Integrand size = 10, antiderivative size = 58 \[ \int e^{-3 i \arctan (a x)} \, dx=\frac {4 i (1-i a x)}{a \sqrt {1+a^2 x^2}}+\frac {i \sqrt {1+a^2 x^2}}{a}-\frac {3 \text {arcsinh}(a x)}{a} \] Output:
4*I*(1-I*a*x)/a/(a^2*x^2+1)^(1/2)+I*(a^2*x^2+1)^(1/2)/a-3*arcsinh(a*x)/a
Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.72 \[ \int e^{-3 i \arctan (a x)} \, dx=\frac {\sqrt {1+a^2 x^2} \left (i+\frac {4}{-i+a x}\right )}{a}-\frac {3 \text {arcsinh}(a x)}{a} \] Input:
Integrate[E^((-3*I)*ArcTan[a*x]),x]
Output:
(Sqrt[1 + a^2*x^2]*(I + 4/(-I + a*x)))/a - (3*ArcSinh[a*x])/a
Time = 0.38 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.03, 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 |
\(\displaystyle \frac {i \sqrt {a^2 x^2+1}}{a}-\frac {\int -\frac {a^4 (1-3 i a x)}{(i a x+1) \sqrt {a^2 x^2+1}}dx}{a^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {a^4 (1-3 i a x)}{(i a x+1) \sqrt {a^2 x^2+1}}dx}{a^4}+\frac {i \sqrt {a^2 x^2+1}}{a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1-3 i a x}{(i a x+1) \sqrt {a^2 x^2+1}}dx+\frac {i \sqrt {a^2 x^2+1}}{a}\) |
\(\Big \downarrow \) 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 |
\(\displaystyle \frac {i \sqrt {a^2 x^2+1}}{a}+\frac {4 i \sqrt {a^2 x^2+1}}{a (1+i a x)}-\frac {3 \text {arcsinh}(a x)}{a}\) |
Input:
Int[E^((-3*I)*ArcTan[a*x]),x]
Output:
(I*Sqrt[1 + a^2*x^2])/a + ((4*I)*Sqrt[1 + a^2*x^2])/(a*(1 + I*a*x)) - (3*A rcSinh[a*x])/a
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 = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.60
method | result | size |
risch | \(\frac {i \sqrt {a^{2} x^{2}+1}}{a}+\frac {4 \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{a^{2} \left (x -\frac {i}{a}\right )}-\frac {3 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}\) | \(93\) |
default | \(\frac {i \left (\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{3}}-2 i a \left (-\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{2}}+3 i a \left (\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{3}+i a \left (\frac {\left (2 \left (x -\frac {i}{a}\right ) a^{2}+2 i a \right ) \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{4 a^{2}}+\frac {\ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{2 \sqrt {a^{2}}}\right )\right )\right )\right )}{a^{3}}\) | \(257\) |
Input:
int(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2),x,method=_RETURNVERBOSE)
Output:
I*(a^2*x^2+1)^(1/2)/a+4/a^2/(x-I/a)*((x-I/a)^2*a^2+2*I*a*(x-I/a))^(1/2)-3* ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/2)
Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.03 \[ \int e^{-3 i \arctan (a x)} \, dx=\frac {4 \, a x + 3 \, {\left (a x - i\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) + \sqrt {a^{2} x^{2} + 1} {\left (i \, a x + 5\right )} - 4 i}{a^{2} x - i \, a} \] Input:
integrate(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2),x, algorithm="fricas")
Output:
(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)
\[ \int e^{-3 i \arctan (a x)} \, dx=i \left (\int \frac {\sqrt {a^{2} x^{2} + 1}}{a^{3} x^{3} - 3 i a^{2} x^{2} - 3 a x + i}\, dx + \int \frac {a^{2} x^{2} \sqrt {a^{2} x^{2} + 1}}{a^{3} x^{3} - 3 i a^{2} x^{2} - 3 a x + i}\, dx\right ) \] Input:
integrate(1/(1+I*a*x)**3*(a**2*x**2+1)**(3/2),x)
Output:
I*(Integral(sqrt(a**2*x**2 + 1)/(a**3*x**3 - 3*I*a**2*x**2 - 3*a*x + I), x ) + Integral(a**2*x**2*sqrt(a**2*x**2 + 1)/(a**3*x**3 - 3*I*a**2*x**2 - 3* a*x + I), x))
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.12 \[ \int e^{-3 i \arctan (a x)} \, dx=\frac {i \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{3} x^{2} - 2 i \, a^{2} x - a} - \frac {3 \, \operatorname {arsinh}\left (a x\right )}{a} + \frac {6 i \, \sqrt {a^{2} x^{2} + 1}}{i \, a^{2} x + a} \] Input:
integrate(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2),x, algorithm="maxima")
Output:
I*(a^2*x^2 + 1)^(3/2)/(a^3*x^2 - 2*I*a^2*x - a) - 3*arcsinh(a*x)/a + 6*I*s qrt(a^2*x^2 + 1)/(I*a^2*x + a)
\[ \int e^{-3 i \arctan (a x)} \, dx=\int { \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (i \, a x + 1\right )}^{3}} \,d x } \] Input:
integrate(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2),x, algorithm="giac")
Output:
undef
Time = 22.71 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.26 \[ \int e^{-3 i \arctan (a x)} \, dx=\frac {\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}}{a}-\frac {3\,\mathrm {asinh}\left (x\,\sqrt {a^2}\right )}{\sqrt {a^2}}-\frac {4\,\sqrt {a^2\,x^2+1}}{\left (-x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \] Input:
int((a^2*x^2 + 1)^(3/2)/(a*x*1i + 1)^3,x)
Output:
((a^2*x^2 + 1)^(1/2)*1i)/a - (3*asinh(x*(a^2)^(1/2)))/(a^2)^(1/2) - (4*(a^ 2*x^2 + 1)^(1/2))/((((a^2)^(1/2)*1i)/a - x*(a^2)^(1/2))*(a^2)^(1/2))
\[ \int e^{-3 i \arctan (a x)} \, dx=-\left (\int \frac {\sqrt {a^{2} x^{2}+1}}{a^{3} i \,x^{3}+3 a^{2} x^{2}-3 a i x -1}d x \right )-\left (\int \frac {\sqrt {a^{2} x^{2}+1}\, x^{2}}{a^{3} i \,x^{3}+3 a^{2} x^{2}-3 a i x -1}d x \right ) a^{2} \] Input:
int(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2),x)
Output:
- (int(sqrt(a**2*x**2 + 1)/(a**3*i*x**3 + 3*a**2*x**2 - 3*a*i*x - 1),x) + int((sqrt(a**2*x**2 + 1)*x**2)/(a**3*i*x**3 + 3*a**2*x**2 - 3*a*i*x - 1), x)*a**2)