\(\int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx\) [85]

Optimal result

Integrand size = 16, antiderivative size = 291 \[ \int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx=-\frac {41 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 a^4}-\frac {15 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{32 a^4}+\frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}+\frac {(1-i a x)^{5/4} (1+i a x)^{7/4}}{8 a^4}+\frac {123 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt {2} a^4}-\frac {123 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt {2} a^4}-\frac {123 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x} \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}\right )}\right )}{64 \sqrt {2} a^4} \] Output:

-41/64*(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4)/a^4-15/32*(1-I*a*x)^(1/4)*(1+I*a*x) 
^(7/4)/a^4+1/4*x^2*(1-I*a*x)^(1/4)*(1+I*a*x)^(7/4)/a^2+1/8*(1-I*a*x)^(5/4) 
*(1+I*a*x)^(7/4)/a^4+123/128*arctan(1-2^(1/2)*(1-I*a*x)^(1/4)/(1+I*a*x)^(1 
/4))*2^(1/2)/a^4-123/128*arctan(1+2^(1/2)*(1-I*a*x)^(1/4)/(1+I*a*x)^(1/4)) 
*2^(1/2)/a^4-123/128*arctanh(2^(1/2)*(1-I*a*x)^(1/4)/(1+I*a*x)^(1/4)/(1+(1 
-I*a*x)^(1/2)/(1+I*a*x)^(1/2)))*2^(1/2)/a^4
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.15 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.51 \[ \int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx=\frac {\sqrt [4]{1-i a x} \left (a^2 x^2 (1+i a x)^{3/4}+i a^3 x^3 (1+i a x)^{3/4}-24\ 2^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},\frac {1}{4},\frac {5}{4},\frac {1}{2} (1-i a x)\right )+8\ 2^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {1}{4},\frac {5}{4},\frac {1}{2} (1-i a x)\right )+2\ 2^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{4},\frac {5}{4},\frac {1}{2} (1-i a x)\right )\right )}{4 a^4} \] Input:

Integrate[E^(((3*I)/2)*ArcTan[a*x])*x^3,x]
 

Output:

((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)
 

Rubi [A] (warning: unable to verify)

Time = 0.79 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.13, 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

\(\displaystyle \frac {x^2 \sqrt [4]{1-i a x} (1+i a x)^{7/4}}{4 a^2}-\frac {\frac {\sqrt [4]{1-i a x} (1+i a x)^{7/4} (11+4 i a x)}{4 a^2}-\frac {41 i \left (\frac {6 i \left (\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{2 \sqrt {2}}\right )\right )}{a}+\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )}{8 a}}{8 a^2}\)

Input:

Int[E^(((3*I)/2)*ArcTan[a*x])*x^3,x]
 

Output:

(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 
)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 60

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]
 

rule 73

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]
 

rule 111

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]
 

rule 164

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]
 

rule 217

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])
 

rule 755

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]]))
 

rule 770

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]
 

rule 1082

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]
 

rule 1103

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]
 

rule 1476

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]
 

rule 1479

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]
 

rule 5585

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]
 

Maple [F]

\[\int {\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {3}{2}} x^{3}d x\]

Input:

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)*x^3,x)
 

Output:

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)*x^3,x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.87 \[ \int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx=\frac {32 \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} \log \left (\frac {64}{123} i \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 32 \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} \log \left (-\frac {64}{123} i \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 32 \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} \log \left (\frac {64}{123} i \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 32 \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} \log \left (-\frac {64}{123} i \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + {\left (16 i \, a^{3} x^{3} + 24 \, a^{2} x^{2} - 30 i \, a x - 63\right )} \sqrt {a^{2} x^{2} + 1} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{64 \, a^{4}} \] Input:

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)*x^3,x, algorithm="fricas")
 

Output:

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
 

Sympy [F]

\[ \int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx=\int x^{3} \left (\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}\, dx \] Input:

integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(3/2)*x**3,x)
 

Output:

Integral(x**3*(I*(a*x - I)/sqrt(a**2*x**2 + 1))**(3/2), x)
 

Maxima [F]

\[ \int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx=\int { x^{3} \left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {3}{2}} \,d x } \] Input:

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)*x^3,x, algorithm="maxima")
 

Output:

integrate(x^3*((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(3/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)*x^3,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Warning, need to choose a branch fo 
r the root of a polynomial with parameters. This might be wrong.The choice 
 was done
 

Mupad [F(-1)]

Timed out. \[ \int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx=\int x^3\,{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{3/2} \,d x \] Input:

int(x^3*((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(3/2),x)
 

Output:

int(x^3*((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(3/2), x)
 

Reduce [F]

\[ \int e^{\frac {3}{2} i \arctan (a x)} x^3 \, dx=\int {\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {3}{2}} x^{3}d x \] Input:

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)*x^3,x)
 

Output:

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)*x^3,x)