\(\int \frac {1}{(a+i a \tan (c+d x))^{2/3}} \, dx\) [297]

Optimal result

Integrand size = 17, antiderivative size = 184 \[ \int \frac {1}{(a+i a \tan (c+d x))^{2/3}} \, dx=-\frac {x}{4\ 2^{2/3} a^{2/3}}-\frac {i \sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2\ 2^{2/3} a^{2/3} d}+\frac {i \log (\cos (c+d x))}{4\ 2^{2/3} a^{2/3} d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4\ 2^{2/3} a^{2/3} d}+\frac {3 i}{4 d (a+i a \tan (c+d x))^{2/3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3560, 3562, 59, 631, 210, 31} \[ \int \frac {1}{(a+i a \tan (c+d x))^{2/3}} \, dx=-\frac {i \sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2\ 2^{2/3} a^{2/3} d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4\ 2^{2/3} a^{2/3} d}+\frac {i \log (\cos (c+d x))}{4\ 2^{2/3} a^{2/3} d}-\frac {x}{4\ 2^{2/3} a^{2/3}}+\frac {3 i}{4 d (a+i a \tan (c+d x))^{2/3}} \]

[In]

Int[(a + I*a*Tan[c + d*x])^(-2/3),x]

[Out]

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

Rule 31

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

Rule 59

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), 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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 631

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 3560

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 3562

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]

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.16 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.27 \[ \int \frac {1}{(a+i a \tan (c+d x))^{2/3}} \, dx=\frac {3 i \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},1,\frac {1}{3},\frac {1}{2} (1+i \tan (c+d x))\right )}{4 d (a+i a \tan (c+d x))^{2/3}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {3 i a \left (\frac {1}{4 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}+\frac {\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}}{2 a}\right )}{d}\)	\(158\)
default	\(\frac {3 i a \left (\frac {1}{4 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}+\frac {\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}}{2 a}\right )}{d}\)	\(158\)

[In]

int(1/(a+I*a*tan(d*x+c))^(2/3),x,method=_RETURNVERBOSE)

[Out]

3*I/d*a*(1/4/a/(a+I*a*tan(d*x+c))^(2/3)+1/2*(1/6*2^(1/3)/a^(2/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))-
1/12*2^(1/3)/a^(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))-1/6
*2^(1/3)/a^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2^(2/3)/a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+1)))/a)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (125) = 250\).

Time = 0.25 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.66 \[ \int \frac {1}{(a+i a \tan (c+d x))^{2/3}} \, dx=\frac {{\left (8 \, a d \left (-\frac {i}{32 \, a^{2} d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (4 i \, a d \left (-\frac {i}{32 \, a^{2} d^{3}}\right )^{\frac {1}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 4 \, {\left (-i \, \sqrt {3} a d + a d\right )} \left (-\frac {i}{32 \, a^{2} d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} - 2 \, {\left (\sqrt {3} a d + i \, a d\right )} \left (-\frac {i}{32 \, a^{2} d^{3}}\right )^{\frac {1}{3}}\right ) - 4 \, {\left (i \, \sqrt {3} a d + a d\right )} \left (-\frac {i}{32 \, a^{2} d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + 2 \, {\left (\sqrt {3} a d - i \, a d\right )} \left (-\frac {i}{32 \, a^{2} d^{3}}\right )^{\frac {1}{3}}\right ) - 3 \cdot 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} {\left (-i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a d} \]

[In]

integrate(1/(a+I*a*tan(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

1/8*(8*a*d*(-1/32*I/(a^2*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log(4*I*a*d*(-1/32*I/(a^2*d^3))^(1/3) + 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)) - 4*(-I*sqrt(3)*a*d + a*d)*(-1/32*I/(a^2*d^3))^(1/3)*e
^(2*I*d*x + 2*I*c)*log(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*(sqrt(3)*a*d +
I*a*d)*(-1/32*I/(a^2*d^3))^(1/3)) - 4*(I*sqrt(3)*a*d + a*d)*(-1/32*I/(a^2*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log(
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*(sqrt(3)*a*d - I*a*d)*(-1/32*I/(a^2*d^
3))^(1/3)) - 3*2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*(-I*e^(2*I*d*x + 2*I*c) - I)*e^(2/3*I*d*x + 2/3*I*c
))*e^(-2*I*d*x - 2*I*c)/(a*d)

Sympy [F]

\[ \int \frac {1}{(a+i a \tan (c+d x))^{2/3}} \, dx=\int \frac {1}{\left (i a \tan {\left (c + d x \right )} + a\right )^{\frac {2}{3}}}\, dx \]

[In]

integrate(1/(a+I*a*tan(d*x+c))**(2/3),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.82 \[ \int \frac {1}{(a+i a \tan (c+d x))^{2/3}} \, dx=-\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + 2^{\frac {1}{3}} a^{\frac {1}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 2 \cdot 2^{\frac {1}{3}} a^{\frac {1}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - \frac {6 \, a}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}\right )}}{8 \, a d} \]

[In]

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

[Out]

-1/8*I*(2*sqrt(3)*2^(1/3)*a^(1/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)*a^(1/3) + 2*(I*a*tan(d*x + c) + a)^(1/3)
)/a^(1/3)) + 2^(1/3)*a^(1/3)*log(2^(2/3)*a^(2/3) + 2^(1/3)*(I*a*tan(d*x + c) + a)^(1/3)*a^(1/3) + (I*a*tan(d*x
+ c) + a)^(2/3)) - 2*2^(1/3)*a^(1/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) - 6*a/(I*a*tan(d*x
+ c) + a)^(2/3))/(a*d)

Giac [F]

\[ \int \frac {1}{(a+i a \tan (c+d x))^{2/3}} \, dx=\int { \frac {1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]

[In]

integrate(1/(a+I*a*tan(d*x+c))^(2/3),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(a+i a \tan (c+d x))^{2/3}} \, dx=\frac {3{}\mathrm {i}}{4\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{2/3}}-\frac {{\left (\frac {1}{32}{}\mathrm {i}\right )}^{1/3}\,\ln \left (d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,36{}\mathrm {i}+144\,{\left (\frac {1}{32}{}\mathrm {i}\right )}^{1/3}\,a^{1/3}\,d^2\right )}{a^{2/3}\,d}-\frac {{\left (\frac {1}{32}{}\mathrm {i}\right )}^{1/3}\,\ln \left (d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,36{}\mathrm {i}+144\,{\left (\frac {1}{32}{}\mathrm {i}\right )}^{1/3}\,a^{1/3}\,d^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{2/3}\,d}+\frac {{\left (\frac {1}{32}{}\mathrm {i}\right )}^{1/3}\,\ln \left (d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,36{}\mathrm {i}-144\,{\left (\frac {1}{32}{}\mathrm {i}\right )}^{1/3}\,a^{1/3}\,d^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{2/3}\,d} \]

[In]

int(1/(a + a*tan(c + d*x)*1i)^(2/3),x)

[Out]

3i/(4*d*(a + a*tan(c + d*x)*1i)^(2/3)) - ((1i/32)^(1/3)*log(d^2*(a + a*tan(c + d*x)*1i)^(1/3)*36i + 144*(1i/32
)^(1/3)*a^(1/3)*d^2))/(a^(2/3)*d) - ((1i/32)^(1/3)*log(d^2*(a + a*tan(c + d*x)*1i)^(1/3)*36i + 144*(1i/32)^(1/
3)*a^(1/3)*d^2*((3^(1/2)*1i)/2 - 1/2))*((3^(1/2)*1i)/2 - 1/2))/(a^(2/3)*d) + ((1i/32)^(1/3)*log(d^2*(a + a*tan
(c + d*x)*1i)^(1/3)*36i - 144*(1i/32)^(1/3)*a^(1/3)*d^2*((3^(1/2)*1i)/2 + 1/2))*((3^(1/2)*1i)/2 + 1/2))/(a^(2/
3)*d)