3.1184 \(\int \frac{(a-i a x)^{3/4}}{(a+i a x)^{3/4}} \, dx\)

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

[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]

________________________________________________________________________________________

Rubi [A]  time = 0.158251, 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, 331, 297, 1162, 617, 204, 1165, 628} \[ -\frac{i (a-i a x)^{3/4} \sqrt [4]{a+i a x}}{a}+\frac{3 i \log \left (\frac{\sqrt{a-i a x}}{\sqrt{a+i a x}}-\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}+1\right )}{2 \sqrt{2}}-\frac{3 i \log \left (\frac{\sqrt{a-i a x}}{\sqrt{a+i a x}}+\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}+1\right )}{2 \sqrt{2}}-\frac{3 i \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt{2}}+\frac{3 i \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{\sqrt{2}} \]

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 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 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 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])

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]

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*}

Mathematica [C]  time = 0.0248626, size = 70, normalized size = 0.27 \[ \frac{2 i \sqrt [4]{2} (1+i x)^{3/4} (a-i a x)^{7/4} \, _2F_1\left (\frac{3}{4},\frac{7}{4};\frac{11}{4};\frac{1}{2}-\frac{i x}{2}\right )}{7 a (a+i a x)^{3/4}} \]

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))

________________________________________________________________________________________

Maple [F]  time = 0.043, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a-iax \right ) ^{{\frac{3}{4}}} \left ( a+iax \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Fricas [A]  time = 1.58805, size = 594, normalized size = 2.32 \begin{align*} \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{align*}

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- a \left (i x - 1\right )\right )^{\frac{3}{4}}}{\left (a \left (i x + 1\right )\right )^{\frac{3}{4}}}\, dx \end{align*}

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((-a*(I*x - 1))**(3/4)/(a*(I*x + 1))**(3/4), x)

________________________________________________________________________________________

Giac [A]  time = 1.18519, size = 242, normalized size = 0.95 \begin{align*} \frac{3}{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{3}{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{3}{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{3}{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 )}^{\frac{3}{4}}}{a} \end{align*}

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]

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