\(\int e^{\frac {5}{2} i \arctan (a x)} x^2 \, dx\) [95]

Optimal result

Integrand size = 16, antiderivative size = 300 \[ \int e^{\frac {5}{2} i \arctan (a x)} x^2 \, dx=\frac {55 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}+\frac {11 i (1-i a x)^{3/4} (1+i a x)^{5/4}}{4 a^3}+\frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}+\frac {i (1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a^3}-\frac {55 i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}+\frac {55 i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}-\frac {55 i \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x} \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}\right )}\right )}{8 \sqrt {2} a^3} \] Output:

55/8*I*(1-I*a*x)^(3/4)*(1+I*a*x)^(1/4)/a^3+11/4*I*(1-I*a*x)^(3/4)*(1+I*a*x 
)^(5/4)/a^3+2*I*(1+I*a*x)^(9/4)/a^3/(1-I*a*x)^(1/4)+1/3*I*(1-I*a*x)^(3/4)* 
(1+I*a*x)^(9/4)/a^3-55/16*I*arctan(1-2^(1/2)*(1-I*a*x)^(1/4)/(1+I*a*x)^(1/ 
4))*2^(1/2)/a^3+55/16*I*arctan(1+2^(1/2)*(1-I*a*x)^(1/4)/(1+I*a*x)^(1/4))* 
2^(1/2)/a^3-55/16*I*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^3
 

Mathematica [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.29 \[ \int e^{\frac {5}{2} i \arctan (a x)} x^2 \, dx=\frac {-\sqrt [4]{1+i a x} (-i+a x)^2 (7 i+a x)+44 \sqrt [4]{2} (i+a x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {3}{4},\frac {7}{4},\frac {1}{2} (1-i a x)\right )}{3 a^3 \sqrt [4]{1-i a x}} \] Input:

Integrate[E^(((5*I)/2)*ArcTan[a*x])*x^2,x]
 

Output:

(-((1 + I*a*x)^(1/4)*(-I + a*x)^2*(7*I + a*x)) + 44*2^(1/4)*(I + a*x)*Hype 
rgeometric2F1[-5/4, 3/4, 7/4, (1 - I*a*x)/2])/(3*a^3*(1 - I*a*x)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.81 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.18, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5585, 100, 27, 90, 60, 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.

\(\displaystyle \int x^2 e^{\frac {5}{2} i \arctan (a x)} \, dx\)

\(\Big \downarrow \) 5585

\(\displaystyle \int \frac {x^2 (1+i a x)^{5/4}}{(1-i a x)^{5/4}}dx\)

\(\Big \downarrow \) 100

\(\displaystyle \frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {2 i \int \frac {a (5 i-a x) (i a x+1)^{5/4}}{2 \sqrt [4]{1-i a x}}dx}{a^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {i \int \frac {(5 i-a x) (i a x+1)^{5/4}}{\sqrt [4]{1-i a x}}dx}{a^2}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {i \left (\frac {11}{2} i \int \frac {(i a x+1)^{5/4}}{\sqrt [4]{1-i a x}}dx-\frac {(1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a}\right )}{a^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {i \left (\frac {11}{2} i \left (\frac {5}{4} \int \frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}dx+\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a}\right )-\frac {(1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a}\right )}{a^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {i \left (\frac {11}{2} i \left (\frac {5}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt [4]{1-i a x} (i a x+1)^{3/4}}dx+\frac {i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}\right )+\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a}\right )-\frac {(1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a}\right )}{a^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {i \left (\frac {11}{2} i \left (\frac {5}{4} \left (\frac {2 i \int \frac {\sqrt {1-i a x}}{(i a x+1)^{3/4}}d\sqrt [4]{1-i a x}}{a}+\frac {i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}\right )+\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a}\right )-\frac {(1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a}\right )}{a^2}\)

\(\Big \downarrow \) 854

\(\displaystyle \frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {i \left (\frac {11}{2} i \left (\frac {5}{4} \left (\frac {2 i \int \frac {\sqrt {1-i a x}}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{a}+\frac {i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}\right )+\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a}\right )-\frac {(1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a}\right )}{a^2}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {i \left (\frac {11}{2} i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \int \frac {\sqrt {1-i a x}+1}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}-\frac {1}{2} \int \frac {1-\sqrt {1-i a x}}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )}{a}+\frac {i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}\right )+\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a}\right )-\frac {(1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a}\right )}{a^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {i \left (\frac {11}{2} i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+\frac {1}{2} \int \frac {1}{\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )-\frac {1}{2} \int \frac {1-\sqrt {1-i a x}}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )}{a}+\frac {i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}\right )+\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a}\right )-\frac {(1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a}\right )}{a^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {i \left (\frac {11}{2} i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\sqrt {1-i a x}-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\sqrt {1-i a x}-1}d\left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\sqrt {1-i a x}}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )}{a}+\frac {i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}\right )+\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a}\right )-\frac {(1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a}\right )}{a^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {i \left (\frac {11}{2} i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\sqrt {1-i a x}}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )}{a}+\frac {i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}\right )+\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a}\right )-\frac {(1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a}\right )}{a^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {i \left (\frac {11}{2} i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-\frac {2 \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1\right )}{\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )\right )}{a}+\frac {i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}\right )+\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a}\right )-\frac {(1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a}\right )}{a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {i \left (\frac {11}{2} i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1\right )}{\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )\right )}{a}+\frac {i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}\right )+\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a}\right )-\frac {(1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a}\right )}{a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {i \left (\frac {11}{2} i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{2 \sqrt {2}}-\frac {1}{2} \int \frac {\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}{\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )\right )}{a}+\frac {i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}\right )+\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a}\right )-\frac {(1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a}\right )}{a^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {i \left (\frac {11}{2} i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{2 \sqrt {2}}\right )\right )}{a}+\frac {i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}\right )+\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a}\right )-\frac {(1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a}\right )}{a^2}\)

Input:

Int[E^(((5*I)/2)*ArcTan[a*x])*x^2,x]
 

Output:

((2*I)*(1 + I*a*x)^(9/4))/(a^3*(1 - I*a*x)^(1/4)) - (I*(-1/3*((1 - I*a*x)^ 
(3/4)*(1 + I*a*x)^(9/4))/a + ((11*I)/2)*(((I/2)*(1 - I*a*x)^(3/4)*(1 + I*a 
*x)^(5/4))/a + (5*((I*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/a + ((2*I)*((-( 
ArcTan[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)]/Sqrt[2]) + ArcTa 
n[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))/4)))/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 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

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]

\[\int {\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {5}{2}} x^{2}d x\]

Input:

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x^2,x)
 

Output:

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x^2,x)
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.81 \[ \int e^{\frac {5}{2} i \arctan (a x)} x^2 \, dx=\frac {12 \, a^{3} \sqrt {\frac {3025 i}{64 \, a^{6}}} \log \left (\frac {8}{55} i \, a^{3} \sqrt {\frac {3025 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 12 \, a^{3} \sqrt {\frac {3025 i}{64 \, a^{6}}} \log \left (-\frac {8}{55} i \, a^{3} \sqrt {\frac {3025 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 12 \, a^{3} \sqrt {-\frac {3025 i}{64 \, a^{6}}} \log \left (\frac {8}{55} i \, a^{3} \sqrt {-\frac {3025 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 12 \, a^{3} \sqrt {-\frac {3025 i}{64 \, a^{6}}} \log \left (-\frac {8}{55} i \, a^{3} \sqrt {-\frac {3025 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - {\left (8 \, a^{3} x^{3} - 26 i \, a^{2} x^{2} - 61 \, a x - 287 i\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{24 \, a^{3}} \] Input:

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x^2,x, algorithm="fricas")
 

Output:

1/24*(12*a^3*sqrt(3025/64*I/a^6)*log(8/55*I*a^3*sqrt(3025/64*I/a^6) + sqrt 
(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 12*a^3*sqrt(3025/64*I/a^6)*log(-8/55*I* 
a^3*sqrt(3025/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + 12*a^3*sq 
rt(-3025/64*I/a^6)*log(8/55*I*a^3*sqrt(-3025/64*I/a^6) + sqrt(I*sqrt(a^2*x 
^2 + 1)/(a*x + I))) - 12*a^3*sqrt(-3025/64*I/a^6)*log(-8/55*I*a^3*sqrt(-30 
25/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - (8*a^3*x^3 - 26*I*a^ 
2*x^2 - 61*a*x - 287*I)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/a^3
 

Sympy [F(-1)]

Timed out. \[ \int e^{\frac {5}{2} i \arctan (a x)} x^2 \, dx=\text {Timed out} \] Input:

integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(5/2)*x**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int e^{\frac {5}{2} i \arctan (a x)} x^2 \, dx=\int { x^{2} \left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}} \,d x } \] Input:

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x^2,x, algorithm="maxima")
 

Output:

integrate(x^2*((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(5/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int e^{\frac {5}{2} i \arctan (a x)} x^2 \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x^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. \[ \int e^{\frac {5}{2} i \arctan (a x)} x^2 \, dx=\int x^2\,{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{5/2} \,d x \] Input:

int(x^2*((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2),x)
 

Output:

int(x^2*((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2), x)
 

Reduce [F]

\[ \int e^{\frac {5}{2} i \arctan (a x)} x^2 \, dx=\int {\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {5}{2}} x^{2}d x \] Input:

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x^2,x)
 

Output:

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x^2,x)