\(\int (a+i a \tan (c+d x))^{2/3} \, dx\) [279]

Optimal result

Integrand size = 17, antiderivative size = 156 \[ \int (a+i a \tan (c+d x))^{2/3} \, dx=-\frac {a^{2/3} x}{2 \sqrt [3]{2}}+\frac {i \sqrt {3} a^{2/3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}+\frac {i a^{2/3} \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac {3 i a^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3562, 57, 631, 210, 31} \[ \int (a+i a \tan (c+d x))^{2/3} \, dx=\frac {i \sqrt {3} a^{2/3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}+\frac {3 i a^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}+\frac {i a^{2/3} \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac {a^{2/3} x}{2 \sqrt [3]{2}} \]

[In]

Int[(a + I*a*Tan[c + d*x])^(2/3),x]

[Out]

-1/2*(a^(2/3)*x)/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 31

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

Rule 57

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 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 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 {(i a) \text {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 ) \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 )}{2 \sqrt [3]{2} d}+\frac {(3 i a) \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 )}{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 ) \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 )}{\sqrt [3]{2} d} \\ & = -\frac {a^{2/3} x}{2 \sqrt [3]{2}}+\frac {i \sqrt {3} a^{2/3} \arctan \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 [A] (verified)

Time = 0.13 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.71 \[ \int (a+i a \tan (c+d x))^{2/3} \, dx=\frac {i a^{2/3} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-\log (i+\tan (c+d x))+3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )\right )}{2 \sqrt [3]{2} d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {3 i a \left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right )}{d}\)	\(133\)
default	\(\frac {3 i a \left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right )}{d}\)	\(133\)

[In]

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

[Out]

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

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. 226 vs. \(2 (107) = 214\).

Time = 0.26 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.45 \[ \int (a+i a \tan (c+d x))^{2/3} \, dx=\frac {1}{2} \, {\left (-i \, \sqrt {3} - 1\right )} \left (-\frac {i \, a^{2}}{2 \, d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} a \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + {\left (i \, \sqrt {3} d^{2} - d^{2}\right )} \left (-\frac {i \, a^{2}}{2 \, d^{3}}\right )^{\frac {2}{3}}}{a}\right ) + \frac {1}{2} \, {\left (i \, \sqrt {3} - 1\right )} \left (-\frac {i \, a^{2}}{2 \, d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} a \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + {\left (-i \, \sqrt {3} d^{2} - d^{2}\right )} \left (-\frac {i \, a^{2}}{2 \, d^{3}}\right )^{\frac {2}{3}}}{a}\right ) + \left (-\frac {i \, a^{2}}{2 \, d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2 \, d^{2} \left (-\frac {i \, a^{2}}{2 \, d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} a \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}}{a}\right ) \]

[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)

Sympy [F]

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

[In]

integrate((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 = 0.31 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.87 \[ \int (a+i a \tan (c+d x))^{2/3} \, dx=\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {5}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} a^{\frac {5}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} a^{\frac {5}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )\right )}}{4 \, a d} \]

[In]

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

[Out]

1/4*I*(2*sqrt(3)*2^(2/3)*a^(5/3)*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^(2/3)*a^(5/3)*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^(2/3)*a^(5/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)))/(a*d)

Giac [F]

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

[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)

Mupad [B] (verification not implemented)

Time = 5.54 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.10 \[ \int (a+i a \tan (c+d x))^{2/3} \, dx=-\frac {{\left (\frac {1}{2}{}\mathrm {i}\right )}^{1/3}\,a^{2/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}+{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}\right )}{d}+\frac {{\left (\frac {1}{2}{}\mathrm {i}\right )}^{1/3}\,a^{2/3}\,\ln \left (-\frac {9\,a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d^2}-\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{7/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d}-\frac {{\left (\frac {1}{2}{}\mathrm {i}\right )}^{1/3}\,a^{2/3}\,\ln \left (-\frac {9\,a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d^2}+\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{7/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,d^2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d} \]

[In]

int((a + a*tan(c + d*x)*1i)^(2/3),x)

[Out]

((1i/2)^(1/3)*a^(2/3)*log(- (9*a^2*(a + a*tan(c + d*x)*1i)^(1/3))/d^2 - (9*(-1)^(1/3)*2^(1/3)*a^(7/3)*(3^(1/2)
*1i - 1))/(2*d^2))*((3^(1/2)*1i)/2 + 1/2))/d - ((1i/2)^(1/3)*a^(2/3)*log((a*(tan(c + d*x)*1i + 1))^(1/3) + (-1
)^(1/3)*2^(1/3)*a^(1/3)))/d - ((1i/2)^(1/3)*a^(2/3)*log((9*(-1)^(1/3)*2^(1/3)*a^(7/3)*(3^(1/2)*1i + 1))/(2*d^2
) - (9*a^2*(a + a*tan(c + d*x)*1i)^(1/3))/d^2)*((3^(1/2)*1i)/2 - 1/2))/d