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

Optimal. Leaf size=111 \[ -\frac {b B-a C}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac {\left (a^2 (-C)+2 a b B+b^2 C\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {x \left (a^2 B+2 a b C-b^2 B\right )}{\left (a^2+b^2\right )^2} \]

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

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Rubi [A]  time = 0.21, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3632, 3529, 3531, 3530} \[ -\frac {b B-a C}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac {\left (a^2 (-C)+2 a b B+b^2 C\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {x \left (a^2 B+2 a b C-b^2 B\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[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[a*Cos[c + d*x] + b*Sin[c + d*x]])
/((a^2 + b^2)^2*d) - (b*B - a*C)/((a^2 + b^2)*d*(a + b*Tan[c + d*x]))

Rule 3529

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

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 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 {\cot (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 {B+C \tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx\\ &=-\frac {b B-a C}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\int \frac {a B+b C-(b B-a C) \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{a^2+b^2}\\ &=\frac {\left (a^2 B-b^2 B+2 a b C\right ) x}{\left (a^2+b^2\right )^2}-\frac {b B-a C}{\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 \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{\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 (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {b B-a C}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 2.22, size = 190, normalized size = 1.71 \[ \frac {\frac {C ((-b-i a) \log (-\tan (c+d x)+i)+i (a+i b) \log (\tan (c+d x)+i)+2 b \log (a+b \tan (c+d x)))}{a^2+b^2}-(b B-a C) \left (\frac {2 b \left (\frac {a^2+b^2}{a+b \tan (c+d x)}-2 a \log (a+b \tan (c+d x))\right )}{\left (a^2+b^2\right )^2}+\frac {i \log (-\tan (c+d x)+i)}{(a+i b)^2}-\frac {i \log (\tan (c+d x)+i)}{(a-i b)^2}\right )}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 222, normalized size = 2.00 \[ \frac {2 \, C a b^{2} - 2 \, B b^{3} + 2 \, {\left (B a^{3} + 2 \, C a^{2} b - B a b^{2}\right )} d x - {\left (C a^{3} - 2 \, B a^{2} b - C a b^{2} + {\left (C a^{2} b - 2 \, B a b^{2} - C b^{3}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (C a^{2} b - B a b^{2} - {\left (B a^{2} b + 2 \, C a b^{2} - B b^{3}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \tan \left (d x + c\right ) + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(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*b^2 - 2*B*b^3 + 2*(B*a^3 + 2*C*a^2*b - B*a*b^2)*d*x - (C*a^3 - 2*B*a^2*b - C*a*b^2 + (C*a^2*b - 2*B
*a*b^2 - C*b^3)*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)) - 2*(C
*a^2*b - B*a*b^2 - (B*a^2*b + 2*C*a*b^2 - B*b^3)*d*x)*tan(d*x + c))/((a^4*b + 2*a^2*b^3 + b^5)*d*tan(d*x + c)
+ (a^5 + 2*a^3*b^2 + a*b^4)*d)

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giac [B]  time = 2.61, size = 234, normalized size = 2.11 \[ \frac {\frac {2 \, {\left (B a^{2} + 2 \, C a b - B b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (C a^{2} - 2 \, B a b - C b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (C a^{2} b - 2 \, B a b^{2} - C b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} + \frac {2 \, {\left (C a^{2} b \tan \left (d x + c\right ) - 2 \, B a b^{2} \tan \left (d x + c\right ) - C b^{3} \tan \left (d x + c\right ) + 2 \, C a^{3} - 3 \, B a^{2} b - B b^{3}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 0.76, size = 301, normalized size = 2.71 \[ -\frac {B b}{d \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}+\frac {a C}{d \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 a b \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {a^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) C}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (a +b \tan \left (d x +c \right )\right ) b^{2} C}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) B a b}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} C}{2 d \left (a^{2}+b^{2}\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{2} C}{2 d \left (a^{2}+b^{2}\right )^{2}}+\frac {B \arctan \left (\tan \left (d x +c \right )\right ) a^{2}}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b^{2}}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {2 C \arctan \left (\tan \left (d x +c \right )\right ) a b}{d \left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(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^2)/(a+b*tan(d*x+c))*B*b+1/d/(a^2+b^2)/(a+b*tan(d*x+c))*a*C+2/d*a/(a^2+b^2)^2*b*ln(a+b*tan(d*x+c))*
B-1/d*a^2/(a^2+b^2)^2*ln(a+b*tan(d*x+c))*C+1/d/(a^2+b^2)^2*ln(a+b*tan(d*x+c))*b^2*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*arctan(tan(d*x+c))*
a*b

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maxima [A]  time = 0.57, size = 177, normalized size = 1.59 \[ \frac {\frac {2 \, {\left (B a^{2} + 2 \, C a b - B b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (C a^{2} - 2 \, B a b - C b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (C a^{2} - 2 \, B a b - C b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (C a - B b\right )}}{a^{3} + a b^{2} + {\left (a^{2} b + b^{3}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 9.09, size = 153, normalized size = 1.38 \[ \frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-C\,a^2+2\,B\,a\,b+C\,b^2\right )}{d\,{\left (a^2+b^2\right )}^2}-\frac {B\,b-C\,a}{d\,\left (a^2+b^2\right )\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (C+B\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^2+a\,b\,2{}\mathrm {i}+b^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 4.99, size = 2895, normalized size = 26.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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