\(\int \frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}} \, dx\) [229]

Optimal result

Integrand size = 25, antiderivative size = 186 \[ \int \frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}} \, dx=-\frac {i \sqrt [4]{a-i a x} (a+i a x)^{3/4}}{a}+\frac {i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a+i a x}}{\sqrt [4]{a-i a x}}\right )}{\sqrt {2}}-\frac {i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{a+i a x}}{\sqrt [4]{a-i a x}}\right )}{\sqrt {2}}+\frac {i \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a+i a x}}{\sqrt [4]{a-i a x} \left (1+\frac {\sqrt {a+i a x}}{\sqrt {a-i a x}}\right )}\right )}{\sqrt {2}} \] Output:

-I*(a-I*a*x)^(1/4)*(a+I*a*x)^(3/4)/a+1/2*I*arctan(1-2^(1/2)*(a+I*a*x)^(1/4 
)/(a-I*a*x)^(1/4))*2^(1/2)-1/2*I*arctan(1+2^(1/2)*(a+I*a*x)^(1/4)/(a-I*a*x 
)^(1/4))*2^(1/2)+1/2*I*arctanh(2^(1/2)*(a+I*a*x)^(1/4)/(a-I*a*x)^(1/4)/(1+ 
(a+I*a*x)^(1/2)/(a-I*a*x)^(1/2)))*2^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 0.34 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.38 \[ \int \frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}} \, dx=\frac {2 i 2^{3/4} \sqrt [4]{1+i x} (a-i a x)^{5/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {5}{4},\frac {9}{4},\frac {1}{2}-\frac {i x}{2}\right )}{5 a \sqrt [4]{a+i a x}} \] Input:

Integrate[(a - I*a*x)^(1/4)/(a + I*a*x)^(1/4),x]
 

Output:

(((2*I)/5)*2^(3/4)*(1 + I*x)^(1/4)*(a - I*a*x)^(5/4)*Hypergeometric2F1[1/4 
, 5/4, 9/4, 1/2 - (I/2)*x])/(a*(a + I*a*x)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.30 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.27, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {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.

\(\displaystyle \int \frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}} \, dx\)

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 770

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[(a - I*a*x)^(1/4)/(a + I*a*x)^(1/4),x]
 

Output:

((-I)*(a - I*a*x)^(1/4)*(a + I*a*x)^(3/4))/a + (2*I)*((-(ArcTan[1 - (Sqrt[ 
2]*(a - I*a*x)^(1/4))/(a + I*a*x)^(1/4)]/Sqrt[2]) + ArcTan[1 + (Sqrt[2]*(a 
 - I*a*x)^(1/4))/(a + I*a*x)^(1/4)]/Sqrt[2])/2 + (-1/2*Log[1 + Sqrt[a - I* 
a*x] - (Sqrt[2]*(a - I*a*x)^(1/4))/(a + I*a*x)^(1/4)]/Sqrt[2] + Log[1 + Sq 
rt[a - I*a*x] + (Sqrt[2]*(a - I*a*x)^(1/4))/(a + I*a*x)^(1/4)]/(2*Sqrt[2]) 
)/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 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]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.28 (sec) , antiderivative size = 477, normalized size of antiderivative = 2.56

method	result	size
risch	\(\frac {i \left (x -i\right ) \left (x +i\right ) \left (-a \left (i x -1\right )\right )^{\frac {1}{4}}}{\left (i x -1\right ) \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}-\frac {\left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-i\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}-i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}} x^{2}-i \operatorname {RootOf}\left (\textit {\_Z}^{2}-i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {3}{4}}-x^{3}-2 i \operatorname {RootOf}\left (\textit {\_Z}^{2}-i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}} x -i \sqrt {-x^{4}-2 i x^{3}-2 i x +1}\, x -2 i x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}}+\sqrt {-x^{4}-2 i x^{3}-2 i x +1}+x}{\left (i x -1\right )^{2}}\right )}{2}-\frac {i \operatorname {RootOf}\left (\textit {\_Z}^{2}-i\right ) \ln \left (-\frac {-i \operatorname {RootOf}\left (\textit {\_Z}^{2}-i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}} x^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}} x +x^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {3}{4}}-i \sqrt {-x^{4}-2 i x^{3}-2 i x +1}\, x +i \operatorname {RootOf}\left (\textit {\_Z}^{2}-i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}}+2 i x^{2}+\sqrt {-x^{4}-2 i x^{3}-2 i x +1}-x}{\left (i x -1\right )^{2}}\right )}{2}\right ) \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (-\left (i x -1\right )^{3} \left (i x +1\right )\right )^{\frac {1}{4}}}{\left (i x -1\right ) \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}\)	\(477\)

Input:

int((a-I*a*x)^(1/4)/(a+I*a*x)^(1/4),x,method=_RETURNVERBOSE)
 

Output:

I*(x-I)*(x+I)*(-a*(I*x-1))^(1/4)/(I*x-1)/(a*(I*x+1))^(1/4)-(1/2*RootOf(_Z^ 
2-I)*ln((-RootOf(_Z^2-I)*(-2*I*x^3-x^4-2*I*x+1)^(1/4)*x^2-I*RootOf(_Z^2-I) 
*(-2*I*x^3-x^4-2*I*x+1)^(3/4)-x^3-2*I*RootOf(_Z^2-I)*(-2*I*x^3-x^4-2*I*x+1 
)^(1/4)*x-I*(-2*I*x^3-x^4-2*I*x+1)^(1/2)*x-2*I*x^2+RootOf(_Z^2-I)*(-2*I*x^ 
3-x^4-2*I*x+1)^(1/4)+(-2*I*x^3-x^4-2*I*x+1)^(1/2)+x)/(I*x-1)^2)-1/2*I*Root 
Of(_Z^2-I)*ln(-(-I*RootOf(_Z^2-I)*(-2*I*x^3-x^4-2*I*x+1)^(1/4)*x^2+2*RootO 
f(_Z^2-I)*(-2*I*x^3-x^4-2*I*x+1)^(1/4)*x+x^3-RootOf(_Z^2-I)*(-2*I*x^3-x^4- 
2*I*x+1)^(3/4)-I*(-2*I*x^3-x^4-2*I*x+1)^(1/2)*x+I*RootOf(_Z^2-I)*(-2*I*x^3 
-x^4-2*I*x+1)^(1/4)+2*I*x^2+(-2*I*x^3-x^4-2*I*x+1)^(1/2)-x)/(I*x-1)^2))*(- 
a*(I*x-1))^(1/4)/(I*x-1)*(-(I*x-1)^3*(I*x+1))^(1/4)/(a*(I*x+1))^(1/4)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.04 \[ \int \frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}} \, dx=\frac {\sqrt {i} a \log \left (\frac {\sqrt {i} {\left (a x - i \, a\right )} + {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{x - i}\right ) - \sqrt {i} a \log \left (-\frac {\sqrt {i} {\left (a x - i \, a\right )} - {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{x - i}\right ) + \sqrt {-i} a \log \left (\frac {\sqrt {-i} {\left (a x - i \, a\right )} + {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{x - i}\right ) - \sqrt {-i} a \log \left (-\frac {\sqrt {-i} {\left (a x - i \, a\right )} - {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{x - i}\right ) - 2 i \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{2 \, a} \] Input:

integrate((a-I*a*x)^(1/4)/(a+I*a*x)^(1/4),x, algorithm="fricas")
 

Output:

1/2*(sqrt(I)*a*log((sqrt(I)*(a*x - I*a) + (I*a*x + a)^(3/4)*(-I*a*x + a)^( 
1/4))/(x - I)) - sqrt(I)*a*log(-(sqrt(I)*(a*x - I*a) - (I*a*x + a)^(3/4)*( 
-I*a*x + a)^(1/4))/(x - I)) + sqrt(-I)*a*log((sqrt(-I)*(a*x - I*a) + (I*a* 
x + a)^(3/4)*(-I*a*x + a)^(1/4))/(x - I)) - sqrt(-I)*a*log(-(sqrt(-I)*(a*x 
 - I*a) - (I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4))/(x - I)) - 2*I*(I*a*x + a) 
^(3/4)*(-I*a*x + a)^(1/4))/a
 

Sympy [F]

\[ \int \frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}} \, dx=\int \frac {\sqrt [4]{- i a \left (x + i\right )}}{\sqrt [4]{i a \left (x - i\right )}}\, dx \] Input:

integrate((a-I*a*x)**(1/4)/(a+I*a*x)**(1/4),x)
 

Output:

Integral((-I*a*(x + I))**(1/4)/(I*a*(x - I))**(1/4), x)
 

Maxima [F]

\[ \int \frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}} \, dx=\int { \frac {{\left (-i \, a x + a\right )}^{\frac {1}{4}}}{{\left (i \, a x + a\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate((a-I*a*x)^(1/4)/(a+I*a*x)^(1/4),x, algorithm="maxima")
 

Output:

integrate((-I*a*x + a)^(1/4)/(I*a*x + a)^(1/4), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((a-I*a*x)^(1/4)/(a+I*a*x)^(1/4),x, algorithm="giac")
 

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:Warning, need to choose a branch for the root of a poly 
nomial with parameters. This might be wrong.The choice was done assuming 0 
=[0,0]ext_re
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}} \, dx=\int \frac {{\left (a-a\,x\,1{}\mathrm {i}\right )}^{1/4}}{{\left (a+a\,x\,1{}\mathrm {i}\right )}^{1/4}} \,d x \] Input:

int((a - a*x*1i)^(1/4)/(a + a*x*1i)^(1/4),x)
 

Output:

int((a - a*x*1i)^(1/4)/(a + a*x*1i)^(1/4), x)
 

Reduce [F]

\[ \int \frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}} \, dx=\int \frac {\left (-i x +1\right )^{\frac {1}{4}}}{\left (i x +1\right )^{\frac {1}{4}}}d x \] Input:

int((a-I*a*x)^(1/4)/(a+I*a*x)^(1/4),x)
 

Output:

int(( - i*x + 1)**(1/4)/(i*x + 1)**(1/4),x)