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

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

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Rubi [A]  time = 0.14, 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 = {47, 63, 240, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \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}+\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} \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 47

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 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 211

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 240

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 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 {\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) \operatorname {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) \operatorname {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) \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 )}{a}-\frac {(2 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 )}{a}\\ &=\frac {4 i \sqrt [4]{a-i a x}}{a \sqrt [4]{a+i a x}}-\frac {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 )}{a}-\frac {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 )}{a}+\frac {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 )}{\sqrt {2} a}+\frac {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 )}{\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 ) \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 )}{a}+\frac {\left (i \sqrt {2}\right ) \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 )}{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 [C]  time = 0.02, size = 70, normalized size = 0.27 \begin {gather*} \frac {i 2^{3/4} \sqrt [4]{1+i x} (a-i a x)^{5/4} \, _2F_1\left (\frac {5}{4},\frac {5}{4};\frac {9}{4};\frac {1}{2}-\frac {i x}{2}\right )}{5 a^2 \sqrt [4]{a+i a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.19, size = 114, normalized size = 0.43 \begin {gather*} \frac {4 i \sqrt [4]{a-i a x}}{a \sqrt [4]{a+i a x}}-\frac {2 \sqrt [4]{-1} \tanh ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{a}+\frac {2 (-1)^{3/4} \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 0.93, size = 303, normalized size = 1.15 \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 \, x - 2 i}\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 \, x - 2 i}\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 \, x - 2 i}\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 \, x - 2 i}\right ) - 8 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{2 \, a^{2} x - 2 i \, a^{2}} \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]

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

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

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:

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maple [C]  time = 2.01, size = 476, normalized size = 1.80 \begin {gather*} -\frac {4 \left (x +i\right ) \left (-\left (i x -1\right ) a \right )^{\frac {1}{4}}}{\left (i x -1\right ) \left (\left (i x +1\right ) a \right )^{\frac {1}{4}} a}+\frac {\left (\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 )+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 )\right ) \left (-\left (i x -1\right ) a \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 (\left (i x +1\right ) a \right )^{\frac {1}{4}} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 {1}{4}}}{{\left (i \, a x + a\right )}^{\frac {5}{4}}}\,{d x} \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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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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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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