3.13.13 \(\int \frac {\sqrt [4]{a-i a x}}{(a+i a x)^{5/4}} \, dx\) [1213]

Optimal. Leaf size=264 \[ \frac {4 i \sqrt [4]{a-i a x}}{a \sqrt [4]{a+i a x}}+\frac {i \sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{a}-\frac {i \sqrt {2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{a}+\frac {i \log \left (1+\frac {\sqrt {a-i a x}}{\sqrt {a+i a x}}-\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2} a}-\frac {i \log \left (1+\frac {\sqrt {a-i a x}}{\sqrt {a+i a x}}+\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt {2} a} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 264, 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 = {49, 65, 246, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {i \sqrt {2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{a}-\frac {i \sqrt {2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{a}+\frac {4 i \sqrt [4]{a-i a x}}{a \sqrt [4]{a+i a x}}+\frac {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 )}{\sqrt {2} a}-\frac {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 )}{\sqrt {2} a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a - I*a*x)^(1/4)/(a + I*a*x)^(5/4),x]

[Out]

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

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 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 {\sqrt [4]{a-i a x}}{(a+i a x)^{5/4}} \, dx &=\frac {4 i \sqrt [4]{a-i a x}}{a \sqrt [4]{a+i a x}}-\int \frac {1}{(a-i a x)^{3/4} \sqrt [4]{a+i a x}} \, dx\\ &=\frac {4 i \sqrt [4]{a-i a x}}{a \sqrt [4]{a+i a x}}-\frac {(4 i) \text {Subst}\left (\int \frac {1}{\sqrt [4]{2 a-x^4}} \, dx,x,\sqrt [4]{a-i a x}\right )}{a}\\ &=\frac {4 i \sqrt [4]{a-i a x}}{a \sqrt [4]{a+i a x}}-\frac {(4 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 )}{a}\\ &=\frac {4 i \sqrt [4]{a-i a x}}{a \sqrt [4]{a+i a x}}-\frac {(2 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 )}{a}-\frac {(2 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 )}{a}\\ &=\frac {4 i \sqrt [4]{a-i a x}}{a \sqrt [4]{a+i a x}}-\frac {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 )}{a}-\frac {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 )}{a}+\frac {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 )}{\sqrt {2} a}+\frac {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 )}{\sqrt {2} a}\\ &=\frac {4 i \sqrt [4]{a-i a x}}{a \sqrt [4]{a+i a x}}+\frac {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 )}{\sqrt {2} a}-\frac {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 )}{\sqrt {2} a}-\frac {\left (i \sqrt {2}\right ) \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 )}{a}+\frac {\left (i \sqrt {2}\right ) \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 )}{a}\\ &=\frac {4 i \sqrt [4]{a-i a x}}{a \sqrt [4]{a+i a x}}+\frac {i \sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{a}-\frac {i \sqrt {2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{a}+\frac {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 )}{\sqrt {2} a}-\frac {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 )}{\sqrt {2} a}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 109, normalized size = 0.41 \begin {gather*} \frac {2 \left (\frac {2 i \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}-\sqrt [4]{-1} \tanh ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )+(-1)^{3/4} \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a - I*a*x)^(1/4)/(a + I*a*x)^(5/4),x]

[Out]

(2*(((2*I)*(a - I*a*x)^(1/4))/(a + I*a*x)^(1/4) - (-1)^(1/4)*ArcTanh[((-1)^(1/4)*(a - I*a*x)^(1/4))/(a + I*a*x
)^(1/4)] + (-1)^(3/4)*ArcTanh[((-1)^(3/4)*(a - I*a*x)^(1/4))/(a + I*a*x)^(1/4)]))/a

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.15, size = 478, normalized size = 1.81

method	result	size
risch	\(-\frac {4 \left (x +i\right ) \left (-a \left (i x -1\right )\right )^{\frac {1}{4}}}{a \left (i x -1\right ) \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}+\frac {\left (\RootOf \left (\textit {\_Z}^{2}-i\right ) \ln \left (-\frac {\left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}-i\right ) x^{2}+i \RootOf \left (\textit {\_Z}^{2}-i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {3}{4}}+x^{3}+i \sqrt {-x^{4}-2 i x^{3}-2 i x +1}\, x +2 i \RootOf \left (\textit {\_Z}^{2}-i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}} x +2 i x^{2}-\sqrt {-x^{4}-2 i x^{3}-2 i x +1}-\RootOf \left (\textit {\_Z}^{2}-i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}}-x}{\left (i x -1\right )^{2}}\right )+i \RootOf \left (\textit {\_Z}^{2}-i\right ) \ln \left (-\frac {i \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}-i\right ) x^{2}-2 \RootOf \left (\textit {\_Z}^{2}-i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}} x +x^{3}+\RootOf \left (\textit {\_Z}^{2}-i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {3}{4}}-i \sqrt {-x^{4}-2 i x^{3}-2 i x +1}\, x -i \RootOf \left (\textit {\_Z}^{2}-i\right ) \left (-x^{4}-2 i x^{3}-2 i x +1\right )^{\frac {1}{4}}+2 i x^{2}+\sqrt {-x^{4}-2 i x^{3}-2 i x +1}-x}{\left (i x -1\right )^{2}}\right )\right ) \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (-\left (i x -1\right )^{3} \left (i x +1\right )\right )^{\frac {1}{4}}}{a \left (i x -1\right ) \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}\)	\(478\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a-I*a*x)^(1/4)/(a+I*a*x)^(5/4),x,method=_RETURNVERBOSE)

[Out]

-4*(x+I)/a*(-a*(-1+I*x))^(1/4)/(-1+I*x)/(a*(1+I*x))^(1/4)+(RootOf(_Z^2-I)*ln(-((1-x^4-2*I*x^3-2*I*x)^(1/4)*Roo
tOf(_Z^2-I)*x^2+I*RootOf(_Z^2-I)*(1-x^4-2*I*x^3-2*I*x)^(3/4)+x^3+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)+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*RootOf(_Z^2-I)*(1
-x^4-2*I*x^3-2*I*x)^(1/4)*x+x^3+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-I*R
ootOf(_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*(-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)^(1/4)/(a+I*a*x)^(5/4),x, algorithm="maxima")

[Out]

integrate((-I*a*x + a)^(1/4)/(I*a*x + a)^(5/4), x)

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Fricas [A]
time = 0.68, size = 296, normalized size = 1.12 \begin {gather*} -\frac {{\left (a^{2} x - i \, a^{2}\right )} \sqrt {\frac {4 i}{a^{2}}} \log \left (\frac {{\left (a^{2} x - i \, a^{2}\right )} \sqrt {\frac {4 i}{a^{2}}} + 2 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{2 \, {\left (x - i\right )}}\right ) - {\left (a^{2} x - i \, a^{2}\right )} \sqrt {\frac {4 i}{a^{2}}} \log \left (-\frac {{\left (a^{2} x - i \, a^{2}\right )} \sqrt {\frac {4 i}{a^{2}}} - 2 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{2 \, {\left (x - i\right )}}\right ) + {\left (a^{2} x - i \, a^{2}\right )} \sqrt {-\frac {4 i}{a^{2}}} \log \left (\frac {{\left (a^{2} x - i \, a^{2}\right )} \sqrt {-\frac {4 i}{a^{2}}} + 2 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{2 \, {\left (x - i\right )}}\right ) - {\left (a^{2} x - i \, a^{2}\right )} \sqrt {-\frac {4 i}{a^{2}}} \log \left (-\frac {{\left (a^{2} x - i \, a^{2}\right )} \sqrt {-\frac {4 i}{a^{2}}} - 2 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{2 \, {\left (x - i\right )}}\right ) - 8 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{2 \, {\left (a^{2} x - i \, a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-I*a*x)^(1/4)/(a+I*a*x)^(5/4),x, algorithm="fricas")

[Out]

-1/2*((a^2*x - I*a^2)*sqrt(4*I/a^2)*log(1/2*((a^2*x - I*a^2)*sqrt(4*I/a^2) + 2*(I*a*x + a)^(3/4)*(-I*a*x + a)^
(1/4))/(x - I)) - (a^2*x - I*a^2)*sqrt(4*I/a^2)*log(-1/2*((a^2*x - I*a^2)*sqrt(4*I/a^2) - 2*(I*a*x + a)^(3/4)*
(-I*a*x + a)^(1/4))/(x - I)) + (a^2*x - I*a^2)*sqrt(-4*I/a^2)*log(1/2*((a^2*x - I*a^2)*sqrt(-4*I/a^2) + 2*(I*a
*x + a)^(3/4)*(-I*a*x + a)^(1/4))/(x - I)) - (a^2*x - I*a^2)*sqrt(-4*I/a^2)*log(-1/2*((a^2*x - I*a^2)*sqrt(-4*
I/a^2) - 2*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4))/(x - I)) - 8*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4))/(a^2*x - I
*a^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [4]{- i a \left (x + i\right )}}{\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)**(1/4)/(a+I*a*x)**(5/4),x)

[Out]

Integral((-I*a*(x + I))**(1/4)/(I*a*(x - I))**(5/4), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-I*a*x)^(1/4)/(a+I*a*x)^(5/4),x, algorithm="giac")

[Out]

integrate((-I*a*x + a)^(1/4)/(I*a*x + a)^(5/4), 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 )}^{1/4}}{{\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)^(1/4)/(a + a*x*1i)^(5/4),x)

[Out]

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

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