\(\int (2 x+e^{2 x+5 x^2} (2+10 x)) \, dx\) [7825]

Optimal result

Integrand size = 21, antiderivative size = 16 \[ \int \left (2 x+e^{2 x+5 x^2} (2+10 x)\right ) \, dx=-2+e^{2 x+5 x^2}+x^2 \]

[Out]

x^2+exp(5*x^2+2*x)-2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2268} \[ \int \left (2 x+e^{2 x+5 x^2} (2+10 x)\right ) \, dx=x^2+e^{5 x^2+2 x} \]

[In]

Int[2*x + E^(2*x + 5*x^2)*(2 + 10*x),x]

[Out]

E^(2*x + 5*x^2) + x^2

Rule 2268

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[e*(F^(a + b*x + c*x^2)/(2
*c*Log[F])), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rubi steps \begin{align*} \text {integral}& = x^2+\int e^{2 x+5 x^2} (2+10 x) \, dx \\ & = e^{2 x+5 x^2}+x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44 \[ \int \left (2 x+e^{2 x+5 x^2} (2+10 x)\right ) \, dx=2 \left (\frac {1}{2} e^{x (2+5 x)}+\frac {x^2}{2}\right ) \]

[In]

Integrate[2*x + E^(2*x + 5*x^2)*(2 + 10*x),x]

[Out]

2*(E^(x*(2 + 5*x))/2 + x^2/2)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
risch	\({\mathrm e}^{x \left (2+5 x \right )}+x^{2}\)	\(13\)
default	\({\mathrm e}^{5 x^{2}+2 x}+x^{2}\)	\(15\)
norman	\({\mathrm e}^{5 x^{2}+2 x}+x^{2}\)	\(15\)
parallelrisch	\({\mathrm e}^{5 x^{2}+2 x}+x^{2}\)	\(15\)
parts	\({\mathrm e}^{5 x^{2}+2 x}+x^{2}\)	\(15\)

[In]

int((10*x+2)*exp(5*x^2+2*x)+2*x,x,method=_RETURNVERBOSE)

[Out]

exp(x*(2+5*x))+x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \left (2 x+e^{2 x+5 x^2} (2+10 x)\right ) \, dx=x^{2} + e^{\left (5 \, x^{2} + 2 \, x\right )} \]

[In]

integrate((10*x+2)*exp(5*x^2+2*x)+2*x,x, algorithm="fricas")

[Out]

x^2 + e^(5*x^2 + 2*x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \left (2 x+e^{2 x+5 x^2} (2+10 x)\right ) \, dx=x^{2} + e^{5 x^{2} + 2 x} \]

[In]

integrate((10*x+2)*exp(5*x**2+2*x)+2*x,x)

[Out]

x**2 + exp(5*x**2 + 2*x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \left (2 x+e^{2 x+5 x^2} (2+10 x)\right ) \, dx=x^{2} + e^{\left (5 \, x^{2} + 2 \, x\right )} \]

[In]

integrate((10*x+2)*exp(5*x^2+2*x)+2*x,x, algorithm="maxima")

[Out]

x^2 + e^(5*x^2 + 2*x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \left (2 x+e^{2 x+5 x^2} (2+10 x)\right ) \, dx=x^{2} + e^{\left (5 \, x^{2} + 2 \, x\right )} \]

[In]

integrate((10*x+2)*exp(5*x^2+2*x)+2*x,x, algorithm="giac")

[Out]

x^2 + e^(5*x^2 + 2*x)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \left (2 x+e^{2 x+5 x^2} (2+10 x)\right ) \, dx={\mathrm {e}}^{5\,x^2+2\,x}+x^2 \]

[In]

int(2*x + exp(2*x + 5*x^2)*(10*x + 2),x)

[Out]

exp(2*x + 5*x^2) + x^2