\(\int \frac {e^{\arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx\) [255]

Optimal result

Integrand size = 21, antiderivative size = 93 \[ \int \frac {e^{\arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\frac {(1+i) 2^{-\frac {1}{2}-\frac {i}{2}} (1-i a x)^{\frac {1}{2}+\frac {i}{2}} \sqrt {1+a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}+\frac {i}{2},\frac {1}{2}+\frac {i}{2},\frac {3}{2}+\frac {i}{2},\frac {1}{2} (1-i a x)\right )}{a \sqrt {c+a^2 c x^2}} \]

[Out]

(1+I)*2^(-1/2-1/2*I)*(1-I*a*x)^(1/2+1/2*I)*hypergeom([1/2+1/2*I, 1/2+1/2*I],[3/2+1/2*I],1/2-1/2*I*a*x)*(a^2*x^
2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5184, 5181, 71} \[ \int \frac {e^{\arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\frac {(1+i) 2^{-\frac {1}{2}-\frac {i}{2}} (1-i a x)^{\frac {1}{2}+\frac {i}{2}} \sqrt {a^2 x^2+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}+\frac {i}{2},\frac {1}{2}+\frac {i}{2},\frac {3}{2}+\frac {i}{2},\frac {1}{2} (1-i a x)\right )}{a \sqrt {a^2 c x^2+c}} \]

[In]

Int[E^ArcTan[a*x]/Sqrt[c + a^2*c*x^2],x]

[Out]

((1 + I)*(1 - I*a*x)^(1/2 + I/2)*Sqrt[1 + a^2*x^2]*Hypergeometric2F1[1/2 + I/2, 1/2 + I/2, 3/2 + I/2, (1 - I*a
*x)/2])/(2^(1/2 + I/2)*a*Sqrt[c + a^2*c*x^2])

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 5181

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

Rule 5184

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[c^IntPart[p]*((c + d*x^2)^FracP
art[p]/(1 + a^2*x^2)^FracPart[p]), Int[(1 + a^2*x^2)^p*E^(n*ArcTan[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x]
&& EqQ[d, a^2*c] &&  !(IntegerQ[p] || GtQ[c, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1+a^2 x^2} \int \frac {e^{\arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}} \\ & = \frac {\sqrt {1+a^2 x^2} \int (1-i a x)^{-\frac {1}{2}+\frac {i}{2}} (1+i a x)^{-\frac {1}{2}-\frac {i}{2}} \, dx}{\sqrt {c+a^2 c x^2}} \\ & = \frac {(1+i) 2^{-\frac {1}{2}-\frac {i}{2}} (1-i a x)^{\frac {1}{2}+\frac {i}{2}} \sqrt {1+a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}+\frac {i}{2},\frac {1}{2}+\frac {i}{2},\frac {3}{2}+\frac {i}{2},\frac {1}{2} (1-i a x)\right )}{a \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\frac {(1+i) 2^{-\frac {1}{2}-\frac {i}{2}} (1-i a x)^{\frac {1}{2}+\frac {i}{2}} \sqrt {1+a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}+\frac {i}{2},\frac {1}{2}+\frac {i}{2},\frac {3}{2}+\frac {i}{2},\frac {1}{2} (1-i a x)\right )}{a \sqrt {c+a^2 c x^2}} \]

[In]

Integrate[E^ArcTan[a*x]/Sqrt[c + a^2*c*x^2],x]

[Out]

((1 + I)*(1 - I*a*x)^(1/2 + I/2)*Sqrt[1 + a^2*x^2]*Hypergeometric2F1[1/2 + I/2, 1/2 + I/2, 3/2 + I/2, (1 - I*a
*x)/2])/(2^(1/2 + I/2)*a*Sqrt[c + a^2*c*x^2])

Maple [F]

\[\int \frac {{\mathrm e}^{\arctan \left (a x \right )}}{\sqrt {a^{2} c \,x^{2}+c}}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(e^(arctan(a*x))/sqrt(a^2*c*x^2 + c), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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