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

Optimal. Leaf size=115 \[ \frac {a (A b-a B)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\left (a^2 A+2 a b B-A b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {x \left (a^2 (-B)+2 a A b+b^2 B\right )}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3591, 3531, 3530} \[ \frac {a (A b-a B)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\left (a^2 A+2 a b B-A b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {x \left (a^2 (-B)+2 a A b+b^2 B\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3591

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

Rubi steps

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

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Mathematica [C]  time = 2.13, size = 140, normalized size = 1.22 \[ \frac {\frac {2 \left (\left (a^2 (-A)-2 a b B+A b^2\right ) \log (a+b \tan (c+d x))-\frac {a \left (a^2+b^2\right ) (a B-A b)}{b (a+b \tan (c+d x))}\right )}{\left (a^2+b^2\right )^2}+\frac {(A+i B) \log (-\tan (c+d x)+i)}{(a+i b)^2}+\frac {(A-i B) \log (\tan (c+d x)+i)}{(a-i b)^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.65, size = 221, normalized size = 1.92 \[ -\frac {2 \, B a^{2} b - 2 \, A a b^{2} + 2 \, {\left (B a^{3} - 2 \, A a^{2} b - B a b^{2}\right )} d x + {\left (A a^{3} + 2 \, B a^{2} b - A a b^{2} + {\left (A a^{2} b + 2 \, B a b^{2} - A b^{3}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (B a^{3} - A a^{2} b - {\left (B a^{2} b - 2 \, A a b^{2} - B b^{3}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \tan \left (d x + c\right ) + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*B*a^2*b - 2*A*a*b^2 + 2*(B*a^3 - 2*A*a^2*b - B*a*b^2)*d*x + (A*a^3 + 2*B*a^2*b - A*a*b^2 + (A*a^2*b +
2*B*a*b^2 - A*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
*(B*a^3 - A*a^2*b - (B*a^2*b - 2*A*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 = 0.44, size = 241, normalized size = 2.10 \[ -\frac {\frac {2 \, {\left (B a^{2} - 2 \, A a b - B b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (A a^{2} + 2 \, B a b - A b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (A a^{2} b + 2 \, B a b^{2} - A b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac {2 \, {\left (A a^{2} b^{2} \tan \left (d x + c\right ) + 2 \, B a b^{3} \tan \left (d x + c\right ) - A b^{4} \tan \left (d x + c\right ) - B a^{4} + 2 \, A a^{3} b + B a^{2} b^{2}\right )}}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(B*a^2 - 2*A*a*b - B*b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) - (A*a^2 + 2*B*a*b - A*b^2)*log(tan(d*x +
c)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) + 2*(A*a^2*b + 2*B*a*b^2 - A*b^3)*log(abs(b*tan(d*x + c) + a))/(a^4*b + 2*a^
2*b^3 + b^5) - 2*(A*a^2*b^2*tan(d*x + c) + 2*B*a*b^3*tan(d*x + c) - A*b^4*tan(d*x + c) - B*a^4 + 2*A*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 [B]  time = 0.27, size = 305, normalized size = 2.65 \[ \frac {a A}{d \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}-\frac {a^{2} B}{d \left (a^{2}+b^{2}\right ) b \left (a +b \tan \left (d x +c \right )\right )}-\frac {a^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) A}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (a +b \tan \left (d x +c \right )\right ) A \,b^{2}}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {2 \ln \left (a +b \tan \left (d x +c \right )\right ) B a b}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} A}{2 d \left (a^{2}+b^{2}\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A \,b^{2}}{2 d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) B a b}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {2 A \arctan \left (\tan \left (d x +c \right )\right ) a b}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {B \arctan \left (\tan \left (d x +c \right )\right ) a^{2}}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b^{2}}{d \left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.70, size = 185, normalized size = 1.61 \[ -\frac {\frac {2 \, {\left (B a^{2} - 2 \, A a b - B b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (A a^{2} + 2 \, B a b - A b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (A a^{2} + 2 \, B a b - A b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (B a^{2} - A a b\right )}}{a^{3} b + a b^{3} + {\left (a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.60, size = 163, normalized size = 1.42 \[ \frac {a\,\left (A\,b-B\,a\right )}{b\,d\,\left (a^2+b^2\right )\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )}{2\,d\,\left (a^2+a\,b\,2{}\mathrm {i}-b^2\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (a^2\,1{}\mathrm {i}+2\,a\,b-b^2\,1{}\mathrm {i}\right )}-\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {A}{a^2+b^2}-\frac {2\,b\,\left (A\,b-B\,a\right )}{{\left (a^2+b^2\right )}^2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.84, size = 2987, normalized size = 25.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*(A + B*tan(c))/tan(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((A*log(tan(c + d*x)**2 + 1)/(2*d) -
B*x + B*tan(c + d*x)/d)/a**2, Eq(b, 0)), (I*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*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) - I
*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) + I*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) + 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), Eq(a, -I*b)), (-I*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*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) + I*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) - I*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) + 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), Eq(a, I*
b)), (x*(A + B*tan(c))*tan(c)/(a + b*tan(c))**2, Eq(d, 0)), (-2*A*a**3*b*log(a/b + tan(c + d*x))/(2*a**5*b*d +
2*a**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x))
+ A*a**3*b*log(tan(c + d*x)**2 + 1)/(2*a**5*b*d + 2*a**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*
tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x)) + 2*A*a**3*b/(2*a**5*b*d + 2*a**4*b**2*d*tan(c + d*x) + 4*a
**3*b**3*d + 4*a**2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x)) + 4*A*a**2*b**2*d*x/(2*a**5*b*d
+ 2*a**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x)
) - 2*A*a**2*b**2*log(a/b + tan(c + d*x))*tan(c + d*x)/(2*a**5*b*d + 2*a**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*
d + 4*a**2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x)) + A*a**2*b**2*log(tan(c + d*x)**2 + 1)*ta
n(c + d*x)/(2*a**5*b*d + 2*a**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c + d*x) + 2*a*b**5*d
+ 2*b**6*d*tan(c + d*x)) + 4*A*a*b**3*d*x*tan(c + d*x)/(2*a**5*b*d + 2*a**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*
d + 4*a**2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x)) + 2*A*a*b**3*log(a/b + tan(c + d*x))/(2*a
**5*b*d + 2*a**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*tan(
c + d*x)) - A*a*b**3*log(tan(c + d*x)**2 + 1)/(2*a**5*b*d + 2*a**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a**
2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x)) + 2*A*a*b**3/(2*a**5*b*d + 2*a**4*b**2*d*tan(c + d
*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x)) + 2*A*b**4*log(a/b + ta
n(c + d*x))*tan(c + d*x)/(2*a**5*b*d + 2*a**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c + d*x)
+ 2*a*b**5*d + 2*b**6*d*tan(c + d*x)) - A*b**4*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*a**5*b*d + 2*a**4*b**
2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x)) - 2*B*a**4
/(2*a**5*b*d + 2*a**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d
*tan(c + d*x)) - 2*B*a**3*b*d*x/(2*a**5*b*d + 2*a**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c
+ d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x)) - 2*B*a**2*b**2*d*x*tan(c + d*x)/(2*a**5*b*d + 2*a**4*b**2*d*tan
(c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x)) - 4*B*a**2*b**2*l
og(a/b + tan(c + d*x))/(2*a**5*b*d + 2*a**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c + d*x) +
2*a*b**5*d + 2*b**6*d*tan(c + d*x)) + 2*B*a**2*b**2*log(tan(c + d*x)**2 + 1)/(2*a**5*b*d + 2*a**4*b**2*d*tan(
c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x)) - 2*B*a**2*b**2/(2
*a**5*b*d + 2*a**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*ta
n(c + d*x)) + 2*B*a*b**3*d*x/(2*a**5*b*d + 2*a**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c +
d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x)) - 4*B*a*b**3*log(a/b + tan(c + d*x))*tan(c + d*x)/(2*a**5*b*d + 2*a
**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x)) + 2
*B*a*b**3*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*a**5*b*d + 2*a**4*b**2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a
**2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x)) + 2*B*b**4*d*x*tan(c + d*x)/(2*a**5*b*d + 2*a**4
*b**2*d*tan(c + d*x) + 4*a**3*b**3*d + 4*a**2*b**4*d*tan(c + d*x) + 2*a*b**5*d + 2*b**6*d*tan(c + d*x)), True)
)

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