\(\int \frac {1}{(a+b \csc ^2(c+d x))^2} \, dx\) [6]

Optimal result

Integrand size = 14, antiderivative size = 92 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^2} \, dx=\frac {x}{a^2}+\frac {\sqrt {b} (3 a+2 b) \arctan \left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 (a+b)^{3/2} d}+\frac {b \cot (c+d x)}{2 a (a+b) d \left (a+b+b \cot ^2(c+d x)\right )} \] Output:

x/a^2+1/2*b^(1/2)*(3*a+2*b)*arctan(b^(1/2)*cot(d*x+c)/(a+b)^(1/2))/a^2/(a+ 
b)^(3/2)/d+1/2*b*cot(d*x+c)/a/(a+b)/d/(a+b+b*cot(d*x+c)^2)
 

Mathematica [A] (verified)

Time = 6.61 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.80 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^2} \, dx=-\frac {(-a-2 b+a \cos (2 (c+d x))) \csc ^4(c+d x) \left (-\sqrt {b} (3 a+2 b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {b}}\right ) (a+2 b-a \cos (2 (c+d x)))+\sqrt {a+b} \left (2 \left (a^2+3 a b+2 b^2\right ) (c+d x)-2 a (a+b) (c+d x) \cos (2 (c+d x))+a b \sin (2 (c+d x))\right )\right )}{8 a^2 (a+b)^{3/2} d \left (a+b \csc ^2(c+d x)\right )^2} \] Input:

Integrate[(a + b*Csc[c + d*x]^2)^(-2),x]
 

Output:

-1/8*((-a - 2*b + a*Cos[2*(c + d*x)])*Csc[c + d*x]^4*(-(Sqrt[b]*(3*a + 2*b 
)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[b]]*(a + 2*b - a*Cos[2*(c + d*x)] 
)) + Sqrt[a + b]*(2*(a^2 + 3*a*b + 2*b^2)*(c + d*x) - 2*a*(a + b)*(c + d*x 
)*Cos[2*(c + d*x)] + a*b*Sin[2*(c + d*x)])))/(a^2*(a + b)^(3/2)*d*(a + b*C 
sc[c + d*x]^2)^2)
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.22, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4616, 316, 397, 216, 218}

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.

\(\displaystyle \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a+b \sec \left (c+d x+\frac {\pi }{2}\right )^2\right )^2}dx\)

\(\Big \downarrow \) 4616

\(\displaystyle -\frac {\int \frac {1}{\left (\cot ^2(c+d x)+1\right ) \left (b \cot ^2(c+d x)+a+b\right )^2}d\cot (c+d x)}{d}\)

\(\Big \downarrow \) 316

\(\displaystyle -\frac {\frac {\int \frac {-b \cot ^2(c+d x)+2 a+b}{\left (\cot ^2(c+d x)+1\right ) \left (b \cot ^2(c+d x)+a+b\right )}d\cot (c+d x)}{2 a (a+b)}-\frac {b \cot (c+d x)}{2 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )}}{d}\)

\(\Big \downarrow \) 397

\(\displaystyle -\frac {\frac {\frac {2 (a+b) \int \frac {1}{\cot ^2(c+d x)+1}d\cot (c+d x)}{a}-\frac {b (3 a+2 b) \int \frac {1}{b \cot ^2(c+d x)+a+b}d\cot (c+d x)}{a}}{2 a (a+b)}-\frac {b \cot (c+d x)}{2 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )}}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {\frac {\frac {2 (a+b) \arctan (\cot (c+d x))}{a}-\frac {b (3 a+2 b) \int \frac {1}{b \cot ^2(c+d x)+a+b}d\cot (c+d x)}{a}}{2 a (a+b)}-\frac {b \cot (c+d x)}{2 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )}}{d}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {\frac {\frac {2 (a+b) \arctan (\cot (c+d x))}{a}-\frac {\sqrt {b} (3 a+2 b) \arctan \left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}}{2 a (a+b)}-\frac {b \cot (c+d x)}{2 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )}}{d}\)

Input:

Int[(a + b*Csc[c + d*x]^2)^(-2),x]
 

Output:

-((((2*(a + b)*ArcTan[Cot[c + d*x]])/a - (Sqrt[b]*(3*a + 2*b)*ArcTan[(Sqrt 
[b]*Cot[c + d*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/(2*a*(a + b)) - (b*Cot[c 
+ d*x])/(2*a*(a + b)*(a + b + b*Cot[c + d*x]^2)))/d)
 

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

rule 3042

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

rule 4616

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = 
FreeFactors[Tan[e + f*x], x]}, Simp[ff/f   Subst[Int[(a + b + b*ff^2*x^2)^p 
/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] 
&& NeQ[a + b, 0] && NeQ[p, -1]
 

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(\frac {-\frac {b \left (-\frac {a \tan \left (d x +c \right )}{2 \left (a +b \right ) \left (\tan \left (d x +c \right )^{2} a +\tan \left (d x +c \right )^{2} b +b \right )}+\frac {\left (3 a +2 b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {b \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {b \left (a +b \right )}}\right )}{a^{2}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{a^{2}}}{d}\)	\(101\)
default	\(\frac {-\frac {b \left (-\frac {a \tan \left (d x +c \right )}{2 \left (a +b \right ) \left (\tan \left (d x +c \right )^{2} a +\tan \left (d x +c \right )^{2} b +b \right )}+\frac {\left (3 a +2 b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {b \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {b \left (a +b \right )}}\right )}{a^{2}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{a^{2}}}{d}\)	\(101\)
risch	\(\frac {x}{a^{2}}-\frac {i b \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a \right )}{a^{2} \left (a +b \right ) d \left (-a \,{\mathrm e}^{4 i \left (d x +c \right )}+2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+4 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a \right )}+\frac {3 \sqrt {-b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-b \left (a +b \right )}+a +2 b}{a}\right )}{4 \left (a +b \right )^{2} d a}+\frac {\sqrt {-b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-b \left (a +b \right )}+a +2 b}{a}\right ) b}{2 \left (a +b \right )^{2} d \,a^{2}}-\frac {3 \sqrt {-b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-b \left (a +b \right )}-a -2 b}{a}\right )}{4 \left (a +b \right )^{2} d a}-\frac {\sqrt {-b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-b \left (a +b \right )}-a -2 b}{a}\right ) b}{2 \left (a +b \right )^{2} d \,a^{2}}\)	\(307\)

Input:

int(1/(a+b*csc(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (80) = 160\).

Time = 0.11 (sec) , antiderivative size = 492, normalized size of antiderivative = 5.35 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^2} \, dx=\left [\frac {8 \, {\left (a^{2} + a b\right )} d x \cos \left (d x + c\right )^{2} - 4 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 8 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + {\left ({\left (3 \, a^{2} + 2 \, a b\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - 5 \, a b - 2 \, b^{2}\right )} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} + 5 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{a^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} + a b\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right )}{8 \, {\left ({\left (a^{4} + a^{3} b\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} d\right )}}, \frac {4 \, {\left (a^{2} + a b\right )} d x \cos \left (d x + c\right )^{2} - 2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + {\left ({\left (3 \, a^{2} + 2 \, a b\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - 5 \, a b - 2 \, b^{2}\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{4 \, {\left ({\left (a^{4} + a^{3} b\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} d\right )}}\right ] \] Input:

integrate(1/(a+b*csc(d*x+c)^2)^2,x, algorithm="fricas")
 

Output:

[1/8*(8*(a^2 + a*b)*d*x*cos(d*x + c)^2 - 4*a*b*cos(d*x + c)*sin(d*x + c) - 
 8*(a^2 + 2*a*b + b^2)*d*x + ((3*a^2 + 2*a*b)*cos(d*x + c)^2 - 3*a^2 - 5*a 
*b - 2*b^2)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(d*x + c)^4 - 2 
*(a^2 + 5*a*b + 4*b^2)*cos(d*x + c)^2 + 4*((a^2 + 3*a*b + 2*b^2)*cos(d*x + 
 c)^3 - (a^2 + 2*a*b + b^2)*cos(d*x + c))*sqrt(-b/(a + b))*sin(d*x + c) + 
a^2 + 2*a*b + b^2)/(a^2*cos(d*x + c)^4 - 2*(a^2 + a*b)*cos(d*x + c)^2 + a^ 
2 + 2*a*b + b^2)))/((a^4 + a^3*b)*d*cos(d*x + c)^2 - (a^4 + 2*a^3*b + a^2* 
b^2)*d), 1/4*(4*(a^2 + a*b)*d*x*cos(d*x + c)^2 - 2*a*b*cos(d*x + c)*sin(d* 
x + c) - 4*(a^2 + 2*a*b + b^2)*d*x + ((3*a^2 + 2*a*b)*cos(d*x + c)^2 - 3*a 
^2 - 5*a*b - 2*b^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(d*x + c)^2 - 
 a - b)*sqrt(b/(a + b))/(b*cos(d*x + c)*sin(d*x + c))))/((a^4 + a^3*b)*d*c 
os(d*x + c)^2 - (a^4 + 2*a^3*b + a^2*b^2)*d)]
 

Sympy [F]

\[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^2} \, dx=\int \frac {1}{\left (a + b \csc ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \] Input:

integrate(1/(a+b*csc(d*x+c)**2)**2,x)
 

Output:

Integral((a + b*csc(c + d*x)**2)**(-2), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^2} \, dx=\frac {\frac {b \tan \left (d x + c\right )}{a^{2} b + a b^{2} + {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \tan \left (d x + c\right )^{2}} - \frac {{\left (3 \, a b + 2 \, b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{3} + a^{2} b\right )} \sqrt {{\left (a + b\right )} b}} + \frac {2 \, {\left (d x + c\right )}}{a^{2}}}{2 \, d} \] Input:

integrate(1/(a+b*csc(d*x+c)^2)^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.52 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^2} \, dx=-\frac {\frac {{\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (3 \, a b + 2 \, b^{2}\right )}}{{\left (a^{3} + a^{2} b\right )} \sqrt {a b + b^{2}}} - \frac {b \tan \left (d x + c\right )}{{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + b\right )} {\left (a^{2} + a b\right )}} - \frac {2 \, {\left (d x + c\right )}}{a^{2}}}{2 \, d} \] Input:

integrate(1/(a+b*csc(d*x+c)^2)^2,x, algorithm="giac")
 

Output:

-1/2*((pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c 
) + b*tan(d*x + c))/sqrt(a*b + b^2)))*(3*a*b + 2*b^2)/((a^3 + a^2*b)*sqrt( 
a*b + b^2)) - b*tan(d*x + c)/((a*tan(d*x + c)^2 + b*tan(d*x + c)^2 + b)*(a 
^2 + a*b)) - 2*(d*x + c)/a^2)/d
 

Mupad [B] (verification not implemented)

Time = 16.43 (sec) , antiderivative size = 1958, normalized size of antiderivative = 21.28 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

int(1/(a + b/sin(c + d*x)^2)^2,x)
 

Output:

(b*tan(c + d*x))/(2*a*d*(b + tan(c + d*x)^2*(a + b))*(a + b)) - (atan((((- 
b*(a + b)^3)^(1/2)*((tan(c + d*x)*(28*a*b^3 + 16*a^3*b + 4*a^4 + 8*b^4 + 3 
3*a^2*b^2))/(2*(a^2*b + a^3)) - (((2*a^6*b + 2*a^4*b^3 + 4*a^5*b^2)/(a^3*b 
 + a^4) - (tan(c + d*x)*(-b*(a + b)^3)^(1/2)*(3*a + 2*b)*(80*a^7*b + 16*a^ 
8 + 32*a^4*b^4 + 112*a^5*b^3 + 144*a^6*b^2))/(8*(a^2*b + a^3)*(3*a^4*b + a 
^5 + a^2*b^3 + 3*a^3*b^2)))*(-b*(a + b)^3)^(1/2)*(3*a + 2*b))/(4*(3*a^4*b 
+ a^5 + a^2*b^3 + 3*a^3*b^2)))*(3*a + 2*b)*1i)/(4*(3*a^4*b + a^5 + a^2*b^3 
 + 3*a^3*b^2)) + ((-b*(a + b)^3)^(1/2)*((tan(c + d*x)*(28*a*b^3 + 16*a^3*b 
 + 4*a^4 + 8*b^4 + 33*a^2*b^2))/(2*(a^2*b + a^3)) + (((2*a^6*b + 2*a^4*b^3 
 + 4*a^5*b^2)/(a^3*b + a^4) + (tan(c + d*x)*(-b*(a + b)^3)^(1/2)*(3*a + 2* 
b)*(80*a^7*b + 16*a^8 + 32*a^4*b^4 + 112*a^5*b^3 + 144*a^6*b^2))/(8*(a^2*b 
 + a^3)*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))*(-b*(a + b)^3)^(1/2)*(3*a 
+ 2*b))/(4*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))*(3*a + 2*b)*1i)/(4*(3*a 
^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))/(((7*a*b^2)/2 + 3*a^2*b + b^3)/(a^3*b 
+ a^4) - ((-b*(a + b)^3)^(1/2)*((tan(c + d*x)*(28*a*b^3 + 16*a^3*b + 4*a^4 
 + 8*b^4 + 33*a^2*b^2))/(2*(a^2*b + a^3)) - (((2*a^6*b + 2*a^4*b^3 + 4*a^5 
*b^2)/(a^3*b + a^4) - (tan(c + d*x)*(-b*(a + b)^3)^(1/2)*(3*a + 2*b)*(80*a 
^7*b + 16*a^8 + 32*a^4*b^4 + 112*a^5*b^3 + 144*a^6*b^2))/(8*(a^2*b + a^3)* 
(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))*(-b*(a + b)^3)^(1/2)*(3*a + 2*b))/ 
(4*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))*(3*a + 2*b))/(4*(3*a^4*b + a...
 

Reduce [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 1989, normalized size of antiderivative = 21.62 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

int(1/(a+b*csc(d*x+c)^2)^2,x)
 

Output:

(6*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan( 
(tan((c + d*x)/2)*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*sin( 
c + d*x)**2*a**2 + 4*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(a)*sqrt(a + b 
) + 2*a + b)*atan((tan((c + d*x)/2)*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) 
 + 2*a + b)))*sin(c + d*x)**2*a*b + 6*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*s 
qrt(a)*sqrt(a + b) + 2*a + b)*atan((tan((c + d*x)/2)*b)/(sqrt(b)*sqrt(2*sq 
rt(a)*sqrt(a + b) + 2*a + b)))*a*b + 4*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2* 
sqrt(a)*sqrt(a + b) + 2*a + b)*atan((tan((c + d*x)/2)*b)/(sqrt(b)*sqrt(2*s 
qrt(a)*sqrt(a + b) + 2*a + b)))*b**2 - 6*sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b 
) + 2*a + b)*atan((tan((c + d*x)/2)*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) 
 + 2*a + b)))*sin(c + d*x)**2*a**3 - 10*sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) 
 + 2*a + b)*atan((tan((c + d*x)/2)*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) 
+ 2*a + b)))*sin(c + d*x)**2*a**2*b - 4*sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) 
 + 2*a + b)*atan((tan((c + d*x)/2)*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) 
+ 2*a + b)))*sin(c + d*x)**2*a*b**2 - 6*sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) 
 + 2*a + b)*atan((tan((c + d*x)/2)*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) 
+ 2*a + b)))*a**2*b - 10*sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*ata 
n((tan((c + d*x)/2)*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*a* 
b**2 - 4*sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((tan((c + d*x) 
/2)*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*b**3 + 2*cos(c ...