\(\int \frac {e^{2 \arctan (a x)}}{(c+a^2 c x^2)^{3/2}} \, dx\) [270]

Optimal result

Integrand size = 23, antiderivative size = 37 \[ \int \frac {e^{2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {e^{2 \arctan (a x)} (2+a x)}{5 a c \sqrt {c+a^2 c x^2}} \]

[Out]

1/5*exp(2*arctan(a*x))*(a*x+2)/a/c/(a^2*c*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {5177} \[ \int \frac {e^{2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {(a x+2) e^{2 \arctan (a x)}}{5 a c \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

(E^(2*ArcTan[a*x])*(2 + a*x))/(5*a*c*Sqrt[c + a^2*c*x^2])

Rule 5177

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(n + a*x)*(E^(n*ArcTan[a*x])/(
a*c*(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^{2 \arctan (a x)} (2+a x)}{5 a c \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {e^{2 \arctan (a x)} (2+a x)}{5 a c \sqrt {c+a^2 c x^2}} \]

[In]

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

[Out]

(E^(2*ArcTan[a*x])*(2 + a*x))/(5*a*c*Sqrt[c + a^2*c*x^2])

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.05

method	result	size
gosper	\(\frac {\left (a^{2} x^{2}+1\right ) \left (a x +2\right ) {\mathrm e}^{2 \arctan \left (a x \right )}}{5 a \left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\)	\(39\)

[In]

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

[Out]

1/5*(a^2*x^2+1)*(a*x+2)*exp(2*arctan(a*x))/a/(a^2*c*x^2+c)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.19 \[ \int \frac {e^{2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {a^{2} c x^{2} + c} {\left (a x + 2\right )} e^{\left (2 \, \arctan \left (a x\right )\right )}}{5 \, {\left (a^{3} c^{2} x^{2} + a c^{2}\right )}} \]

[In]

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

[Out]

1/5*sqrt(a^2*c*x^2 + c)*(a*x + 2)*e^(2*arctan(a*x))/(a^3*c^2*x^2 + a*c^2)

Sympy [F]

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

[In]

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

[Out]

Integral(exp(2*atan(a*x))/(c*(a**2*x**2 + 1))**(3/2), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.68 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int \frac {e^{2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {{\mathrm {e}}^{2\,\mathrm {atan}\left (a\,x\right )}\,\left (\frac {x}{5\,c}+\frac {2}{5\,a\,c}\right )}{\sqrt {c\,a^2\,x^2+c}} \]

[In]

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

[Out]

(exp(2*atan(a*x))*(x/(5*c) + 2/(5*a*c)))/(c + a^2*c*x^2)^(1/2)