Integrand size = 14, antiderivative size = 137 \[ \int e^{-3 i \arctan (a x)} x^3 \, dx=\frac {(1-i a x)^3}{a^4 \sqrt {1+a^2 x^2}}+\frac {27 \sqrt {1+a^2 x^2}}{4 a^4}-\frac {x^2 \sqrt {1+a^2 x^2}}{a^2}+\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}-\frac {9 i (2 i+3 a x) \sqrt {1+a^2 x^2}}{8 a^4}+\frac {51 i \text {arcsinh}(a x)}{8 a^4} \]
51/8*I*arcsinh(a*x)/a^4+(1-I*a*x)^3/a^4/(a^2*x^2+1)^(1/2)+27/4*(a^2*x^2+1) ^(1/2)/a^4-x^2*(a^2*x^2+1)^(1/2)/a^2+1/4*I*x^3*(a^2*x^2+1)^(1/2)/a-9/8*I*( 2*I+3*a*x)*(a^2*x^2+1)^(1/2)/a^4
Time = 0.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.58 \[ \int e^{-3 i \arctan (a x)} x^3 \, dx=\sqrt {1+a^2 x^2} \left (\frac {6}{a^4}-\frac {19 i x}{8 a^3}-\frac {x^2}{a^2}+\frac {i x^3}{4 a}-\frac {4 i}{a^4 (-i+a x)}\right )+\frac {51 i \text {arcsinh}(a x)}{8 a^4} \]
Sqrt[1 + a^2*x^2]*(6/a^4 - (((19*I)/8)*x)/a^3 - x^2/a^2 + ((I/4)*x^3)/a - (4*I)/(a^4*(-I + a*x))) + (((51*I)/8)*ArcSinh[a*x])/a^4
Time = 0.80 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.12, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.071, Rules used = {5583, 2164, 25, 2027, 2164, 27, 563, 25, 2346, 2346, 27, 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.
| \(\displaystyle \int x^3 e^{-3 i \arctan (a x)} \, dx\) |
| \(\Big \downarrow \) 5583 |
| \(\displaystyle \int \frac {x^3 (1-i a x)^2}{(1+i a x) \sqrt {a^2 x^2+1}}dx\) |
| \(\Big \downarrow \) 2164 |
| \(\displaystyle i a \int -\frac {\sqrt {a^2 x^2+1} \left (x^4+\frac {i x^3}{a}\right )}{(i a x+1)^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -i a \int \frac {\sqrt {a^2 x^2+1} \left (x^4+\frac {i x^3}{a}\right )}{(i a x+1)^2}dx\) |
| \(\Big \downarrow \) 2027 |
| \(\displaystyle -i a \int \frac {x^3 \left (x+\frac {i}{a}\right ) \sqrt {a^2 x^2+1}}{(i a x+1)^2}dx\) |
| \(\Big \downarrow \) 2164 |
| \(\displaystyle a^2 \int \frac {x^3 \left (a^2 x^2+1\right )^{3/2}}{a^2 (i a x+1)^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {x^3 \left (a^2 x^2+1\right )^{3/2}}{(1+i a x)^3}dx\) |
| \(\Big \downarrow \) 563 |
| \(\displaystyle \frac {4 \sqrt {a^2 x^2+1}}{a^4 (1+i a x)}-\frac {i \int -\frac {a^4 x^4+3 i a^3 x^3-4 a^2 x^2-4 i a x+4}{\sqrt {a^2 x^2+1}}dx}{a^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {i \int \frac {a^4 x^4+3 i a^3 x^3-4 a^2 x^2-4 i a x+4}{\sqrt {a^2 x^2+1}}dx}{a^3}+\frac {4 \sqrt {a^2 x^2+1}}{a^4 (1+i a x)}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {i \left (\frac {1}{4} a^2 x^3 \sqrt {a^2 x^2+1}+\frac {\int \frac {12 i x^3 a^5-19 x^2 a^4-16 i x a^3+16 a^2}{\sqrt {a^2 x^2+1}}dx}{4 a^2}\right )}{a^3}+\frac {4 \sqrt {a^2 x^2+1}}{a^4 (1+i a x)}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {i \left (\frac {1}{4} a^2 x^3 \sqrt {a^2 x^2+1}+\frac {\frac {\int \frac {3 \left (-19 x^2 a^6-24 i x a^5+16 a^4\right )}{\sqrt {a^2 x^2+1}}dx}{3 a^2}+4 i a^3 x^2 \sqrt {a^2 x^2+1}}{4 a^2}\right )}{a^3}+\frac {4 \sqrt {a^2 x^2+1}}{a^4 (1+i a x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {i \left (\frac {1}{4} a^2 x^3 \sqrt {a^2 x^2+1}+\frac {\frac {\int \frac {-19 x^2 a^6-24 i x a^5+16 a^4}{\sqrt {a^2 x^2+1}}dx}{a^2}+4 i a^3 x^2 \sqrt {a^2 x^2+1}}{4 a^2}\right )}{a^3}+\frac {4 \sqrt {a^2 x^2+1}}{a^4 (1+i a x)}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {i \left (\frac {1}{4} a^2 x^3 \sqrt {a^2 x^2+1}+\frac {\frac {-\frac {19}{2} a^4 x \sqrt {a^2 x^2+1}+\frac {\int \frac {3 a^6 (17-16 i a x)}{\sqrt {a^2 x^2+1}}dx}{2 a^2}}{a^2}+4 i a^3 x^2 \sqrt {a^2 x^2+1}}{4 a^2}\right )}{a^3}+\frac {4 \sqrt {a^2 x^2+1}}{a^4 (1+i a x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {i \left (\frac {1}{4} a^2 x^3 \sqrt {a^2 x^2+1}+\frac {\frac {-\frac {19}{2} a^4 x \sqrt {a^2 x^2+1}+\frac {3}{2} a^4 \int \frac {17-16 i a x}{\sqrt {a^2 x^2+1}}dx}{a^2}+4 i a^3 x^2 \sqrt {a^2 x^2+1}}{4 a^2}\right )}{a^3}+\frac {4 \sqrt {a^2 x^2+1}}{a^4 (1+i a x)}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle \frac {i \left (\frac {1}{4} a^2 x^3 \sqrt {a^2 x^2+1}+\frac {\frac {-\frac {19}{2} a^4 x \sqrt {a^2 x^2+1}+\frac {3}{2} a^4 \left (17 \int \frac {1}{\sqrt {a^2 x^2+1}}dx-\frac {16 i \sqrt {a^2 x^2+1}}{a}\right )}{a^2}+4 i a^3 x^2 \sqrt {a^2 x^2+1}}{4 a^2}\right )}{a^3}+\frac {4 \sqrt {a^2 x^2+1}}{a^4 (1+i a x)}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {4 \sqrt {a^2 x^2+1}}{a^4 (1+i a x)}+\frac {i \left (\frac {1}{4} a^2 x^3 \sqrt {a^2 x^2+1}+\frac {\frac {-\frac {19}{2} a^4 x \sqrt {a^2 x^2+1}+\frac {3}{2} a^4 \left (\frac {17 \text {arcsinh}(a x)}{a}-\frac {16 i \sqrt {a^2 x^2+1}}{a}\right )}{a^2}+4 i a^3 x^2 \sqrt {a^2 x^2+1}}{4 a^2}\right )}{a^3}\) |
(4*Sqrt[1 + a^2*x^2])/(a^4*(1 + I*a*x)) + (I*((a^2*x^3*Sqrt[1 + a^2*x^2])/ 4 + ((4*I)*a^3*x^2*Sqrt[1 + a^2*x^2] + ((-19*a^4*x*Sqrt[1 + a^2*x^2])/2 + (3*a^4*(((-16*I)*Sqrt[1 + a^2*x^2])/a + (17*ArcSinh[a*x])/a))/2)/a^2)/(4*a ^2)))/a^3
3.1.52.3.1 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 = 0.37 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91
| method | result | size |
| risch | \(\frac {i \left (2 a^{3} x^{3}+8 i a^{2} x^{2}-19 a x -48 i\right ) \sqrt {a^{2} x^{2}+1}}{8 a^{4}}+\frac {i \left (\frac {51 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}-\frac {32 \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{a^{2} \left (x -\frac {i}{a}\right )}\right )}{8 a^{3}}\) | \(124\) |
| default | \(\frac {i \left (\frac {x \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{4}+\frac {3 \sqrt {a^{2} x^{2}+1}\, x}{8}+\frac {3 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{8 \sqrt {a^{2}}}\right )}{a^{3}}+\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^{6}}-\frac {3 \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 )}{a^{4}}-\frac {3 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^{5}}\) | \(682\) |
1/8*I*(2*a^3*x^3+8*I*a^2*x^2-19*a*x-48*I)*(a^2*x^2+1)^(1/2)/a^4+1/8*I/a^3* (51*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/2)-32/a^2/(x-I/a)*((x -I/a)^2*a^2+2*I*a*(x-I/a))^(1/2))
Time = 0.25 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.64 \[ \int e^{-3 i \arctan (a x)} x^3 \, dx=\frac {-32 i \, a x - 51 \, {\left (i \, a x + 1\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) + {\left (2 i \, a^{4} x^{4} - 6 \, a^{3} x^{3} - 11 i \, a^{2} x^{2} + 29 \, a x - 80 i\right )} \sqrt {a^{2} x^{2} + 1} - 32}{8 \, {\left (a^{5} x - i \, a^{4}\right )}} \]
1/8*(-32*I*a*x - 51*(I*a*x + 1)*log(-a*x + sqrt(a^2*x^2 + 1)) + (2*I*a^4*x ^4 - 6*a^3*x^3 - 11*I*a^2*x^2 + 29*a*x - 80*I)*sqrt(a^2*x^2 + 1) - 32)/(a^ 5*x - I*a^4)
\[ \int e^{-3 i \arctan (a x)} x^3 \, dx=i \left (\int \frac {x^{3} \sqrt {a^{2} x^{2} + 1}}{a^{3} x^{3} - 3 i a^{2} x^{2} - 3 a x + i}\, dx + \int \frac {a^{2} x^{5} \sqrt {a^{2} x^{2} + 1}}{a^{3} x^{3} - 3 i a^{2} x^{2} - 3 a x + i}\, dx\right ) \]
I*(Integral(x**3*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**5*sqrt(a**2*x**2 + 1)/(a**3*x**3 - 3*I*a**2*x**2 - 3*a*x + I), x))
Time = 0.28 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.58 \[ \int e^{-3 i \arctan (a x)} x^3 \, dx=\frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{6} x^{2} - 2 i \, a^{5} x - a^{4}} + \frac {3 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{2 i \, a^{5} x + 2 \, a^{4}} + \frac {6 \, \sqrt {a^{2} x^{2} + 1}}{i \, a^{5} x + a^{4}} + \frac {i \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{4 \, a^{3}} + \frac {3 i \, \sqrt {a^{2} x^{2} + 1} x}{8 \, a^{3}} - \frac {3 i \, \sqrt {-a^{2} x^{2} + 4 i \, a x + 3} x}{2 \, a^{3}} - \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{4}} + \frac {3 i \, \arcsin \left (i \, a x + 2\right )}{2 \, a^{4}} + \frac {63 i \, \operatorname {arsinh}\left (a x\right )}{8 \, a^{4}} + \frac {9 \, \sqrt {a^{2} x^{2} + 1}}{2 \, a^{4}} - \frac {3 \, \sqrt {-a^{2} x^{2} + 4 i \, a x + 3}}{a^{4}} \]
(a^2*x^2 + 1)^(3/2)/(a^6*x^2 - 2*I*a^5*x - a^4) + 3*(a^2*x^2 + 1)^(3/2)/(2 *I*a^5*x + 2*a^4) + 6*sqrt(a^2*x^2 + 1)/(I*a^5*x + a^4) + 1/4*I*(a^2*x^2 + 1)^(3/2)*x/a^3 + 3/8*I*sqrt(a^2*x^2 + 1)*x/a^3 - 3/2*I*sqrt(-a^2*x^2 + 4* I*a*x + 3)*x/a^3 - (a^2*x^2 + 1)^(3/2)/a^4 + 3/2*I*arcsin(I*a*x + 2)/a^4 + 63/8*I*arcsinh(a*x)/a^4 + 9/2*sqrt(a^2*x^2 + 1)/a^4 - 3*sqrt(-a^2*x^2 + 4 *I*a*x + 3)/a^4
Exception generated. \[ \int e^{-3 i \arctan (a x)} x^3 \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Time = 0.51 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.01 \[ \int e^{-3 i \arctan (a x)} x^3 \, dx=\frac {\sqrt {a^2\,x^2+1}\,\left (\frac {4}{{\left (a^2\right )}^{3/2}}+\frac {2\,\sqrt {a^2}}{a^4}-\frac {x^2\,\sqrt {a^2}}{a^2}+\frac {x^3\,{\left (a^2\right )}^{3/2}\,1{}\mathrm {i}}{4\,a^3}-\frac {x\,\sqrt {a^2}\,19{}\mathrm {i}}{8\,a^3}\right )}{\sqrt {a^2}}+\frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )\,51{}\mathrm {i}}{8\,a^3\,\sqrt {a^2}}+\frac {\sqrt {a^2\,x^2+1}\,4{}\mathrm {i}}{a^3\,\left (-x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]