3.276 \(\int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx\)
Optimal. Leaf size=157 \[ -\frac {a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\left (a^2 (-B)+2 a A b+b^2 B\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac {x \left (a^2 A+2 a b B-A b^2\right )}{\left (a^2+b^2\right )^2}-\frac {a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right ) \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )^2} \]
[Out]
-(A*a^2-A*b^2+2*B*a*b)*x/(a^2+b^2)^2-(2*A*a*b-B*a^2+B*b^2)*ln(cos(d*x+c))/(a^2+b^2)^2/d-a*(2*A*b^3-a*(a^2+3*b^
2)*B)*ln(a+b*tan(d*x+c))/b^2/(a^2+b^2)^2/d-a^2*(A*b-B*a)/b^2/(a^2+b^2)/d/(a+b*tan(d*x+c))
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Rubi [A] time = 0.27, antiderivative size = 157, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used =
{3604, 3626, 3617, 31, 3475} \[ -\frac {a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right ) \log (a+b \tan (c+d x))}{b^2 d \left (a^2+b^2\right )^2}-\frac {\left (a^2 (-B)+2 a A b+b^2 B\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac {x \left (a^2 A+2 a b B-A b^2\right )}{\left (a^2+b^2\right )^2} \]
Antiderivative was successfully verified.
[In]
Int[(Tan[c + d*x]^2*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x]
[Out]
-(((a^2*A - A*b^2 + 2*a*b*B)*x)/(a^2 + b^2)^2) - ((2*a*A*b - a^2*B + b^2*B)*Log[Cos[c + d*x]])/((a^2 + b^2)^2*
d) - (a*(2*A*b^3 - a*(a^2 + 3*b^2)*B)*Log[a + b*Tan[c + d*x]])/(b^2*(a^2 + b^2)^2*d) - (a^2*(A*b - a*B))/(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]
Rubi steps
\begin {align*} \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx &=-\frac {a^2 (A b-a B)}{b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\int \frac {-a (A b-a B)+b (A b-a B) \tan (c+d x)+\left (a^2+b^2\right ) B \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )}\\ &=-\frac {\left (a^2 A-A b^2+2 a b B\right ) x}{\left (a^2+b^2\right )^2}-\frac {a^2 (A b-a B)}{b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\left (2 a A b-a^2 B+b^2 B\right ) \int \tan (c+d x) \, dx}{\left (a^2+b^2\right )^2}-\frac {\left (a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\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 A-A b^2+2 a b B\right ) x}{\left (a^2+b^2\right )^2}-\frac {\left (2 a A b-a^2 B+b^2 B\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {a^2 (A b-a B)}{b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\left (a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\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 A-A b^2+2 a b B\right ) x}{\left (a^2+b^2\right )^2}-\frac {\left (2 a A b-a^2 B+b^2 B\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right ) \log (a+b \tan (c+d x))}{b^2 \left (a^2+b^2\right )^2 d}-\frac {a^2 (A b-a B)}{b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end {align*}
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Mathematica [C] time = 2.14, size = 323, normalized size = 2.06 \[ \frac {-2 i a \left (a B \left (a^2+3 b^2\right )-2 A b^3\right ) \tan ^{-1}(\tan (c+d x)) (a+b \tan (c+d x))+a \left (a \left (a B \left (a^2+3 b^2\right )-2 A b^3\right ) \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )-2 B \left (a^2+b^2\right )^2 \log (\cos (c+d x))+2 (a+i b)^2 (c+d x) \left (-A b^2+a B (2 b+i a)\right )\right )+b \tan (c+d x) \left (a \left (a B \left (a^2+3 b^2\right )-2 A b^3\right ) \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )-2 B \left (a^2+b^2\right )^2 \log (\cos (c+d x))+2 (a+i b) \left (i a^3 B (c+d x+i)+a^2 b (A+B (c+d x+i))-a b^2 (A (c+d x+i)-2 i B (c+d x))-i A b^3 (c+d x)\right )\right )}{2 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))} \]
Antiderivative was successfully verified.
[In]
Integrate[(Tan[c + d*x]^2*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x]
[Out]
(a*(2*(a + I*b)^2*(-(A*b^2) + a*(I*a + 2*b)*B)*(c + d*x) - 2*(a^2 + b^2)^2*B*Log[Cos[c + d*x]] + a*(-2*A*b^3 +
a*(a^2 + 3*b^2)*B)*Log[(a*Cos[c + d*x] + b*Sin[c + d*x])^2]) + b*(2*(a + I*b)*((-I)*A*b^3*(c + d*x) + I*a^3*B
*(I + c + d*x) - a*b^2*((-2*I)*B*(c + d*x) + A*(I + c + d*x)) + a^2*b*(A + B*(I + c + d*x))) - 2*(a^2 + b^2)^2
*B*Log[Cos[c + d*x]] + a*(-2*A*b^3 + a*(a^2 + 3*b^2)*B)*Log[(a*Cos[c + d*x] + b*Sin[c + d*x])^2])*Tan[c + d*x]
- (2*I)*a*(-2*A*b^3 + a*(a^2 + 3*b^2)*B)*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 = 0.80, size = 311, normalized size = 1.98 \[ \frac {2 \, B a^{3} b^{2} - 2 \, A a^{2} b^{3} - 2 \, {\left (A a^{3} b^{2} + 2 \, B a^{2} b^{3} - A a b^{4}\right )} d x + {\left (B a^{5} + 3 \, B a^{3} b^{2} - 2 \, A a^{2} b^{3} + {\left (B a^{4} b + 3 \, B a^{2} b^{3} - 2 \, A a b^{4}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - {\left (B a^{5} + 2 \, B a^{3} b^{2} + B a b^{4} + {\left (B a^{4} b + 2 \, B a^{2} b^{3} + B b^{5}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (B a^{4} b - A a^{3} b^{2} + {\left (A a^{2} b^{3} + 2 \, B a b^{4} - A b^{5}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} d \tan \left (d x + c\right ) + {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="fricas")
[Out]
1/2*(2*B*a^3*b^2 - 2*A*a^2*b^3 - 2*(A*a^3*b^2 + 2*B*a^2*b^3 - A*a*b^4)*d*x + (B*a^5 + 3*B*a^3*b^2 - 2*A*a^2*b^
3 + (B*a^4*b + 3*B*a^2*b^3 - 2*A*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)) - (B*a^5 + 2*B*a^3*b^2 + B*a*b^4 + (B*a^4*b + 2*B*a^2*b^3 + B*b^5)*tan(d*x + c))*log(1/(tan(
d*x + c)^2 + 1)) - 2*(B*a^4*b - A*a^3*b^2 + (A*a^2*b^3 + 2*B*a*b^4 - A*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 = 0.74, size = 244, normalized size = 1.55 \[ -\frac {\frac {2 \, {\left (A a^{2} + 2 \, B a b - A b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (B a^{2} - 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (B a^{4} + 3 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (B a^{4} \tan \left (d x + c\right ) + 3 \, B a^{2} b^{2} \tan \left (d x + c\right ) - 2 \, A a b^{3} \tan \left (d x + c\right ) + A a^{4} + 2 \, B a^{3} b - A a^{2} b^{2}\right )}}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="giac")
[Out]
-1/2*(2*(A*a^2 + 2*B*a*b - A*b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) + (B*a^2 - 2*A*a*b - B*b^2)*log(tan(d*x +
c)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) - 2*(B*a^4 + 3*B*a^2*b^2 - 2*A*a*b^3)*log(abs(b*tan(d*x + c) + a))/(a^4*b^2
+ 2*a^2*b^4 + b^6) + 2*(B*a^4*tan(d*x + c) + 3*B*a^2*b^2*tan(d*x + c) - 2*A*a*b^3*tan(d*x + c) + A*a^4 + 2*B*a
^3*b - A*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 = 0.28, size = 313, normalized size = 1.99 \[ -\frac {a^{2} A}{d b \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}+\frac {a^{3} B}{d \,b^{2} \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}-\frac {2 a b \ln \left (a +b \tan \left (d x +c \right )\right ) A}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {a^{4} \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )^{2} b^{2}}+\frac {3 a^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A a b}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} B}{2 d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{2} B}{2 d \left (a^{2}+b^{2}\right )^{2}}-\frac {A \arctan \left (\tan \left (d x +c \right )\right ) a^{2}}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {A \arctan \left (\tan \left (d x +c \right )\right ) b^{2}}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {2 B \arctan \left (\tan \left (d x +c \right )\right ) a b}{d \left (a^{2}+b^{2}\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x)
[Out]
-1/d*a^2/b/(a^2+b^2)/(a+b*tan(d*x+c))*A+1/d*a^3/b^2/(a^2+b^2)/(a+b*tan(d*x+c))*B-2/d*a/(a^2+b^2)^2*b*ln(a+b*ta
n(d*x+c))*A+1/d*a^4/(a^2+b^2)^2/b^2*ln(a+b*tan(d*x+c))*B+3/d*a^2/(a^2+b^2)^2*ln(a+b*tan(d*x+c))*B+1/d/(a^2+b^2
)^2*ln(1+tan(d*x+c)^2)*A*a*b-1/2/d/(a^2+b^2)^2*ln(1+tan(d*x+c)^2)*a^2*B+1/2/d/(a^2+b^2)^2*ln(1+tan(d*x+c)^2)*b
^2*B-1/d/(a^2+b^2)^2*A*arctan(tan(d*x+c))*a^2+1/d/(a^2+b^2)^2*A*arctan(tan(d*x+c))*b^2-2/d/(a^2+b^2)^2*B*arcta
n(tan(d*x+c))*a*b
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maxima [A] time = 0.86, size = 197, normalized size = 1.25 \[ -\frac {\frac {2 \, {\left (A a^{2} + 2 \, B a b - A b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (B a^{4} + 3 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}} + \frac {{\left (B a^{2} - 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (B a^{3} - A a^{2} b\right )}}{a^{3} b^{2} + a b^{4} + {\left (a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="maxima")
[Out]
-1/2*(2*(A*a^2 + 2*B*a*b - A*b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) - 2*(B*a^4 + 3*B*a^2*b^2 - 2*A*a*b^3)*log(
b*tan(d*x + c) + a)/(a^4*b^2 + 2*a^2*b^4 + b^6) + (B*a^2 - 2*A*a*b - B*b^2)*log(tan(d*x + c)^2 + 1)/(a^4 + 2*a
^2*b^2 + b^4) - 2*(B*a^3 - A*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 = 6.56, size = 165, normalized size = 1.05 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^2+a\,b\,2{}\mathrm {i}+b^2\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )}-\frac {a^2\,\left (A\,b-B\,a\right )}{b^2\,d\,\left (a^2+b^2\right )\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}+\frac {a\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a^3+3\,B\,a\,b^2-2\,A\,b^3\right )}{b^2\,d\,{\left (a^2+b^2\right )}^2} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((tan(c + d*x)^2*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x))^2,x)
[Out]
(log(tan(c + d*x) + 1i)*(A*1i + B))/(2*d*(a*b*2i - a^2 + b^2)) + (log(tan(c + d*x) - 1i)*(A + B*1i))/(2*d*(2*a
*b - a^2*1i + b^2*1i)) - (a^2*(A*b - B*a))/(b^2*d*(a^2 + b^2)*(a + b*tan(c + d*x))) + (a*log(a + b*tan(c + d*x
))*(B*a^3 - 2*A*b^3 + 3*B*a*b^2))/(b^2*d*(a^2 + b^2)^2)
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sympy [A] time = 2.30, size = 3485, normalized size = 22.20 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)**2*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**2,x)
[Out]
Piecewise((zoo*x*(A + B*tan(c)), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((-A*x + A*tan(c + d*x)/d - B*log(tan(c + d*
x)**2 + 1)/(2*d) + B*tan(c + d*x)**2/(2*d))/a**2, Eq(b, 0)), (-A*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*A*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) + A*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*A*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*A/(-4*b**2*d*tan(c + d*x)**2 + 8*I*b*
*2*d*tan(c + d*x) + 4*b**2*d) - 3*I*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) - 6*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) + 3*I*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) - 2*B*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*B*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*B*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*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) + 4*B/(-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)), (-A*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*A*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) + A*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*A*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*A/(-4*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) + 4*b**2*d) + 3*I*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) - 6*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) - 3*I*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) - 2*B*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*B*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*B*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*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) + 4*B/(-4*b**2*d*ta
n(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) + 4*b**2*d), Eq(a, I*b)), (x*(A + B*tan(c))*tan(c)**2/(a + b*tan(c))**
2, Eq(d, 0)), (-2*A*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*A*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*A*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*A*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*A*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*tan(c + d*x) + 2*a*b**6*d
+ 2*b**7*d*tan(c + d*x)) - 2*A*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*A*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*A*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*t
an(c + d*x) + 2*a*b**6*d + 2*b**7*d*tan(c + d*x)) + 2*A*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*A*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*B*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*B
*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*B*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*B*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)) - B*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*B*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*B*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*B*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*t
an(c + d*x)) - B*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*B*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)) + B*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)) + B*b**5*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)), True))
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