3.1185 \(\int \frac{1}{\sqrt [4]{a-i a x} (a+i a x)^{3/4}} \, dx\)

Optimal. Leaf size=233 \[ \frac{i \log \left (\frac{\sqrt{a-i a x}}{\sqrt{a+i a x}}-\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}+1\right )}{\sqrt{2} a}-\frac{i \log \left (\frac{\sqrt{a-i a x}}{\sqrt{a+i a x}}+\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}+1\right )}{\sqrt{2} a}-\frac{i \sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{a}+\frac{i \sqrt{2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{a} \]

[Out]

((-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)

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Rubi [A]  time = 0.135098, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {63, 331, 297, 1162, 617, 204, 1165, 628} \[ \frac{i \log \left (\frac{\sqrt{a-i a x}}{\sqrt{a+i a x}}-\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}+1\right )}{\sqrt{2} a}-\frac{i \log \left (\frac{\sqrt{a-i a x}}{\sqrt{a+i a x}}+\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}+1\right )}{\sqrt{2} a}-\frac{i \sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{a}+\frac{i \sqrt{2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{a-i a x}}{\sqrt [4]{a+i a x}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Int[1/((a - I*a*x)^(1/4)*(a + I*a*x)^(3/4)),x]

[Out]

((-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 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{1}{\sqrt [4]{a-i a x} (a+i a x)^{3/4}} \, dx &=\frac{(4 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 )}{a}\\ &=\frac{(4 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 )}{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{(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{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{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{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*}

Mathematica [C]  time = 0.0220721, size = 70, normalized size = 0.3 \[ \frac{2 i \sqrt [4]{2} (1+i x)^{3/4} (a-i a x)^{3/4} \, _2F_1\left (\frac{3}{4},\frac{3}{4};\frac{7}{4};\frac{1}{2}-\frac{i x}{2}\right )}{3 a (a+i a x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a - I*a*x)^(1/4)*(a + I*a*x)^(3/4)),x]

[Out]

(((2*I)/3)*2^(1/4)*(1 + I*x)^(3/4)*(a - I*a*x)^(3/4)*Hypergeometric2F1[3/4, 3/4, 7/4, 1/2 - (I/2)*x])/(a*(a +
I*a*x)^(3/4))

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Maple [F]  time = 0.039, size = 0, normalized size = 0. \begin{align*} \int{{\frac{1}{\sqrt [4]{a-iax}}} \left ( a+iax \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (i \, a x + a\right )}^{\frac{3}{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(1/(a-I*a*x)^(1/4)/(a+I*a*x)^(3/4),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.62714, size = 601, normalized size = 2.58 \begin{align*} \frac{1}{2} \, \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{1}{4}}{\left (-i \, a x + a\right )}^{\frac{3}{4}}}{2 \, x + 2 i}\right ) - \frac{1}{2} \, \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{1}{4}}{\left (-i \, a x + a\right )}^{\frac{3}{4}}}{2 \, x + 2 i}\right ) + \frac{1}{2} \, \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{1}{4}}{\left (-i \, a x + a\right )}^{\frac{3}{4}}}{2 \, x + 2 i}\right ) - \frac{1}{2} \, \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{1}{4}}{\left (-i \, a x + a\right )}^{\frac{3}{4}}}{2 \, x + 2 i}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(1/4)/(a+I*a*x)^(3/4),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(1/4)/(a+I*a*x)^(3/4),x, algorithm="giac")

[Out]

Exception raised: TypeError