3.1178 \(\int \frac{\sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.172916, antiderivative size = 256, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {50, 63, 240, 211, 1165, 628, 1162, 617, 204} \[ -\frac{i \sqrt [4]{a-i a x} (a+i a x)^{3/4}}{a}-\frac{i \log \left (\frac{\sqrt{a-i a x}}{\sqrt{a+i a x}}-\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}+1\right )}{2 \sqrt{2}}+\frac{i \log \left (\frac{\sqrt{a-i a x}}{\sqrt{a+i a x}}+\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}+1\right )}{2 \sqrt{2}}-\frac{i \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt{2}}+\frac{i \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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])

Rubi steps

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

Mathematica [C]  time = 0.0208058, size = 70, normalized size = 0.27 \[ \frac{2 i 2^{3/4} \sqrt [4]{1+i x} (a-i a x)^{5/4} \, _2F_1\left (\frac{1}{4},\frac{5}{4};\frac{9}{4};\frac{1}{2}-\frac{i x}{2}\right )}{5 a \sqrt [4]{a+i a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((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))

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Maple [F]  time = 0.049, size = 0, normalized size = 0. \begin{align*} \int{\sqrt [4]{a-iax}{\frac{1}{\sqrt [4]{a+iax}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-i \, a x + a\right )}^{\frac{1}{4}}}{{\left (i \, a x + a\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.67944, size = 540, normalized size = 2.11 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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(-(sqr
t(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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt [4]{- a \left (i x - 1\right )}}{\sqrt [4]{a \left (i x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.25306, size = 252, normalized size = 0.98 \begin{align*} \frac{1}{2} i \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + \frac{2 \,{\left (-i \, a x + a\right )}^{\frac{1}{4}}}{{\left (i \, a x + a\right )}^{\frac{1}{4}}}\right )}\right ) + \frac{1}{2} i \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - \frac{2 \,{\left (-i \, a x + a\right )}^{\frac{1}{4}}}{{\left (i \, a x + a\right )}^{\frac{1}{4}}}\right )}\right ) + \frac{1}{4} i \, \sqrt{2} \log \left (\frac{\sqrt{2}{\left (-i \, a x + a\right )}^{\frac{1}{4}}}{{\left (i \, a x + a\right )}^{\frac{1}{4}}} + \frac{\sqrt{-i \, a x + a}}{\sqrt{i \, a x + a}} + 1\right ) - \frac{1}{4} i \, \sqrt{2} \log \left (-\frac{\sqrt{2}{\left (-i \, a x + a\right )}^{\frac{1}{4}}}{{\left (i \, a x + a\right )}^{\frac{1}{4}}} + \frac{\sqrt{-i \, a x + a}}{\sqrt{i \, a x + a}} + 1\right ) + \frac{i \,{\left (-i \, a x + a\right )}^{\frac{1}{4}}{\left (-i \, a x - a\right )}}{{\left (i \, a x + a\right )}^{\frac{1}{4}} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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