3.12.8 \(\int \frac {(a-i a x)^{3/4}}{(a+i a x)^{3/4}} \, dx\)

Optimal. Leaf size=256 \[ -\frac {i (a-i a x)^{3/4} \sqrt [4]{a+i a x}}{a}+\frac {3 i \log \left (\frac {\sqrt {a-i a x}}{\sqrt {a+i a x}}-\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}+1\right )}{2 \sqrt {2}}-\frac {3 i \log \left (\frac {\sqrt {a-i a x}}{\sqrt {a+i a x}}+\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}+1\right )}{2 \sqrt {2}}-\frac {3 i \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2}}+\frac {3 i \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2}} \]

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Rubi [A]  time = 0.16, antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {50, 63, 331, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {i (a-i a x)^{3/4} \sqrt [4]{a+i a x}}{a}+\frac {3 i \log \left (\frac {\sqrt {a-i a x}}{\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 {3 i \log \left (\frac {\sqrt {a-i a x}}{\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 {3 i \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2}}+\frac {3 i \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a - I*a*x)^(3/4)/(a + I*a*x)^(3/4),x]

[Out]

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

Rule 50

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] + Dist[(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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(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 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{3/4}} \, dx &=-\frac {i (a-i a x)^{3/4} \sqrt [4]{a+i a x}}{a}+\frac {1}{2} (3 a) \int \frac {1}{\sqrt [4]{a-i a x} (a+i a x)^{3/4}} \, dx\\ &=-\frac {i (a-i a x)^{3/4} \sqrt [4]{a+i a x}}{a}+6 i \operatorname {Subst}\left (\int \frac {x^2}{\left (2 a-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{a-i a x}\right )\\ &=-\frac {i (a-i a x)^{3/4} \sqrt [4]{a+i a x}}{a}+6 i \operatorname {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )\\ &=-\frac {i (a-i a x)^{3/4} \sqrt [4]{a+i a x}}{a}-3 i \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )+3 i \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )\\ &=-\frac {i (a-i a x)^{3/4} \sqrt [4]{a+i a x}}{a}+\frac {3}{2} i \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )+\frac {3}{2} i \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )+\frac {(3 i) \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{2 \sqrt {2}}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{2 \sqrt {2}}\\ &=-\frac {i (a-i a x)^{3/4} \sqrt [4]{a+i a x}}{a}+\frac {3 i \log \left (1+\frac {\sqrt {a-i a x}}{\sqrt {a+i a x}}-\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{2 \sqrt {2}}-\frac {3 i \log \left (1+\frac {\sqrt {a-i a x}}{\sqrt {a+i a x}}+\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{2 \sqrt {2}}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2}}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2}}\\ &=-\frac {i (a-i a x)^{3/4} \sqrt [4]{a+i a x}}{a}-\frac {3 i \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2}}+\frac {3 i \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2}}+\frac {3 i \log \left (1+\frac {\sqrt {a-i a x}}{\sqrt {a+i a x}}-\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{2 \sqrt {2}}-\frac {3 i \log \left (1+\frac {\sqrt {a-i a x}}{\sqrt {a+i a x}}+\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{2 \sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 70, normalized size = 0.27 \begin {gather*} \frac {2 i \sqrt [4]{2} (1+i x)^{3/4} (a-i a x)^{7/4} \, _2F_1\left (\frac {3}{4},\frac {7}{4};\frac {11}{4};\frac {1}{2}-\frac {i x}{2}\right )}{7 a (a+i a x)^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a - I*a*x)^(3/4)/(a + I*a*x)^(3/4),x]

[Out]

(((2*I)/7)*2^(1/4)*(1 + I*x)^(3/4)*(a - I*a*x)^(7/4)*Hypergeometric2F1[3/4, 7/4, 11/4, 1/2 - (I/2)*x])/(a*(a +
I*a*x)^(3/4))

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IntegrateAlgebraic [A]  time = 0.60, size = 128, normalized size = 0.50 \begin {gather*} \frac {(-1)^{3/4} (x-i)^{3/4} (a-i a x)^{3/4} \left (-\sqrt [4]{-1} \sqrt [4]{x-i} (x+i)^{3/4}+3 (-1)^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{x+i}}{\sqrt [4]{x-i}}\right )-3 (-1)^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{x+i}}{\sqrt [4]{x-i}}\right )\right )}{(x+i)^{3/4} (a+i a x)^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a - I*a*x)^(3/4)/(a + I*a*x)^(3/4),x]

[Out]

((-1)^(3/4)*(-I + x)^(3/4)*(a - I*a*x)^(3/4)*(-((-1)^(1/4)*(-I + x)^(1/4)*(I + x)^(3/4)) + 3*(-1)^(3/4)*ArcTan
[(I + x)^(1/4)/(-I + x)^(1/4)] - 3*(-1)^(3/4)*ArcTanh[(I + x)^(1/4)/(-I + x)^(1/4)]))/((I + x)^(3/4)*(a + I*a*
x)^(3/4))

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fricas [A]  time = 1.48, size = 204, normalized size = 0.80 \begin {gather*} \frac {\sqrt {9 i} a \log \left (\frac {\sqrt {9 i} {\left (a x + i \, a\right )} + 3 \, {\left (i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, a x + a\right )}^{\frac {3}{4}}}{3 \, x + 3 i}\right ) - \sqrt {9 i} a \log \left (-\frac {\sqrt {9 i} {\left (a x + i \, a\right )} - 3 \, {\left (i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, a x + a\right )}^{\frac {3}{4}}}{3 \, x + 3 i}\right ) + \sqrt {-9 i} a \log \left (\frac {\sqrt {-9 i} {\left (a x + i \, a\right )} + 3 \, {\left (i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, a x + a\right )}^{\frac {3}{4}}}{3 \, x + 3 i}\right ) - \sqrt {-9 i} a \log \left (-\frac {\sqrt {-9 i} {\left (a x + i \, a\right )} - 3 \, {\left (i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, a x + a\right )}^{\frac {3}{4}}}{3 \, x + 3 i}\right ) - 2 i \, {\left (i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, a x + a\right )}^{\frac {3}{4}}}{2 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-I*a*x)^(3/4)/(a+I*a*x)^(3/4),x, algorithm="fricas")

[Out]

1/2*(sqrt(9*I)*a*log((sqrt(9*I)*(a*x + I*a) + 3*(I*a*x + a)^(1/4)*(-I*a*x + a)^(3/4))/(3*x + 3*I)) - sqrt(9*I)
*a*log(-(sqrt(9*I)*(a*x + I*a) - 3*(I*a*x + a)^(1/4)*(-I*a*x + a)^(3/4))/(3*x + 3*I)) + sqrt(-9*I)*a*log((sqrt
(-9*I)*(a*x + I*a) + 3*(I*a*x + a)^(1/4)*(-I*a*x + a)^(3/4))/(3*x + 3*I)) - sqrt(-9*I)*a*log(-(sqrt(-9*I)*(a*x
+ I*a) - 3*(I*a*x + a)^(1/4)*(-I*a*x + a)^(3/4))/(3*x + 3*I)) - 2*I*(I*a*x + a)^(1/4)*(-I*a*x + a)^(3/4))/a

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (-i \, a x + a\right )}^{\frac {3}{4}}}{{\left (i \, a x + a\right )}^{\frac {3}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-I*a*x)^(3/4)/(a+I*a*x)^(3/4),x, algorithm="giac")

[Out]

integrate((-I*a*x + a)^(3/4)/(I*a*x + a)^(3/4), x)

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maple [C]  time = 2.14, size = 464, normalized size = 1.81 \begin {gather*} -\frac {i \left (x -i\right ) \left (x +i\right ) a}{\left (\left (i x +1\right ) a \right )^{\frac {3}{4}} \left (-\left (i x -1\right ) a \right )^{\frac {1}{4}}}+\frac {\left (-\frac {3 \RootOf \left (\textit {\_Z}^{2}+i\right ) \ln \left (-\frac {x^{3}-\left (-x^{4}+2 i x^{3}+2 i x +1\right )^{\frac {1}{4}} x^{2} \RootOf \left (\textit {\_Z}^{2}+i\right )-2 i x^{2}+2 i \left (-x^{4}+2 i x^{3}+2 i x +1\right )^{\frac {1}{4}} x \RootOf \left (\textit {\_Z}^{2}+i\right )-i \sqrt {-x^{4}+2 i x^{3}+2 i x +1}\, x -x +i \left (-x^{4}+2 i x^{3}+2 i x +1\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{2}+i\right )+\left (-x^{4}+2 i x^{3}+2 i x +1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+i\right )-\sqrt {-x^{4}+2 i x^{3}+2 i x +1}}{\left (i x +1\right )^{2}}\right )}{2}-\frac {3 i \RootOf \left (\textit {\_Z}^{2}+i\right ) \ln \left (-\frac {x^{3}-i \left (-x^{4}+2 i x^{3}+2 i x +1\right )^{\frac {1}{4}} x^{2} \RootOf \left (\textit {\_Z}^{2}+i\right )-2 i x^{2}-2 \left (-x^{4}+2 i x^{3}+2 i x +1\right )^{\frac {1}{4}} x \RootOf \left (\textit {\_Z}^{2}+i\right )+i \sqrt {-x^{4}+2 i x^{3}+2 i x +1}\, x -x +\left (-x^{4}+2 i x^{3}+2 i x +1\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{2}+i\right )+i \left (-x^{4}+2 i x^{3}+2 i x +1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+i\right )+\sqrt {-x^{4}+2 i x^{3}+2 i x +1}}{\left (i x +1\right )^{2}}\right )}{2}\right ) \left (-\left (i x -1\right ) \left (i x +1\right )^{3}\right )^{\frac {1}{4}} a}{\left (\left (i x +1\right ) a \right )^{\frac {3}{4}} \left (-\left (i x -1\right ) a \right )^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-I*a*x+a)^(3/4)/(I*a*x+a)^(3/4),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (-i \, a x + a\right )}^{\frac {3}{4}}}{{\left (i \, a x + a\right )}^{\frac {3}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-I*a*x)^(3/4)/(a+I*a*x)^(3/4),x, algorithm="maxima")

[Out]

integrate((-I*a*x + a)^(3/4)/(I*a*x + a)^(3/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a-a\,x\,1{}\mathrm {i}\right )}^{3/4}}{{\left (a+a\,x\,1{}\mathrm {i}\right )}^{3/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - a*x*1i)^(3/4)/(a + a*x*1i)^(3/4),x)

[Out]

int((a - a*x*1i)^(3/4)/(a + a*x*1i)^(3/4), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- i a \left (x + i\right )\right )^{\frac {3}{4}}}{\left (i a \left (x - i\right )\right )^{\frac {3}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-I*a*x)**(3/4)/(a+I*a*x)**(3/4),x)

[Out]

Integral((-I*a*(x + I))**(3/4)/(I*a*(x - I))**(3/4), x)

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