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

Optimal. Leaf size=200 \[ \frac{(-1)^{3/4} \sqrt{a} (7 B+4 i A) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}+\frac{(4 A-i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{(1+i) \sqrt{a} (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d} \]

[Out]

((-1)^(3/4)*Sqrt[a]*((4*I)*A + 7*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]
)/(4*d) + ((1 + I)*Sqrt[a]*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])
/d + ((4*A - I*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(4*d) + (B*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Ta
n[c + d*x]])/(2*d)

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Rubi [A]  time = 0.680083, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac{(-1)^{3/4} \sqrt{a} (7 B+4 i A) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}+\frac{(4 A-i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{(1+i) \sqrt{a} (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{B \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1)^(3/4)*Sqrt[a]*((4*I)*A + 7*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]
)/(4*d) + ((1 + I)*Sqrt[a]*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])
/d + ((4*A - I*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(4*d) + (B*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Ta
n[c + d*x]])/(2*d)

Rule 3597

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

Rule 3601

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

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]

Rule 3599

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [F]  time = 10.1308, size = 0, normalized size = 0. \[ \int \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [B]  time = 0.073, size = 838, 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+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^(3/2)*(A+B*tan(d*x+c)),x)

[Out]

-1/8/d*(a*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^(1/2)*(4*I*A*2^(1/2)*(I*a)^(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-4*I*B*2^(1/2)*(I*a)^(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))*a-6*I*B
*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)*tan(d*x+c)-7*I*B*(-I*a)^(1/2)*ln(1/2*(2*I*a*ta
n(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))*tan(d*x+c)*a+4*B*2^(1/2)*(I*a)^(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+4*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)*tan(d*x+c)^2-8*I*A*(a*tan(d*x+c)*
(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)-4*I*A*(-I*a)^(1/2)*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))*a+4*A*2^(1/2)*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*t
an(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a+8*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1
/2)*(-I*a)^(1/2)*(I*a)^(1/2)*tan(d*x+c)+4*A*(-I*a)^(1/2)*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))*tan(d*x+c)*a-2*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I
*a)^(1/2)-7*B*(-I*a)^(1/2)*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))*a)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-tan(d*x+c)+I)/(-I*a)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt{i \, a \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.96178, size = 2155, normalized size = 10.78 \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))^(1/2)*tan(d*x+c)^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/8*(2*sqrt(2)*((4*A - 3*I*B)*e^(2*I*d*x + 2*I*c) + 4*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) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-16*I*A^2
- 56*A*B + 49*I*B^2)*a/d^2)*log((sqrt(2)*((4*I*A + 7*B)*e^(2*I*d*x + 2*I*c) + 4*I*A + 7*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) + 2*I*d*sqrt((-16*
I*A^2 - 56*A*B + 49*I*B^2)*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(4*I*A + 7*B)) - (d*e^(2*I*d*x + 2
*I*c) + d)*sqrt((-16*I*A^2 - 56*A*B + 49*I*B^2)*a/d^2)*log((sqrt(2)*((4*I*A + 7*B)*e^(2*I*d*x + 2*I*c) + 4*I*A
+ 7*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) - 2*I*d*sqrt((-16*I*A^2 - 56*A*B + 49*I*B^2)*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(4*I*A
+ 7*B)) - 4*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)*a/d^2)*log((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) + I*d*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2
*I*c)/(I*A + B)) + 4*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)*a/d^2)*log((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) - I*d*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*
I*d*x - 2*I*c)/(I*A + B)))/(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))**(1/2)*tan(d*x+c)**(3/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.35899, size = 378, normalized size = 1.89 \begin{align*} -\frac{2 \,{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} + 2 \,{\left (-i \, a \tan \left (d x + c\right ) - a\right )} a + a^{2}\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 (\frac{-i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + i \, a^{2}}{\sqrt{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} - 2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} + a^{4}}} + 1\right )} -{\left ({\left (a \tan \left (d x + c\right ) - i \, a\right )} a + i \, a^{2}\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 )}{\left (\frac{-i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + i \, a^{2}}{\sqrt{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} - 2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} + a^{4}}} + 1\right )}}{2 \,{\left ({\left (a \tan \left (d x + c\right ) - i \, a\right )} a^{2} + 2 i \, a^{3}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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