3.13.12 \(\int \frac {(a-i a x)^{5/4}}{(a+i a x)^{5/4}} \, dx\) [1212]

Optimal. Leaf size=287 \[ \frac {4 i (a-i a x)^{5/4}}{a \sqrt [4]{a+i a x}}+\frac {5 i \sqrt [4]{a-i a x} (a+i a x)^{3/4}}{a}+\frac {5 i \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2}}-\frac {5 i \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2}}+\frac {5 i \log \left (1+\frac {\sqrt {a-i a x}}{\sqrt {a+i a x}}-\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{2 \sqrt {2}}-\frac {5 i \log \left (1+\frac {\sqrt {a-i a x}}{\sqrt {a+i a x}}+\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{2 \sqrt {2}} \]

[Out]

4*I*(a-I*a*x)^(5/4)/a/(a+I*a*x)^(1/4)+5*I*(a-I*a*x)^(1/4)*(a+I*a*x)^(3/4)/a+5/2*I*arctan(1-(a-I*a*x)^(1/4)*2^(
1/2)/(a+I*a*x)^(1/4))*2^(1/2)-5/2*I*arctan(1+(a-I*a*x)^(1/4)*2^(1/2)/(a+I*a*x)^(1/4))*2^(1/2)+5/4*I*ln(1-(a-I*
a*x)^(1/4)*2^(1/2)/(a+I*a*x)^(1/4)+(a-I*a*x)^(1/2)/(a+I*a*x)^(1/2))*2^(1/2)-5/4*I*ln(1+(a-I*a*x)^(1/4)*2^(1/2)
/(a+I*a*x)^(1/4)+(a-I*a*x)^(1/2)/(a+I*a*x)^(1/2))*2^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {49, 52, 65, 246, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {5 i \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2}}-\frac {5 i \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2}}+\frac {4 i (a-i a x)^{5/4}}{a \sqrt [4]{a+i a x}}+\frac {5 i (a+i a x)^{3/4} \sqrt [4]{a-i a x}}{a}+\frac {5 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 {5 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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a - I*a*x)^(5/4)/(a + I*a*x)^(5/4),x]

[Out]

((4*I)*(a - I*a*x)^(5/4))/(a*(a + I*a*x)^(1/4)) + ((5*I)*(a - I*a*x)^(1/4)*(a + I*a*x)^(3/4))/a + ((5*I)*ArcTa
n[1 - (Sqrt[2]*(a - I*a*x)^(1/4))/(a + I*a*x)^(1/4)])/Sqrt[2] - ((5*I)*ArcTan[1 + (Sqrt[2]*(a - I*a*x)^(1/4))/
(a + I*a*x)^(1/4)])/Sqrt[2] + (((5*I)/2)*Log[1 + Sqrt[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] - (((5*I)/2)*Log[1 + Sqrt[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 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 52

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 65

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 210

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 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[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 631

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 642

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 1176

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 1179

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)^{5/4}}{(a+i a x)^{5/4}} \, dx &=\frac {4 i (a-i a x)^{5/4}}{a \sqrt [4]{a+i a x}}-5 \int \frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}} \, dx\\ &=\frac {4 i (a-i a x)^{5/4}}{a \sqrt [4]{a+i a x}}+\frac {5 i \sqrt [4]{a-i a x} (a+i a x)^{3/4}}{a}-\frac {1}{2} (5 a) \int \frac {1}{(a-i a x)^{3/4} \sqrt [4]{a+i a x}} \, dx\\ &=\frac {4 i (a-i a x)^{5/4}}{a \sqrt [4]{a+i a x}}+\frac {5 i \sqrt [4]{a-i a x} (a+i a x)^{3/4}}{a}-10 i \text {Subst}\left (\int \frac {1}{\sqrt [4]{2 a-x^4}} \, dx,x,\sqrt [4]{a-i a x}\right )\\ &=\frac {4 i (a-i a x)^{5/4}}{a \sqrt [4]{a+i a x}}+\frac {5 i \sqrt [4]{a-i a x} (a+i a x)^{3/4}}{a}-10 i \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\frac {\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )\\ &=\frac {4 i (a-i a x)^{5/4}}{a \sqrt [4]{a+i a x}}+\frac {5 i \sqrt [4]{a-i a x} (a+i a x)^{3/4}}{a}-5 i \text {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 )-5 i \text {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 {4 i (a-i a x)^{5/4}}{a \sqrt [4]{a+i a x}}+\frac {5 i \sqrt [4]{a-i a x} (a+i a x)^{3/4}}{a}-\frac {5}{2} i \text {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 {5}{2} i \text {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 {(5 i) \text {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 {(5 i) \text {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 {4 i (a-i a x)^{5/4}}{a \sqrt [4]{a+i a x}}+\frac {5 i \sqrt [4]{a-i a x} (a+i a x)^{3/4}}{a}+\frac {5 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 {5 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 {(5 i) \text {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 {(5 i) \text {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 {4 i (a-i a x)^{5/4}}{a \sqrt [4]{a+i a x}}+\frac {5 i \sqrt [4]{a-i a x} (a+i a x)^{3/4}}{a}+\frac {5 i \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2}}-\frac {5 i \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2}}+\frac {5 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 {5 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 [A]
time = 0.58, size = 117, normalized size = 0.41 \begin {gather*} -\frac {\sqrt [4]{a-i a x} \left (\sqrt [4]{i+x} (-9 i+x)+5 i \sqrt [4]{-i+x} \tan ^{-1}\left (\frac {\sqrt [4]{i+x}}{\sqrt [4]{-i+x}}\right )+5 i \sqrt [4]{-i+x} \tanh ^{-1}\left (\frac {\sqrt [4]{i+x}}{\sqrt [4]{-i+x}}\right )\right )}{\sqrt [4]{i+x} \sqrt [4]{a+i a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a - I*a*x)^(5/4)/(a + I*a*x)^(5/4),x]

[Out]

-(((a - I*a*x)^(1/4)*((I + x)^(1/4)*(-9*I + x) + (5*I)*(-I + x)^(1/4)*ArcTan[(I + x)^(1/4)/(-I + x)^(1/4)] + (
5*I)*(-I + x)^(1/4)*ArcTanh[(I + x)^(1/4)/(-I + x)^(1/4)]))/((I + x)^(1/4)*(a + I*a*x)^(1/4)))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.57, size = 480, normalized size = 1.67

method	result	size
risch	\(-\frac {i \left (x^{2}-8 i x +9\right ) \left (-a \left (i x -1\right )\right )^{\frac {1}{4}}}{\left (i x -1\right ) \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}-\frac {\left (\frac {5 \RootOf \left (\textit {\_Z}^{2}+i\right ) \ln \left (-\frac {-\left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+i\right ) x^{2}+x^{3}+i \RootOf \left (\textit {\_Z}^{2}+i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {3}{4}}-i \sqrt {-x^{4}-2 i x^{3}-2 i x +1}\, x -2 i \RootOf \left (\textit {\_Z}^{2}+i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}} x +2 i x^{2}+\sqrt {-x^{4}-2 i x^{3}-2 i x +1}+\RootOf \left (\textit {\_Z}^{2}+i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}}-x}{\left (i x -1\right )^{2}}\right )}{2}+\frac {5 i \RootOf \left (\textit {\_Z}^{2}+i\right ) \ln \left (-\frac {-i \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+i\right ) x^{2}+2 \RootOf \left (\textit {\_Z}^{2}+i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}} x +x^{3}+i \sqrt {-x^{4}-2 i x^{3}-2 i x +1}\, x +\RootOf \left (\textit {\_Z}^{2}+i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {3}{4}}+i \RootOf \left (\textit {\_Z}^{2}+i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}}+2 i x^{2}-\sqrt {-x^{4}-2 i x^{3}-2 i x +1}-x}{\left (i x -1\right )^{2}}\right )}{2}\right ) \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (-\left (i x -1\right )^{3} \left (i x +1\right )\right )^{\frac {1}{4}}}{\left (i x -1\right ) \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}\)	\(480\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a-I*a*x)^(5/4)/(a+I*a*x)^(5/4),x,method=_RETURNVERBOSE)

[Out]

-I*(x^2+9-8*I*x)*(-a*(-1+I*x))^(1/4)/(-1+I*x)/(a*(1+I*x))^(1/4)-(5/2*RootOf(_Z^2+I)*ln(-(-(1-x^4-2*I*x^3-2*I*x
)^(1/4)*RootOf(_Z^2+I)*x^2+x^3+I*RootOf(_Z^2+I)*(1-x^4-2*I*x^3-2*I*x)^(3/4)-I*(1-x^4-2*I*x^3-2*I*x)^(1/2)*x-2*
I*RootOf(_Z^2+I)*(1-x^4-2*I*x^3-2*I*x)^(1/4)*x+2*I*x^2+(1-x^4-2*I*x^3-2*I*x)^(1/2)+RootOf(_Z^2+I)*(1-x^4-2*I*x
^3-2*I*x)^(1/4)-x)/(-1+I*x)^2)+5/2*I*RootOf(_Z^2+I)*ln(-(-I*(1-x^4-2*I*x^3-2*I*x)^(1/4)*RootOf(_Z^2+I)*x^2+2*R
ootOf(_Z^2+I)*(1-x^4-2*I*x^3-2*I*x)^(1/4)*x+x^3+I*(1-x^4-2*I*x^3-2*I*x)^(1/2)*x+RootOf(_Z^2+I)*(1-x^4-2*I*x^3-
2*I*x)^(3/4)+I*RootOf(_Z^2+I)*(1-x^4-2*I*x^3-2*I*x)^(1/4)+2*I*x^2-(1-x^4-2*I*x^3-2*I*x)^(1/2)-x)/(-1+I*x)^2))*
(-a*(-1+I*x))^(1/4)/(-1+I*x)*(-(-1+I*x)^3*(1+I*x))^(1/4)/(a*(1+I*x))^(1/4)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-I*a*x)^(5/4)/(a+I*a*x)^(5/4),x, algorithm="maxima")

[Out]

integrate((-I*a*x + a)^(5/4)/(I*a*x + a)^(5/4), x)

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Fricas [A]
time = 0.82, size = 233, normalized size = 0.81 \begin {gather*} -\frac {\sqrt {25 i} {\left (a x - i \, a\right )} \log \left (\frac {\sqrt {25 i} {\left (a x - i \, a\right )} + 5 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{5 \, {\left (x - i\right )}}\right ) - \sqrt {25 i} {\left (a x - i \, a\right )} \log \left (-\frac {\sqrt {25 i} {\left (a x - i \, a\right )} - 5 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{5 \, {\left (x - i\right )}}\right ) + \sqrt {-25 i} {\left (a x - i \, a\right )} \log \left (\frac {\sqrt {-25 i} {\left (a x - i \, a\right )} + 5 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{5 \, {\left (x - i\right )}}\right ) - \sqrt {-25 i} {\left (a x - i \, a\right )} \log \left (-\frac {\sqrt {-25 i} {\left (a x - i \, a\right )} - 5 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{5 \, {\left (x - i\right )}}\right ) + 2 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, x - 9\right )}}{2 \, {\left (a x - i \, a\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-I*a*x)^(5/4)/(a+I*a*x)^(5/4),x, algorithm="fricas")

[Out]

-1/2*(sqrt(25*I)*(a*x - I*a)*log(1/5*(sqrt(25*I)*(a*x - I*a) + 5*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4))/(x - I)
) - sqrt(25*I)*(a*x - I*a)*log(-1/5*(sqrt(25*I)*(a*x - I*a) - 5*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4))/(x - I))
+ sqrt(-25*I)*(a*x - I*a)*log(1/5*(sqrt(-25*I)*(a*x - I*a) + 5*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4))/(x - I))
- sqrt(-25*I)*(a*x - I*a)*log(-1/5*(sqrt(-25*I)*(a*x - I*a) - 5*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4))/(x - I)
) + 2*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4)*(-I*x - 9))/(a*x - I*a)

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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 {5}{4}}}{\left (i a \left (x - i\right )\right )^{\frac {5}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-I*a*x)**(5/4)/(a+I*a*x)**(5/4),x)

[Out]

Integral((-I*a*(x + I))**(5/4)/(I*a*(x - I))**(5/4), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-I*a*x)^(5/4)/(a+I*a*x)^(5/4),x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:ext_reduce Error:
Bad Argument Typeintegrate(-(-i)/4*16*((sageVARa+(-i)*sageVARa*sageVARx)^(1/4))^8/(-((sageVARa+(-i)*sageVARa*
sageVARx)^(1/4

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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 )}^{5/4}}{{\left (a+a\,x\,1{}\mathrm {i}\right )}^{5/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - a*x*1i)^(5/4)/(a + a*x*1i)^(5/4),x)

[Out]

int((a - a*x*1i)^(5/4)/(a + a*x*1i)^(5/4), x)

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