\(\int \frac {e^{\arctan (x)} x}{(1+x^2)^{3/2}} \, dx\) [98]

Optimal result

Integrand size = 15, antiderivative size = 22 \[ \int \frac {e^{\arctan (x)} x}{\left (1+x^2\right )^{3/2}} \, dx=-\frac {e^{\arctan (x)} (1-x)}{2 \sqrt {1+x^2}} \]

[Out]

-1/2*exp(arctan(x))*(1-x)/(x^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {5185} \[ \int \frac {e^{\arctan (x)} x}{\left (1+x^2\right )^{3/2}} \, dx=-\frac {(1-x) e^{\arctan (x)}}{2 \sqrt {x^2+1}} \]

[In]

Int[(E^ArcTan[x]*x)/(1 + x^2)^(3/2),x]

[Out]

-1/2*(E^ArcTan[x]*(1 - x))/Sqrt[1 + x^2]

Rule 5185

Int[(E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_))/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(1 - a*n*x))*(E^(n*Ar
cTan[a*x])/(d*(n^2 + 1)*Sqrt[c + d*x^2])), x] /; FreeQ[{a, c, d, n}, x] && EqQ[d, a^2*c] &&  !IntegerQ[I*n]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {e^{\arctan (x)} x}{\left (1+x^2\right )^{3/2}} \, dx=\frac {1}{2} (1-i x)^{-\frac {1}{2}+\frac {i}{2}} (1+i x)^{-\frac {1}{2}-\frac {i}{2}} (-1+x) \]

[In]

Integrate[(E^ArcTan[x]*x)/(1 + x^2)^(3/2),x]

[Out]

(-1 + x)/(2*(1 - I*x)^(1/2 - I/2)*(1 + I*x)^(1/2 + I/2))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73

method	result	size
gosper	\(\frac {\left (-1+x \right ) {\mathrm e}^{\arctan \left (x \right )}}{2 \sqrt {x^{2}+1}}\)	\(16\)

[In]

int(exp(arctan(x))*x/(x^2+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/2*(-1+x)*exp(arctan(x))/(x^2+1)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\arctan (x)} x}{\left (1+x^2\right )^{3/2}} \, dx=\frac {{\left (x - 1\right )} e^{\arctan \left (x\right )}}{2 \, \sqrt {x^{2} + 1}} \]

[In]

integrate(exp(arctan(x))*x/(x^2+1)^(3/2),x, algorithm="fricas")

[Out]

1/2*(x - 1)*e^arctan(x)/sqrt(x^2 + 1)

Sympy [A] (verification not implemented)

Time = 11.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {e^{\arctan (x)} x}{\left (1+x^2\right )^{3/2}} \, dx=\frac {x e^{\operatorname {atan}{\left (x \right )}}}{2 \sqrt {x^{2} + 1}} - \frac {e^{\operatorname {atan}{\left (x \right )}}}{2 \sqrt {x^{2} + 1}} \]

[In]

integrate(exp(atan(x))*x/(x**2+1)**(3/2),x)

[Out]

x*exp(atan(x))/(2*sqrt(x**2 + 1)) - exp(atan(x))/(2*sqrt(x**2 + 1))

Maxima [F]

\[ \int \frac {e^{\arctan (x)} x}{\left (1+x^2\right )^{3/2}} \, dx=\int { \frac {x e^{\arctan \left (x\right )}}{{\left (x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(exp(arctan(x))*x/(x^2+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(x*e^arctan(x)/(x^2 + 1)^(3/2), x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\arctan (x)} x}{\left (1+x^2\right )^{3/2}} \, dx=\frac {1}{2} \, {\left (\frac {x}{\sqrt {x^{2} + 1}} - \frac {1}{\sqrt {x^{2} + 1}}\right )} e^{\arctan \left (x\right )} \]

[In]

integrate(exp(arctan(x))*x/(x^2+1)^(3/2),x, algorithm="giac")

[Out]

1/2*(x/sqrt(x^2 + 1) - 1/sqrt(x^2 + 1))*e^arctan(x)

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\arctan (x)} x}{\left (1+x^2\right )^{3/2}} \, dx=\int \frac {x\,{\mathrm {e}}^{\mathrm {atan}\left (x\right )}}{{\left (x^2+1\right )}^{3/2}} \,d x \]

[In]

int((x*exp(atan(x)))/(x^2 + 1)^(3/2),x)

[Out]

int((x*exp(atan(x)))/(x^2 + 1)^(3/2), x)