Integrand size = 12, antiderivative size = 92 \[ \int e^{-3 i \arctan (a x)} x \, dx=-\frac {9 \sqrt {1+a^2 x^2}}{2 a^2}-\frac {3 \left (1+a^2 x^2\right )^{3/2}}{2 a^2 (1+i a x)}-\frac {\left (1+a^2 x^2\right )^{5/2}}{a^2 (1+i a x)^3}-\frac {9 i \text {arcsinh}(a x)}{2 a^2} \]
-3/2*(a^2*x^2+1)^(3/2)/a^2/(1+I*a*x)-(a^2*x^2+1)^(5/2)/a^2/(1+I*a*x)^3-9/2 *I*arcsinh(a*x)/a^2-9/2*(a^2*x^2+1)^(1/2)/a^2
Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.65 \[ \int e^{-3 i \arctan (a x)} x \, dx=\sqrt {1+a^2 x^2} \left (-\frac {3}{a^2}+\frac {i x}{2 a}+\frac {4 i}{a^2 (-i+a x)}\right )-\frac {9 i \text {arcsinh}(a x)}{2 a^2} \]
Sqrt[1 + a^2*x^2]*(-3/a^2 + ((I/2)*x)/a + (4*I)/(a^2*(-I + a*x))) - (((9*I )/2)*ArcSinh[a*x])/a^2
Time = 0.55 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.97, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5583, 2164, 25, 2027, 2164, 27, 563, 25, 2346, 27, 455, 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.
\(\displaystyle \int x e^{-3 i \arctan (a x)} \, dx\) |
\(\Big \downarrow \) 5583 |
\(\displaystyle \int \frac {x (1-i a x)^2}{(1+i a x) \sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 2164 |
\(\displaystyle i a \int -\frac {\left (x^2+\frac {i x}{a}\right ) \sqrt {a^2 x^2+1}}{(i a x+1)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -i a \int \frac {\left (x^2+\frac {i x}{a}\right ) \sqrt {a^2 x^2+1}}{(i a x+1)^2}dx\) |
\(\Big \downarrow \) 2027 |
\(\displaystyle -i a \int \frac {x \left (x+\frac {i}{a}\right ) \sqrt {a^2 x^2+1}}{(i a x+1)^2}dx\) |
\(\Big \downarrow \) 2164 |
\(\displaystyle a^2 \int \frac {x \left (a^2 x^2+1\right )^{3/2}}{a^2 (i a x+1)^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x \left (a^2 x^2+1\right )^{3/2}}{(1+i a x)^3}dx\) |
\(\Big \downarrow \) 563 |
\(\displaystyle \frac {i \int -\frac {-a^2 x^2-3 i a x+4}{\sqrt {a^2 x^2+1}}dx}{a}-\frac {4 \sqrt {a^2 x^2+1}}{a^2 (1+i a x)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {i \int \frac {-a^2 x^2-3 i a x+4}{\sqrt {a^2 x^2+1}}dx}{a}-\frac {4 \sqrt {a^2 x^2+1}}{a^2 (1+i a x)}\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle -\frac {i \left (-\frac {1}{2} x \sqrt {a^2 x^2+1}+\frac {\int \frac {3 a^2 (3-2 i a x)}{\sqrt {a^2 x^2+1}}dx}{2 a^2}\right )}{a}-\frac {4 \sqrt {a^2 x^2+1}}{a^2 (1+i a x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {i \left (-\frac {1}{2} x \sqrt {a^2 x^2+1}+\frac {3}{2} \int \frac {3-2 i a x}{\sqrt {a^2 x^2+1}}dx\right )}{a}-\frac {4 \sqrt {a^2 x^2+1}}{a^2 (1+i a x)}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle -\frac {i \left (-\frac {1}{2} x \sqrt {a^2 x^2+1}+\frac {3}{2} \left (3 \int \frac {1}{\sqrt {a^2 x^2+1}}dx-\frac {2 i \sqrt {a^2 x^2+1}}{a}\right )\right )}{a}-\frac {4 \sqrt {a^2 x^2+1}}{a^2 (1+i a x)}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle -\frac {i \left (-\frac {1}{2} x \sqrt {a^2 x^2+1}+\frac {3}{2} \left (\frac {3 \text {arcsinh}(a x)}{a}-\frac {2 i \sqrt {a^2 x^2+1}}{a}\right )\right )}{a}-\frac {4 \sqrt {a^2 x^2+1}}{a^2 (1+i a x)}\) |
(-4*Sqrt[1 + a^2*x^2])/(a^2*(1 + I*a*x)) - (I*(-1/2*(x*Sqrt[1 + a^2*x^2]) + (3*(((-2*I)*Sqrt[1 + a^2*x^2])/a + (3*ArcSinh[a*x])/a))/2))/a
3.1.54.3.1 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[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[(-(-c)^(m - n - 2))*d^(2*n - m + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)* b^(n + 2)*(c + d*x))), x] - Simp[d^(2*n - m + 2)/b^(n + 1) Int[(1/Sqrt[a + b*x^2])*ExpandToSum[(2^(-n - 1)*(-c)^(m - n - 1) - d^m*x^m*(-c + d*x)^(-n - 1))/(c + d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2 , 0] && IGtQ[m, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[p, 1] && EqQ[u, 1])
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*e Int[(d + e*x)^(m - 1)*PolynomialQuotient[Pq, a*e + b*d*x, x]* (a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && EqQ[PolynomialRemainder[Pq, a*e + b*d*x, x], 0 ]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1)) Int[(a + b*x^2)^p*ExpandToS um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && !LeQ[p, -1]
Int[E^(ArcTan[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a* x)^((I*n + 1)/2)/((1 + I*a*x)^((I*n - 1)/2)*Sqrt[1 + a^2*x^2])), x] /; Free Q[{a, m}, x] && IntegerQ[(I*n - 1)/2]
Time = 0.33 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.15
method | result | size |
risch | \(\frac {i \left (a x +6 i\right ) \sqrt {a^{2} x^{2}+1}}{2 a^{2}}-\frac {i \left (\frac {9 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}-\frac {8 \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{a^{2} \left (x -\frac {i}{a}\right )}\right )}{2 a}\) | \(106\) |
default | \(-\frac {\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{3}}-2 i a \left (-\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{2}}+3 i a \left (\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{3}+i a \left (\frac {\left (2 \left (x -\frac {i}{a}\right ) a^{2}+2 i a \right ) \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{4 a^{2}}+\frac {\ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{2 \sqrt {a^{2}}}\right )\right )\right )}{a^{4}}+\frac {i \left (-\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{2}}+3 i a \left (\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{3}+i a \left (\frac {\left (2 \left (x -\frac {i}{a}\right ) a^{2}+2 i a \right ) \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{4 a^{2}}+\frac {\ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{2 \sqrt {a^{2}}}\right )\right )\right )}{a^{3}}\) | \(463\) |
1/2*I*(a*x+6*I)*(a^2*x^2+1)^(1/2)/a^2-1/2*I/a*(9*ln(a^2*x/(a^2)^(1/2)+(a^2 *x^2+1)^(1/2))/(a^2)^(1/2)-8/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 = 72, normalized size of antiderivative = 0.78 \[ \int e^{-3 i \arctan (a x)} x \, dx=\frac {8 i \, a x - 9 \, {\left (-i \, a x - 1\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) + \sqrt {a^{2} x^{2} + 1} {\left (i \, a^{2} x^{2} - 5 \, a x + 14 i\right )} + 8}{2 \, {\left (a^{3} x - i \, a^{2}\right )}} \]
1/2*(8*I*a*x - 9*(-I*a*x - 1)*log(-a*x + sqrt(a^2*x^2 + 1)) + sqrt(a^2*x^2 + 1)*(I*a^2*x^2 - 5*a*x + 14*I) + 8)/(a^3*x - I*a^2)
\[ \int e^{-3 i \arctan (a x)} x \, dx=i \left (\int \frac {x \sqrt {a^{2} x^{2} + 1}}{a^{3} x^{3} - 3 i a^{2} x^{2} - 3 a x + i}\, dx + \int \frac {a^{2} x^{3} \sqrt {a^{2} x^{2} + 1}}{a^{3} x^{3} - 3 i a^{2} x^{2} - 3 a x + i}\, dx\right ) \]
I*(Integral(x*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**3*sqrt(a**2*x**2 + 1)/(a**3*x**3 - 3*I*a**2*x**2 - 3*a*x + I), x))
Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.22 \[ \int e^{-3 i \arctan (a x)} x \, dx=-\frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{4} x^{2} - 2 i \, a^{3} x - a^{2}} - \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{2 i \, a^{3} x + 2 \, a^{2}} - \frac {6 \, \sqrt {a^{2} x^{2} + 1}}{i \, a^{3} x + a^{2}} - \frac {9 i \, \operatorname {arsinh}\left (a x\right )}{2 \, a^{2}} - \frac {3 \, \sqrt {a^{2} x^{2} + 1}}{2 \, a^{2}} \]
-(a^2*x^2 + 1)^(3/2)/(a^4*x^2 - 2*I*a^3*x - a^2) - (a^2*x^2 + 1)^(3/2)/(2* I*a^3*x + 2*a^2) - 6*sqrt(a^2*x^2 + 1)/(I*a^3*x + a^2) - 9/2*I*arcsinh(a*x )/a^2 - 3/2*sqrt(a^2*x^2 + 1)/a^2
\[ \int e^{-3 i \arctan (a x)} x \, dx=\int { \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{{\left (i \, a x + 1\right )}^{3}} \,d x } \]
Time = 0.48 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.14 \[ \int e^{-3 i \arctan (a x)} x \, dx=-\frac {\sqrt {a^2\,x^2+1}\,\left (\frac {3\,\sqrt {a^2}}{a^2}-\frac {x\,\sqrt {a^2}\,1{}\mathrm {i}}{2\,a}\right )}{\sqrt {a^2}}-\frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )\,9{}\mathrm {i}}{2\,a\,\sqrt {a^2}}-\frac {\sqrt {a^2\,x^2+1}\,4{}\mathrm {i}}{a\,\left (-x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]