3.3.91 \(\int \frac {\tan ^3(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\) [291]

Optimal. Leaf size=237 \[ -\frac {i x}{4 \sqrt [3]{2} \sqrt [3]{a}}-\frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {\log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {21}{10 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 (a+i a \tan (c+d x))^{2/3}}{10 a d} \]

[Out]

-1/8*I*x*2^(2/3)/a^(1/3)-1/8*ln(cos(d*x+c))*2^(2/3)/a^(1/3)/d-3/8*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(1/3))
*2^(2/3)/a^(1/3)/d-1/4*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)/
a^(1/3)/d+21/10/d/(a+I*a*tan(d*x+c))^(1/3)+3/5*tan(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(1/3)+3/10*(a+I*a*tan(d*x+c))
^(2/3)/a/d

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Rubi [A]
time = 0.20, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3641, 3673, 3607, 3562, 57, 631, 210, 31} \begin {gather*} -\frac {\sqrt {3} \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 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 (a+i a \tan (c+d x))^{2/3}}{10 a d}+\frac {21}{10 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 \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 {\log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {i x}{4 \sqrt [3]{2} \sqrt [3]{a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/4*I)*x)/(2^(1/3)*a^(1/3)) - (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^(1/3)*a^(1/3)*d) - Log[Cos[c + d*x]]/(4*2^(1/3)*a^(1/3)*d) - (3*Log[2^(1/3)*a^(1/3) - (a + I*a*Ta
n[c + d*x])^(1/3)])/(4*2^(1/3)*a^(1/3)*d) + 21/(10*d*(a + I*a*Tan[c + d*x])^(1/3)) + (3*Tan[c + d*x]^2)/(5*d*(
a + I*a*Tan[c + d*x])^(1/3)) + (3*(a + I*a*Tan[c + d*x])^(2/3))/(10*a*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]

Rule 3607

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
b*c - a*d))*((a + b*Tan[e + f*x])^m/(2*a*f*m)), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1
), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rule 3641

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[d*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[1/(a*(m + n - 1)), Int[(a
+ b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 2)*Simp[d*(b*c*m + a*d*(-1 + n)) - a*c^2*(m + n - 1) + d*(b*d*m
- a*c*(m + 2*n - 2))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[
a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[n, 1] && NeQ[m + n - 1, 0] && (IntegerQ[n] || IntegersQ[2*m, 2*n])

Rule 3673

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\tan ^3(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx &=\frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 \int \frac {\tan (c+d x) \left (2 a-\frac {1}{3} i a \tan (c+d x)\right )}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx}{5 a}\\ &=\frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 (a+i a \tan (c+d x))^{2/3}}{10 a d}-\frac {3 \int \frac {\frac {i a}{3}+2 a \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx}{5 a}\\ &=\frac {21}{10 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 (a+i a \tan (c+d x))^{2/3}}{10 a d}+\frac {i \int (a+i a \tan (c+d x))^{2/3} \, dx}{2 a}\\ &=\frac {21}{10 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 (a+i a \tan (c+d x))^{2/3}}{10 a d}+\frac {\text {Subst}\left (\int \frac {1}{(a-x) \sqrt [3]{a+x}} \, dx,x,i a \tan (c+d x)\right )}{2 d}\\ &=-\frac {i x}{4 \sqrt [3]{2} \sqrt [3]{a}}-\frac {\log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {21}{10 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 (a+i a \tan (c+d x))^{2/3}}{10 a d}-\frac {3 \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 d}+\frac {3 \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 \sqrt [3]{2} \sqrt [3]{a} d}\\ &=-\frac {i x}{4 \sqrt [3]{2} \sqrt [3]{a}}-\frac {\log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {3 \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 {21}{10 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 (a+i a \tan (c+d x))^{2/3}}{10 a d}+\frac {3 \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 \sqrt [3]{2} \sqrt [3]{a} d}\\ &=-\frac {i x}{4 \sqrt [3]{2} \sqrt [3]{a}}-\frac {\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 {\log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {3 \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 {21}{10 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 (a+i a \tan (c+d x))^{2/3}}{10 a d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.90, size = 115, normalized size = 0.49 \begin {gather*} \frac {3 \sec ^2(c+d x) \left (40+24 \cos (2 (c+d x))+5 \, _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 ) (1+\cos (2 (c+d x))+i \sin (2 (c+d x)))+4 i \sin (2 (c+d x))\right )}{80 d \sqrt [3]{a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sec[c + d*x]^2*(40 + 24*Cos[2*(c + d*x)] + 5*Hypergeometric2F1[2/3, 1, 5/3, E^((2*I)*(c + d*x))/(1 + E^((2*
I)*(c + d*x)))]*(1 + Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) + (4*I)*Sin[2*(c + d*x)]))/(80*d*(a + I*a*Tan[c +
d*x])^(1/3))

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Maple [A]
time = 0.10, size = 192, normalized size = 0.81

method	result	size
derivativedivides	\(-\frac {3 \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}-\frac {a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}+\frac {\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) a^{2}}{2}-\frac {a^{2}}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d \,a^{2}}\)	\(192\)
default	\(-\frac {3 \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}-\frac {a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}+\frac {\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) a^{2}}{2}-\frac {a^{2}}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d \,a^{2}}\)	\(192\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.54, size = 190, normalized size = 0.80 \begin {gather*} -\frac {10 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {11}{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 ) - 5 \cdot 2^{\frac {2}{3}} a^{\frac {11}{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 ) + 10 \cdot 2^{\frac {2}{3}} a^{\frac {11}{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 ) + 24 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}} a^{2} - 60 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} a^{3} - \frac {60 \, a^{4}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}}{40 \, a^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/40*(10*sqrt(3)*2^(2/3)*a^(11/3)*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)) - 5*2^(2/3)*a^(11/3)*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)) + 10*2^(2/3)*a^(11/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) + 24*(I*a*tan
(d*x + c) + a)^(5/3)*a^2 - 60*(I*a*tan(d*x + c) + a)^(2/3)*a^3 - 60*a^4/(I*a*tan(d*x + c) + a)^(1/3))/(a^4*d)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (172) = 344\).
time = 1.49, size = 418, normalized size = 1.76 \begin {gather*} \frac {3 \cdot 2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (7 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 20 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )} + 10 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \left (-\frac {1}{a d^{3}}\right )^{\frac {1}{3}} \log \left (-2 \, \left (\frac {1}{2}\right )^{\frac {2}{3}} a d^{2} \left (-\frac {1}{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 ) - 5 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left ({\left (i \, \sqrt {3} a d + a d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (i \, \sqrt {3} a d + a d\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \left (-\frac {1}{a d^{3}}\right )^{\frac {1}{3}} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (i \, \sqrt {3} a d^{2} - a d^{2}\right )} \left (-\frac {1}{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 ) - 5 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left ({\left (-i \, \sqrt {3} a d + a d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-i \, \sqrt {3} a d + a d\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \left (-\frac {1}{a d^{3}}\right )^{\frac {1}{3}} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (-i \, \sqrt {3} a d^{2} - a d^{2}\right )} \left (-\frac {1}{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 )}{20 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(3*2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(7*e^(4*I*d*x + 4*I*c) + 20*e^(2*I*d*x + 2*I*c) + 5)*e^(4/
3*I*d*x + 4/3*I*c) + 10*(1/2)^(1/3)*(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*(-1/(a*d^3))^(1/3)*log
(-2*(1/2)^(2/3)*a*d^2*(-1/(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)
) - 5*(1/2)^(1/3)*((I*sqrt(3)*a*d + a*d)*e^(4*I*d*x + 4*I*c) + (I*sqrt(3)*a*d + a*d)*e^(2*I*d*x + 2*I*c))*(-1/
(a*d^3))^(1/3)*log(-(1/2)^(2/3)*(I*sqrt(3)*a*d^2 - a*d^2)*(-1/(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)) - 5*(1/2)^(1/3)*((-I*sqrt(3)*a*d + a*d)*e^(4*I*d*x + 4*I*c) + (-I*sqrt(3
)*a*d + a*d)*e^(2*I*d*x + 2*I*c))*(-1/(a*d^3))^(1/3)*log(-(1/2)^(2/3)*(-I*sqrt(3)*a*d^2 - a*d^2)*(-1/(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)))/(a*d*e^(4*I*d*x + 4*I*c) + a*d*e
^(2*I*d*x + 2*I*c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [B]
time = 0.36, size = 228, normalized size = 0.96 \begin {gather*} \frac {3}{2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}+\frac {3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{2/3}}{2\,a\,d}-\frac {3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/3}}{5\,a^2\,d}+\frac {4^{1/3}\,\ln \left (18\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}+9\,4^{2/3}\,{\left (-a\right )}^{1/3}\,d\right )}{4\,{\left (-a\right )}^{1/3}\,d}+\frac {4^{1/3}\,\ln \left (18\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}+144\,4^{2/3}\,{\left (-a\right )}^{1/3}\,d\,{\left (-\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}^2\right )\,\left (-\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}{{\left (-a\right )}^{1/3}\,d}-\frac {4^{1/3}\,\ln \left (18\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}+144\,4^{2/3}\,{\left (-a\right )}^{1/3}\,d\,{\left (\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}^2\right )\,\left (\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}{{\left (-a\right )}^{1/3}\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/(2*d*(a + a*tan(c + d*x)*1i)^(1/3)) + (3*(a + a*tan(c + d*x)*1i)^(2/3))/(2*a*d) - (3*(a + a*tan(c + d*x)*1i)
^(5/3))/(5*a^2*d) + (4^(1/3)*log(18*d*(a + a*tan(c + d*x)*1i)^(1/3) + 9*4^(2/3)*(-a)^(1/3)*d))/(4*(-a)^(1/3)*d
) + (4^(1/3)*log(18*d*(a + a*tan(c + d*x)*1i)^(1/3) + 144*4^(2/3)*(-a)^(1/3)*d*((3^(1/2)*1i)/8 - 1/8)^2)*((3^(
1/2)*1i)/8 - 1/8))/((-a)^(1/3)*d) - (4^(1/3)*log(18*d*(a + a*tan(c + d*x)*1i)^(1/3) + 144*4^(2/3)*(-a)^(1/3)*d
*((3^(1/2)*1i)/8 + 1/8)^2)*((3^(1/2)*1i)/8 + 1/8))/((-a)^(1/3)*d)

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