3.484 \(\int \frac{\cot (c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=168 \[ \frac{b^2 \left (3 a^2+b^2\right )}{a^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{b^2}{2 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{b^2 \left (3 a^2 b^2+6 a^4+b^4\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{a^3 d \left (a^2+b^2\right )^3}-\frac{b x \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^3}+\frac{\log (\sin (c+d x))}{a^3 d} \]

[Out]

-((b*(3*a^2 - b^2)*x)/(a^2 + b^2)^3) + Log[Sin[c + d*x]]/(a^3*d) - (b^2*(6*a^4 + 3*a^2*b^2 + b^4)*Log[a*Cos[c
+ d*x] + b*Sin[c + d*x]])/(a^3*(a^2 + b^2)^3*d) + b^2/(2*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (b^2*(3*a^2
+ b^2))/(a^2*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

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Rubi [A]  time = 0.41004, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3569, 3649, 3651, 3530, 3475} \[ \frac{b^2 \left (3 a^2+b^2\right )}{a^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{b^2}{2 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{b^2 \left (3 a^2 b^2+6 a^4+b^4\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{a^3 d \left (a^2+b^2\right )^3}-\frac{b x \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^3}+\frac{\log (\sin (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]/(a + b*Tan[c + d*x])^3,x]

[Out]

-((b*(3*a^2 - b^2)*x)/(a^2 + b^2)^3) + Log[Sin[c + d*x]]/(a^3*d) - (b^2*(6*a^4 + 3*a^2*b^2 + b^4)*Log[a*Cos[c
+ d*x] + b*Sin[c + d*x]])/(a^3*(a^2 + b^2)^3*d) + b^2/(2*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (b^2*(3*a^2
+ b^2))/(a^2*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3651

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

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 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\cot (c+d x)}{(a+b \tan (c+d x))^3} \, dx &=\frac{b^2}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\int \frac{\cot (c+d x) \left (2 \left (a^2+b^2\right )-2 a b \tan (c+d x)+2 b^2 \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^2} \, dx}{2 a \left (a^2+b^2\right )}\\ &=\frac{b^2}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{b^2 \left (3 a^2+b^2\right )}{a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{\cot (c+d x) \left (2 \left (a^2+b^2\right )^2-4 a^3 b \tan (c+d x)+2 b^2 \left (3 a^2+b^2\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 a^2 \left (a^2+b^2\right )^2}\\ &=-\frac{b \left (3 a^2-b^2\right ) x}{\left (a^2+b^2\right )^3}+\frac{b^2}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{b^2 \left (3 a^2+b^2\right )}{a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \cot (c+d x) \, dx}{a^3}-\frac{\left (b^2 \left (6 a^4+3 a^2 b^2+b^4\right )\right ) \int \frac{b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{a^3 \left (a^2+b^2\right )^3}\\ &=-\frac{b \left (3 a^2-b^2\right ) x}{\left (a^2+b^2\right )^3}+\frac{\log (\sin (c+d x))}{a^3 d}-\frac{b^2 \left (6 a^4+3 a^2 b^2+b^4\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{a^3 \left (a^2+b^2\right )^3 d}+\frac{b^2}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{b^2 \left (3 a^2+b^2\right )}{a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 3.85454, size = 209, normalized size = 1.24 \[ \frac{\frac{4 a b^2}{\left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{2 b^2}{a^2+a b \tan (c+d x)}-\frac{2 b^2 \left (3 a^2 b^2+6 a^4+b^4\right ) \log (a+b \tan (c+d x))}{a^2 \left (a^2+b^2\right )^2}+\frac{2 \left (a^2+b^2\right ) \log (\tan (c+d x))}{a^2}+\frac{b^2}{(a+b \tan (c+d x))^2}-\frac{a (a-i b) \log (-\tan (c+d x)+i)}{(a+i b)^2}-\frac{a (a+i b) \log (\tan (c+d x)+i)}{(a-i b)^2}}{2 a d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]/(a + b*Tan[c + d*x])^3,x]

[Out]

(-((a*(a - I*b)*Log[I - Tan[c + d*x]])/(a + I*b)^2) + (2*(a^2 + b^2)*Log[Tan[c + d*x]])/a^2 - (a*(a + I*b)*Log
[I + Tan[c + d*x]])/(a - I*b)^2 - (2*b^2*(6*a^4 + 3*a^2*b^2 + b^4)*Log[a + b*Tan[c + d*x]])/(a^2*(a^2 + b^2)^2
) + b^2/(a + b*Tan[c + d*x])^2 + (4*a*b^2)/((a^2 + b^2)*(a + b*Tan[c + d*x])) + (2*b^2)/(a^2 + a*b*Tan[c + d*x
]))/(2*a*(a^2 + b^2)*d)

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Maple [A]  time = 0.106, size = 304, normalized size = 1.8 \begin{align*} -{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){a}^{3}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) a{b}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-3\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}b}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{\ln \left ( \tan \left ( dx+c \right ) \right ) }{d{a}^{3}}}+{\frac{{b}^{2}}{2\, \left ({a}^{2}+{b}^{2} \right ) ad \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) }}+{\frac{{b}^{4}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}{a}^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) }}-6\,{\frac{a\ln \left ( a+b\tan \left ( dx+c \right ) \right ){b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-3\,{\frac{{b}^{4}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{3}a}}-{\frac{{b}^{6}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{3}{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)/(a+b*tan(d*x+c))^3,x)

[Out]

-1/2/d/(a^2+b^2)^3*ln(1+tan(d*x+c)^2)*a^3+3/2/d/(a^2+b^2)^3*ln(1+tan(d*x+c)^2)*a*b^2-3/d/(a^2+b^2)^3*arctan(ta
n(d*x+c))*a^2*b+1/d/(a^2+b^2)^3*arctan(tan(d*x+c))*b^3+1/d/a^3*ln(tan(d*x+c))+1/2*b^2/a/(a^2+b^2)/d/(a+b*tan(d
*x+c))^2+3/d/(a^2+b^2)^2/(a+b*tan(d*x+c))*b^2+1/d*b^4/(a^2+b^2)^2/a^2/(a+b*tan(d*x+c))-6/d*a/(a^2+b^2)^3*ln(a+
b*tan(d*x+c))*b^2-3/d*b^4/(a^2+b^2)^3/a*ln(a+b*tan(d*x+c))-1/d*b^6/(a^2+b^2)^3/a^3*ln(a+b*tan(d*x+c))

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Maxima [A]  time = 1.64328, size = 392, normalized size = 2.33 \begin{align*} -\frac{\frac{2 \,{\left (3 \, a^{2} b - b^{3}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{2 \,{\left (6 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{9} + 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} + a^{3} b^{6}} + \frac{{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{7 \, a^{3} b^{2} + 3 \, a b^{4} + 2 \,{\left (3 \, a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )}{a^{8} + 2 \, a^{6} b^{2} + a^{4} b^{4} +{\left (a^{6} b^{2} + 2 \, a^{4} b^{4} + a^{2} b^{6}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (a^{7} b + 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \tan \left (d x + c\right )} - \frac{2 \, \log \left (\tan \left (d x + c\right )\right )}{a^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*(2*(3*a^2*b - b^3)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 2*(6*a^4*b^2 + 3*a^2*b^4 + b^6)*log(b*
tan(d*x + c) + a)/(a^9 + 3*a^7*b^2 + 3*a^5*b^4 + a^3*b^6) + (a^3 - 3*a*b^2)*log(tan(d*x + c)^2 + 1)/(a^6 + 3*a
^4*b^2 + 3*a^2*b^4 + b^6) - (7*a^3*b^2 + 3*a*b^4 + 2*(3*a^2*b^3 + b^5)*tan(d*x + c))/(a^8 + 2*a^6*b^2 + a^4*b^
4 + (a^6*b^2 + 2*a^4*b^4 + a^2*b^6)*tan(d*x + c)^2 + 2*(a^7*b + 2*a^5*b^3 + a^3*b^5)*tan(d*x + c)) - 2*log(tan
(d*x + c))/a^3)/d

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Fricas [B]  time = 2.22766, size = 1050, normalized size = 6.25 \begin{align*} \frac{9 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - 2 \,{\left (3 \, a^{7} b - a^{5} b^{3}\right )} d x -{\left (7 \, a^{4} b^{4} + a^{2} b^{6} + 2 \,{\left (3 \, a^{5} b^{3} - a^{3} b^{5}\right )} d x\right )} \tan \left (d x + c\right )^{2} +{\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6} +{\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac{\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) -{\left (6 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6} +{\left (6 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (6 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\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 (4 \, a^{5} b^{3} - 3 \, a^{3} b^{5} - a b^{7} + 2 \,{\left (3 \, a^{6} b^{2} - a^{4} b^{4}\right )} d x\right )} \tan \left (d x + c\right )}{2 \,{\left ({\left (a^{9} b^{2} + 3 \, a^{7} b^{4} + 3 \, a^{5} b^{6} + a^{3} b^{8}\right )} d \tan \left (d x + c\right )^{2} + 2 \,{\left (a^{10} b + 3 \, a^{8} b^{3} + 3 \, a^{6} b^{5} + a^{4} b^{7}\right )} d \tan \left (d x + c\right ) +{\left (a^{11} + 3 \, a^{9} b^{2} + 3 \, a^{7} b^{4} + a^{5} b^{6}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(9*a^4*b^4 + 3*a^2*b^6 - 2*(3*a^7*b - a^5*b^3)*d*x - (7*a^4*b^4 + a^2*b^6 + 2*(3*a^5*b^3 - a^3*b^5)*d*x)*t
an(d*x + c)^2 + (a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6 + (a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*tan(d*x + c)^
2 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*tan(d*x + c))*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1)) - (6*a^6*
b^2 + 3*a^4*b^4 + a^2*b^6 + (6*a^4*b^4 + 3*a^2*b^6 + b^8)*tan(d*x + c)^2 + 2*(6*a^5*b^3 + 3*a^3*b^5 + a*b^7)*t
an(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*(4*a^5*b^3 - 3*a^3*
b^5 - a*b^7 + 2*(3*a^6*b^2 - a^4*b^4)*d*x)*tan(d*x + c))/((a^9*b^2 + 3*a^7*b^4 + 3*a^5*b^6 + a^3*b^8)*d*tan(d*
x + c)^2 + 2*(a^10*b + 3*a^8*b^3 + 3*a^6*b^5 + a^4*b^7)*d*tan(d*x + c) + (a^11 + 3*a^9*b^2 + 3*a^7*b^4 + a^5*b
^6)*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*tan(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.35725, size = 443, normalized size = 2.64 \begin{align*} -\frac{\frac{2 \,{\left (3 \, a^{2} b - b^{3}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{2 \,{\left (6 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{9} b + 3 \, a^{7} b^{3} + 3 \, a^{5} b^{5} + a^{3} b^{7}} - \frac{18 \, a^{4} b^{4} \tan \left (d x + c\right )^{2} + 9 \, a^{2} b^{6} \tan \left (d x + c\right )^{2} + 3 \, b^{8} \tan \left (d x + c\right )^{2} + 42 \, a^{5} b^{3} \tan \left (d x + c\right ) + 26 \, a^{3} b^{5} \tan \left (d x + c\right ) + 8 \, a b^{7} \tan \left (d x + c\right ) + 25 \, a^{6} b^{2} + 19 \, a^{4} b^{4} + 6 \, a^{2} b^{6}}{{\left (a^{9} + 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} + a^{3} b^{6}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{2}} - \frac{2 \, \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(2*(3*a^2*b - b^3)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (a^3 - 3*a*b^2)*log(tan(d*x + c)^2 + 1
)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 2*(6*a^4*b^3 + 3*a^2*b^5 + b^7)*log(abs(b*tan(d*x + c) + a))/(a^9*b +
3*a^7*b^3 + 3*a^5*b^5 + a^3*b^7) - (18*a^4*b^4*tan(d*x + c)^2 + 9*a^2*b^6*tan(d*x + c)^2 + 3*b^8*tan(d*x + c)^
2 + 42*a^5*b^3*tan(d*x + c) + 26*a^3*b^5*tan(d*x + c) + 8*a*b^7*tan(d*x + c) + 25*a^6*b^2 + 19*a^4*b^4 + 6*a^2
*b^6)/((a^9 + 3*a^7*b^2 + 3*a^5*b^4 + a^3*b^6)*(b*tan(d*x + c) + a)^2) - 2*log(abs(tan(d*x + c)))/a^3)/d