3.167 \(\int \frac{(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac{9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=225 \[ \frac{(2-2 i) a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{4 a (19 A-21 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a (7 B+8 i A) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 a (63 B+67 i A) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)} \]

[Out]

((2 - 2*I)*a^(3/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*
a*A*Sqrt[a + I*a*Tan[c + d*x]])/(7*d*Tan[c + d*x]^(7/2)) - (2*a*((8*I)*A + 7*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3
5*d*Tan[c + d*x]^(5/2)) + (4*a*(19*A - (21*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(3/2)) + (4*a
*((67*I)*A + 63*B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d*Sqrt[Tan[c + d*x]])

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Rubi [A]  time = 0.735221, antiderivative size = 225, 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 = {3593, 3598, 12, 3544, 205} \[ \frac{(2-2 i) a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{4 a (19 A-21 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a (7 B+8 i A) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 a (63 B+67 i A) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(9/2),x]

[Out]

((2 - 2*I)*a^(3/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*
a*A*Sqrt[a + I*a*Tan[c + d*x]])/(7*d*Tan[c + d*x]^(7/2)) - (2*a*((8*I)*A + 7*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3
5*d*Tan[c + d*x]^(5/2)) + (4*a*(19*A - (21*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(3/2)) + (4*a
*((67*I)*A + 63*B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d*Sqrt[Tan[c + d*x]])

Rule 3593

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^2*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^
(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m -
1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ
[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

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+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac{9}{2}}(c+d x)} \, dx &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2}{7} \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{1}{2} a (8 i A+7 B)-\frac{1}{2} a (6 A-7 i B) \tan (c+d x)\right )}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\\ &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 a (8 i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{2} a^2 (19 A-21 i B)-a^2 (8 i A+7 B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{35 a}\\ &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 a (8 i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 a (19 A-21 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{8 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{4} a^3 (67 i A+63 B)+\frac{1}{2} a^3 (19 A-21 i B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{105 a^2}\\ &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 a (8 i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 a (19 A-21 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a (67 i A+63 B) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}+\frac{16 \int \frac{105 a^4 (A-i B) \sqrt{a+i a \tan (c+d x)}}{8 \sqrt{\tan (c+d x)}} \, dx}{105 a^3}\\ &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 a (8 i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 a (19 A-21 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a (67 i A+63 B) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}+(2 a (A-i B)) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 a (8 i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 a (19 A-21 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a (67 i A+63 B) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}-\frac{\left (4 a^3 (i A+B)\right ) \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 )}{d}\\ &=-\frac{(2+2 i) a^{3/2} (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 )}{d}-\frac{2 a A \sqrt{a+i a \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 a (8 i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 a (19 A-21 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a (67 i A+63 B) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 11.0162, size = 261, normalized size = 1.16 \[ \frac{a (B+i A) e^{-3 i (c+d x)} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right )^2 (\tan (c+d x)-i) \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right ) \sqrt{a+i a \tan (c+d x)}}{d \sqrt{-1+e^{2 i (c+d x)}}}-\frac{a \csc ^3(c+d x) \sqrt{a+i a \tan (c+d x)} (7 (A+6 i B) \cos (c+d x)+(53 A-42 i B) \cos (3 (c+d x))+2 \sin (c+d x) ((147 B+158 i A) \cos (2 (c+d x))-110 i A-105 B))}{210 d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(9/2),x]

[Out]

-(a*Csc[c + d*x]^3*(7*(A + (6*I)*B)*Cos[c + d*x] + (53*A - (42*I)*B)*Cos[3*(c + d*x)] + 2*((-110*I)*A - 105*B
+ ((158*I)*A + 147*B)*Cos[2*(c + d*x)])*Sin[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(210*d*Sqrt[Tan[c + d*x]]) +
(a*(I*A + B)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))^2*Ar
cTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]]*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(d*E^((3
*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))])

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Maple [B]  time = 0.044, size = 796, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x)

[Out]

1/210/d*(a*(1+I*tan(d*x+c)))^(1/2)*a/tan(d*x+c)^(7/2)*(504*B*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3*(a*tan(d*x+
c)*(1+I*tan(d*x+c)))^(1/2)+536*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)
-420*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(
1/2)*tan(d*x+c)^4*a+152*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+105*I*(I
*a)^(1/2)*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+420*A*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a
)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^4*a-168*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*ta
n(d*x+c)))^(1/2)-105*(I*a)^(1/2)*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-96*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I
*tan(d*x+c)))^(1/2)+210*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^
(1/2))*(-I*a)^(1/2)*tan(d*x+c)^4*a+210*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a
)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^4*a-84*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c
)))^(1/2)*tan(d*x+c)-60*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2))/(I*a)^(1/2)/(-I*a)^(
1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)

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Maxima [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+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.84498, size = 1774, normalized size = 7.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

-1/210*(4*sqrt(2)*((211*A - 189*I*B)*a*e^(8*I*d*x + 8*I*c) - 10*(16*A - 21*I*B)*a*e^(6*I*d*x + 6*I*c) + 14*(A
+ 6*I*B)*a*e^(4*I*d*x + 4*I*c) + 70*(4*A - 3*I*B)*a*e^(2*I*d*x + 2*I*c) - 105*(A - I*B)*a)*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) - 105*sqrt((-8*I*A
^2 - 16*A*B + 8*I*B^2)*a^3/d^2)*(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4
*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2*B)*a)*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) + I*sqrt((-8
*I*A^2 - 16*A*B + 8*I*B^2)*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((2*I*A + 2*B)*a)) + 105*sqrt(
(-8*I*A^2 - 16*A*B + 8*I*B^2)*a^3/d^2)*(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I
*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2*B)*a)*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) - I*s
qrt((-8*I*A^2 - 16*A*B + 8*I*B^2)*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((2*I*A + 2*B)*a)))/(d*
e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)

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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+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.48651, size = 309, normalized size = 1.37 \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 )}^{2} a^{6} +{\left (\left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} - \left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6}\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}{{\left (-2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} a + 14 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{2} - 40 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} + 60 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} - 50 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} + 22 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6} - 4 i \, a^{7}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x, algorithm="giac")

[Out]

((I - 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)^2*a^6 + ((2*I + 2)*(I*a*tan(d*x + c)
+ a)^2*a^5 - (2*I + 2)*(I*a*tan(d*x + c) + a)*a^6)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*tan(d*x
+ c) + a)*B)/((-2*I*(I*a*tan(d*x + c) + a)^6*a + 14*I*(I*a*tan(d*x + c) + a)^5*a^2 - 40*I*(I*a*tan(d*x + c) +
a)^4*a^3 + 60*I*(I*a*tan(d*x + c) + a)^3*a^4 - 50*I*(I*a*tan(d*x + c) + a)^2*a^5 + 22*I*(I*a*tan(d*x + c) + a
)*a^6 - 4*I*a^7)*d)