\(\int e^{n \arctan (a+b x)} \, dx\) [240]

Optimal result

Integrand size = 10, antiderivative size = 91 \[ \int e^{n \arctan (a+b x)} \, dx=-\frac {2^{1-\frac {i n}{2}} (1-i a-i b x)^{1+\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (1+\frac {i n}{2},\frac {i n}{2},2+\frac {i n}{2},\frac {1}{2} (1-i a-i b x)\right )}{b (2 i-n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5201, 71} \[ \int e^{n \arctan (a+b x)} \, dx=-\frac {2^{1-\frac {i n}{2}} (-i a-i b x+1)^{1+\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {i n}{2}+1,\frac {i n}{2},\frac {i n}{2}+2,\frac {1}{2} (-i a-i b x+1)\right )}{b (-n+2 i)} \]

[In]

Int[E^(n*ArcTan[a + b*x]),x]

[Out]

-((2^(1 - (I/2)*n)*(1 - I*a - I*b*x)^(1 + (I/2)*n)*Hypergeometric2F1[1 + (I/2)*n, (I/2)*n, 2 + (I/2)*n, (1 - I
*a - I*b*x)/2])/(b*(2*I - n)))

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 5201

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.)), x_Symbol] :> Int[(1 - I*a*c - I*b*c*x)^(I*(n/2))/(1 + I*a*c +
I*b*c*x)^(I*(n/2)), x] /; FreeQ[{a, b, c, n}, x]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.66 \[ \int e^{n \arctan (a+b x)} \, dx=\frac {4 e^{(2 i+n) \arctan (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i n}{2},2-\frac {i n}{2},-e^{2 i \arctan (a+b x)}\right )}{b (2 i+n)} \]

[In]

Integrate[E^(n*ArcTan[a + b*x]),x]

[Out]

(4*E^((2*I + n)*ArcTan[a + b*x])*Hypergeometric2F1[2, 1 - (I/2)*n, 2 - (I/2)*n, -E^((2*I)*ArcTan[a + b*x])])/(
b*(2*I + n))

Maple [F]

\[\int {\mathrm e}^{n \arctan \left (b x +a \right )}d x\]

[In]

int(exp(n*arctan(b*x+a)),x)

[Out]

int(exp(n*arctan(b*x+a)),x)

Fricas [F]

\[ \int e^{n \arctan (a+b x)} \, dx=\int { e^{\left (n \arctan \left (b x + a\right )\right )} \,d x } \]

[In]

integrate(exp(n*arctan(b*x+a)),x, algorithm="fricas")

[Out]

integral(e^(n*arctan(b*x + a)), x)

Sympy [F]

\[ \int e^{n \arctan (a+b x)} \, dx=\int e^{n \operatorname {atan}{\left (a + b x \right )}}\, dx \]

[In]

integrate(exp(n*atan(b*x+a)),x)

[Out]

Integral(exp(n*atan(a + b*x)), x)

Maxima [F]

\[ \int e^{n \arctan (a+b x)} \, dx=\int { e^{\left (n \arctan \left (b x + a\right )\right )} \,d x } \]

[In]

integrate(exp(n*arctan(b*x+a)),x, algorithm="maxima")

[Out]

integrate(e^(n*arctan(b*x + a)), x)

Giac [F]

\[ \int e^{n \arctan (a+b x)} \, dx=\int { e^{\left (n \arctan \left (b x + a\right )\right )} \,d x } \]

[In]

integrate(exp(n*arctan(b*x+a)),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

int(exp(n*atan(a + b*x)),x)

[Out]

int(exp(n*atan(a + b*x)), x)