3.189 \(\int \frac{A+B \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=240 \[ -\frac{(21 A+11 i B) \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(-25 B+39 i A) \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \sqrt{\tan (c+d x)}}+\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{5 A+3 i B}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{A+i B}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \]

[Out]

((1/4 + I/4)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) +
(A + I*B)/(3*d*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)) + (5*A + (3*I)*B)/(2*a*d*Tan[c + d*x]^(3/2)*S
qrt[a + I*a*Tan[c + d*x]]) - ((21*A + (11*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(6*a^2*d*Tan[c + d*x]^(3/2)) + (((
39*I)*A - 25*B)*Sqrt[a + I*a*Tan[c + d*x]])/(6*a^2*d*Sqrt[Tan[c + d*x]])

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Rubi [A]  time = 0.769485, antiderivative size = 240, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {3596, 3598, 12, 3544, 205} \[ -\frac{(21 A+11 i B) \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(-25 B+39 i A) \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \sqrt{\tan (c+d x)}}+\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{5 A+3 i B}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{A+i B}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

((1/4 + I/4)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) +
(A + I*B)/(3*d*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)) + (5*A + (3*I)*B)/(2*a*d*Tan[c + d*x]^(3/2)*S
qrt[a + I*a*Tan[c + d*x]]) - ((21*A + (11*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(6*a^2*d*Tan[c + d*x]^(3/2)) + (((
39*I)*A - 25*B)*Sqrt[a + I*a*Tan[c + d*x]])/(6*a^2*d*Sqrt[Tan[c + d*x]])

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{A+B \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx &=\frac{A+i B}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{\int \frac{\frac{3}{2} a (3 A+i B)-3 a (i A-B) \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac{A+i B}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{5 A+3 i B}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{3}{4} a^2 (21 A+11 i B)-3 a^2 (5 i A-3 B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{3 a^4}\\ &=\frac{A+i B}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{5 A+3 i B}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(21 A+11 i B) \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{8} a^3 (39 i A-25 B)-\frac{3}{4} a^3 (21 A+11 i B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{9 a^5}\\ &=\frac{A+i B}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{5 A+3 i B}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(21 A+11 i B) \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(39 i A-25 B) \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \sqrt{\tan (c+d x)}}+\frac{4 \int -\frac{9 a^4 (A-i B) \sqrt{a+i a \tan (c+d x)}}{16 \sqrt{\tan (c+d x)}} \, dx}{9 a^6}\\ &=\frac{A+i B}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{5 A+3 i B}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(21 A+11 i B) \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(39 i A-25 B) \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \sqrt{\tan (c+d x)}}-\frac{(A-i B) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{4 a^2}\\ &=\frac{A+i B}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{5 A+3 i B}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(21 A+11 i B) \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(39 i A-25 B) \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \sqrt{\tan (c+d x)}}+\frac{(i A+B) \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 ) (i A+B) \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{A+i B}{3 d \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{5 A+3 i B}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(21 A+11 i B) \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(39 i A-25 B) \sqrt{a+i a \tan (c+d x)}}{6 a^2 d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 6.64076, size = 244, normalized size = 1.02 \[ \frac{i e^{-2 i (c+d x)} \sec ^2(c+d x) \left (-3 i (A-i B) e^{3 i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )-i A \left (18 e^{2 i (c+d x)}-87 e^{4 i (c+d x)}+52 e^{6 i (c+d x)}+1\right )+B \left (12 e^{2 i (c+d x)}-51 e^{4 i (c+d x)}+38 e^{6 i (c+d x)}+1\right )\right )}{12 a d \left (-1+e^{2 i (c+d x)}\right ) \sqrt{\tan (c+d x)} (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

((I/12)*(B*(1 + 12*E^((2*I)*(c + d*x)) - 51*E^((4*I)*(c + d*x)) + 38*E^((6*I)*(c + d*x))) - I*A*(1 + 18*E^((2*
I)*(c + d*x)) - 87*E^((4*I)*(c + d*x)) + 52*E^((6*I)*(c + d*x))) - (3*I)*(A - I*B)*E^((3*I)*(c + d*x))*(-1 + E
^((2*I)*(c + d*x)))^(3/2)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sec[c + d*x]^2)/(a*d*E^((2*
I)*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))*Sqrt[Tan[c + d*x]]*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [B]  time = 0.08, size = 1012, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(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/tan(d*x+c)^(3/2)*(-48*I*B*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1
/2)*(-I*a)^(1/2)-100*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^4-9*I*A*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
*B*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)^5*a+384*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+156*I*A*(a*tan(d*x+c)
*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4+3*I*B*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+9*A*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+204*B*(-I*a)^(1/2
)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2-9*I*B*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-276*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*
(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2-9*B*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-32*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/
2)*tan(d*x+c)+256*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3-3*A*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-16*I
*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)+3*I*A*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)^5*a)/(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((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.22245, size = 1656, normalized size = 6.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/12*(sqrt(2)*(2*(26*A + 19*I*B)*e^(8*I*d*x + 8*I*c) - (35*A + 13*I*B)*e^(6*I*d*x + 6*I*c) - 3*(23*A + 13*I*B
)*e^(4*I*d*x + 4*I*c) + (19*A + 13*I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*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^(I*d*x + I*c) + 3*sqrt(1/2)*(a^2*d*e^(8*I*d*x + 8*I*
c) - 2*a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c))*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*log((2*
I*sqrt(1/2)*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x
+ 2*I*c) + I*A + B)*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^(I*d*x + I*c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) - 3*sqrt(1/2)*(a^2*d*e^(8*I*d*x + 8*I*c) - 2*a^2*d*e^(6
*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c))*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*log((-2*I*sqrt(1/2)*a^2*
d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A
+ B)*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^(I*d*x +
I*c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)))/(a^2*d*e^(8*I*d*x + 8*I*c) - 2*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((A+B*tan(d*x+c))/tan(d*x+c)**(5/2)/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.47054, size = 261, normalized size = 1.09 \begin{align*} \frac{\left (i + 1\right ) \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} +{\left (-\left (2 i - 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} + \left (2 i - 2\right ) \, a^{4}\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} \sqrt{i \, a \tan \left (d x + c\right ) + a} B}{2 \,{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} a - 5 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{2} + 9 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} - 7 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} + 2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

1/2*((I + 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)*a^4 + (-(2*I - 2)*(I*a*tan(d*x +
c) + a)*a^3 + (2*I - 2)*a^4)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*tan(d*x + c) + a)*B)/(((I*a*t
an(d*x + c) + a)^6*a - 5*(I*a*tan(d*x + c) + a)^5*a^2 + 9*(I*a*tan(d*x + c) + a)^4*a^3 - 7*(I*a*tan(d*x + c) +
a)^3*a^4 + 2*(I*a*tan(d*x + c) + a)^2*a^5)*d)