\(\int \frac {\csc ^3(c+d x)}{a+b \tan (c+d x)} \, dx\) [55]

Optimal result

Integrand size = 21, antiderivative size = 122 \[ \int \frac {\csc ^3(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{2 a d}-\frac {b^2 \text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {b \sqrt {a^2+b^2} \text {arctanh}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{a^3 d}+\frac {b \csc (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \] Output:

-1/2*arctanh(cos(d*x+c))/a/d-b^2*arctanh(cos(d*x+c))/a^3/d+b*(a^2+b^2)^(1/ 
2)*arctanh((b*cos(d*x+c)-a*sin(d*x+c))/(a^2+b^2)^(1/2))/a^3/d+b*csc(d*x+c) 
/a^2/d-1/2*cot(d*x+c)*csc(d*x+c)/a/d
 

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.47 \[ \int \frac {\csc ^3(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {-16 b \sqrt {a^2+b^2} \text {arctanh}\left (\frac {-b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )+4 a b \cot \left (\frac {1}{2} (c+d x)\right )-a^2 \csc ^2\left (\frac {1}{2} (c+d x)\right )-4 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-8 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )+4 a b \tan \left (\frac {1}{2} (c+d x)\right )}{8 a^3 d} \] Input:

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

Output:

(-16*b*Sqrt[a^2 + b^2]*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]] 
+ 4*a*b*Cot[(c + d*x)/2] - a^2*Csc[(c + d*x)/2]^2 - 4*a^2*Log[Cos[(c + d*x 
)/2]] - 8*b^2*Log[Cos[(c + d*x)/2]] + 4*a^2*Log[Sin[(c + d*x)/2]] + 8*b^2* 
Log[Sin[(c + d*x)/2]] + a^2*Sec[(c + d*x)/2]^2 + 4*a*b*Tan[(c + d*x)/2])/( 
8*a^3*d)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4001, 3042, 3589, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (c+d x)^3 (a+b \tan (c+d x))}dx\)

\(\Big \downarrow \) 4001

\(\displaystyle \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a \cos (c+d x)+b \sin (c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)}{\sin (c+d x)^3 (a \cos (c+d x)+b \sin (c+d x))}dx\)

\(\Big \downarrow \) 3589

\(\displaystyle \int \left (-\frac {b^3 \sec ^2(c+d x)}{a^3 (a \cos (c+d x)+b \sin (c+d x))}+\frac {b^2 \csc (c+d x) \sec ^2(c+d x)}{a^3}-\frac {b \csc ^2(c+d x) \sec (c+d x)}{a^2}+\frac {\csc ^3(c+d x)}{a}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {b^2 \text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {b \csc (c+d x)}{a^2 d}+\frac {b \sqrt {a^2+b^2} \text {arctanh}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{a^3 d}-\frac {\text {arctanh}(\cos (c+d x))}{2 a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}\)

Input:

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

Output:

-1/2*ArcTanh[Cos[c + d*x]]/(a*d) - (b^2*ArcTanh[Cos[c + d*x]])/(a^3*d) + ( 
b*Sqrt[a^2 + b^2]*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2 
]])/(a^3*d) + (b*Csc[c + d*x])/(a^2*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a* 
d)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3589

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_. 
) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[Ex 
pandTrig[cos[c + d*x]^m*(sin[c + d*x]^n/(a*cos[c + d*x] + b*sin[c + d*x])), 
 x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegersQ[m, n]
 

rule 4001

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n 
_.), x_Symbol] :> Int[Sin[e + f*x]^m*((a*Cos[e + f*x] + b*Sin[e + f*x])^n/C 
os[e + f*x]^n), x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && ILtQ 
[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))
 

Maple [A] (verified)

Time = 1.88 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {\frac {\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {1}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (2 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \sqrt {a^{2}+b^{2}}\, \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{3}}}{d}\)	\(140\)
default	\(\frac {\frac {\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {1}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (2 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \sqrt {a^{2}+b^{2}}\, \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{3}}}{d}\)	\(140\)
risch	\(\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (-i a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-i a -2 b \right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{a^{3} d}-\frac {i \sqrt {-a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {\sqrt {-a^{2}-b^{2}}\, \left (i b +a \right )}{a^{2}+b^{2}}\right )}{d \,a^{3}}+\frac {i \sqrt {-a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {\sqrt {-a^{2}-b^{2}}\, \left (i b +a \right )}{a^{2}+b^{2}}\right )}{d \,a^{3}}\)	\(284\)

Input:

int(csc(d*x+c)^3/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/4/a^2*(1/2*a*tan(1/2*d*x+1/2*c)^2+2*b*tan(1/2*d*x+1/2*c))-1/8/a/tan 
(1/2*d*x+1/2*c)^2+1/4/a^3*(2*a^2+4*b^2)*ln(tan(1/2*d*x+1/2*c))+1/2*b/a^2/t 
an(1/2*d*x+1/2*c)-2*b*(a^2+b^2)^(1/2)/a^3*arctanh(1/2*(2*a*tan(1/2*d*x+1/2 
*c)-2*b)/(a^2+b^2)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (114) = 228\).

Time = 0.13 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.21 \[ \int \frac {\csc ^3(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {2 \, a^{2} \cos \left (d x + c\right ) - 4 \, a b \sin \left (d x + c\right ) + 2 \, {\left (b \cos \left (d x + c\right )^{2} - b\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - {\left ({\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )}} \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {\csc ^3(c+d x)}{a+b \tan (c+d x)} \, dx=\int \frac {\csc ^{3}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \] Input:

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

Output:

Integral(csc(c + d*x)**3/(a + b*tan(c + d*x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.76 \[ \int \frac {\csc ^3(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {\frac {4 \, b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}} + \frac {4 \, {\left (a^{2} + 2 \, b^{2}\right )} \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {{\left (a - \frac {4 \, b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a^{2} \sin \left (d x + c\right )^{2}} + \frac {8 \, {\left (a^{2} b + b^{3}\right )} \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}}}{8 \, d} \] Input:

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

Output:

1/8*((4*b*sin(d*x + c)/(cos(d*x + c) + 1) + a*sin(d*x + c)^2/(cos(d*x + c) 
 + 1)^2)/a^2 + 4*(a^2 + 2*b^2)*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - 
(a - 4*b*sin(d*x + c)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)^2/(a^2*sin(d* 
x + c)^2) + 8*(a^2*b + b^3)*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + s 
qrt(a^2 + b^2))/(b - a*sin(d*x + c)/(cos(d*x + c) + 1) - sqrt(a^2 + b^2))) 
/(sqrt(a^2 + b^2)*a^3))/d
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.71 \[ \int \frac {\csc ^3(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} + \frac {4 \, {\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {8 \, {\left (a^{2} b + b^{3}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} - \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \] Input:

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

Output:

1/8*((a*tan(1/2*d*x + 1/2*c)^2 + 4*b*tan(1/2*d*x + 1/2*c))/a^2 + 4*(a^2 + 
2*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 8*(a^2*b + b^3)*log(abs(2*a*ta 
n(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/2*c) 
 - 2*b + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^3) - (6*a^2*tan(1/2*d*x + 
1/2*c)^2 + 12*b^2*tan(1/2*d*x + 1/2*c)^2 - 4*a*b*tan(1/2*d*x + 1/2*c) + a^ 
2)/(a^3*tan(1/2*d*x + 1/2*c)^2))/d
 

Mupad [B] (verification not implemented)

Time = 1.68 (sec) , antiderivative size = 764, normalized size of antiderivative = 6.26 \[ \int \frac {\csc ^3(c+d x)}{a+b \tan (c+d x)} \, dx =\text {Too large to display} \] Input:

int(1/(sin(c + d*x)^3*(a + b*tan(c + d*x))),x)
 

Output:

(b^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(2*((a^3*d)/2 - (a^3*d*co 
s(2*c + 2*d*x))/2)) - (a^2*(cos(c + d*x)/2 - log(sin(c/2 + (d*x)/2)/cos(c/ 
2 + (d*x)/2))/4 + (log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2* 
d*x))/4))/((a^3*d)/2 - (a^3*d*cos(2*c + 2*d*x))/2) + (a*b*sin(c + d*x))/(( 
a^3*d)/2 - (a^3*d*cos(2*c + 2*d*x))/2) - (b^2*log(sin(c/2 + (d*x)/2)/cos(c 
/2 + (d*x)/2))*cos(2*c + 2*d*x))/(2*((a^3*d)/2 - (a^3*d*cos(2*c + 2*d*x))/ 
2)) + (b*atan((a^4*sin(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2)*1i + b^4*sin(c/2 + 
 (d*x)/2)*(a^2 + b^2)^(1/2)*8i + a*b^3*cos(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2 
)*4i + a^3*b*cos(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2)*3i + a^2*b^2*sin(c/2 + ( 
d*x)/2)*(a^2 + b^2)^(1/2)*8i)/(a^5*cos(c/2 + (d*x)/2) + 8*b^5*sin(c/2 + (d 
*x)/2) + 4*a*b^4*cos(c/2 + (d*x)/2) + 4*a^4*b*sin(c/2 + (d*x)/2) + 5*a^3*b 
^2*cos(c/2 + (d*x)/2) + 12*a^2*b^3*sin(c/2 + (d*x)/2)))*(a^2 + b^2)^(1/2)* 
1i)/((a^3*d)/2 - (a^3*d*cos(2*c + 2*d*x))/2) - (b*cos(2*c + 2*d*x)*atan((a 
^4*sin(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2)*1i + b^4*sin(c/2 + (d*x)/2)*(a^2 + 
 b^2)^(1/2)*8i + a*b^3*cos(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2)*4i + a^3*b*cos 
(c/2 + (d*x)/2)*(a^2 + b^2)^(1/2)*3i + a^2*b^2*sin(c/2 + (d*x)/2)*(a^2 + b 
^2)^(1/2)*8i)/(a^5*cos(c/2 + (d*x)/2) + 8*b^5*sin(c/2 + (d*x)/2) + 4*a*b^4 
*cos(c/2 + (d*x)/2) + 4*a^4*b*sin(c/2 + (d*x)/2) + 5*a^3*b^2*cos(c/2 + (d* 
x)/2) + 12*a^2*b^3*sin(c/2 + (d*x)/2)))*(a^2 + b^2)^(1/2)*1i)/((a^3*d)/2 - 
 (a^3*d*cos(2*c + 2*d*x))/2)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.08 \[ \int \frac {\csc ^3(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {4 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a i -b i}{\sqrt {a^{2}+b^{2}}}\right ) \sin \left (d x +c \right )^{2} b i -\cos \left (d x +c \right ) a^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b^{2}+2 \sin \left (d x +c \right ) a b}{2 \sin \left (d x +c \right )^{2} a^{3} d} \] Input:

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

Output:

(4*sqrt(a**2 + b**2)*atan((tan((c + d*x)/2)*a*i - b*i)/sqrt(a**2 + b**2))* 
sin(c + d*x)**2*b*i - cos(c + d*x)*a**2 + log(tan((c + d*x)/2))*sin(c + d* 
x)**2*a**2 + 2*log(tan((c + d*x)/2))*sin(c + d*x)**2*b**2 + 2*sin(c + d*x) 
*a*b)/(2*sin(c + d*x)**2*a**3*d)