\(\int \frac {e^{a x} x}{(1+a x)^2} \, dx\) [164]

Optimal result

Integrand size = 14, antiderivative size = 16 \[ \int \frac {e^{a x} x}{(1+a x)^2} \, dx=\frac {e^{a x}}{a^2 (1+a x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2228} \[ \int \frac {e^{a x} x}{(1+a x)^2} \, dx=\frac {e^{a x}}{a^2 (a x+1)} \]

[In]

Int[(E^(a*x)*x)/(1 + a*x)^2,x]

[Out]

E^(a*x)/(a^2*(1 + a*x))

Rule 2228

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[g*u^(m + 1)*(F^(c*v)/(b*c*
e*Log[F])), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps \begin{align*} \text {integral}& = \frac {e^{a x}}{a^2 (1+a x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {e^{a x} x}{(1+a x)^2} \, dx=\frac {e^{a x}}{a^2 (1+a x)} \]

[In]

Integrate[(E^(a*x)*x)/(1 + a*x)^2,x]

[Out]

E^(a*x)/(a^2*(1 + a*x))

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00

method	result	size
gosper	\(\frac {{\mathrm e}^{a x}}{a^{2} \left (a x +1\right )}\)	\(16\)
derivativedivides	\(\frac {{\mathrm e}^{a x}}{a^{2} \left (a x +1\right )}\)	\(16\)
default	\(\frac {{\mathrm e}^{a x}}{a^{2} \left (a x +1\right )}\)	\(16\)
norman	\(\frac {{\mathrm e}^{a x}}{a^{2} \left (a x +1\right )}\)	\(16\)
risch	\(\frac {{\mathrm e}^{a x}}{a^{2} \left (a x +1\right )}\)	\(16\)
parallelrisch	\(\frac {{\mathrm e}^{a x}}{a^{2} \left (a x +1\right )}\)	\(16\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {e^{a x} x}{(1+a x)^2} \, dx=\frac {e^{\left (a x\right )}}{a^{3} x + a^{2}} \]

[In]

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

[Out]

e^(a*x)/(a^3*x + a^2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {e^{a x} x}{(1+a x)^2} \, dx=\frac {e^{a x}}{a^{3} x + a^{2}} \]

[In]

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

[Out]

exp(a*x)/(a**3*x + a**2)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {e^{a x} x}{(1+a x)^2} \, dx=\frac {e^{\left (a x\right )}}{a^{3} x + a^{2}} \]

[In]

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

[Out]

e^(a*x)/(a^3*x + a^2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (15) = 30\).

Time = 0.31 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.81 \[ \int \frac {e^{a x} x}{(1+a x)^2} \, dx=-\frac {e^{\left (-{\left (a x + 1\right )} {\left (\frac {1}{a x + 1} - 1\right )}\right )}}{{\left (a x + 1\right )} a^{2} {\left (\frac {1}{a x + 1} - 1\right )} - a^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {e^{a x} x}{(1+a x)^2} \, dx=\frac {{\mathrm {e}}^{a\,x}}{a^2\,\left (a\,x+1\right )} \]

[In]

int((x*exp(a*x))/(a*x + 1)^2,x)

[Out]

exp(a*x)/(a^2*(a*x + 1))