\(\int \frac {e^{\arctan (a x)}}{(c+a^2 c x^2)^{5/2}} \, dx\) [257]

Optimal result

Integrand size = 21, antiderivative size = 72 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {e^{\arctan (a x)} (1+3 a x)}{10 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {3 e^{\arctan (a x)} (1+a x)}{10 a c^2 \sqrt {c+a^2 c x^2}} \]

[Out]

1/10*exp(arctan(a*x))*(3*a*x+1)/a/c/(a^2*c*x^2+c)^(3/2)+3/10*exp(arctan(a*x))*(a*x+1)/a/c^2/(a^2*c*x^2+c)^(1/2
)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {5178, 5177} \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {3 (a x+1) e^{\arctan (a x)}}{10 a c^2 \sqrt {a^2 c x^2+c}}+\frac {(3 a x+1) e^{\arctan (a x)}}{10 a c \left (a^2 c x^2+c\right )^{3/2}} \]

[In]

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

[Out]

(E^ArcTan[a*x]*(1 + 3*a*x))/(10*a*c*(c + a^2*c*x^2)^(3/2)) + (3*E^ArcTan[a*x]*(1 + a*x))/(10*a*c^2*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]

Rule 5178

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(n - 2*a*(p + 1)*x)*(c + d*x^2)
^(p + 1)*(E^(n*ArcTan[a*x])/(a*c*(n^2 + 4*(p + 1)^2))), x] + Dist[2*(p + 1)*((2*p + 3)/(c*(n^2 + 4*(p + 1)^2))
), Int[(c + d*x^2)^(p + 1)*E^(n*ArcTan[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[d, a^2*c] && LtQ[p, -1]
&&  !IntegerQ[I*n] && NeQ[n^2 + 4*(p + 1)^2, 0] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = \frac {e^{\arctan (a x)} (1+3 a x)}{10 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {3 \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{5 c} \\ & = \frac {e^{\arctan (a x)} (1+3 a x)}{10 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {3 e^{\arctan (a x)} (1+a x)}{10 a c^2 \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {e^{\arctan (a x)} \left (4+6 a x+3 a^2 x^2+3 a^3 x^3\right )}{10 c^2 \left (a+a^3 x^2\right ) \sqrt {c+a^2 c x^2}} \]

[In]

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

[Out]

(E^ArcTan[a*x]*(4 + 6*a*x + 3*a^2*x^2 + 3*a^3*x^3))/(10*c^2*(a + a^3*x^2)*Sqrt[c + a^2*c*x^2])

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.75

method	result	size
gosper	\(\frac {\left (a^{2} x^{2}+1\right ) \left (3 a^{3} x^{3}+3 a^{2} x^{2}+6 a x +4\right ) {\mathrm e}^{\arctan \left (a x \right )}}{10 a \left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\)	\(54\)

[In]

int(exp(arctan(a*x))/(a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/10*(a^2*x^2+1)*(3*a^3*x^3+3*a^2*x^2+6*a*x+4)*exp(arctan(a*x))/a/(a^2*c*x^2+c)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {{\left (3 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + 6 \, a x + 4\right )} \sqrt {a^{2} c x^{2} + c} e^{\left (\arctan \left (a x\right )\right )}}{10 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \]

[In]

integrate(exp(arctan(a*x))/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

1/10*(3*a^3*x^3 + 3*a^2*x^2 + 6*a*x + 4)*sqrt(a^2*c*x^2 + c)*e^(arctan(a*x))/(a^5*c^3*x^4 + 2*a^3*c^3*x^2 + a*
c^3)

Sympy [F]

\[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {e^{\operatorname {atan}{\left (a x \right )}}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(exp(atan(a*x))/(a**2*c*x**2+c)**(5/2),x)

[Out]

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

Maxima [F]

\[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {e^{\left (\arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(exp(arctan(a*x))/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(e^(arctan(a*x))/(a^2*c*x^2 + c)^(5/2), x)

Giac [F]

\[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {e^{\left (\arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(exp(arctan(a*x))/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {{\mathrm {e}}^{\mathrm {atan}\left (a\,x\right )}\,\left (\frac {2}{5\,a^3\,c^2}+\frac {3\,x^3}{10\,c^2}+\frac {3\,x}{5\,a^2\,c^2}+\frac {3\,x^2}{10\,a\,c^2}\right )}{\frac {\sqrt {c\,a^2\,x^2+c}}{a^2}+x^2\,\sqrt {c\,a^2\,x^2+c}} \]

[In]

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

[Out]

(exp(atan(a*x))*(2/(5*a^3*c^2) + (3*x^3)/(10*c^2) + (3*x)/(5*a^2*c^2) + (3*x^2)/(10*a*c^2)))/((c + a^2*c*x^2)^
(1/2)/a^2 + x^2*(c + a^2*c*x^2)^(1/2))