3.3.74 \(\int \tan (c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx\) [274]

Optimal. Leaf size=174 \[ \frac {i \sqrt [3]{a} x}{2\ 2^{2/3}}-\frac {\sqrt {3} \sqrt [3]{a} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2^{2/3} d}+\frac {\sqrt [3]{a} \log (\cos (c+d x))}{2\ 2^{2/3} d}+\frac {3 \sqrt [3]{a} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2\ 2^{2/3} d}+\frac {3 \sqrt [3]{a+i a \tan (c+d x)}}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3608, 3562, 59, 631, 210, 31} \begin {gather*} -\frac {\sqrt {3} \sqrt [3]{a} \text {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^{2/3} d}+\frac {3 \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac {3 \sqrt [3]{a} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2\ 2^{2/3} d}+\frac {\sqrt [3]{a} \log (\cos (c+d x))}{2\ 2^{2/3} d}+\frac {i \sqrt [3]{a} x}{2\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3608

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*(
(a + b*Tan[e + f*x])^m/(f*m)), x] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [F]
time = 180.00, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

________________________________________________________________________________________

Maple [A]
time = 0.10, size = 150, normalized size = 0.86

method	result	size
derivativedivides	\(\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+3 \left (\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}\right ) a}{d}\)	\(150\)
default	\(\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+3 \left (\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}\right ) a}{d}\)	\(150\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.52, size = 153, normalized size = 0.88 \begin {gather*} -\frac {2 \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {7}{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 {7}{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 {7}{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 ) - 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{2}}{4 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(3)*2^(1/3)*a^(7/3)*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^(7/3)*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^(7/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) - 12*(I*a*tan(d*x + c
) + a)^(1/3)*a^2)/(a^2*d)

________________________________________________________________________________________

Fricas [A]
time = 0.55, size = 239, normalized size = 1.37 \begin {gather*} \frac {\left (\frac {1}{4}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} d - d\right )} \left (\frac {a}{d^{3}}\right )^{\frac {1}{3}} \log \left (\left (\frac {1}{4}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} d + d\right )} \left (\frac {a}{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 ) + \left (\frac {1}{4}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} d - d\right )} \left (\frac {a}{d^{3}}\right )^{\frac {1}{3}} \log \left (\left (\frac {1}{4}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} d + d\right )} \left (\frac {a}{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 ) + 2 \, \left (\frac {1}{4}\right )^{\frac {1}{3}} d \left (\frac {a}{d^{3}}\right )^{\frac {1}{3}} \log \left (-2 \, \left (\frac {1}{4}\right )^{\frac {1}{3}} d \left (\frac {a}{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 ) + 6 \cdot 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 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((1/4)^(1/3)*(-I*sqrt(3)*d - d)*(a/d^3)^(1/3)*log((1/4)^(1/3)*(I*sqrt(3)*d + d)*(a/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)) + (1/4)^(1/3)*(I*sqrt(3)*d - d)*(a/d^3)^(1/3)*log((
1/4)^(1/3)*(-I*sqrt(3)*d + d)*(a/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)) + 2*(1/4)^(1/3)*d*(a/d^3)^(1/3)*log(-2*(1/4)^(1/3)*d*(a/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)) + 6*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))/d

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt [3]{i a \left (\tan {\left (c + d x \right )} - i\right )} \tan {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 4.28, size = 176, normalized size = 1.01 \begin {gather*} \frac {3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}+\frac {2^{1/3}\,a^{1/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-2^{1/3}\,a^{1/3}\right )}{2\,d}+\frac {4^{2/3}\,a^{1/3}\,\ln \left (\frac {9\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}-\frac {9\,2^{1/3}\,a^{4/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,d}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,d}-\frac {4^{2/3}\,a^{1/3}\,\ln \left (\frac {9\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}+\frac {9\,2^{1/3}\,a^{4/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,d}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*(a + a*tan(c + d*x)*1i)^(1/3))/d + (2^(1/3)*a^(1/3)*log((a*(tan(c + d*x)*1i + 1))^(1/3) - 2^(1/3)*a^(1/3)))
/(2*d) + (4^(2/3)*a^(1/3)*log((9*a*(a + a*tan(c + d*x)*1i)^(1/3))/d - (9*2^(1/3)*a^(4/3)*(3^(1/2)*1i - 1))/(2*
d))*((3^(1/2)*1i)/2 - 1/2))/(4*d) - (4^(2/3)*a^(1/3)*log((9*a*(a + a*tan(c + d*x)*1i)^(1/3))/d + (9*2^(1/3)*a^
(4/3)*(3^(1/2)*1i + 1))/(2*d))*((3^(1/2)*1i)/2 + 1/2))/(4*d)

________________________________________________________________________________________