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 = 0.81 (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, 2027, 2164, 25, 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 |
| \(\displaystyle -i a \int \frac {\sqrt {a^2 x^2+1} \left (\frac {i x^3}{a}-x^4\right )}{(1-i a x)^2}dx\) |
| \(\Big \downarrow \) 2027 |
| \(\displaystyle -i a \int \frac {\left (\frac {i}{a}-x\right ) x^3 \sqrt {a^2 x^2+1}}{(1-i a x)^2}dx\) |
| \(\Big \downarrow \) 2164 |
| \(\displaystyle -a^2 \int -\frac {x^3 \left (a^2 x^2+1\right )^{3/2}}{a^2 (1-i a x)^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle a^2 \int \frac {x^3 \left (a^2 x^2+1\right )^{3/2}}{a^2 (1-i a x)^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {x^3 \left (a^2 x^2+1\right )^{3/2}}{(1-i a x)^3}dx\) |
| \(\Big \downarrow \) 563 |
| \(\displaystyle \frac {i \int -\frac {a^4 x^4-3 i a^3 x^3-4 a^2 x^2+4 i a x+4}{\sqrt {a^2 x^2+1}}dx}{a^3}+\frac {4 \sqrt {a^2 x^2+1}}{a^4 (1-i a x)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {4 \sqrt {a^2 x^2+1}}{a^4 (1-i a x)}-\frac {i \int \frac {a^4 x^4-3 i a^3 x^3-4 a^2 x^2+4 i a x+4}{\sqrt {a^2 x^2+1}}dx}{a^3}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {4 \sqrt {a^2 x^2+1}}{a^4 (1-i a x)}-\frac {i \left (\frac {1}{4} a^2 x^3 \sqrt {a^2 x^2+1}+\frac {\int \frac {-12 i x^3 a^5-19 x^2 a^4+16 i x a^3+16 a^2}{\sqrt {a^2 x^2+1}}dx}{4 a^2}\right )}{a^3}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {4 \sqrt {a^2 x^2+1}}{a^4 (1-i a x)}-\frac {i \left (\frac {1}{4} a^2 x^3 \sqrt {a^2 x^2+1}+\frac {\frac {\int \frac {3 \left (-19 x^2 a^6+24 i x a^5+16 a^4\right )}{\sqrt {a^2 x^2+1}}dx}{3 a^2}-4 i a^3 x^2 \sqrt {a^2 x^2+1}}{4 a^2}\right )}{a^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4 \sqrt {a^2 x^2+1}}{a^4 (1-i a x)}-\frac {i \left (\frac {1}{4} a^2 x^3 \sqrt {a^2 x^2+1}+\frac {\frac {\int \frac {-19 x^2 a^6+24 i x a^5+16 a^4}{\sqrt {a^2 x^2+1}}dx}{a^2}-4 i a^3 x^2 \sqrt {a^2 x^2+1}}{4 a^2}\right )}{a^3}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {4 \sqrt {a^2 x^2+1}}{a^4 (1-i a x)}-\frac {i \left (\frac {1}{4} a^2 x^3 \sqrt {a^2 x^2+1}+\frac {\frac {-\frac {19}{2} a^4 x \sqrt {a^2 x^2+1}+\frac {\int \frac {3 a^6 (16 i a x+17)}{\sqrt {a^2 x^2+1}}dx}{2 a^2}}{a^2}-4 i a^3 x^2 \sqrt {a^2 x^2+1}}{4 a^2}\right )}{a^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4 \sqrt {a^2 x^2+1}}{a^4 (1-i a x)}-\frac {i \left (\frac {1}{4} a^2 x^3 \sqrt {a^2 x^2+1}+\frac {\frac {-\frac {19}{2} a^4 x \sqrt {a^2 x^2+1}+\frac {3}{2} a^4 \int \frac {16 i a x+17}{\sqrt {a^2 x^2+1}}dx}{a^2}-4 i a^3 x^2 \sqrt {a^2 x^2+1}}{4 a^2}\right )}{a^3}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle \frac {4 \sqrt {a^2 x^2+1}}{a^4 (1-i a x)}-\frac {i \left (\frac {1}{4} a^2 x^3 \sqrt {a^2 x^2+1}+\frac {\frac {-\frac {19}{2} a^4 x \sqrt {a^2 x^2+1}+\frac {3}{2} a^4 \left (17 \int \frac {1}{\sqrt {a^2 x^2+1}}dx+\frac {16 i \sqrt {a^2 x^2+1}}{a}\right )}{a^2}-4 i a^3 x^2 \sqrt {a^2 x^2+1}}{4 a^2}\right )}{a^3}\) |
| \(\Big \downarrow \) 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
3.1.19.3.1 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 = 0.42 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91
| method | result | size |
| risch | \(-\frac {i \left (2 a^{3} x^{3}-8 i a^{2} x^{2}-19 a x +48 i\right ) \sqrt {a^{2} x^{2}+1}}{8 a^{4}}-\frac {i \left (\frac {51 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}-\frac {32 \sqrt {\left (x +\frac {i}{a}\right )^{2} a^{2}-2 i a \left (x +\frac {i}{a}\right )}}{a^{2} \left (x +\frac {i}{a}\right )}\right )}{8 a^{3}}\) | \(124\) |
| meijerg | \(\frac {-2 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right )}{4 \sqrt {a^{2} x^{2}+1}}}{a^{4} \sqrt {\pi }}+\frac {3 i \left (\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {5}{2}} \left (5 a^{2} x^{2}+15\right )}{10 a^{4} \sqrt {a^{2} x^{2}+1}}-\frac {3 \sqrt {\pi }\, \left (a^{2}\right )^{\frac {5}{2}} \operatorname {arcsinh}\left (a x \right )}{2 a^{5}}\right )}{a^{3} \sqrt {\pi }\, \sqrt {a^{2}}}-\frac {3 \left (\frac {8 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (-2 a^{4} x^{4}+8 a^{2} x^{2}+16\right )}{6 \sqrt {a^{2} x^{2}+1}}\right )}{a^{4} \sqrt {\pi }}-\frac {i \left (-\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {7}{2}} \left (-14 a^{4} x^{4}+35 a^{2} x^{2}+105\right )}{56 a^{6} \sqrt {a^{2} x^{2}+1}}+\frac {15 \sqrt {\pi }\, \left (a^{2}\right )^{\frac {7}{2}} \operatorname {arcsinh}\left (a x \right )}{8 a^{7}}\right )}{a^{3} \sqrt {\pi }\, \sqrt {a^{2}}}\) | \(231\) |
| default | \(\frac {x^{2}}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {a^{2} x^{2}+1}}-3 a^{2} \left (\frac {x^{4}}{3 a^{2} \sqrt {a^{2} x^{2}+1}}-\frac {4 \left (\frac {x^{2}}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {a^{2} x^{2}+1}}\right )}{3 a^{2}}\right )-i a^{3} \left (\frac {x^{5}}{4 a^{2} \sqrt {a^{2} x^{2}+1}}-\frac {5 \left (\frac {x^{3}}{2 a^{2} \sqrt {a^{2} x^{2}+1}}-\frac {3 \left (-\frac {x}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{a^{2} \sqrt {a^{2}}}\right )}{2 a^{2}}\right )}{4 a^{2}}\right )+3 i a \left (\frac {x^{3}}{2 a^{2} \sqrt {a^{2} x^{2}+1}}-\frac {3 \left (-\frac {x}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{a^{2} \sqrt {a^{2}}}\right )}{2 a^{2}}\right )\) | \(286\) |
-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 = 0.26 (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)
\[ \int e^{3 i \arctan (a x)} x^3 \, dx=- i \left (\int \frac {i x^{3}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x^{4}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{6}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{5}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \]
-I*(Integral(I*x**3/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x) + Integral(-3*a*x**4/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x) + Integral(a**3*x**6/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x **2 + 1)), x) + Integral(-3*I*a**2*x**5/(a**2*x**2*sqrt(a**2*x**2 + 1) + s qrt(a**2*x**2 + 1)), x))
Time = 0.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.83 \[ \int e^{3 i \arctan (a x)} x^3 \, dx=-\frac {i \, a x^{5}}{4 \, \sqrt {a^{2} x^{2} + 1}} - \frac {x^{4}}{\sqrt {a^{2} x^{2} + 1}} + \frac {17 i \, x^{3}}{8 \, \sqrt {a^{2} x^{2} + 1} a} + \frac {5 \, x^{2}}{\sqrt {a^{2} x^{2} + 1} a^{2}} + \frac {51 i \, x}{8 \, \sqrt {a^{2} x^{2} + 1} a^{3}} - \frac {51 i \, \operatorname {arsinh}\left (a x\right )}{8 \, a^{4}} + \frac {10}{\sqrt {a^{2} x^{2} + 1} a^{4}} \]
-1/4*I*a*x^5/sqrt(a^2*x^2 + 1) - x^4/sqrt(a^2*x^2 + 1) + 17/8*I*x^3/(sqrt( a^2*x^2 + 1)*a) + 5*x^2/(sqrt(a^2*x^2 + 1)*a^2) + 51/8*I*x/(sqrt(a^2*x^2 + 1)*a^3) - 51/8*I*arcsinh(a*x)/a^4 + 10/(sqrt(a^2*x^2 + 1)*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 = 0.50 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00 \[ \int e^{3 i \arctan (a x)} x^3 \, dx=\frac {\sqrt {a^2\,x^2+1}\,\left (\frac {4}{{\left (a^2\right )}^{3/2}}+\frac {2\,\sqrt {a^2}}{a^4}-\frac {x^2\,\sqrt {a^2}}{a^2}-\frac {x^3\,{\left (a^2\right )}^{3/2}\,1{}\mathrm {i}}{4\,a^3}+\frac {x\,\sqrt {a^2}\,19{}\mathrm {i}}{8\,a^3}\right )}{\sqrt {a^2}}-\frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )\,51{}\mathrm {i}}{8\,a^3\,\sqrt {a^2}}+\frac {\sqrt {a^2\,x^2+1}\,4{}\mathrm {i}}{a^3\,\left (x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]