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

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

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

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Rubi [A]  time = 0.14, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3529, 3531, 3530} \[ -\frac {A b-a B}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac {\left (a^2 (-B)+2 a A b+b^2 B\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {x \left (a^2 A+2 a b B-A b^2\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

Rubi steps

\begin {align*} \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx &=-\frac {A b-a B}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\int \frac {a A+b B-(A b-a B) \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{a^2+b^2}\\ &=\frac {\left (a^2 A-A b^2+2 a b B\right ) x}{\left (a^2+b^2\right )^2}-\frac {A b-a B}{\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 \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 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 (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {A b-a 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 = 190, normalized size = 1.71 \[ \frac {\frac {B ((-b-i a) \log (-\tan (c+d x)+i)+i (a+i b) \log (\tan (c+d x)+i)+2 b \log (a+b \tan (c+d x)))}{a^2+b^2}-(A b-a B) \left (\frac {2 b \left (\frac {a^2+b^2}{a+b \tan (c+d x)}-2 a \log (a+b \tan (c+d x))\right )}{\left (a^2+b^2\right )^2}+\frac {i \log (-\tan (c+d x)+i)}{(a+i b)^2}-\frac {i \log (\tan (c+d x)+i)}{(a-i b)^2}\right )}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((B*(((-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) - (A*b - a*B)*((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 = 0.79, size = 222, normalized size = 2.00 \[ \frac {2 \, B a b^{2} - 2 \, A b^{3} + 2 \, {\left (A a^{3} + 2 \, B a^{2} b - A a b^{2}\right )} d x - {\left (B a^{3} - 2 \, A a^{2} b - B a b^{2} + {\left (B a^{2} b - 2 \, A a b^{2} - B b^{3}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (B a^{2} b - A a b^{2} - {\left (A a^{2} b + 2 \, B a b^{2} - A b^{3}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \tan \left (d x + c\right ) + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*B*a*b^2 - 2*A*b^3 + 2*(A*a^3 + 2*B*a^2*b - A*a*b^2)*d*x - (B*a^3 - 2*A*a^2*b - B*a*b^2 + (B*a^2*b - 2*A
*a*b^2 - B*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^2*b - A*a*b^2 - (A*a^2*b + 2*B*a*b^2 - A*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 = 0.42, size = 234, normalized size = 2.11 \[ \frac {\frac {2 \, {\left (A a^{2} + 2 \, B a b - A b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (B a^{2} - 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (B a^{2} b - 2 \, A a b^{2} - B b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} + \frac {2 \, {\left (B a^{2} b \tan \left (d x + c\right ) - 2 \, A a b^{2} \tan \left (d x + c\right ) - B b^{3} \tan \left (d x + c\right ) + 2 \, B a^{3} - 3 \, A a^{2} b - A b^{3}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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^2*b - 2*A*a*b^2 - B*b^3)*log(abs(b*tan(d*x + c) + a))/(a^4*b + 2*a^2
*b^3 + b^5) + 2*(B*a^2*b*tan(d*x + c) - 2*A*a*b^2*tan(d*x + c) - B*b^3*tan(d*x + c) + 2*B*a^3 - 3*A*a^2*b - A*
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 = 0.26, size = 301, normalized size = 2.71 \[ -\frac {A b}{d \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}+\frac {a B}{d \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 a b \ln \left (a +b \tan \left (d x +c \right )\right ) A}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {a^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (a +b \tan \left (d x +c \right )\right ) b^{2} B}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A a b}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} B}{2 d \left (a^{2}+b^{2}\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{2} B}{2 d \left (a^{2}+b^{2}\right )^{2}}+\frac {A \arctan \left (\tan \left (d x +c \right )\right ) a^{2}}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {A \arctan \left (\tan \left (d x +c \right )\right ) b^{2}}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {2 B \arctan \left (\tan \left (d x +c \right )\right ) a b}{d \left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.86, size = 177, normalized size = 1.59 \[ \frac {\frac {2 \, {\left (A a^{2} + 2 \, B a b - A b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (B a^{2} - 2 \, A a b - B b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (B a^{2} - 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (B a - A b\right )}}{a^{3} + a b^{2} + {\left (a^{2} b + b^{3}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 6.48, size = 153, normalized size = 1.38 \[ \frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-B\,a^2+2\,A\,a\,b+B\,b^2\right )}{d\,{\left (a^2+b^2\right )}^2}-\frac {A\,b-B\,a}{d\,\left (a^2+b^2\right )\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^2+a\,b\,2{}\mathrm {i}+b^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.80, size = 2878, normalized size = 25.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*(A + B*tan(c))/tan(c)**2, Eq(a, 0) & Eq(b, 0) & Eq(d, 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) + 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) - I*B*d*x*tan(c + d*x)**2/(-4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*ta
n(c + d*x) + 4*b**2*d) - 2*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)
+ 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) - 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), 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*ta
n(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) + 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) + 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) - 2*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) - 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) + 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), Eq(a, I*b)), (x*(A + B*tan(c))/(a + b*tan(c))**2, Eq(d, 0)), ((A*x + B
*log(tan(c + d*x)**2 + 1)/(2*d))/a**2, Eq(b, 0)), (2*A*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*A*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*A*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*A*a**2*b*log(tan(c + d*x)**2 + 1)/(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*A*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*A*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*A*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*A*a*b**2*log(t
an(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*A*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*A*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*B*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)) + B*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*B*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*B*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*ta
n(c + d*x)) - 2*B*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)) + B*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*B*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*B*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)) - B*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*B*a*b**2/(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*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)) - B*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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