3.3.82 \(\int \tan (c+d x) (a+i a \tan (c+d x))^{4/3} \, dx\) [282]

Optimal. Leaf size=192 \[ \frac {i a^{4/3} x}{2^{2/3}}-\frac {\sqrt [3]{2} \sqrt {3} a^{4/3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{d}+\frac {a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}+\frac {3 a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}+\frac {3 a \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac {3 (a+i a \tan (c+d x))^{4/3}}{4 d} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3608, 3559, 3562, 59, 631, 210, 31} \begin {gather*} -\frac {\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 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 {a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}+\frac {i a^{4/3} x}{2^{2/3}}+\frac {3 a \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac {3 (a+i a \tan (c+d x))^{4/3}}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(I*a^(4/3)*x)/2^(2/3) - (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 + (a^(4/3)*Log[Cos[c + d*x]])/(2^(2/3)*d) + (3*a^(4/3)*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c +
d*x])^(1/3)])/(2^(2/3)*d) + (3*a*(a + I*a*Tan[c + d*x])^(1/3))/d + (3*(a + I*a*Tan[c + d*x])^(4/3))/(4*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 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) (a+i a \tan (c+d x))^{4/3} \, dx &=\frac {3 (a+i a \tan (c+d x))^{4/3}}{4 d}-i \int (a+i a \tan (c+d x))^{4/3} \, dx\\ &=\frac {3 a \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac {3 (a+i a \tan (c+d x))^{4/3}}{4 d}-(2 i a) \int \sqrt [3]{a+i a \tan (c+d x)} \, dx\\ &=\frac {3 a \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac {3 (a+i a \tan (c+d x))^{4/3}}{4 d}-\frac {\left (2 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 {i a^{4/3} x}{2^{2/3}}+\frac {a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}+\frac {3 a \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac {3 (a+i a \tan (c+d x))^{4/3}}{4 d}-\frac {\left (3 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 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 {i a^{4/3} x}{2^{2/3}}+\frac {a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}+\frac {3 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 a \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac {3 (a+i a \tan (c+d x))^{4/3}}{4 d}+\frac {\left (3 \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 {i a^{4/3} x}{2^{2/3}}-\frac {\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 {a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}+\frac {3 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 a \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac {3 (a+i a \tan (c+d x))^{4/3}}{4 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]*(a + I*a*Tan[c + d*x])^(4/3),x]

[Out]

$Aborted

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Maple [A]
time = 0.10, size = 169, normalized size = 0.88

method	result	size
derivativedivides	\(\frac {\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}{4}+3 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+6 \left (\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}\right ) a^{2}}{d}\)	\(169\)
default	\(\frac {\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}{4}+3 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+6 \left (\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}\right ) a^{2}}{d}\)	\(169\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(3/4*(a+I*a*tan(d*x+c))^(4/3)+3*a*(a+I*a*tan(d*x+c))^(1/3)+6*(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^2)

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Maxima [A]
time = 0.53, size = 172, normalized size = 0.90 \begin {gather*} -\frac {4 \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {10}{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 \cdot 2^{\frac {1}{3}} a^{\frac {10}{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 ) - 4 \cdot 2^{\frac {1}{3}} a^{\frac {10}{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 ) - 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}} a^{2} - 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{3}}{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))^(4/3),x, algorithm="maxima")

[Out]

-1/4*(4*sqrt(3)*2^(1/3)*a^(10/3)*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^(1/3)*a^(10/3)*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)) - 4*2^(1/3)*a^(10/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) - 3*(I*a*tan(d*x
+ c) + a)^(4/3)*a^2 - 12*(I*a*tan(d*x + c) + a)^(1/3)*a^3)/(a^2*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. 350 vs. \(2 (144) = 288\).
time = 0.60, size = 350, normalized size = 1.82 \begin {gather*} \frac {3 \cdot 2^{\frac {1}{3}} {\left (3 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, 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 )} + 2^{\frac {1}{3}} {\left ({\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 {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 )} + 2^{\frac {1}{3}} {\left (i \, \sqrt {3} d + d\right )} \left (\frac {a^{4}}{d^{3}}\right )^{\frac {1}{3}}}{2 \, a}\right ) + 2^{\frac {1}{3}} {\left ({\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 {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 )} + 2^{\frac {1}{3}} {\left (-i \, \sqrt {3} d + d\right )} \left (\frac {a^{4}}{d^{3}}\right )^{\frac {1}{3}}}{2 \, a}\right ) + 2 \cdot 2^{\frac {1}{3}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \left (\frac {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 )} - 2^{\frac {1}{3}} \left (\frac {a^{4}}{d^{3}}\right )^{\frac {1}{3}} d}{a}\right )}{2 \, {\left (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)*(a+I*a*tan(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

1/2*(3*2^(1/3)*(3*a*e^(2*I*d*x + 2*I*c) + 2*a)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + 2
^(1/3)*((-I*sqrt(3)*d - d)*e^(2*I*d*x + 2*I*c) - I*sqrt(3)*d - d)*(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) + 2^(1/3)*(I*sqrt(3)*d + d)*(a^4/d^3)^(1/3))/a) + 2^(1/3)
*((I*sqrt(3)*d - d)*e^(2*I*d*x + 2*I*c) + I*sqrt(3)*d - d)*(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) + 2^(1/3)*(-I*sqrt(3)*d + d)*(a^4/d^3)^(1/3))/a) + 2*2^(1/3)*(d*
e^(2*I*d*x + 2*I*c) + d)*(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) - 2^(1/3)*(a^4/d^3)^(1/3)*d)/a))/(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 {\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))**(4/3),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**(4/3)*tan(c + d*x), 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)*(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), x)

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Mupad [B]
time = 4.32, size = 198, normalized size = 1.03 \begin {gather*} \frac {3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}}{4\,d}+\frac {3\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}+\frac {2^{1/3}\,a^{4/3}\,\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 )}{d}+\frac {2^{1/3}\,a^{4/3}\,\ln \left (\frac {18\,a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}-\frac {9\,2^{1/3}\,a^{7/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{d}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d}-\frac {2^{1/3}\,a^{4/3}\,\ln \left (\frac {18\,a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}+\frac {9\,2^{1/3}\,a^{7/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{d}\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)*(a + a*tan(c + d*x)*1i)^(4/3),x)

[Out]

(3*(a + a*tan(c + d*x)*1i)^(4/3))/(4*d) + (3*a*(a + a*tan(c + d*x)*1i)^(1/3))/d + (2^(1/3)*a^(4/3)*log((a*(tan
(c + d*x)*1i + 1))^(1/3) - 2^(1/3)*a^(1/3)))/d + (2^(1/3)*a^(4/3)*log((18*a^2*(a + a*tan(c + d*x)*1i)^(1/3))/d
- (9*2^(1/3)*a^(7/3)*(3^(1/2)*1i - 1))/d)*((3^(1/2)*1i)/2 - 1/2))/d - (2^(1/3)*a^(4/3)*log((18*a^2*(a + a*tan
(c + d*x)*1i)^(1/3))/d + (9*2^(1/3)*a^(7/3)*(3^(1/2)*1i + 1))/d)*((3^(1/2)*1i)/2 + 1/2))/d

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