\(\int \frac {\tan ^4(c+d x)}{a+b \tan (c+d x)} \, dx\) [458]

Optimal result

Integrand size = 21, antiderivative size = 97 \[ \int \frac {\tan ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {a x}{a^2+b^2}+\frac {b \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}+\frac {a^4 \log (a+b \tan (c+d x))}{b^3 \left (a^2+b^2\right ) d}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan ^2(c+d x)}{2 b d} \]

[Out]

a*x/(a^2+b^2)+b*ln(cos(d*x+c))/(a^2+b^2)/d+a^4*ln(a+b*tan(d*x+c))/b^3/(a^2+b^2)/d-a*tan(d*x+c)/b^2/d+1/2*tan(d
*x+c)^2/b/d

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3647, 3728, 3708, 3698, 31, 3556} \[ \int \frac {\tan ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {b \log (\cos (c+d x))}{d \left (a^2+b^2\right )}+\frac {a x}{a^2+b^2}+\frac {a^4 \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan ^2(c+d x)}{2 b d} \]

[In]

Int[Tan[c + d*x]^4/(a + b*Tan[c + d*x]),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3556

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

Rule 3647

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 - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*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]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3698

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A/(b*f), Subst[Int[(a + x)^m, x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A, C]

Rule 3708

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

Rule 3728

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] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d
*Tan[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps \begin{align*} \text {integral}& = \frac {\tan ^2(c+d x)}{2 b d}+\frac {\int \frac {\tan (c+d x) \left (-2 a-2 b \tan (c+d x)-2 a \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 b} \\ & = -\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan ^2(c+d x)}{2 b d}+\frac {\int \frac {2 a^2+2 \left (a^2-b^2\right ) \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{2 b^2} \\ & = \frac {a x}{a^2+b^2}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan ^2(c+d x)}{2 b d}+\frac {a^4 \int \frac {1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b^2 \left (a^2+b^2\right )}-\frac {b \int \tan (c+d x) \, dx}{a^2+b^2} \\ & = \frac {a x}{a^2+b^2}+\frac {b \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan ^2(c+d x)}{2 b d}+\frac {a^4 \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^3 \left (a^2+b^2\right ) d} \\ & = \frac {a x}{a^2+b^2}+\frac {b \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}+\frac {a^4 \log (a+b \tan (c+d x))}{b^3 \left (a^2+b^2\right ) d}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan ^2(c+d x)}{2 b d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.46 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.10 \[ \int \frac {\tan ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {\log (i-\tan (c+d x))}{i a-b}-\frac {\log (i+\tan (c+d x))}{i a+b}+\frac {2 a^4 \log (a+b \tan (c+d x))}{b^3 \left (a^2+b^2\right )}-\frac {2 a \tan (c+d x)}{b^2}+\frac {\tan ^2(c+d x)}{b}}{2 d} \]

[In]

Integrate[Tan[c + d*x]^4/(a + b*Tan[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {-\frac {-\frac {b \left (\tan ^{2}\left (d x +c \right )\right )}{2}+a \tan \left (d x +c \right )}{b^{2}}+\frac {a^{4} \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{3} \left (a^{2}+b^{2}\right )}+\frac {-\frac {b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+a \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}}{d}\)	\(92\)
default	\(\frac {-\frac {-\frac {b \left (\tan ^{2}\left (d x +c \right )\right )}{2}+a \tan \left (d x +c \right )}{b^{2}}+\frac {a^{4} \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{3} \left (a^{2}+b^{2}\right )}+\frac {-\frac {b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+a \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}}{d}\)	\(92\)
norman	\(\frac {a x}{a^{2}+b^{2}}+\frac {\tan ^{2}\left (d x +c \right )}{2 b d}-\frac {a \tan \left (d x +c \right )}{b^{2} d}+\frac {a^{4} \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{3} \left (a^{2}+b^{2}\right ) d}-\frac {b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \left (a^{2}+b^{2}\right )}\)	\(101\)
parallelrisch	\(-\frac {-2 a \,b^{3} d x -a^{2} b^{2} \left (\tan ^{2}\left (d x +c \right )\right )-b^{4} \left (\tan ^{2}\left (d x +c \right )\right )+\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{4}-2 a^{4} \ln \left (a +b \tan \left (d x +c \right )\right )+2 a^{3} b \tan \left (d x +c \right )+2 a \,b^{3} \tan \left (d x +c \right )}{2 \left (a^{2}+b^{2}\right ) b^{3} d}\)	\(111\)
risch	\(-\frac {x}{i b -a}+\frac {2 i a^{2} x}{b^{3}}+\frac {2 i a^{2} c}{b^{3} d}-\frac {2 i x}{b}-\frac {2 i c}{b d}-\frac {2 i a^{4} x}{\left (a^{2}+b^{2}\right ) b^{3}}-\frac {2 i a^{4} c}{\left (a^{2}+b^{2}\right ) b^{3} d}+\frac {-2 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 i a}{b^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}{b^{3} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b d}+\frac {a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{\left (a^{2}+b^{2}\right ) b^{3} d}\)	\(236\)

[In]

int(tan(d*x+c)^4/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.38 \[ \int \frac {\tan ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {2 \, a b^{3} d x + a^{4} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + {\left (a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{2} - {\left (a^{4} - b^{4}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} + b^{5}\right )} d} \]

[In]

integrate(tan(d*x+c)^4/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.77 (sec) , antiderivative size = 677, normalized size of antiderivative = 6.98 \[ \int \frac {\tan ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\begin {cases} \tilde {\infty } x \tan ^{3}{\left (c \right )} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {x + \frac {\tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {\tan {\left (c + d x \right )}}{d}}{a} & \text {for}\: b = 0 \\- \frac {3 i d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {3 d x}{2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {2 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {2 i \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {\tan ^{3}{\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {i \tan ^{2}{\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {3 i}{2 b d \tan {\left (c + d x \right )} - 2 i b d} & \text {for}\: a = - i b \\\frac {3 i d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {3 d x}{2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {2 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {2 i \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {\tan ^{3}{\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {i \tan ^{2}{\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {3 i}{2 b d \tan {\left (c + d x \right )} + 2 i b d} & \text {for}\: a = i b \\\frac {x \tan ^{4}{\left (c \right )}}{a + b \tan {\left (c \right )}} & \text {for}\: d = 0 \\\frac {2 a^{4} \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{2 a^{2} b^{3} d + 2 b^{5} d} - \frac {2 a^{3} b \tan {\left (c + d x \right )}}{2 a^{2} b^{3} d + 2 b^{5} d} + \frac {a^{2} b^{2} \tan ^{2}{\left (c + d x \right )}}{2 a^{2} b^{3} d + 2 b^{5} d} + \frac {2 a b^{3} d x}{2 a^{2} b^{3} d + 2 b^{5} d} - \frac {2 a b^{3} \tan {\left (c + d x \right )}}{2 a^{2} b^{3} d + 2 b^{5} d} - \frac {b^{4} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} b^{3} d + 2 b^{5} d} + \frac {b^{4} \tan ^{2}{\left (c + d x \right )}}{2 a^{2} b^{3} d + 2 b^{5} d} & \text {otherwise} \end {cases} \]

[In]

integrate(tan(d*x+c)**4/(a+b*tan(d*x+c)),x)

[Out]

Piecewise((zoo*x*tan(c)**3, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((x + tan(c + d*x)**3/(3*d) - tan(c + d*x)/d)/a,
Eq(b, 0)), (-3*I*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) - 2*I*b*d) - 3*d*x/(2*b*d*tan(c + d*x) - 2*I*b*d) - 2*lo
g(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x) - 2*I*b*d) + 2*I*log(tan(c + d*x)**2 + 1)/(2*b*d*tan(c
+ d*x) - 2*I*b*d) + tan(c + d*x)**3/(2*b*d*tan(c + d*x) - 2*I*b*d) + I*tan(c + d*x)**2/(2*b*d*tan(c + d*x) -
2*I*b*d) + 3*I/(2*b*d*tan(c + d*x) - 2*I*b*d), Eq(a, -I*b)), (3*I*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b
*d) - 3*d*x/(2*b*d*tan(c + d*x) + 2*I*b*d) - 2*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I
*b*d) - 2*I*log(tan(c + d*x)**2 + 1)/(2*b*d*tan(c + d*x) + 2*I*b*d) + tan(c + d*x)**3/(2*b*d*tan(c + d*x) + 2*
I*b*d) - I*tan(c + d*x)**2/(2*b*d*tan(c + d*x) + 2*I*b*d) - 3*I/(2*b*d*tan(c + d*x) + 2*I*b*d), Eq(a, I*b)), (
x*tan(c)**4/(a + b*tan(c)), Eq(d, 0)), (2*a**4*log(a/b + tan(c + d*x))/(2*a**2*b**3*d + 2*b**5*d) - 2*a**3*b*t
an(c + d*x)/(2*a**2*b**3*d + 2*b**5*d) + a**2*b**2*tan(c + d*x)**2/(2*a**2*b**3*d + 2*b**5*d) + 2*a*b**3*d*x/(
2*a**2*b**3*d + 2*b**5*d) - 2*a*b**3*tan(c + d*x)/(2*a**2*b**3*d + 2*b**5*d) - b**4*log(tan(c + d*x)**2 + 1)/(
2*a**2*b**3*d + 2*b**5*d) + b**4*tan(c + d*x)**2/(2*a**2*b**3*d + 2*b**5*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.02 \[ \int \frac {\tan ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {2 \, a^{4} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b^{3} + b^{5}} + \frac {2 \, {\left (d x + c\right )} a}{a^{2} + b^{2}} - \frac {b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {b \tan \left (d x + c\right )^{2} - 2 \, a \tan \left (d x + c\right )}{b^{2}}}{2 \, d} \]

[In]

integrate(tan(d*x+c)^4/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.90 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03 \[ \int \frac {\tan ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {2 \, a^{4} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{3} + b^{5}} + \frac {2 \, {\left (d x + c\right )} a}{a^{2} + b^{2}} - \frac {b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {b \tan \left (d x + c\right )^{2} - 2 \, a \tan \left (d x + c\right )}{b^{2}}}{2 \, d} \]

[In]

integrate(tan(d*x+c)^4/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

1/2*(2*a^4*log(abs(b*tan(d*x + c) + a))/(a^2*b^3 + b^5) + 2*(d*x + c)*a/(a^2 + b^2) - b*log(tan(d*x + c)^2 + 1
)/(a^2 + b^2) + (b*tan(d*x + c)^2 - 2*a*tan(d*x + c))/b^2)/d

Mupad [B] (verification not implemented)

Time = 5.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.12 \[ \int \frac {\tan ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,b\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (b+a\,1{}\mathrm {i}\right )}-\frac {a\,\mathrm {tan}\left (c+d\,x\right )}{b^2\,d}+\frac {a^4\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{b^3\,d\,\left (a^2+b^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (a+b\,1{}\mathrm {i}\right )} \]

[In]

int(tan(c + d*x)^4/(a + b*tan(c + d*x)),x)

[Out]

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