\(\int \sqrt {e^{a+b x}} x \, dx\) [93]

Optimal result

Integrand size = 13, antiderivative size = 34 \[ \int \sqrt {e^{a+b x}} x \, dx=-\frac {4 \sqrt {e^{a+b x}}}{b^2}+\frac {2 \sqrt {e^{a+b x}} x}{b} \]

[Out]

-4*exp(b*x+a)^(1/2)/b^2+2*x*exp(b*x+a)^(1/2)/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2207, 2225} \[ \int \sqrt {e^{a+b x}} x \, dx=\frac {2 x \sqrt {e^{a+b x}}}{b}-\frac {4 \sqrt {e^{a+b x}}}{b^2} \]

[In]

Int[Sqrt[E^(a + b*x)]*x,x]

[Out]

(-4*Sqrt[E^(a + b*x)])/b^2 + (2*Sqrt[E^(a + b*x)]*x)/b

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sqrt {e^{a+b x}} x}{b}-\frac {2 \int \sqrt {e^{a+b x}} \, dx}{b} \\ & = -\frac {4 \sqrt {e^{a+b x}}}{b^2}+\frac {2 \sqrt {e^{a+b x}} x}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.62 \[ \int \sqrt {e^{a+b x}} x \, dx=\frac {2 \sqrt {e^{a+b x}} (-2+b x)}{b^2} \]

[In]

Integrate[Sqrt[E^(a + b*x)]*x,x]

[Out]

(2*Sqrt[E^(a + b*x)]*(-2 + b*x))/b^2

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.56

method	result	size
gosper	\(\frac {2 \left (b x -2\right ) \sqrt {{\mathrm e}^{b x +a}}}{b^{2}}\)	\(19\)
risch	\(\frac {2 \left (b x -2\right ) \sqrt {{\mathrm e}^{b x +a}}}{b^{2}}\)	\(19\)
parallelrisch	\(\frac {2 b \sqrt {{\mathrm e}^{b x +a}}\, x -4 \sqrt {{\mathrm e}^{b x +a}}}{b^{2}}\)	\(28\)
meijerg	\(\frac {4 \sqrt {{\mathrm e}^{b x +a}}\, {\mathrm e}^{-a -\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}} \left (1-\frac {\left (-b x \,{\mathrm e}^{\frac {a}{2}}+2\right ) {\mathrm e}^{\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}}}{2}\right )}{b^{2}}\)	\(50\)

[In]

int(x*exp(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*(b*x-2)*exp(b*x+a)^(1/2)/b^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.56 \[ \int \sqrt {e^{a+b x}} x \, dx=\frac {2 \, {\left (b x - 2\right )} e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )}}{b^{2}} \]

[In]

integrate(x*exp(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

2*(b*x - 2)*e^(1/2*b*x + 1/2*a)/b^2

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \sqrt {e^{a+b x}} x \, dx=\begin {cases} \frac {\left (2 b x - 4\right ) \sqrt {e^{a + b x}}}{b^{2}} & \text {for}\: b^{2} \neq 0 \\\frac {x^{2}}{2} & \text {otherwise} \end {cases} \]

[In]

integrate(x*exp(b*x+a)**(1/2),x)

[Out]

Piecewise(((2*b*x - 4)*sqrt(exp(a + b*x))/b**2, Ne(b**2, 0)), (x**2/2, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \sqrt {e^{a+b x}} x \, dx=\frac {2 \, {\left (b x e^{\left (\frac {1}{2} \, a\right )} - 2 \, e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (\frac {1}{2} \, b x\right )}}{b^{2}} \]

[In]

integrate(x*exp(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

2*(b*x*e^(1/2*a) - 2*e^(1/2*a))*e^(1/2*b*x)/b^2

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.56 \[ \int \sqrt {e^{a+b x}} x \, dx=\frac {2 \, {\left (b x - 2\right )} e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )}}{b^{2}} \]

[In]

integrate(x*exp(b*x+a)^(1/2),x, algorithm="giac")

[Out]

2*(b*x - 2)*e^(1/2*b*x + 1/2*a)/b^2

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.53 \[ \int \sqrt {e^{a+b x}} x \, dx=\frac {2\,\sqrt {{\mathrm {e}}^{a+b\,x}}\,\left (b\,x-2\right )}{b^2} \]

[In]

int(x*exp(a + b*x)^(1/2),x)

[Out]

(2*exp(a + b*x)^(1/2)*(b*x - 2))/b^2