\(\int e^{-\frac {3}{2} i \arctan (a x)} x^3 \, dx\) [112]

Optimal result

Integrand size = 16, antiderivative size = 291 \[ \int e^{-\frac {3}{2} i \arctan (a x)} x^3 \, dx=-\frac {41 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{64 a^4}-\frac {7 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{32 a^4}+\frac {x^2 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {(1-i a x)^{11/4} \sqrt [4]{1+i a x}}{8 a^4}-\frac {123 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt {2} a^4}+\frac {123 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt {2} a^4}-\frac {123 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x} \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}\right )}\right )}{64 \sqrt {2} a^4} \] Output:

-41/64*(1-I*a*x)^(3/4)*(1+I*a*x)^(1/4)/a^4-7/32*(1-I*a*x)^(7/4)*(1+I*a*x)^ 
(1/4)/a^4+1/4*x^2*(1-I*a*x)^(7/4)*(1+I*a*x)^(1/4)/a^2-1/8*(1-I*a*x)^(11/4) 
*(1+I*a*x)^(1/4)/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 = 0.14 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.44 \[ \int e^{-\frac {3}{2} i \arctan (a x)} x^3 \, dx=\frac {(1-i a x)^{7/4} \left (7 a^2 x^2 \sqrt [4]{1+i a x}+12 \sqrt [4]{2} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {7}{4},\frac {11}{4},\frac {1}{2} (1-i a x)\right )-20 \sqrt [4]{2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {7}{4},\frac {11}{4},\frac {1}{2} (1-i a x)\right )+7 \sqrt [4]{2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{4},\frac {11}{4},\frac {1}{2} (1-i a x)\right )\right )}{28 a^4} \] Input:

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

Output:

((1 - I*a*x)^(7/4)*(7*a^2*x^2*(1 + I*a*x)^(1/4) + 12*2^(1/4)*Hypergeometri 
c2F1[-5/4, 7/4, 11/4, (1 - I*a*x)/2] - 20*2^(1/4)*Hypergeometric2F1[-1/4, 
7/4, 11/4, (1 - I*a*x)/2] + 7*2^(1/4)*Hypergeometric2F1[3/4, 7/4, 11/4, (1 
 - I*a*x)/2]))/(28*a^4)
 

Rubi [A] (warning: unable to verify)

Time = 0.81 (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, 854, 826, 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

\(\displaystyle \frac {\int -\frac {x (1-i a x)^{3/4} (4-3 i a x)}{2 (i a x+1)^{3/4}}dx}{4 a^2}+\frac {x^2 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^2 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {\int \frac {x (1-i a x)^{3/4} (4-3 i a x)}{(i a x+1)^{3/4}}dx}{8 a^2}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {x^2 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {\frac {41 i \int \frac {(1-i a x)^{3/4}}{(i a x+1)^{3/4}}dx}{8 a}+\frac {\sqrt [4]{1+i a x} (11-4 i a x) (1-i a x)^{7/4}}{4 a^2}}{8 a^2}\)

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 854

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(x^2*(1 - I*a*x)^(7/4)*(1 + I*a*x)^(1/4))/(4*a^2) - (((1 - I*a*x)^(7/4)*(1 
 + I*a*x)^(1/4)*(11 - (4*I)*a*x))/(4*a^2) + (((41*I)/8)*(((-I)*(1 - I*a*x) 
^(3/4)*(1 + I*a*x)^(1/4))/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 + (Log[1 + Sqrt[1 - I*a*x] - (Sqrt[2]*(1 - 
I*a*x)^(1/4))/(1 + I*a*x)^(1/4)]/(2*Sqrt[2]) - Log[1 + Sqrt[1 - I*a*x] + ( 
Sqrt[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 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   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 854

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 
 1)/n)   Subst[Int[x^m/(1 - b*x^n)^(p + (m + 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)] && IntegersQ[m, p + (m + 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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.86 \[ \int e^{-\frac {3}{2} i \arctan (a x)} x^3 \, dx=-\frac {32 \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} \log \left (\frac {64}{123} \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 32 \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} \log \left (-\frac {64}{123} \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 32 \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} \log \left (\frac {64}{123} \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 32 \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} \log \left (-\frac {64}{123} \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + {\left (16 \, a^{4} x^{4} + 40 i \, a^{3} x^{3} - 54 \, a^{2} x^{2} - 93 i \, a x + 63\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{64 \, a^{4}} \] Input:

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

Output:

-1/64*(32*a^4*sqrt(15129/4096*I/a^8)*log(64/123*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)*lo 
g(-64/123*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*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)*lo 
g(-64/123*a^4*sqrt(-15129/4096*I/a^8) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I) 
)) + (16*a^4*x^4 + 40*I*a^3*x^3 - 54*a^2*x^2 - 93*I*a*x + 63)*sqrt(I*sqrt( 
a^2*x^2 + 1)/(a*x + I)))/a^4
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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
 

Mupad [F(-1)]

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

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

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

Output:

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