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

Optimal. Leaf size=234 \[ \frac {\sqrt {3} \sqrt [3]{a} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2^{2/3} d}+\frac {3 \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)}}{7 d}-\frac {3 (a+i a \tan (c+d x))^{4/3}}{28 a d}-\frac {18 \sqrt [3]{a+i a \tan (c+d x)}}{7 d}-\frac {3 \sqrt [3]{a} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2\ 2^{2/3} d}-\frac {\sqrt [3]{a} \log (\cos (c+d x))}{2\ 2^{2/3} d}-\frac {i \sqrt [3]{a} x}{2\ 2^{2/3}} \]

[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-18/7*(a+I*a*tan(d*x+c))^(1/3)/d+3/7*tan(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/3)/d-3/28*(a+I*a*tan(d*x+c))^
(4/3)/a/d

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Rubi [A]  time = 0.29, antiderivative size = 234, 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 = {3560, 3592, 3527, 3481, 57, 617, 204, 31} \[ \frac {3 \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)}}{7 d}+\frac {\sqrt {3} \sqrt [3]{a} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2^{2/3} d}-\frac {3 (a+i a \tan (c+d x))^{4/3}}{28 a d}-\frac {18 \sqrt [3]{a+i a \tan (c+d x)}}{7 d}-\frac {3 \sqrt [3]{a} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2\ 2^{2/3} d}-\frac {\sqrt [3]{a} \log (\cos (c+d x))}{2\ 2^{2/3} d}-\frac {i \sqrt [3]{a} x}{2\ 2^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3*(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) - (18*(a + I*a*Tan[c + d*x])^(1/3))/(7*d) + (3*Tan[c + d*x]^2*(a + I*a*T
an[c + d*x])^(1/3))/(7*d) - (3*(a + I*a*Tan[c + d*x])^(4/3))/(28*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_))^(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 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 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 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 3527

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]

Rule 3560

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 3592

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 \tan ^3(c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx &=\frac {3 \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)}}{7 d}-\frac {3 \int \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}{7 a}\\ &=\frac {3 \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)}}{7 d}-\frac {3 (a+i a \tan (c+d x))^{4/3}}{28 a d}-\frac {3 \int \sqrt [3]{a+i a \tan (c+d x)} \left (-\frac {i a}{3}+2 a \tan (c+d x)\right ) \, dx}{7 a}\\ &=-\frac {18 \sqrt [3]{a+i a \tan (c+d x)}}{7 d}+\frac {3 \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)}}{7 d}-\frac {3 (a+i a \tan (c+d x))^{4/3}}{28 a d}+i \int \sqrt [3]{a+i a \tan (c+d x)} \, dx\\ &=-\frac {18 \sqrt [3]{a+i a \tan (c+d x)}}{7 d}+\frac {3 \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)}}{7 d}-\frac {3 (a+i a \tan (c+d x))^{4/3}}{28 a d}+\frac {a \operatorname {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 {18 \sqrt [3]{a+i a \tan (c+d x)}}{7 d}+\frac {3 \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)}}{7 d}-\frac {3 (a+i a \tan (c+d x))^{4/3}}{28 a d}+\frac {\left (3 \sqrt [3]{a}\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\ 2^{2/3} d}+\frac {\left (3 a^{2/3}\right ) \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 \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 {18 \sqrt [3]{a+i a \tan (c+d x)}}{7 d}+\frac {3 \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)}}{7 d}-\frac {3 (a+i a \tan (c+d x))^{4/3}}{28 a d}-\frac {\left (3 \sqrt [3]{a}\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 )}{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 {18 \sqrt [3]{a+i a \tan (c+d x)}}{7 d}+\frac {3 \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)}}{7 d}-\frac {3 (a+i a \tan (c+d x))^{4/3}}{28 a d}\\ \end {align*}

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

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [B]  time = 0.66, size = 403, normalized size = 1.72 \[ -\frac {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 (15 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 21 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 14\right )} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} - \left (\frac {1}{4}\right )^{\frac {1}{3}} {\left ({\left (7 i \, \sqrt {3} d - 7 \, d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (14 i \, \sqrt {3} d - 14 \, d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i \, \sqrt {3} d - 7 \, 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 ({\left (-7 i \, \sqrt {3} d - 7 \, d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-14 i \, \sqrt {3} d - 14 \, d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 7 i \, \sqrt {3} d - 7 \, 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 ) - 14 \, \left (\frac {1}{4}\right )^{\frac {1}{3}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \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 )}{14 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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]

-1/14*(3*2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*(15*e^(4*I*d*x + 4*I*c) + 21*e^(2*I*d*x + 2*I*c) + 14)*e^
(2/3*I*d*x + 2/3*I*c) - (1/4)^(1/3)*((7*I*sqrt(3)*d - 7*d)*e^(4*I*d*x + 4*I*c) + (14*I*sqrt(3)*d - 14*d)*e^(2*
I*d*x + 2*I*c) + 7*I*sqrt(3)*d - 7*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)*((-7*I*sqrt(3)*d - 7*d)*e^(4*I*d*
x + 4*I*c) + (-14*I*sqrt(3)*d - 14*d)*e^(2*I*d*x + 2*I*c) - 7*I*sqrt(3)*d - 7*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*(1/4)^(1/3)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + 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)))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e
^(2*I*d*x + 2*I*c) + d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \tan \left (d x + c\right )^{3}\,{d x} \]

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

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maple [A]  time = 0.16, size = 198, normalized size = 0.85 \[ -\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{3}}}{7 d \,a^{2}}+\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}{4 d a}-\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{d}-\frac {a^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{2 d}+\frac {a^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{4 d}+\frac {a^{\frac {1}{3}} 2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{2 d} \]

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)

[Out]

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

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maxima [A]  time = 0.81, size = 190, normalized size = 0.81 \[ \frac {14 \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {13}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + 7 \cdot 2^{\frac {1}{3}} a^{\frac {13}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 14 \cdot 2^{\frac {1}{3}} a^{\frac {13}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{3}} a^{2} + 21 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}} a^{3} - 84 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{4}}{28 \, a^{4} d} \]

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]

1/28*(14*sqrt(3)*2^(1/3)*a^(13/3)*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)) + 7*2^(1/3)*a^(13/3)*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)) - 14*2^(1/3)*a^(13/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) - 12*(I*a*tan(
d*x + c) + a)^(7/3)*a^2 + 21*(I*a*tan(d*x + c) + a)^(4/3)*a^3 - 84*(I*a*tan(d*x + c) + a)^(1/3)*a^4)/(a^4*d)

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mupad [B]  time = 0.63, size = 235, normalized size = 1.00 \[ -\frac {3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}+\frac {3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}}{4\,a\,d}-\frac {3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/3}}{7\,a^2\,d}+\frac {2^{1/3}\,{\left (-a\right )}^{1/3}\,\ln \left (9\,a\,{\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-9\,2^{1/3}\,{\left (-a\right )}^{4/3}\right )}{2\,d}+\frac {4^{2/3}\,{\left (-a\right )}^{1/3}\,\ln \left (\frac {9\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}-\frac {9\,2^{1/3}\,{\left (-a\right )}^{4/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,d}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,d}-\frac {4^{2/3}\,{\left (-a\right )}^{1/3}\,\ln \left (\frac {9\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}+\frac {9\,2^{1/3}\,{\left (-a\right )}^{4/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,d}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,d} \]

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*(a + a*tan(c + d*x)*1i)^(4/3))/(4*a*d) - (3*(a + a*tan(c + d*x)*1i)^(1/3))/d - (3*(a + a*tan(c + d*x)*1i)^(
7/3))/(7*a^2*d) + (2^(1/3)*(-a)^(1/3)*log(9*a*(a*(tan(c + d*x)*1i + 1))^(1/3) - 9*2^(1/3)*(-a)^(4/3)))/(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)

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

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

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