Integrand size = 16, antiderivative size = 337 \[ \int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx=-\frac {41 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 a^4}+\frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}-\frac {\sqrt [4]{1-i a x} (1+i a x)^{7/4} (11+4 i a x)}{32 a^4}+\frac {123 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt {2} a^4}-\frac {123 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt {2} a^4}+\frac {123 \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{128 \sqrt {2} a^4}-\frac {123 \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{128 \sqrt {2} a^4} \]
-41/64*(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4)/a^4+1/4*x^2*(1-I*a*x)^(1/4)*(1+I*a* x)^(7/4)/a^2-1/32*(1-I*a*x)^(1/4)*(1+I*a*x)^(7/4)*(11+4*I*a*x)/a^4+123/128 *arctan(1-(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4))/a^4*2^(1/2)-123/128*arc tan(1+(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4))/a^4*2^(1/2)+123/256*ln(1-(1 -I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4)+(1-I*a*x)^(1/2)/(1+I*a*x)^(1/2))/a^4 *2^(1/2)-123/256*ln(1+(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4)+(1-I*a*x)^(1 /2)/(1+I*a*x)^(1/2))/a^4*2^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.08 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.44 \[ \int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx=\frac {\sqrt [4]{1-i a x} \left (a^2 x^2 (1+i a x)^{3/4}+i a^3 x^3 (1+i a x)^{3/4}-24\ 2^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},\frac {1}{4},\frac {5}{4},\frac {1}{2} (1-i a x)\right )+8\ 2^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {1}{4},\frac {5}{4},\frac {1}{2} (1-i a x)\right )+2\ 2^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{4},\frac {5}{4},\frac {1}{2} (1-i a x)\right )\right )}{4 a^4} \]
((1 - I*a*x)^(1/4)*(a^2*x^2*(1 + I*a*x)^(3/4) + I*a^3*x^3*(1 + I*a*x)^(3/4 ) - 24*2^(3/4)*Hypergeometric2F1[-11/4, 1/4, 5/4, (1 - I*a*x)/2] + 8*2^(3/ 4)*Hypergeometric2F1[-7/4, 1/4, 5/4, (1 - I*a*x)/2] + 2*2^(3/4)*Hypergeome tric2F1[-3/4, 1/4, 5/4, (1 - I*a*x)/2]))/(4*a^4)
Time = 0.45 (sec) , antiderivative size = 330, normalized size of antiderivative = 0.98, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.938, Rules used = {5585, 111, 27, 164, 60, 73, 770, 755, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
\(\displaystyle \int x^3 e^{\frac {3}{2} i \arctan (a x)} \, dx\) |
\(\Big \downarrow \) 5585 |
\(\displaystyle \int \frac {x^3 (1+i a x)^{3/4}}{(1-i a x)^{3/4}}dx\) |
\(\Big \downarrow \) 111 |
\(\displaystyle \frac {\int -\frac {x (i a x+1)^{3/4} (3 i a x+4)}{2 (1-i a x)^{3/4}}dx}{4 a^2}+\frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}-\frac {\int \frac {x (i a x+1)^{3/4} (3 i a x+4)}{(1-i a x)^{3/4}}dx}{8 a^2}\) |
\(\Big \downarrow \) 164 |
\(\displaystyle \frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}-\frac {\frac {\sqrt [4]{1-i a x} (1+i a x)^{7/4} (11+4 i a x)}{4 a^2}-\frac {41 i \int \frac {(i a x+1)^{3/4}}{(1-i a x)^{3/4}}dx}{8 a}}{8 a^2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}-\frac {\frac {\sqrt [4]{1-i a x} (1+i a x)^{7/4} (11+4 i a x)}{4 a^2}-\frac {41 i \left (\frac {3}{2} \int \frac {1}{(1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx+\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )}{8 a}}{8 a^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}-\frac {\frac {\sqrt [4]{1-i a x} (1+i a x)^{7/4} (11+4 i a x)}{4 a^2}-\frac {41 i \left (\frac {6 i \int \frac {1}{\sqrt [4]{i a x+1}}d\sqrt [4]{1-i a x}}{a}+\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )}{8 a}}{8 a^2}\) |
\(\Big \downarrow \) 770 |
\(\displaystyle \frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}-\frac {\frac {\sqrt [4]{1-i a x} (1+i a x)^{7/4} (11+4 i a x)}{4 a^2}-\frac {41 i \left (\frac {6 i \int \frac {1}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{a}+\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )}{8 a}}{8 a^2}\) |
\(\Big \downarrow \) 755 |
\(\displaystyle \frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}-\frac {\frac {\sqrt [4]{1-i a x} (1+i a x)^{7/4} (11+4 i a x)}{4 a^2}-\frac {41 i \left (\frac {6 i \left (\frac {1}{2} \int \frac {1-\sqrt {1-i a x}}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+\frac {1}{2} \int \frac {\sqrt {1-i a x}+1}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )}{a}+\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )}{8 a}}{8 a^2}\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}-\frac {\frac {\sqrt [4]{1-i a x} (1+i a x)^{7/4} (11+4 i a x)}{4 a^2}-\frac {41 i \left (\frac {6 i \left (\frac {1}{2} \int \frac {1-\sqrt {1-i a x}}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+\frac {1}{2} \int \frac {1}{\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )\right )}{a}+\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )}{8 a}}{8 a^2}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}-\frac {\frac {\sqrt [4]{1-i a x} (1+i a x)^{7/4} (11+4 i a x)}{4 a^2}-\frac {41 i \left (\frac {6 i \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\sqrt {1-i a x}-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\sqrt {1-i a x}-1}d\left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1\right )}{\sqrt {2}}\right )+\frac {1}{2} \int \frac {1-\sqrt {1-i a x}}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )}{a}+\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )}{8 a}}{8 a^2}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}-\frac {\frac {\sqrt [4]{1-i a x} (1+i a x)^{7/4} (11+4 i a x)}{4 a^2}-\frac {41 i \left (\frac {6 i \left (\frac {1}{2} \int \frac {1-\sqrt {1-i a x}}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )\right )}{a}+\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )}{8 a}}{8 a^2}\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}-\frac {\frac {\sqrt [4]{1-i a x} (1+i a x)^{7/4} (11+4 i a x)}{4 a^2}-\frac {41 i \left (\frac {6 i \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-\frac {2 \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1\right )}{\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )\right )}{a}+\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )}{8 a}}{8 a^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}-\frac {\frac {\sqrt [4]{1-i a x} (1+i a x)^{7/4} (11+4 i a x)}{4 a^2}-\frac {41 i \left (\frac {6 i \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1\right )}{\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )\right )}{a}+\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )}{8 a}}{8 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}-\frac {\frac {\sqrt [4]{1-i a x} (1+i a x)^{7/4} (11+4 i a x)}{4 a^2}-\frac {41 i \left (\frac {6 i \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}{\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )\right )}{a}+\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )}{8 a}}{8 a^2}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}-\frac {\frac {\sqrt [4]{1-i a x} (1+i a x)^{7/4} (11+4 i a x)}{4 a^2}-\frac {41 i \left (\frac {6 i \left (\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{2 \sqrt {2}}\right )\right )}{a}+\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )}{8 a}}{8 a^2}\) |
(x^2*(1 - I*a*x)^(1/4)*(1 + I*a*x)^(7/4))/(4*a^2) - (((1 - I*a*x)^(1/4)*(1 + I*a*x)^(7/4)*(11 + (4*I)*a*x))/(4*a^2) - (((41*I)/8)*((I*(1 - I*a*x)^(1 /4)*(1 + I*a*x)^(3/4))/a + ((6*I)*((-(ArcTan[1 - (Sqrt[2]*(1 - I*a*x)^(1/4 ))/(1 + I*a*x)^(1/4)]/Sqrt[2]) + ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)]/Sqrt[2])/2 + (-1/2*Log[1 + Sqrt[1 - I*a*x] - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)]/Sqrt[2] + Log[1 + Sqrt[1 - I*a*x] + (Sq rt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)]/(2*Sqrt[2]))/2))/a))/a)/(8*a^2 )
3.1.70.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & & AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n) Subst[In t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 /n]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a *x)^(I*(n/2))/(1 + I*a*x)^(I*(n/2))), x] /; FreeQ[{a, m, n}, x] && !Intege rQ[(I*n - 1)/2]
\[\int {\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {3}{2}} x^{3}d x\]
Time = 0.28 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.75 \[ \int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx=\frac {32 \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} \log \left (\frac {64}{123} i \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 32 \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} \log \left (-\frac {64}{123} i \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 32 \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} \log \left (\frac {64}{123} i \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 32 \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} \log \left (-\frac {64}{123} i \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + {\left (16 i \, a^{3} x^{3} + 24 \, a^{2} x^{2} - 30 i \, a x - 63\right )} \sqrt {a^{2} x^{2} + 1} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{64 \, a^{4}} \]
1/64*(32*a^4*sqrt(15129/4096*I/a^8)*log(64/123*I*a^4*sqrt(15129/4096*I/a^8 ) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 32*a^4*sqrt(15129/4096*I/a^8)*l og(-64/123*I*a^4*sqrt(15129/4096*I/a^8) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 32*a^4*sqrt(-15129/4096*I/a^8)*log(64/123*I*a^4*sqrt(-15129/4096*I/ a^8) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + 32*a^4*sqrt(-15129/4096*I/a^ 8)*log(-64/123*I*a^4*sqrt(-15129/4096*I/a^8) + sqrt(I*sqrt(a^2*x^2 + 1)/(a *x + I))) + (16*I*a^3*x^3 + 24*a^2*x^2 - 30*I*a*x - 63)*sqrt(a^2*x^2 + 1)* sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/a^4
\[ \int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx=\int x^{3} \left (\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}\, dx \]
\[ \int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx=\int { x^{3} \left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {3}{2}} \,d x } \]
Exception generated. \[ \int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:The choice was done assuming 0=[0,0 ]Warning, replacing 0 by 23, a substitution variable should perhaps be pur ged.Warni
Timed out. \[ \int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx=\int x^3\,{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{3/2} \,d x \]