3.3.81 \(\int \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3} \, dx\) [281]

Optimal. Leaf size=203 \[ \frac {a^{4/3} x}{2^{2/3}}+\frac {i \sqrt [3]{2} \sqrt {3} a^{4/3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{d}-\frac {i a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}-\frac {3 i a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}-\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {3 i (a+i a \tan (c+d x))^{7/3}}{7 a d} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3624, 3559, 3562, 59, 631, 210, 31} \begin {gather*} \frac {i \sqrt [3]{2} \sqrt {3} a^{4/3} \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 )}{d}-\frac {3 i a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}-\frac {i a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}+\frac {a^{4/3} x}{2^{2/3}}-\frac {3 i (a+i a \tan (c+d x))^{7/3}}{7 a d}-\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(4/3)*x)/2^(2/3) + (I*2^(1/3)*Sqrt[3]*a^(4/3)*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt
[3]*a^(1/3))])/d - (I*a^(4/3)*Log[Cos[c + d*x]])/(2^(2/3)*d) - ((3*I)*a^(4/3)*Log[2^(1/3)*a^(1/3) - (a + I*a*T
an[c + d*x])^(1/3)])/(2^(2/3)*d) - ((3*I)*a*(a + I*a*Tan[c + d*x])^(1/3))/d - (((3*I)/7)*(a + I*a*Tan[c + d*x]
)^(7/3))/(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 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 3559

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))
), x] + Dist[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] && G
tQ[n, 1]

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 3624

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

Rubi steps

\begin {align*} \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3} \, dx &=-\frac {3 i (a+i a \tan (c+d x))^{7/3}}{7 a d}-\int (a+i a \tan (c+d x))^{4/3} \, dx\\ &=-\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {3 i (a+i a \tan (c+d x))^{7/3}}{7 a d}-(2 a) \int \sqrt [3]{a+i a \tan (c+d x)} \, dx\\ &=-\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {3 i (a+i a \tan (c+d x))^{7/3}}{7 a d}+\frac {\left (2 i a^2\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{2/3}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=\frac {a^{4/3} x}{2^{2/3}}-\frac {i a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}-\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {3 i (a+i a \tan (c+d x))^{7/3}}{7 a d}+\frac {\left (3 i a^{4/3}\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^{2/3} d}+\frac {\left (3 i a^{5/3}\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 )}{\sqrt [3]{2} d}\\ &=\frac {a^{4/3} x}{2^{2/3}}-\frac {i a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}-\frac {3 i a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}-\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {3 i (a+i a \tan (c+d x))^{7/3}}{7 a d}-\frac {\left (3 i \sqrt [3]{2} a^{4/3}\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 )}{d}\\ &=\frac {a^{4/3} x}{2^{2/3}}+\frac {i \sqrt [3]{2} \sqrt {3} a^{4/3} \tan ^{-1}\left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{d}-\frac {i a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}-\frac {3 i a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}-\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {3 i (a+i a \tan (c+d x))^{7/3}}{7 a d}\\ \end {align*}

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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]^2*(a + I*a*Tan[c + d*x])^(4/3),x]

[Out]

$Aborted

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Maple [A]
time = 0.10, size = 175, normalized size = 0.86

method	result	size
derivativedivides	\(-\frac {3 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{3}}}{7}+a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2 \left (\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}\right ) a^{3}\right )}{d a}\)	\(175\)
default	\(-\frac {3 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{3}}}{7}+a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2 \left (\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}\right ) a^{3}\right )}{d a}\)	\(175\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^(4/3),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.53, size = 172, normalized size = 0.85 \begin {gather*} \frac {i \, {\left (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 ) - 6 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{3}} a^{2} - 42 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{4}\right )}}{14 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

1/14*I*(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*ta
n(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)) - 6*(I*a*tan
(d*x + c) + a)^(7/3)*a^2 - 42*(I*a*tan(d*x + c) + a)^(1/3)*a^4)/(a^3*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. 409 vs. \(2 (147) = 294\).
time = 0.57, size = 409, normalized size = 2.01 \begin {gather*} -\frac {6 \cdot 2^{\frac {1}{3}} {\left (11 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 14 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i \, a\right )} \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 )} + 7 \, {\left ({\left (-i \, \sqrt {3} d + d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (-i \, \sqrt {3} d + d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, \sqrt {3} d + d\right )} \left (\frac {2 i \, a^{4}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2 \cdot 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 (\sqrt {3} d + i \, d\right )} \left (\frac {2 i \, a^{4}}{d^{3}}\right )^{\frac {1}{3}}}{2 \, a}\right ) + 7 \, {\left ({\left (i \, \sqrt {3} d + d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (i \, \sqrt {3} d + d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, \sqrt {3} d + d\right )} \left (\frac {2 i \, a^{4}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2 \cdot 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 (\sqrt {3} d - i \, d\right )} \left (\frac {2 i \, a^{4}}{d^{3}}\right )^{\frac {1}{3}}}{2 \, a}\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 )} \left (\frac {2 i \, a^{4}}{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 )} - i \, \left (\frac {2 i \, a^{4}}{d^{3}}\right )^{\frac {1}{3}} d}{a}\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

-1/14*(6*2^(1/3)*(11*I*a*e^(4*I*d*x + 4*I*c) + 14*I*a*e^(2*I*d*x + 2*I*c) + 7*I*a)*(a/(e^(2*I*d*x + 2*I*c) + 1
))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + 7*((-I*sqrt(3)*d + d)*e^(4*I*d*x + 4*I*c) + 2*(-I*sqrt(3)*d + d)*e^(2*I*d*x
+ 2*I*c) - I*sqrt(3)*d + d)*(2*I*a^4/d^3)^(1/3)*log(1/2*(2*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) + (sqrt(3)*d + I*d)*(2*I*a^4/d^3)^(1/3))/a) + 7*((I*sqrt(3)*d + d)*e^(4*I*d*x + 4*I*c) + 2
*(I*sqrt(3)*d + d)*e^(2*I*d*x + 2*I*c) + I*sqrt(3)*d + d)*(2*I*a^4/d^3)^(1/3)*log(1/2*(2*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) - (sqrt(3)*d - I*d)*(2*I*a^4/d^3)^(1/3))/a) - 14*(d*e^(4*I*d*
x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*(2*I*a^4/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*(2*I*a^4/d^3)^(1/3)*d)/a))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d
)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {4}{3}} \tan ^{2}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2*(a+I*a*tan(d*x+c))**(4/3),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**(4/3)*tan(c + d*x)**2, 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)^2*(a+I*a*tan(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(4/3)*tan(d*x + c)^2, x)

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Mupad [B]
time = 4.07, size = 218, normalized size = 1.07 \begin {gather*} -\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/3}\,3{}\mathrm {i}}{7\,a\,d}-\frac {a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,3{}\mathrm {i}}{d}+\frac {{\left (2{}\mathrm {i}\right )}^{1/3}\,a^{4/3}\,\ln \left (a^2\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,18{}\mathrm {i}+18\,{\left (2{}\mathrm {i}\right )}^{1/3}\,a^{7/3}\,d^2\right )}{d}+\frac {{\left (2{}\mathrm {i}\right )}^{1/3}\,a^{4/3}\,\ln \left (a^2\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,18{}\mathrm {i}+18\,{\left (2{}\mathrm {i}\right )}^{1/3}\,a^{7/3}\,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 )}{d}-\frac {{\left (2{}\mathrm {i}\right )}^{1/3}\,a^{4/3}\,\ln \left (a^2\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,18{}\mathrm {i}-18\,{\left (2{}\mathrm {i}\right )}^{1/3}\,a^{7/3}\,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 )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^2*(a + a*tan(c + d*x)*1i)^(4/3),x)

[Out]

(2i^(1/3)*a^(4/3)*log(a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/3)*18i + 18*2i^(1/3)*a^(7/3)*d^2))/d - (a*(a + a*tan(
c + d*x)*1i)^(1/3)*3i)/d - ((a + a*tan(c + d*x)*1i)^(7/3)*3i)/(7*a*d) + (2i^(1/3)*a^(4/3)*log(a^2*d^2*(a + a*t
an(c + d*x)*1i)^(1/3)*18i + 18*2i^(1/3)*a^(7/3)*d^2*((3^(1/2)*1i)/2 - 1/2))*((3^(1/2)*1i)/2 - 1/2))/d - (2i^(1
/3)*a^(4/3)*log(a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/3)*18i - 18*2i^(1/3)*a^(7/3)*d^2*((3^(1/2)*1i)/2 + 1/2))*((
3^(1/2)*1i)/2 + 1/2))/d

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