\(\int \csc (c+d x) (a+b \tan (c+d x))^4 \, dx\) [44]

Optimal result

Integrand size = 19, antiderivative size = 118 \[ \int \csc (c+d x) (a+b \tan (c+d x))^4 \, dx=-\frac {a^4 \text {arctanh}(\cos (c+d x))}{d}+\frac {4 a^3 b \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b^3 \text {arctanh}(\sin (c+d x))}{d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {b^4 \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}+\frac {2 a b^3 \sec (c+d x) \tan (c+d x)}{d} \]

[Out]

-a^4*arctanh(cos(d*x+c))/d+4*a^3*b*arctanh(sin(d*x+c))/d-2*a*b^3*arctanh(sin(d*x+c))/d+6*a^2*b^2*sec(d*x+c)/d-
b^4*sec(d*x+c)/d+1/3*b^4*sec(d*x+c)^3/d+2*a*b^3*sec(d*x+c)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3598, 3855, 2686, 8, 2691} \[ \int \csc (c+d x) (a+b \tan (c+d x))^4 \, dx=-\frac {a^4 \text {arctanh}(\cos (c+d x))}{d}+\frac {4 a^3 b \text {arctanh}(\sin (c+d x))}{d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {2 a b^3 \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a b^3 \tan (c+d x) \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}-\frac {b^4 \sec (c+d x)}{d} \]

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3598

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[Expand[Sin[e
+ f*x]^m*(a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IGtQ[n, 0]

Rule 3855

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

Rubi steps \begin{align*} \text {integral}& = \int \left (a^4 \csc (c+d x)+4 a^3 b \sec (c+d x)+6 a^2 b^2 \sec (c+d x) \tan (c+d x)+4 a b^3 \sec (c+d x) \tan ^2(c+d x)+b^4 \sec (c+d x) \tan ^3(c+d x)\right ) \, dx \\ & = a^4 \int \csc (c+d x) \, dx+\left (4 a^3 b\right ) \int \sec (c+d x) \, dx+\left (6 a^2 b^2\right ) \int \sec (c+d x) \tan (c+d x) \, dx+\left (4 a b^3\right ) \int \sec (c+d x) \tan ^2(c+d x) \, dx+b^4 \int \sec (c+d x) \tan ^3(c+d x) \, dx \\ & = -\frac {a^4 \text {arctanh}(\cos (c+d x))}{d}+\frac {4 a^3 b \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a b^3 \sec (c+d x) \tan (c+d x)}{d}-\left (2 a b^3\right ) \int \sec (c+d x) \, dx+\frac {\left (6 a^2 b^2\right ) \text {Subst}(\int 1 \, dx,x,\sec (c+d x))}{d}+\frac {b^4 \text {Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{d} \\ & = -\frac {a^4 \text {arctanh}(\cos (c+d x))}{d}+\frac {4 a^3 b \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b^3 \text {arctanh}(\sin (c+d x))}{d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {b^4 \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}+\frac {2 a b^3 \sec (c+d x) \tan (c+d x)}{d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(352\) vs. \(2(118)=236\).

Time = 6.85 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.98 \[ \int \csc (c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {72 a^2 b^2-10 b^4-12 a^4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-48 a^3 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+24 a b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 a^4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+48 a^3 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-24 a b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {12 a b^3}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {b^4}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+2 b^2 \left (36 a^2-b^2+2 b^2 \cos (c+d x)+\left (36 a^2-5 b^2\right ) \cos (2 (c+d x))\right ) \sec ^3(c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )-\frac {12 a b^3}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {b^4}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}}{12 d} \]

[In]

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

[Out]

(72*a^2*b^2 - 10*b^4 - 12*a^4*Log[Cos[(c + d*x)/2]] - 48*a^3*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 24*a
*b^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*a^4*Log[Sin[(c + d*x)/2]] + 48*a^3*b*Log[Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]] - 24*a*b^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (12*a*b^3)/(Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2])^2 + b^4/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + 2*b^2*(36*a^2 - b^2 + 2*b^2*Cos[c + d*x] + (36*a
^2 - 5*b^2)*Cos[2*(c + d*x)])*Sec[c + d*x]^3*Sin[(c + d*x)/2]^2 - (12*a*b^3)/(Cos[(c + d*x)/2] + Sin[(c + d*x)
/2])^2 + b^4/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(12*d)

Maple [A] (verified)

Time = 1.37 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.44

method	result	size
derivativedivides	\(\frac {b^{4} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+4 a \,b^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {6 a^{2} b^{2}}{\cos \left (d x +c \right )}+4 a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}\)	\(170\)
default	\(\frac {b^{4} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+4 a \,b^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {6 a^{2} b^{2}}{\cos \left (d x +c \right )}+4 a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}\)	\(170\)
risch	\(-\frac {2 b^{2} {\mathrm e}^{i \left (d x +c \right )} \left (-18 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-36 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-18 a^{2}+3 b^{2}-6 i a b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {4 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {2 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {4 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {2 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\)	\(257\)

[In]

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

[Out]

1/d*(b^4*(1/3*sin(d*x+c)^4/cos(d*x+c)^3-1/3*sin(d*x+c)^4/cos(d*x+c)-1/3*(2+sin(d*x+c)^2)*cos(d*x+c))+4*a*b^3*(
1/2*sin(d*x+c)^3/cos(d*x+c)^2+1/2*sin(d*x+c)-1/2*ln(sec(d*x+c)+tan(d*x+c)))+6*a^2*b^2/cos(d*x+c)+4*a^3*b*ln(se
c(d*x+c)+tan(d*x+c))+a^4*ln(csc(d*x+c)-cot(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.48 \[ \int \csc (c+d x) (a+b \tan (c+d x))^4 \, dx=-\frac {3 \, a^{4} \cos \left (d x + c\right )^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, a^{4} \cos \left (d x + c\right )^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 12 \, a b^{3} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 6 \, {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, b^{4} - 6 \, {\left (6 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2}}{6 \, d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

-1/6*(3*a^4*cos(d*x + c)^3*log(1/2*cos(d*x + c) + 1/2) - 3*a^4*cos(d*x + c)^3*log(-1/2*cos(d*x + c) + 1/2) - 1
2*a*b^3*cos(d*x + c)*sin(d*x + c) - 6*(2*a^3*b - a*b^3)*cos(d*x + c)^3*log(sin(d*x + c) + 1) + 6*(2*a^3*b - a*
b^3)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) - 2*b^4 - 6*(6*a^2*b^2 - b^4)*cos(d*x + c)^2)/(d*cos(d*x + c)^3)

Sympy [F]

\[ \int \csc (c+d x) (a+b \tan (c+d x))^4 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{4} \csc {\left (c + d x \right )}\, dx \]

[In]

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

[Out]

Integral((a + b*tan(c + d*x))**4*csc(c + d*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.18 \[ \int \csc (c+d x) (a+b \tan (c+d x))^4 \, dx=-\frac {3 \, a b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, a^{3} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 3 \, a^{4} \log \left (\cot \left (d x + c\right ) + \csc \left (d x + c\right )\right ) - \frac {18 \, a^{2} b^{2}}{\cos \left (d x + c\right )} + \frac {{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} b^{4}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

[In]

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

[Out]

-1/3*(3*a*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 6*a^3*b*
(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 3*a^4*log(cot(d*x + c) + csc(d*x + c)) - 18*a^2*b^2/cos(d*x
+ c) + (3*cos(d*x + c)^2 - 1)*b^4/cos(d*x + c)^3)/d

Giac [A] (verification not implemented)

none

Time = 1.04 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.64 \[ \int \csc (c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {3 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 6 \, {\left (2 \, a^{3} b - a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, {\left (2 \, a^{3} b - a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {4 \, {\left (3 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, a^{2} b^{2} + b^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \]

[In]

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

[Out]

1/3*(3*a^4*log(abs(tan(1/2*d*x + 1/2*c))) + 6*(2*a^3*b - a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*(2*a^3*
b - a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 4*(3*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 9*a^2*b^2*tan(1/2*d*x + 1/
2*c)^4 + 18*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*b^4*tan(1/2*d*x + 1/2*c)^2 - 3*a*b^3*tan(1/2*d*x + 1/2*c) - 9*a
^2*b^2 + b^4)/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 6.07 (sec) , antiderivative size = 496, normalized size of antiderivative = 4.20 \[ \int \csc (c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {12\,a^2\,b^2-\frac {4\,b^4}{3}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (4\,b^4-24\,a^2\,b^2\right )+12\,a^2\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-4\,a\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {a\,b\,\mathrm {atan}\left (\frac {a\,b\,\left (2\,a^2-b^2\right )\,\left (4\,a\,b^3-8\,a^3\,b+2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-12\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2-b^2\right )\right )\,2{}\mathrm {i}+a\,b\,\left (2\,a^2-b^2\right )\,\left (4\,a\,b^3-8\,a^3\,b+2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+12\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2-b^2\right )\right )\,2{}\mathrm {i}}{2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (64\,a^6\,b^2-64\,a^4\,b^4+16\,a^2\,b^6\right )+16\,a^7\,b-8\,a^5\,b^3+2\,a\,b\,\left (2\,a^2-b^2\right )\,\left (4\,a\,b^3-8\,a^3\,b+2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-12\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2-b^2\right )\right )-2\,a\,b\,\left (2\,a^2-b^2\right )\,\left (4\,a\,b^3-8\,a^3\,b+2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+12\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2-b^2\right )\right )}\right )\,\left (2\,a^2-b^2\right )\,4{}\mathrm {i}}{d} \]

[In]

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

[Out]

(a^4*log(tan(c/2 + (d*x)/2)))/d - (12*a^2*b^2 - (4*b^4)/3 + tan(c/2 + (d*x)/2)^2*(4*b^4 - 24*a^2*b^2) + 12*a^2
*b^2*tan(c/2 + (d*x)/2)^4 + 4*a*b^3*tan(c/2 + (d*x)/2) - 4*a*b^3*tan(c/2 + (d*x)/2)^5)/(d*(3*tan(c/2 + (d*x)/2
)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1)) - (a*b*atan((a*b*(2*a^2 - b^2)*(4*a*b^3 - 8*a^3*b +
2*a^4*tan(c/2 + (d*x)/2) - 12*a*b*tan(c/2 + (d*x)/2)*(2*a^2 - b^2))*2i + a*b*(2*a^2 - b^2)*(4*a*b^3 - 8*a^3*b
+ 2*a^4*tan(c/2 + (d*x)/2) + 12*a*b*tan(c/2 + (d*x)/2)*(2*a^2 - b^2))*2i)/(2*tan(c/2 + (d*x)/2)*(16*a^2*b^6 -
64*a^4*b^4 + 64*a^6*b^2) + 16*a^7*b - 8*a^5*b^3 + 2*a*b*(2*a^2 - b^2)*(4*a*b^3 - 8*a^3*b + 2*a^4*tan(c/2 + (d*
x)/2) - 12*a*b*tan(c/2 + (d*x)/2)*(2*a^2 - b^2)) - 2*a*b*(2*a^2 - b^2)*(4*a*b^3 - 8*a^3*b + 2*a^4*tan(c/2 + (d
*x)/2) + 12*a*b*tan(c/2 + (d*x)/2)*(2*a^2 - b^2))))*(2*a^2 - b^2)*4i)/d