3.223 \(\int \frac{1}{\tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=162 \[ -\frac{25 \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \sqrt{\tan (c+d x)}}+\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{11}{6 a d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{1}{3 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \]

[Out]

((1/4 + I/4)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) + 1/(3*d*Sq
rt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) + 11/(6*a*d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (2
5*Sqrt[a + I*a*Tan[c + d*x]])/(6*a^2*d*Sqrt[Tan[c + d*x]])

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Rubi [A]  time = 0.418833, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3559, 3596, 3598, 12, 3544, 205} \[ -\frac{25 \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \sqrt{\tan (c+d x)}}+\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{11}{6 a d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{1}{3 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

((1/4 + I/4)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) + 1/(3*d*Sq
rt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) + 11/(6*a*d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (2
5*Sqrt[a + I*a*Tan[c + d*x]])/(6*a^2*d*Sqrt[Tan[c + d*x]])

Rule 3559

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

Rule 3596

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

Rule 3598

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3544

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{\tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx &=\frac{1}{3 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{\int \frac{\frac{7 a}{2}-2 i a \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac{1}{3 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{11}{6 a d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{25 a^2}{4}-\frac{11}{2} i a^2 \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{3 a^4}\\ &=\frac{1}{3 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{11}{6 a d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{25 \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \sqrt{\tan (c+d x)}}+\frac{2 \int \frac{3 i a^3 \sqrt{a+i a \tan (c+d x)}}{8 \sqrt{\tan (c+d x)}} \, dx}{3 a^5}\\ &=\frac{1}{3 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{11}{6 a d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{25 \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \sqrt{\tan (c+d x)}}+\frac{i \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{4 a^2}\\ &=\frac{1}{3 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{11}{6 a d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{25 \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \sqrt{\tan (c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{2 d}\\ &=\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{1}{3 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{11}{6 a d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{25 \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.82417, size = 190, normalized size = 1.17 \[ \frac{i e^{-2 i (c+d x)} \sqrt{\tan (c+d x)} \left (\sqrt{-1+e^{2 i (c+d x)}} \left (13 e^{2 i (c+d x)}-38 e^{4 i (c+d x)}+1\right )+3 e^{3 i (c+d x)} \left (-1+e^{2 i (c+d x)}\right ) \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{6 \sqrt{2} a d \left (-1+e^{2 i (c+d x)}\right )^{3/2} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

((I/6)*(Sqrt[-1 + E^((2*I)*(c + d*x))]*(1 + 13*E^((2*I)*(c + d*x)) - 38*E^((4*I)*(c + d*x))) + 3*E^((3*I)*(c +
d*x))*(-1 + E^((2*I)*(c + d*x)))*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Tan[c + d*x]])
/(Sqrt[2]*a*d*E^((2*I)*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))^(3/2)*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*
(c + d*x)))])

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Maple [B]  time = 0.046, size = 505, normalized size = 3.1 \begin{align*}{\frac{1}{24\,{a}^{2}d \left ( -\tan \left ( dx+c \right ) +i \right ) ^{3}}\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( 9\,i\sqrt{2}\ln \left ({\frac{1}{\tan \left ( dx+c \right ) +i} \left ( 2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) \right ) } \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{3}a-3\,\sqrt{2}\ln \left ({\frac{2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) }{\tan \left ( dx+c \right ) +i}} \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{4}a+100\,\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( \tan \left ( dx+c \right ) \right ) ^{3}-3\,i\sqrt{2}\ln \left ({\frac{1}{\tan \left ( dx+c \right ) +i} \left ( 2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) \right ) } \right ) \tan \left ( dx+c \right ) a+9\,\sqrt{2}\ln \left ({\frac{2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) }{\tan \left ( dx+c \right ) +i}} \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{2}a-204\,\tan \left ( dx+c \right ) \sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}-256\,i\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( \tan \left ( dx+c \right ) \right ) ^{2}+48\,i\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) } \right ){\frac{1}{\sqrt{\tan \left ( dx+c \right ) }}}{\frac{1}{\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{-ia}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24/d*(a*(1+I*tan(d*x+c)))^(1/2)/a^2*(9*I*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^
(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^3*a-3*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1
+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^4*a+100*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*t
an(d*x+c)))^(1/2)*tan(d*x+c)^3-3*I*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*
a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)*a+9*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x
+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a-204*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^
(1/2)*(-I*a)^(1/2)-256*I*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2+48*I*(-I*a)^(1/2)*(a*
tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2))/tan(d*x+c)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+I)^3/(
-I*a)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 3.06977, size = 1199, normalized size = 7.4 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-38 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 25 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 14 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (i \, d x + i \, c\right )} - 3 \,{\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt{\frac{i}{2 \, a^{3} d^{2}}} \log \left (\frac{1}{4} \,{\left (2 i \, a^{2} d \sqrt{\frac{i}{2 \, a^{3} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 3 \,{\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt{\frac{i}{2 \, a^{3} d^{2}}} \log \left (\frac{1}{4} \,{\left (-2 i \, a^{2} d \sqrt{\frac{i}{2 \, a^{3} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{12 \,{\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/12*(sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(
-38*I*e^(6*I*d*x + 6*I*c) - 25*I*e^(4*I*d*x + 4*I*c) + 14*I*e^(2*I*d*x + 2*I*c) + I)*e^(I*d*x + I*c) - 3*(a^2*
d*e^(6*I*d*x + 6*I*c) - a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/2*I/(a^3*d^2))*log(1/4*(2*I*a^2*d*sqrt(1/2*I/(a^3*d^
2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*
d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 3*(a^2*d*e^(6*I*d*x + 6*I*c)
- a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/2*I/(a^3*d^2))*log(1/4*(-2*I*a^2*d*sqrt(1/2*I/(a^3*d^2))*e^(2*I*d*x + 2*I
*c) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(
e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)))/(a^2*d*e^(6*I*d*x + 6*I*c) - a^2*d*e^(4*I*d*x + 4
*I*c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)**(3/2)/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.34516, size = 150, normalized size = 0.93 \begin{align*} -\frac{2 \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} a^{2} \log \left (\sqrt{i \, a \tan \left (d x + c\right ) + a}\right )}{-\left (i + 1\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} + \left (4 i + 4\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a - \left (5 i + 5\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} + \left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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