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

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

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3628, 3531, 3530} \[ \frac {a (b B-a C)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\left (a^2 B+2 a b C-b^2 B\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {x \left (a^2 (-C)+2 a b B+b^2 C\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*b*B - a^2*C + b^2*C)*x)/(a^2 + b^2)^2 - ((a^2*B - b^2*B + 2*a*b*C)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])
/((a^2 + b^2)^2*d) + (a*(b*B - a*C))/(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 3628

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

Rubi steps

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

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Mathematica [C]  time = 2.19, size = 140, normalized size = 1.22 \[ \frac {\frac {2 \left (\left (a^2 (-B)-2 a b C+b^2 B\right ) \log (a+b \tan (c+d x))-\frac {a \left (a^2+b^2\right ) (a C-b B)}{b (a+b \tan (c+d x))}\right )}{\left (a^2+b^2\right )^2}+\frac {(B+i C) \log (-\tan (c+d x)+i)}{(a+i b)^2}+\frac {(B-i C) \log (\tan (c+d x)+i)}{(a-i b)^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 221, normalized size = 1.92 \[ -\frac {2 \, C a^{2} b - 2 \, B a b^{2} + 2 \, {\left (C a^{3} - 2 \, B a^{2} b - C a b^{2}\right )} d x + {\left (B a^{3} + 2 \, C a^{2} b - B a b^{2} + {\left (B a^{2} b + 2 \, C a b^{2} - B b^{3}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (C a^{3} - B a^{2} b - {\left (C a^{2} b - 2 \, B a b^{2} - C b^{3}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \tan \left (d x + c\right ) + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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^2*b - 2*B*a*b^2 + 2*(C*a^3 - 2*B*a^2*b - C*a*b^2)*d*x + (B*a^3 + 2*C*a^2*b - B*a*b^2 + (B*a^2*b +
2*C*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
*(C*a^3 - B*a^2*b - (C*a^2*b - 2*B*a*b^2 - C*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 = 1.99, size = 241, normalized size = 2.10 \[ -\frac {\frac {2 \, {\left (C a^{2} - 2 \, B a b - C b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (B a^{2} + 2 \, C a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (B a^{2} b + 2 \, C a b^{2} - B b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac {2 \, {\left (B a^{2} b^{2} \tan \left (d x + c\right ) + 2 \, C a b^{3} \tan \left (d x + c\right ) - B b^{4} \tan \left (d x + c\right ) - C a^{4} + 2 \, B a^{3} b + C a^{2} b^{2}\right )}}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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*(C*a^2 - 2*B*a*b - C*b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) - (B*a^2 + 2*C*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*C*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^2*tan(d*x + c) + 2*C*a*b^3*tan(d*x + c) - B*b^4*tan(d*x + c) - C*a^4 + 2*B*a^3*b + C
*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 = 0.31, size = 305, normalized size = 2.65 \[ \frac {a B}{d \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}-\frac {a^{2} C}{d \left (a^{2}+b^{2}\right ) b \left (a +b \tan \left (d x +c \right )\right )}-\frac {a^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (a +b \tan \left (d x +c \right )\right ) b^{2} B}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {2 \ln \left (a +b \tan \left (d x +c \right )\right ) C a b}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} B}{2 d \left (a^{2}+b^{2}\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{2} B}{2 d \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) C a b}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {2 B \arctan \left (\tan \left (d x +c \right )\right ) a b}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {C \arctan \left (\tan \left (d x +c \right )\right ) a^{2}}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {C \arctan \left (\tan \left (d x +c \right )\right ) b^{2}}{d \left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.64, size = 185, normalized size = 1.61 \[ -\frac {\frac {2 \, {\left (C a^{2} - 2 \, B a b - C b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (B a^{2} + 2 \, C a b - B b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (B a^{2} + 2 \, C a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (C a^{2} - B a b\right )}}{a^{3} b + a b^{3} + {\left (a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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*(C*a^2 - 2*B*a*b - C*b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) + 2*(B*a^2 + 2*C*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*C*a*b - B*b^2)*log(tan(d*x + c)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4
) + 2*(C*a^2 - B*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 = 9.01, size = 163, normalized size = 1.42 \[ \frac {a\,\left (B\,b-C\,a\right )}{b\,d\,\left (a^2+b^2\right )\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )}{2\,d\,\left (a^2+a\,b\,2{}\mathrm {i}-b^2\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (C+B\,1{}\mathrm {i}\right )}{2\,d\,\left (a^2\,1{}\mathrm {i}+2\,a\,b-b^2\,1{}\mathrm {i}\right )}-\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {B}{a^2+b^2}-\frac {2\,b\,\left (B\,b-C\,a\right )}{{\left (a^2+b^2\right )}^2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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sympy [A]  time = 1.84, size = 2995, normalized size = 26.04 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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