3.279 \(\int (a+i a \tan (c+d x))^{2/3} \, dx\)

Optimal. Leaf size=156 \[ \frac{i \sqrt{3} a^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}+\frac{3 i a^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}+\frac{i a^{2/3} \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac{a^{2/3} x}{2 \sqrt [3]{2}} \]

[Out]

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

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Rubi [A]  time = 0.081721, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {3481, 55, 617, 204, 31} \[ \frac{i \sqrt{3} a^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}+\frac{3 i a^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}+\frac{i a^{2/3} \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac{a^{2/3} x}{2 \sqrt [3]{2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(2/3),x]

[Out]

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

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

Mathematica [C]  time = 0.355125, size = 81, normalized size = 0.52 \[ -\frac{3 i \left (\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )}{2 \sqrt [3]{2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(2/3),x]

[Out]

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

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Maple [A]  time = 0.013, size = 138, normalized size = 0.9 \begin{align*}{\frac{{\frac{i}{2}}{2}^{{\frac{2}{3}}}}{d}{a}^{{\frac{2}{3}}}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ) }-{\frac{{\frac{i}{4}}{2}^{{\frac{2}{3}}}}{d}{a}^{{\frac{2}{3}}}\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{{\frac{i}{2}}\sqrt{3}{2}^{{\frac{2}{3}}}}{d}{a}^{{\frac{2}{3}}}\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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(2/3),x)

[Out]

1/2*I/d*a^(2/3)*2^(2/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))-1/4*I/d*a^(2/3)*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))+1/2*I/d*a^(2/3)*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((a+I*a*tan(d*x+c))^(2/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.6505, size = 645, normalized size = 4.13 \begin{align*} \frac{1}{2} \,{\left (-i \, \sqrt{3} - 1\right )} \left (-\frac{i \, a^{2}}{2 \, d^{3}}\right )^{\frac{1}{3}} \log \left (\frac{2^{\frac{1}{3}} a \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 )} +{\left (i \, \sqrt{3} d^{2} - d^{2}\right )} \left (-\frac{i \, a^{2}}{2 \, d^{3}}\right )^{\frac{2}{3}}}{a}\right ) + \frac{1}{2} \,{\left (i \, \sqrt{3} - 1\right )} \left (-\frac{i \, a^{2}}{2 \, d^{3}}\right )^{\frac{1}{3}} \log \left (\frac{2^{\frac{1}{3}} a \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 )} +{\left (-i \, \sqrt{3} d^{2} - d^{2}\right )} \left (-\frac{i \, a^{2}}{2 \, d^{3}}\right )^{\frac{2}{3}}}{a}\right ) + \left (-\frac{i \, a^{2}}{2 \, d^{3}}\right )^{\frac{1}{3}} \log \left (\frac{2 \, d^{2} \left (-\frac{i \, a^{2}}{2 \, d^{3}}\right )^{\frac{2}{3}} + 2^{\frac{1}{3}} a \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 )}}{a}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

1/2*(-I*sqrt(3) - 1)*(-1/2*I*a^2/d^3)^(1/3)*log((2^(1/3)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x +
2/3*I*c) + (I*sqrt(3)*d^2 - d^2)*(-1/2*I*a^2/d^3)^(2/3))/a) + 1/2*(I*sqrt(3) - 1)*(-1/2*I*a^2/d^3)^(1/3)*log((
2^(1/3)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + (-I*sqrt(3)*d^2 - d^2)*(-1/2*I*a^2/d^3
)^(2/3))/a) + (-1/2*I*a^2/d^3)^(1/3)*log((2*d^2*(-1/2*I*a^2/d^3)^(2/3) + 2^(1/3)*a*(a/(e^(2*I*d*x + 2*I*c) + 1
))^(1/3)*e^(2/3*I*d*x + 2/3*I*c))/a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (i a \tan{\left (c + d x \right )} + a\right )^{\frac{2}{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(2/3),x)

[Out]

Integral((I*a*tan(c + d*x) + a)**(2/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{2}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(2/3),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(2/3), x)