3.33 \(\int \frac {\tan (c+d x) (B \tan (c+d x)+C \tan ^2(c+d x))}{(a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=157 \[ -\frac {a^2 (b B-a C)}{b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\left (a^2 (-C)+2 a b B+b^2 C\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac {x \left (a^2 B+2 a b C-b^2 B\right )}{\left (a^2+b^2\right )^2}-\frac {a \left (a^3 (-C)-3 a b^2 C+2 b^3 B\right ) \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )^2} \]

[Out]

-(B*a^2-B*b^2+2*C*a*b)*x/(a^2+b^2)^2-(2*B*a*b-C*a^2+C*b^2)*ln(cos(d*x+c))/(a^2+b^2)^2/d-a*(2*B*b^3-C*a^3-3*C*a
*b^2)*ln(a+b*tan(d*x+c))/b^2/(a^2+b^2)^2/d-a^2*(B*b-C*a)/b^2/(a^2+b^2)/d/(a+b*tan(d*x+c))

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Rubi [A]  time = 0.31, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3632, 3604, 3626, 3617, 31, 3475} \[ -\frac {a^2 (b B-a C)}{b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {a \left (a^3 (-C)-3 a b^2 C+2 b^3 B\right ) \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )^2}-\frac {\left (a^2 (-C)+2 a b B+b^2 C\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac {x \left (a^2 B+2 a b C-b^2 B\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]*(B*Tan[c + d*x] + C*Tan[c + d*x]^2))/(a + b*Tan[c + d*x])^2,x]

[Out]

-(((a^2*B - b^2*B + 2*a*b*C)*x)/(a^2 + b^2)^2) - ((2*a*b*B - a^2*C + b^2*C)*Log[Cos[c + d*x]])/((a^2 + b^2)^2*
d) - (a*(2*b^3*B - a^3*C - 3*a*b^2*C)*Log[a + b*Tan[c + d*x]])/(b^2*(a^2 + b^2)^2*d) - (a^2*(b*B - a*C))/(b^2*
(a^2 + b^2)*d*(a + b*Tan[c + d*x]))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3604

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

Rule 3617

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A/(b*f), Subst[Int[(a + x)^m, x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A, C]

Rule 3626

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/((a_.) + (b_.)*tan[(e_.) + (f_.)*
(x_)]), x_Symbol] :> Simp[((a*A + b*B - a*C)*x)/(a^2 + b^2), x] + (Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2), I
nt[(1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x]), x], x] - Dist[(A*b - a*B - b*C)/(a^2 + b^2), Int[Tan[e + f*x], x
], x]) /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && NeQ[a^2 + b^2, 0] && NeQ[A*b - a
*B - b*C, 0]

Rule 3632

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

Rubi steps

\begin {align*} \int \frac {\tan (c+d x) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^2} \, dx &=\int \frac {\tan ^2(c+d x) (B+C \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx\\ &=-\frac {a^2 (b B-a C)}{b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\int \frac {-a (b B-a C)+b (b B-a C) \tan (c+d x)+\left (a^2+b^2\right ) C \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )}\\ &=-\frac {\left (a^2 B-b^2 B+2 a b C\right ) x}{\left (a^2+b^2\right )^2}-\frac {a^2 (b B-a C)}{b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\left (2 a b B-a^2 C+b^2 C\right ) \int \tan (c+d x) \, dx}{\left (a^2+b^2\right )^2}-\frac {\left (a \left (2 b^3 B-a^3 C-3 a b^2 C\right )\right ) \int \frac {1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )^2}\\ &=-\frac {\left (a^2 B-b^2 B+2 a b C\right ) x}{\left (a^2+b^2\right )^2}-\frac {\left (2 a b B-a^2 C+b^2 C\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {a^2 (b B-a C)}{b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\left (a \left (2 b^3 B-a^3 C-3 a b^2 C\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^2 \left (a^2+b^2\right )^2 d}\\ &=-\frac {\left (a^2 B-b^2 B+2 a b C\right ) x}{\left (a^2+b^2\right )^2}-\frac {\left (2 a b B-a^2 C+b^2 C\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {a \left (2 b^3 B-a^3 C-3 a b^2 C\right ) \log (a+b \tan (c+d x))}{b^2 \left (a^2+b^2\right )^2 d}-\frac {a^2 (b B-a C)}{b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 2.18, size = 324, normalized size = 2.06 \[ \frac {-2 i a \left (a^3 C+3 a b^2 C-2 b^3 B\right ) \tan ^{-1}(\tan (c+d x)) (a+b \tan (c+d x))+a \left (a \left (a^3 C+3 a b^2 C-2 b^3 B\right ) \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )+2 (a+i b)^2 (c+d x) \left (i a^2 C+2 a b C-b^2 B\right )-2 C \left (a^2+b^2\right )^2 \log (\cos (c+d x))\right )+b \tan (c+d x) \left (a \left (a^3 C+3 a b^2 C-2 b^3 B\right ) \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )-2 C \left (a^2+b^2\right )^2 \log (\cos (c+d x))+2 (a+i b) \left (i a^3 C (c+d x+i)+a^2 b (B+C (c+d x+i))-a b^2 (B (c+d x+i)-2 i C (c+d x))-i b^3 B (c+d x)\right )\right )}{2 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]*(B*Tan[c + d*x] + C*Tan[c + d*x]^2))/(a + b*Tan[c + d*x])^2,x]

[Out]

(a*(2*(a + I*b)^2*(-(b^2*B) + I*a^2*C + 2*a*b*C)*(c + d*x) - 2*(a^2 + b^2)^2*C*Log[Cos[c + d*x]] + a*(-2*b^3*B
+ a^3*C + 3*a*b^2*C)*Log[(a*Cos[c + d*x] + b*Sin[c + d*x])^2]) + b*(2*(a + I*b)*((-I)*b^3*B*(c + d*x) + I*a^3
*C*(I + c + d*x) - a*b^2*((-2*I)*C*(c + d*x) + B*(I + c + d*x)) + a^2*b*(B + C*(I + c + d*x))) - 2*(a^2 + b^2)
^2*C*Log[Cos[c + d*x]] + a*(-2*b^3*B + a^3*C + 3*a*b^2*C)*Log[(a*Cos[c + d*x] + b*Sin[c + d*x])^2])*Tan[c + d*
x] - (2*I)*a*(-2*b^3*B + a^3*C + 3*a*b^2*C)*ArcTan[Tan[c + d*x]]*(a + b*Tan[c + d*x]))/(2*b^2*(a^2 + b^2)^2*d*
(a + b*Tan[c + d*x]))

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fricas [B]  time = 0.82, size = 311, normalized size = 1.98 \[ \frac {2 \, C a^{3} b^{2} - 2 \, B a^{2} b^{3} - 2 \, {\left (B a^{3} b^{2} + 2 \, C a^{2} b^{3} - B a b^{4}\right )} d x + {\left (C a^{5} + 3 \, C a^{3} b^{2} - 2 \, B a^{2} b^{3} + {\left (C a^{4} b + 3 \, C a^{2} b^{3} - 2 \, B a b^{4}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - {\left (C a^{5} + 2 \, C a^{3} b^{2} + C a b^{4} + {\left (C a^{4} b + 2 \, C a^{2} b^{3} + C b^{5}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (C a^{4} b - B a^{3} b^{2} + {\left (B a^{2} b^{3} + 2 \, C a b^{4} - B b^{5}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} d \tan \left (d x + c\right ) + {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*(2*C*a^3*b^2 - 2*B*a^2*b^3 - 2*(B*a^3*b^2 + 2*C*a^2*b^3 - B*a*b^4)*d*x + (C*a^5 + 3*C*a^3*b^2 - 2*B*a^2*b^
3 + (C*a^4*b + 3*C*a^2*b^3 - 2*B*a*b^4)*tan(d*x + c))*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan
(d*x + c)^2 + 1)) - (C*a^5 + 2*C*a^3*b^2 + C*a*b^4 + (C*a^4*b + 2*C*a^2*b^3 + C*b^5)*tan(d*x + c))*log(1/(tan(
d*x + c)^2 + 1)) - 2*(C*a^4*b - B*a^3*b^2 + (B*a^2*b^3 + 2*C*a*b^4 - B*b^5)*d*x)*tan(d*x + c))/((a^4*b^3 + 2*a
^2*b^5 + b^7)*d*tan(d*x + c) + (a^5*b^2 + 2*a^3*b^4 + a*b^6)*d)

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giac [A]  time = 2.02, size = 244, normalized size = 1.55 \[ -\frac {\frac {2 \, {\left (B a^{2} + 2 \, C a b - B b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (C a^{2} - 2 \, B a b - C b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (C a^{4} + 3 \, C a^{2} b^{2} - 2 \, B a b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (C a^{4} \tan \left (d x + c\right ) + 3 \, C a^{2} b^{2} \tan \left (d x + c\right ) - 2 \, B a b^{3} \tan \left (d x + c\right ) + B a^{4} + 2 \, C a^{3} b - B a^{2} b^{2}\right )}}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/2*(2*(B*a^2 + 2*C*a*b - B*b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) + (C*a^2 - 2*B*a*b - C*b^2)*log(tan(d*x +
c)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) - 2*(C*a^4 + 3*C*a^2*b^2 - 2*B*a*b^3)*log(abs(b*tan(d*x + c) + a))/(a^4*b^2
+ 2*a^2*b^4 + b^6) + 2*(C*a^4*tan(d*x + c) + 3*C*a^2*b^2*tan(d*x + c) - 2*B*a*b^3*tan(d*x + c) + B*a^4 + 2*C*a
^3*b - B*a^2*b^2)/((a^4*b + 2*a^2*b^3 + b^5)*(b*tan(d*x + c) + a)))/d

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maple [A]  time = 0.32, size = 313, normalized size = 1.99 \[ -\frac {a^{2} B}{d b \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}+\frac {a^{3} C}{d \,b^{2} \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}-\frac {2 a b \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {a^{4} \ln \left (a +b \tan \left (d x +c \right )\right ) C}{d \left (a^{2}+b^{2}\right )^{2} b^{2}}+\frac {3 a^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) C}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) B a b}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} C}{2 d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{2} C}{2 d \left (a^{2}+b^{2}\right )^{2}}-\frac {B \arctan \left (\tan \left (d x +c \right )\right ) a^{2}}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b^{2}}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {2 C \arctan \left (\tan \left (d x +c \right )\right ) a b}{d \left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)*(B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c))^2,x)

[Out]

-1/d*a^2/b/(a^2+b^2)/(a+b*tan(d*x+c))*B+1/d*a^3/b^2/(a^2+b^2)/(a+b*tan(d*x+c))*C-2/d*a/(a^2+b^2)^2*b*ln(a+b*ta
n(d*x+c))*B+1/d*a^4/(a^2+b^2)^2/b^2*ln(a+b*tan(d*x+c))*C+3/d*a^2/(a^2+b^2)^2*ln(a+b*tan(d*x+c))*C+1/d/(a^2+b^2
)^2*ln(1+tan(d*x+c)^2)*B*a*b-1/2/d/(a^2+b^2)^2*ln(1+tan(d*x+c)^2)*a^2*C+1/2/d/(a^2+b^2)^2*ln(1+tan(d*x+c)^2)*b
^2*C-1/d/(a^2+b^2)^2*B*arctan(tan(d*x+c))*a^2+1/d/(a^2+b^2)^2*B*arctan(tan(d*x+c))*b^2-2/d/(a^2+b^2)^2*C*arcta
n(tan(d*x+c))*a*b

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maxima [A]  time = 0.61, size = 197, normalized size = 1.25 \[ -\frac {\frac {2 \, {\left (B a^{2} + 2 \, C a b - B b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (C a^{4} + 3 \, C a^{2} b^{2} - 2 \, B a b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}} + \frac {{\left (C a^{2} - 2 \, B a b - C b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (C a^{3} - B a^{2} b\right )}}{a^{3} b^{2} + a b^{4} + {\left (a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/2*(2*(B*a^2 + 2*C*a*b - B*b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) - 2*(C*a^4 + 3*C*a^2*b^2 - 2*B*a*b^3)*log(
b*tan(d*x + c) + a)/(a^4*b^2 + 2*a^2*b^4 + b^6) + (C*a^2 - 2*B*a*b - C*b^2)*log(tan(d*x + c)^2 + 1)/(a^4 + 2*a
^2*b^2 + b^4) - 2*(C*a^3 - B*a^2*b)/(a^3*b^2 + a*b^4 + (a^2*b^3 + b^5)*tan(d*x + c)))/d

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mupad [B]  time = 9.11, size = 165, normalized size = 1.05 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (C+B\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^2+a\,b\,2{}\mathrm {i}+b^2\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )}-\frac {a^2\,\left (B\,b-C\,a\right )}{b^2\,d\,\left (a^2+b^2\right )\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}+\frac {a\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (C\,a^3+3\,C\,a\,b^2-2\,B\,b^3\right )}{b^2\,d\,{\left (a^2+b^2\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(c + d*x)*(B*tan(c + d*x) + C*tan(c + d*x)^2))/(a + b*tan(c + d*x))^2,x)

[Out]

(log(tan(c + d*x) + 1i)*(B*1i + C))/(2*d*(a*b*2i - a^2 + b^2)) + (log(tan(c + d*x) - 1i)*(B + C*1i))/(2*d*(2*a
*b - a^2*1i + b^2*1i)) - (a^2*(B*b - C*a))/(b^2*d*(a^2 + b^2)*(a + b*tan(c + d*x))) + (a*log(a + b*tan(c + d*x
))*(C*a^3 - 2*B*b^3 + 3*C*a*b^2))/(b^2*d*(a^2 + b^2)^2)

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sympy [A]  time = 2.29, size = 3497, normalized size = 22.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(B*tan(d*x+c)+C*tan(d*x+c)**2)/(a+b*tan(d*x+c))**2,x)

[Out]

Piecewise((zoo*x*(B*tan(c) + C*tan(c)**2)/tan(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((-B*x + B*tan(c + d*x)/d -
C*log(tan(c + d*x)**2 + 1)/(2*d) + C*tan(c + d*x)**2/(2*d))/a**2, Eq(b, 0)), (-B*d*x*tan(c + d*x)**2/(-4*b**2
*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) + 4*b**2*d) + 2*I*B*d*x*tan(c + d*x)/(-4*b**2*d*tan(c + d*x)**2 +
8*I*b**2*d*tan(c + d*x) + 4*b**2*d) + B*d*x/(-4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) + 4*b**2*d)
+ 3*B*tan(c + d*x)/(-4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) + 4*b**2*d) - 2*I*B/(-4*b**2*d*tan(c +
d*x)**2 + 8*I*b**2*d*tan(c + d*x) + 4*b**2*d) - 3*I*C*d*x*tan(c + d*x)**2/(-4*b**2*d*tan(c + d*x)**2 + 8*I*b*
*2*d*tan(c + d*x) + 4*b**2*d) - 6*C*d*x*tan(c + d*x)/(-4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) + 4*
b**2*d) + 3*I*C*d*x/(-4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) + 4*b**2*d) - 2*C*log(tan(c + d*x)**2
+ 1)*tan(c + d*x)**2/(-4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) + 4*b**2*d) + 4*I*C*log(tan(c + d*x
)**2 + 1)*tan(c + d*x)/(-4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) + 4*b**2*d) + 2*C*log(tan(c + d*x)
**2 + 1)/(-4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) + 4*b**2*d) + 5*I*C*tan(c + d*x)/(-4*b**2*d*tan(
c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) + 4*b**2*d) + 4*C/(-4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + d*x) +
4*b**2*d), Eq(a, -I*b)), (-B*d*x*tan(c + d*x)**2/(-4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) + 4*b**
2*d) - 2*I*B*d*x*tan(c + d*x)/(-4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) + 4*b**2*d) + B*d*x/(-4*b**
2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) + 4*b**2*d) + 3*B*tan(c + d*x)/(-4*b**2*d*tan(c + d*x)**2 - 8*I*
b**2*d*tan(c + d*x) + 4*b**2*d) + 2*I*B/(-4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) + 4*b**2*d) + 3*I
*C*d*x*tan(c + d*x)**2/(-4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) + 4*b**2*d) - 6*C*d*x*tan(c + d*x)
/(-4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) + 4*b**2*d) - 3*I*C*d*x/(-4*b**2*d*tan(c + d*x)**2 - 8*I
*b**2*d*tan(c + d*x) + 4*b**2*d) - 2*C*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**2/(-4*b**2*d*tan(c + d*x)**2 - 8
*I*b**2*d*tan(c + d*x) + 4*b**2*d) - 4*I*C*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(-4*b**2*d*tan(c + d*x)**2 -
8*I*b**2*d*tan(c + d*x) + 4*b**2*d) + 2*C*log(tan(c + d*x)**2 + 1)/(-4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan
(c + d*x) + 4*b**2*d) - 5*I*C*tan(c + d*x)/(-4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) + 4*b**2*d) +
4*C/(-4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) + 4*b**2*d), Eq(a, I*b)), (x*(B*tan(c) + C*tan(c)**2)
*tan(c)/(a + b*tan(c))**2, Eq(d, 0)), (-2*B*a**4*b/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d
+ 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) - 2*B*a**3*b**2*d*x/(2*a**5*b**2*d + 2*a**
4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) - 2*B
*a**2*b**3*d*x*tan(c + d*x)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c
+ d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) - 4*B*a**2*b**3*log(a/b + tan(c + d*x))/(2*a**5*b**2*d + 2*a**4*b
**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + 2*B*a*
*2*b**3*log(tan(c + d*x)**2 + 1)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*t
an(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) - 2*B*a**2*b**3/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x)
+ 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + 2*B*a*b**4*d*x/(2*a**5*b*
*2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c +
d*x)) - 4*B*a*b**4*log(a/b + tan(c + d*x))*tan(c + d*x)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*
b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + 2*B*a*b**4*log(tan(c + d*x)**2 + 1
)*tan(c + d*x)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*
b**6*d + 2*b**7*d*tan(c + d*x)) + 2*B*b**5*d*x*tan(c + d*x)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a*
*3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + 2*C*a**5*log(a/b + tan(c + d*x)
)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b*
*7*d*tan(c + d*x)) + 2*C*a**5/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(
c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + 2*C*a**4*b*log(a/b + tan(c + d*x))*tan(c + d*x)/(2*a**5*b**2*
d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*
x)) + 6*C*a**3*b**2*log(a/b + tan(c + d*x))/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a*
*2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) - C*a**3*b**2*log(tan(c + d*x)**2 + 1)/(2*a**5*b*
*2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c +
d*x)) + 2*C*a**3*b**2/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x
) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) - 4*C*a**2*b**3*d*x/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a*
*3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + 6*C*a**2*b**3*log(a/b + tan(c +
d*x))*tan(c + d*x)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) +
2*a*b**6*d + 2*b**7*d*tan(c + d*x)) - C*a**2*b**3*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*a**5*b**2*d + 2*a*
*4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) - 4*
C*a*b**4*d*x*tan(c + d*x)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c +
d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + C*a*b**4*log(tan(c + d*x)**2 + 1)/(2*a**5*b**2*d + 2*a**4*b**3*d*
tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*tan(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + C*b**5*log(t
an(c + d*x)**2 + 1)*tan(c + d*x)/(2*a**5*b**2*d + 2*a**4*b**3*d*tan(c + d*x) + 4*a**3*b**4*d + 4*a**2*b**5*d*t
an(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)), True))

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