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

Optimal result

Integrand size = 42, antiderivative size = 14 \[ \int \frac {e^x (2+4 x)+(1+2 x)^x (2 x+(1+2 x) \log (1+2 x))}{1+2 x} \, dx=3+2 e^x+(1+2 x)^x \]

[Out]

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

Rubi [F(-1)]

Timed out. \[ \int \frac {e^x (2+4 x)+(1+2 x)^x (2 x+(1+2 x) \log (1+2 x))}{1+2 x} \, dx=\text {\$Aborted} \]

[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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {e^x (2+4 x)+(1+2 x)^x (2 x+(1+2 x) \log (1+2 x))}{1+2 x} \, dx=2 e^x+(1+2 x)^x \]

[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

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^x (2+4 x)+(1+2 x)^x (2 x+(1+2 x) \log (1+2 x))}{1+2 x} \, dx={\left (2 \, x + 1\right )}^{x} + 2 \, e^{x} \]

[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

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {e^x (2+4 x)+(1+2 x)^x (2 x+(1+2 x) \log (1+2 x))}{1+2 x} \, dx=2 e^{x} + e^{x \log {\left (2 x + 1 \right )}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^x (2+4 x)+(1+2 x)^x (2 x+(1+2 x) \log (1+2 x))}{1+2 x} \, dx={\left (2 \, x + 1\right )}^{x} + 2 \, e^{x} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^x (2+4 x)+(1+2 x)^x (2 x+(1+2 x) \log (1+2 x))}{1+2 x} \, dx={\left (2 \, x + 1\right )}^{x} + 2 \, e^{x} \]

[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

Mupad [B] (verification not implemented)

Time = 9.18 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^x (2+4 x)+(1+2 x)^x (2 x+(1+2 x) \log (1+2 x))}{1+2 x} \, dx=2\,{\mathrm {e}}^x+{\left (2\,x+1\right )}^x \]

[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