3.294 \(\int \frac{1}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=184 \[ \frac{i \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac{3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac{i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac{x}{4 \sqrt [3]{2} \sqrt [3]{a}} \]

[Out]

-x/(4*2^(1/3)*a^(1/3)) + ((I/2)*Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/
3))])/(2^(1/3)*a^(1/3)*d) + ((I/4)*Log[Cos[c + d*x]])/(2^(1/3)*a^(1/3)*d) + (((3*I)/4)*Log[2^(1/3)*a^(1/3) - (
a + I*a*Tan[c + d*x])^(1/3)])/(2^(1/3)*a^(1/3)*d) + ((3*I)/2)/(d*(a + I*a*Tan[c + d*x])^(1/3))

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Rubi [A]  time = 0.0982843, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {3479, 3481, 55, 617, 204, 31} \[ \frac{i \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac{3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac{i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac{x}{4 \sqrt [3]{2} \sqrt [3]{a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(-1/3),x]

[Out]

-x/(4*2^(1/3)*a^(1/3)) + ((I/2)*Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/
3))])/(2^(1/3)*a^(1/3)*d) + ((I/4)*Log[Cos[c + d*x]])/(2^(1/3)*a^(1/3)*d) + (((3*I)/4)*Log[2^(1/3)*a^(1/3) - (
a + I*a*Tan[c + d*x])^(1/3)])/(2^(1/3)*a^(1/3)*d) + ((3*I)/2)/(d*(a + I*a*Tan[c + d*x])^(1/3))

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 3481

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Dist[b/d, Subst[Int[(a + x)^(n - 1)/(a - x), x]
, x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 + b^2, 0]

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [C]  time = 0.385705, size = 141, normalized size = 0.77 \[ \frac{3 \left (e^{2 i d x} (\cos (c)+i \sin (c)) \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )-2 i \sin (c) \left (-1+e^{2 i d x}\right )-2 \cos (c) \left (1+e^{2 i d x}\right )\right )}{4 d \sqrt [3]{a+i a \tan (c+d x)} \left (i \cos (c) \left (1+e^{2 i d x}\right )-\sin (c) \left (-1+e^{2 i d x}\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-2*(1 + E^((2*I)*d*x))*Cos[c] + E^((2*I)*d*x)*Hypergeometric2F1[2/3, 1, 5/3, E^((2*I)*(c + d*x))/(1 + E^((
2*I)*(c + d*x)))]*(Cos[c] + I*Sin[c]) - (2*I)*(-1 + E^((2*I)*d*x))*Sin[c]))/(4*d*(I*(1 + E^((2*I)*d*x))*Cos[c]
- (-1 + E^((2*I)*d*x))*Sin[c])*(a + I*a*Tan[c + d*x])^(1/3))

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Maple [A]  time = 0.012, size = 158, normalized size = 0.9 \begin{align*}{\frac{{\frac{3\,i}{2}}}{d}{\frac{1}{\sqrt [3]{a+ia\tan \left ( dx+c \right ) }}}}+{\frac{{\frac{i}{4}}{2}^{{\frac{2}{3}}}}{d}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ){\frac{1}{\sqrt [3]{a}}}}-{\frac{{\frac{i}{8}}{2}^{{\frac{2}{3}}}}{d}\ln \left ( \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}+\sqrt [3]{2}\sqrt [3]{a}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }+{2}^{{\frac{2}{3}}}{a}^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{a}}}}+{\frac{{\frac{i}{4}}\sqrt{3}{2}^{{\frac{2}{3}}}}{d}\arctan \left ({\frac{\sqrt{3}}{3} \left ({{2}^{{\frac{2}{3}}}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [3]{a}}}}+1 \right ) } \right ){\frac{1}{\sqrt [3]{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(d*x+c))^(1/3),x)

[Out]

3/2*I/d/(a+I*a*tan(d*x+c))^(1/3)+1/4*I/d/a^(1/3)*2^(2/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))-1/8*I/d/
a^(1/3)*2^(2/3)*ln((a+I*a*tan(d*x+c))^(2/3)+2^(1/3)*a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+2^(2/3)*a^(2/3))+1/4*I/d/
a^(1/3)*3^(1/2)*2^(2/3)*arctan(1/3*3^(1/2)*(2^(2/3)/a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.70187, size = 941, normalized size = 5.11 \begin{align*} \frac{{\left (4 \, a d \left (-\frac{i}{16 \, a d^{3}}\right )^{\frac{1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (8 \, a d^{2} \left (-\frac{i}{16 \, a d^{3}}\right )^{\frac{2}{3}} + 2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )}\right ) + 2^{\frac{2}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{2}{3}}{\left (3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (\frac{4}{3} i \, d x + \frac{4}{3} i \, c\right )} +{\left (-2 i \, \sqrt{3} a d - 2 \, a d\right )} \left (-\frac{i}{16 \, a d^{3}}\right )^{\frac{1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left ({\left (4 i \, \sqrt{3} a d^{2} - 4 \, a d^{2}\right )} \left (-\frac{i}{16 \, a d^{3}}\right )^{\frac{2}{3}} + 2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )}\right ) +{\left (2 i \, \sqrt{3} a d - 2 \, a d\right )} \left (-\frac{i}{16 \, a d^{3}}\right )^{\frac{1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left ({\left (-4 i \, \sqrt{3} a d^{2} - 4 \, a d^{2}\right )} \left (-\frac{i}{16 \, a d^{3}}\right )^{\frac{2}{3}} + 2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )}\right )\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

1/4*(4*a*d*(-1/16*I/(a*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log(8*a*d^2*(-1/16*I/(a*d^3))^(2/3) + 2^(1/3)*(a/(e^(2*
I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c)) + 2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(3*I*e^(2*I*
d*x + 2*I*c) + 3*I)*e^(4/3*I*d*x + 4/3*I*c) + (-2*I*sqrt(3)*a*d - 2*a*d)*(-1/16*I/(a*d^3))^(1/3)*e^(2*I*d*x +
2*I*c)*log((4*I*sqrt(3)*a*d^2 - 4*a*d^2)*(-1/16*I/(a*d^3))^(2/3) + 2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)
*e^(2/3*I*d*x + 2/3*I*c)) + (2*I*sqrt(3)*a*d - 2*a*d)*(-1/16*I/(a*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log((-4*I*sq
rt(3)*a*d^2 - 4*a*d^2)*(-1/16*I/(a*d^3))^(2/3) + 2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/
3*I*c)))*e^(-2*I*d*x - 2*I*c)/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))**(1/3),x)

[Out]

Integral((I*a*tan(c + d*x) + a)**(-1/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(-1/3), x)