Integrand size = 10, antiderivative size = 60 \[ \int e^{3 i \arctan (a x)} \, dx=-\frac {2 i (1+i a x)^2}{a \sqrt {1+a^2 x^2}}-\frac {3 i \sqrt {1+a^2 x^2}}{a}-\frac {3 \text {arcsinh}(a x)}{a} \]
Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70 \[ \int e^{3 i \arctan (a x)} \, dx=\frac {\sqrt {1+a^2 x^2} \left (-i+\frac {4}{i+a x}\right )}{a}-\frac {3 \text {arcsinh}(a x)}{a} \]
Time = 0.23 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5582, 711, 25, 27, 671, 222}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int e^{3 i \arctan (a x)} \, dx\) |
\(\Big \downarrow \) 5582 |
\(\displaystyle \int \frac {(1+i a x)^2}{(1-i a x) \sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 711 |
\(\displaystyle -\frac {\int -\frac {a^4 (3 i a x+1)}{(1-i a x) \sqrt {a^2 x^2+1}}dx}{a^4}-\frac {i \sqrt {a^2 x^2+1}}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {a^4 (3 i a x+1)}{(1-i a x) \sqrt {a^2 x^2+1}}dx}{a^4}-\frac {i \sqrt {a^2 x^2+1}}{a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {3 i a x+1}{(1-i a x) \sqrt {a^2 x^2+1}}dx-\frac {i \sqrt {a^2 x^2+1}}{a}\) |
\(\Big \downarrow \) 671 |
\(\displaystyle -3 \int \frac {1}{\sqrt {a^2 x^2+1}}dx-\frac {i \sqrt {a^2 x^2+1}}{a}-\frac {4 i \sqrt {a^2 x^2+1}}{a (1-i a x)}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle -\frac {i \sqrt {a^2 x^2+1}}{a}-\frac {4 i \sqrt {a^2 x^2+1}}{a (1-i a x)}-\frac {3 \text {arcsinh}(a x)}{a}\) |
3.1.22.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ ), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m + p + 1)) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))^(n_.)*((a_.) + (c_.)*(x_ )^2)^(p_), x_Symbol] :> Simp[g^n*(d + e*x)^(m + n - 1)*((a + c*x^2)^(p + 1) /(c*e^(n - 1)*(m + n + 2*p + 1))), x] + Simp[1/(c*e^n*(m + n + 2*p + 1)) Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^n*(m + n + 2*p + 1)*(f + g*x) ^n - c*g^n*(m + n + 2*p + 1)*(d + e*x)^n - 2*e*g^n*(m + p + n)*(d + e*x)^(n - 2)*(a*e - c*d*x), x], x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && Eq Q[c*d^2 + a*e^2, 0] && IGtQ[n, 0] && NeQ[m + n + 2*p + 1, 0]
Int[E^(ArcTan[(a_.)*(x_)]*(n_)), x_Symbol] :> Int[(1 - I*a*x)^((I*n + 1)/2) /((1 + I*a*x)^((I*n - 1)/2)*Sqrt[1 + a^2*x^2]), x] /; FreeQ[a, x] && Intege rQ[(I*n - 1)/2]
Time = 0.35 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.55
method | result | size |
risch | \(-\frac {i \sqrt {a^{2} x^{2}+1}}{a}-\frac {3 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}+\frac {4 \sqrt {\left (x +\frac {i}{a}\right )^{2} a^{2}-2 i a \left (x +\frac {i}{a}\right )}}{a^{2} \left (x +\frac {i}{a}\right )}\) | \(93\) |
default | \(\frac {x}{\sqrt {a^{2} x^{2}+1}}-3 a^{2} \left (-\frac {x}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{a^{2} \sqrt {a^{2}}}\right )-i a^{3} \left (\frac {x^{2}}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {a^{2} x^{2}+1}}\right )-\frac {3 i}{a \sqrt {a^{2} x^{2}+1}}\) | \(128\) |
meijerg | \(\frac {x}{\sqrt {a^{2} x^{2}+1}}+\frac {3 i \left (\sqrt {\pi }-\frac {\sqrt {\pi }}{\sqrt {a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}-\frac {3 \left (-\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\sqrt {\pi }\, \left (a^{2}\right )^{\frac {3}{2}} \operatorname {arcsinh}\left (a x \right )}{a^{3}}\right )}{\sqrt {\pi }\, \sqrt {a^{2}}}-\frac {i \left (-2 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right )}{4 \sqrt {a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}\) | \(137\) |
-I*(a^2*x^2+1)^(1/2)/a-3*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/ 2)+4/a^2/(x+I/a)*((x+I/a)^2*a^2-2*I*a*(x+I/a))^(1/2)
Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int e^{3 i \arctan (a x)} \, dx=\frac {4 \, a x + 3 \, {\left (a x + i\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) + \sqrt {a^{2} x^{2} + 1} {\left (-i \, a x + 5\right )} + 4 i}{a^{2} x + i \, a} \]
(4*a*x + 3*(a*x + I)*log(-a*x + sqrt(a^2*x^2 + 1)) + sqrt(a^2*x^2 + 1)*(-I *a*x + 5) + 4*I)/(a^2*x + I*a)
\[ \int e^{3 i \arctan (a x)} \, dx=- i \left (\int \frac {i}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{3}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{2}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \]
-I*(Integral(I/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x) + Integral(-3*a*x/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x) + Integral(a**3*x**3/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1) ), x) + Integral(-3*I*a**2*x**2/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2 *x**2 + 1)), x))
Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95 \[ \int e^{3 i \arctan (a x)} \, dx=-\frac {i \, a x^{2}}{\sqrt {a^{2} x^{2} + 1}} + \frac {4 \, x}{\sqrt {a^{2} x^{2} + 1}} - \frac {3 \, \operatorname {arsinh}\left (a x\right )}{a} - \frac {5 i}{\sqrt {a^{2} x^{2} + 1} a} \]
-I*a*x^2/sqrt(a^2*x^2 + 1) + 4*x/sqrt(a^2*x^2 + 1) - 3*arcsinh(a*x)/a - 5* I/(sqrt(a^2*x^2 + 1)*a)
\[ \int e^{3 i \arctan (a x)} \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{3}}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.47 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.20 \[ \int e^{3 i \arctan (a x)} \, dx=-\frac {\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}}{a}-\frac {3\,\mathrm {asinh}\left (x\,\sqrt {a^2}\right )}{\sqrt {a^2}}+\frac {4\,\sqrt {a^2\,x^2+1}}{\left (x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]