3.1.98 \(\int \frac {\csc (c+d x)}{(a+b \sin ^2(c+d x))^2} \, dx\) [98]

3.1.98.1 Optimal result
3.1.98.2 Mathematica [C] (verified)
3.1.98.3 Rubi [A] (verified)
3.1.98.4 Maple [A] (verified)
3.1.98.5 Fricas [B] (verification not implemented)
3.1.98.6 Sympy [F]
3.1.98.7 Maxima [A] (verification not implemented)
3.1.98.8 Giac [B] (verification not implemented)
3.1.98.9 Mupad [B] (verification not implemented)

3.1.98.1 Optimal result

Integrand size = 21, antiderivative size = 103 \[ \int \frac {\csc (c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {\sqrt {b} (3 a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 (a+b)^{3/2} d}+\frac {b \cos (c+d x)}{2 a (a+b) d \left (a+b-b \cos ^2(c+d x)\right )} \]

output
-arctanh(cos(d*x+c))/a^2/d+1/2*b*cos(d*x+c)/a/(a+b)/d/(a+b-b*cos(d*x+c)^2) 
+1/2*(3*a+2*b)*arctanh(cos(d*x+c)*b^(1/2)/(a+b)^(1/2))*b^(1/2)/a^2/(a+b)^( 
3/2)/d
 
3.1.98.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.15 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.88 \[ \int \frac {\csc (c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {\sqrt {b} (3 a+2 b) \arctan \left (\frac {\sqrt {b}-i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )}{(-a-b)^{3/2}}+\frac {\sqrt {b} (3 a+2 b) \arctan \left (\frac {\sqrt {b}+i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )}{(-a-b)^{3/2}}+2 \left (\frac {a b \cos (c+d x)}{(a+b) (2 a+b-b \cos (2 (c+d x)))}-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{2 a^2 d} \]

input
Integrate[Csc[c + d*x]/(a + b*Sin[c + d*x]^2)^2,x]
 
output
((Sqrt[b]*(3*a + 2*b)*ArcTan[(Sqrt[b] - I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[- 
a - b]])/(-a - b)^(3/2) + (Sqrt[b]*(3*a + 2*b)*ArcTan[(Sqrt[b] + I*Sqrt[a] 
*Tan[(c + d*x)/2])/Sqrt[-a - b]])/(-a - b)^(3/2) + 2*((a*b*Cos[c + d*x])/( 
(a + b)*(2*a + b - b*Cos[2*(c + d*x)])) - Log[Cos[(c + d*x)/2]] + Log[Sin[ 
(c + d*x)/2]]))/(2*a^2*d)
 
3.1.98.3 Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3665, 316, 25, 397, 219, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3665

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 221

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

input
Int[Csc[c + d*x]/(a + b*Sin[c + d*x]^2)^2,x]
 
output
-((((2*(a + b)*ArcTanh[Cos[c + d*x]])/a - (Sqrt[b]*(3*a + 2*b)*ArcTanh[(Sq 
rt[b]*Cos[c + d*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/(2*a*(a + b)) - (b*Cos[ 
c + d*x])/(2*a*(a + b)*(a + b - b*Cos[c + d*x]^2)))/d)
 

3.1.98.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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 3665
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f 
Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + 
 f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
 
3.1.98.4 Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.04

method result size
derivativedivides \(\frac {-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{2 a^{2}}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a^{2}}+\frac {b \left (\frac {a \cos \left (d x +c \right )}{2 \left (a +b \right ) \left (a +b -b \left (\cos ^{2}\left (d x +c \right )\right )\right )}+\frac {\left (3 a +2 b \right ) \operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{2}}}{d}\) \(107\)
default \(\frac {-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{2 a^{2}}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a^{2}}+\frac {b \left (\frac {a \cos \left (d x +c \right )}{2 \left (a +b \right ) \left (a +b -b \left (\cos ^{2}\left (d x +c \right )\right )\right )}+\frac {\left (3 a +2 b \right ) \operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{2}}}{d}\) \(107\)
risch \(-\frac {b \left ({\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}\right )}{a \left (a +b \right ) d \left (b \,{\mathrm e}^{4 i \left (d x +c \right )}-4 a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{2}}+\frac {3 i \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}\, {\mathrm e}^{i \left (d x +c \right )}}{b}+1\right )}{4 \left (a +b \right )^{2} d a}+\frac {i \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}\, {\mathrm e}^{i \left (d x +c \right )}}{b}+1\right ) b}{2 \left (a +b \right )^{2} d \,a^{2}}-\frac {3 i \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}\, {\mathrm e}^{i \left (d x +c \right )}}{b}+1\right )}{4 \left (a +b \right )^{2} d a}-\frac {i \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}\, {\mathrm e}^{i \left (d x +c \right )}}{b}+1\right ) b}{2 \left (a +b \right )^{2} d \,a^{2}}\) \(343\)

input
int(csc(d*x+c)/(a+b*sin(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
 
output
1/d*(-1/2/a^2*ln(1+cos(d*x+c))+1/2/a^2*ln(cos(d*x+c)-1)+1/a^2*b*(1/2*a/(a+ 
b)*cos(d*x+c)/(a+b-b*cos(d*x+c)^2)+1/2*(3*a+2*b)/(a+b)/((a+b)*b)^(1/2)*arc 
tanh(b*cos(d*x+c)/((a+b)*b)^(1/2))))
 
3.1.98.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 208 vs. \(2 (94) = 188\).

Time = 0.31 (sec) , antiderivative size = 455, normalized size of antiderivative = 4.42 \[ \int \frac {\csc (c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\left [-\frac {2 \, a b \cos \left (d x + c\right ) - {\left ({\left (3 \, a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - 5 \, a b - 2 \, b^{2}\right )} \sqrt {\frac {b}{a + b}} \log \left (\frac {b \cos \left (d x + c\right )^{2} + 2 \, {\left (a + b\right )} \sqrt {\frac {b}{a + b}} \cos \left (d x + c\right ) + a + b}{b \cos \left (d x + c\right )^{2} - a - b}\right ) + 2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{3} b + a^{2} b^{2}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} d\right )}}, -\frac {a b \cos \left (d x + c\right ) + {\left ({\left (3 \, a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - 5 \, a b - 2 \, b^{2}\right )} \sqrt {-\frac {b}{a + b}} \arctan \left (\sqrt {-\frac {b}{a + b}} \cos \left (d x + c\right )\right ) + {\left ({\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left ({\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left ({\left (a^{3} b + a^{2} b^{2}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} d\right )}}\right ] \]

input
integrate(csc(d*x+c)/(a+b*sin(d*x+c)^2)^2,x, algorithm="fricas")
 
output
[-1/4*(2*a*b*cos(d*x + c) - ((3*a*b + 2*b^2)*cos(d*x + c)^2 - 3*a^2 - 5*a* 
b - 2*b^2)*sqrt(b/(a + b))*log((b*cos(d*x + c)^2 + 2*(a + b)*sqrt(b/(a + b 
))*cos(d*x + c) + a + b)/(b*cos(d*x + c)^2 - a - b)) + 2*((a*b + b^2)*cos( 
d*x + c)^2 - a^2 - 2*a*b - b^2)*log(1/2*cos(d*x + c) + 1/2) - 2*((a*b + b^ 
2)*cos(d*x + c)^2 - a^2 - 2*a*b - b^2)*log(-1/2*cos(d*x + c) + 1/2))/((a^3 
*b + a^2*b^2)*d*cos(d*x + c)^2 - (a^4 + 2*a^3*b + a^2*b^2)*d), -1/2*(a*b*c 
os(d*x + c) + ((3*a*b + 2*b^2)*cos(d*x + c)^2 - 3*a^2 - 5*a*b - 2*b^2)*sqr 
t(-b/(a + b))*arctan(sqrt(-b/(a + b))*cos(d*x + c)) + ((a*b + b^2)*cos(d*x 
 + c)^2 - a^2 - 2*a*b - b^2)*log(1/2*cos(d*x + c) + 1/2) - ((a*b + b^2)*co 
s(d*x + c)^2 - a^2 - 2*a*b - b^2)*log(-1/2*cos(d*x + c) + 1/2))/((a^3*b + 
a^2*b^2)*d*cos(d*x + c)^2 - (a^4 + 2*a^3*b + a^2*b^2)*d)]
 
3.1.98.6 Sympy [F]

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

input
integrate(csc(d*x+c)/(a+b*sin(d*x+c)**2)**2,x)
 
output
Integral(csc(c + d*x)/(a + b*sin(c + d*x)**2)**2, x)
 
3.1.98.7 Maxima [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.45 \[ \int \frac {\csc (c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {2 \, b \cos \left (d x + c\right )}{a^{3} + 2 \, a^{2} b + a b^{2} - {\left (a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{2}} - \frac {{\left (3 \, a b + 2 \, b^{2}\right )} \log \left (\frac {b \cos \left (d x + c\right ) - \sqrt {{\left (a + b\right )} b}}{b \cos \left (d x + c\right ) + \sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{3} + a^{2} b\right )} \sqrt {{\left (a + b\right )} b}} - \frac {2 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {2 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2}}}{4 \, d} \]

input
integrate(csc(d*x+c)/(a+b*sin(d*x+c)^2)^2,x, algorithm="maxima")
 
output
1/4*(2*b*cos(d*x + c)/(a^3 + 2*a^2*b + a*b^2 - (a^2*b + a*b^2)*cos(d*x + c 
)^2) - (3*a*b + 2*b^2)*log((b*cos(d*x + c) - sqrt((a + b)*b))/(b*cos(d*x + 
 c) + sqrt((a + b)*b)))/((a^3 + a^2*b)*sqrt((a + b)*b)) - 2*log(cos(d*x + 
c) + 1)/a^2 + 2*log(cos(d*x + c) - 1)/a^2)/d
 
3.1.98.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (94) = 188\).

Time = 0.43 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.39 \[ \int \frac {\csc (c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {\frac {{\left (3 \, a b + 2 \, b^{2}\right )} \arctan \left (\frac {b \cos \left (d x + c\right ) + a + b}{\sqrt {-a b - b^{2}} \cos \left (d x + c\right ) + \sqrt {-a b - b^{2}}}\right )}{{\left (a^{3} + a^{2} b\right )} \sqrt {-a b - b^{2}}} - \frac {2 \, {\left (a b - \frac {a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {2 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{{\left (a^{3} + a^{2} b\right )} {\left (a - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {4 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}} - \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2}}}{2 \, d} \]

input
integrate(csc(d*x+c)/(a+b*sin(d*x+c)^2)^2,x, algorithm="giac")
 
output
-1/2*((3*a*b + 2*b^2)*arctan((b*cos(d*x + c) + a + b)/(sqrt(-a*b - b^2)*co 
s(d*x + c) + sqrt(-a*b - b^2)))/((a^3 + a^2*b)*sqrt(-a*b - b^2)) - 2*(a*b 
- a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*b^2*(cos(d*x + c) - 1)/(co 
s(d*x + c) + 1))/((a^3 + a^2*b)*(a - 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) 
+ 1) - 4*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + a*(cos(d*x + c) - 1)^2/ 
(cos(d*x + c) + 1)^2)) - log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) 
/a^2)/d
 
3.1.98.9 Mupad [B] (verification not implemented)

Time = 14.48 (sec) , antiderivative size = 2039, normalized size of antiderivative = 19.80 \[ \int \frac {\csc (c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]

input
int(1/(sin(c + d*x)*(a + b*sin(c + d*x)^2)^2),x)
 
output
(atan(((((cos(c + d*x)*(20*a*b^4 + 8*b^5 + 13*a^2*b^3))/(2*(2*a^3*b + a^4 
+ a^2*b^2)) + ((b*(a + b)^3)^(1/2)*((2*a^4*b^4 + 6*a^5*b^3 + 4*a^6*b^2)/(2 
*a^4*b + a^5 + a^3*b^2) - (cos(c + d*x)*(b*(a + b)^3)^(1/2)*(3*a + 2*b)*(3 
2*a^4*b^5 + 80*a^5*b^4 + 64*a^6*b^3 + 16*a^7*b^2))/(8*(2*a^3*b + a^4 + a^2 
*b^2)*(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 
^5 + a^2*b^3 + 3*a^3*b^2)))*(b*(a + b)^3)^(1/2)*(3*a + 2*b)*1i)/(4*(3*a^4* 
b + a^5 + a^2*b^3 + 3*a^3*b^2)) + (((cos(c + d*x)*(20*a*b^4 + 8*b^5 + 13*a 
^2*b^3))/(2*(2*a^3*b + a^4 + a^2*b^2)) - ((b*(a + b)^3)^(1/2)*((2*a^4*b^4 
+ 6*a^5*b^3 + 4*a^6*b^2)/(2*a^4*b + a^5 + a^3*b^2) + (cos(c + d*x)*(b*(a + 
 b)^3)^(1/2)*(3*a + 2*b)*(32*a^4*b^5 + 80*a^5*b^4 + 64*a^6*b^3 + 16*a^7*b^ 
2))/(8*(2*a^3*b + a^4 + a^2*b^2)*(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^5 + a^2*b^3 + 3*a^3*b^2)))*(b*(a + b)^3)^(1/2) 
*(3*a + 2*b)*1i)/(4*(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2)))/(((3*a*b^3)/2 
+ b^4)/(2*a^4*b + a^5 + a^3*b^2) - (((cos(c + d*x)*(20*a*b^4 + 8*b^5 + 13* 
a^2*b^3))/(2*(2*a^3*b + a^4 + a^2*b^2)) + ((b*(a + b)^3)^(1/2)*((2*a^4*b^4 
 + 6*a^5*b^3 + 4*a^6*b^2)/(2*a^4*b + a^5 + a^3*b^2) - (cos(c + d*x)*(b*(a 
+ b)^3)^(1/2)*(3*a + 2*b)*(32*a^4*b^5 + 80*a^5*b^4 + 64*a^6*b^3 + 16*a^7*b 
^2))/(8*(2*a^3*b + a^4 + a^2*b^2)*(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^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)) + (((cos(c + d...