ln(2*x+1))+2*exp(x)

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Maxima [A]
time = 0.29, size = 12, normalized size = 0.86 \begin {gather*} {\left (2 \, x + 1\right )}^{x} + 2 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1+2*x)*log(1+2*x)+2*x)*exp(x*log(1+2*x))+(4*x+2)*exp(x))/(1+2*x),x, algorithm="maxima")

[Out]

(2*x + 1)^x + 2*e^x

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Fricas [A]
time = 0.37, size = 12, normalized size = 0.86 \begin {gather*} {\left (2 \, x + 1\right )}^{x} + 2 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1+2*x)*log(1+2*x)+2*x)*exp(x*log(1+2*x))+(4*x+2)*exp(x))/(1+2*x),x, algorithm="fricas")

[Out]

(2*x + 1)^x + 2*e^x

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Sympy [A]
time = 0.16, size = 14, normalized size = 1.00 \begin {gather*} 2 e^{x} + e^{x \log {\left (2 x + 1 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1+2*x)*ln(1+2*x)+2*x)*exp(x*ln(1+2*x))+(4*x+2)*exp(x))/(1+2*x),x)

[Out]

2*exp(x) + exp(x*log(2*x + 1))

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Giac [A]
time = 0.42, size = 12, normalized size = 0.86 \begin {gather*} {\left (2 \, x + 1\right )}^{x} + 2 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1+2*x)*log(1+2*x)+2*x)*exp(x*log(1+2*x))+(4*x+2)*exp(x))/(1+2*x),x, algorithm="giac")

[Out]

(2*x + 1)^x + 2*e^x

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Mupad [B]
time = 0.27, size = 12, normalized size = 0.86 \begin {gather*} 2\,{\mathrm {e}}^x+{\left (2\,x+1\right )}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(4*x + 2) + exp(x*log(2*x + 1))*(2*x + log(2*x + 1)*(2*x + 1)))/(2*x + 1),x)

[Out]

2*exp(x) + (2*x + 1)^x

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