3.14.40 \(\int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [1340]

3.14.40.1 Optimal result

Integrand size = 27, antiderivative size = 133 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^2 x}{b \left (a^2-b^2\right )}+\frac {b x}{a^2-b^2}+\frac {2 a^3 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}+\frac {a \sec (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \tan (c+d x)}{\left (a^2-b^2\right ) d} \]

output

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

3.14.40.2 Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.14 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 a^3 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {-\left (\left (a^2-b^2\right ) (c+d x) \cos (c+d x)\right )+b (a-b \sin (c+d x))}{(a-b) (a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}}{b d} \]

input

Integrate[(Sin[c + d*x]*Tan[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
 

output

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

3.14.40.3 Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 3381, 3042, 3086, 24, 3214, 3042, 3139, 1083, 217, 3954, 24}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3381

\(\displaystyle -\frac {a^2 \int \frac {\sin (c+d x)}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {b \int \tan ^2(c+d x)dx}{a^2-b^2}+\frac {a \int \sec (c+d x) \tan (c+d x)dx}{a^2-b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a^2 \int \frac {\sin (c+d x)}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {b \int \tan (c+d x)^2dx}{a^2-b^2}+\frac {a \int \sec (c+d x) \tan (c+d x)dx}{a^2-b^2}\)

\(\Big \downarrow \) 3086

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

\(\displaystyle -\frac {a^2 \left (\frac {4 a \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d}+\frac {x}{b}\right )}{a^2-b^2}-\frac {b \int \tan (c+d x)^2dx}{a^2-b^2}+\frac {a \sec (c+d x)}{d \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 3954

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

\(\Big \downarrow \) 24

\(\displaystyle -\frac {a^2 \left (\frac {x}{b}-\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}\right )}{a^2-b^2}-\frac {b \left (\frac {\tan (c+d x)}{d}-x\right )}{a^2-b^2}+\frac {a \sec (c+d x)}{d \left (a^2-b^2\right )}\)

input

Int[(Sin[c + d*x]*Tan[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
 

output

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

3.14.40.3.1 Defintions of rubi rules used

rule 24

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

rule 217

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

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

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

rule 3086

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_.), x_Symbol] :> Simp[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 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3381

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( 
n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a*(d^2/(a^2 
- b^2))   Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-Simp[ 
b*(d/(a^2 - b^2))   Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] 
 - Simp[a^2*(d^2/(g^2*(a^2 - b^2)))   Int[(g*Cos[e + f*x])^(p + 2)*((d*Sin[ 
e + f*x])^(n - 2)/(a + b*Sin[e + f*x])), x], x]) /; FreeQ[{a, b, d, e, f, g 
}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1 
]
 

rule 3954

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d 
*x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2   Int[(b*Tan[c + d*x])^(n - 2), x] 
, x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
 

3.14.40.4 Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(\frac {-\frac {16}{\left (16 a +16 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {16}{\left (16 a -16 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}+\frac {2 a^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right ) \left (a +b \right ) b \sqrt {a^{2}-b^{2}}}}{d}\)	\(130\)
default	\(\frac {-\frac {16}{\left (16 a +16 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {16}{\left (16 a -16 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}+\frac {2 a^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right ) \left (a +b \right ) b \sqrt {a^{2}-b^{2}}}}{d}\)	\(130\)
risch	\(-\frac {x}{b}+\frac {2 i \left (i a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{d \left (-a^{2}+b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {i a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}-\frac {i a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}\)	\(223\)

input

int(sec(d*x+c)^2*sin(d*x+c)^3/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
 

output

1/d*(-16/(16*a+16*b)/(tan(1/2*d*x+1/2*c)-1)+16/(16*a-16*b)/(tan(1/2*d*x+1/ 
2*c)+1)-2/b*arctan(tan(1/2*d*x+1/2*c))+2/(a-b)/(a+b)*a^3/b/(a^2-b^2)^(1/2) 
*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2)))
 

3.14.40.5 Fricas [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.77 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {\sqrt {-a^{2} + b^{2}} a^{3} \cos \left (d x + c\right ) \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{3} b - 2 \, a b^{3} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d x \cos \left (d x + c\right ) - 2 \, {\left (a^{2} b^{2} - b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right )}, -\frac {\sqrt {a^{2} - b^{2}} a^{3} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - a^{3} b + a b^{3} + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (a^{2} b^{2} - b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right )}\right ] \]

input

integrate(sec(d*x+c)^2*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas" 
)
 

output

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

3.14.40.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]

input

integrate(sec(d*x+c)**2*sin(d*x+c)**3/(a+b*sin(d*x+c)),x)
 

output

Timed out
 

3.14.40.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]

input

integrate(sec(d*x+c)^2*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima" 
)
 

output

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

3.14.40.8 Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.98 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{3}}{{\left (a^{2} b - b^{3}\right )} \sqrt {a^{2} - b^{2}}} - \frac {d x + c}{b} + \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}}{d} \]

input

integrate(sec(d*x+c)^2*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")
 

output

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

3.14.40.9 Mupad [B] (verification not implemented)

Time = 14.03 (sec) , antiderivative size = 1538, normalized size of antiderivative = 11.56 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]

input

int(sin(c + d*x)^3/(cos(c + d*x)^2*(a + b*sin(c + d*x))),x)
 

output

(a^5*cos(c + d*x) + a^5)/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^ 
2)) - (2*a^6*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b*d*(a^2 - b^2) 
*(a^4 + b^4 - 2*a^2*b^2)) - (b*(a^4*sin(c + d*x) - 6*a^4*cos(c + d*x)*atan 
(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))))/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 
 + b^4 - 2*a^2*b^2)) + (b^4*(a + a*cos(c + d*x)))/(d*cos(c + d*x)*(a^2 - b 
^2)*(a^4 + b^4 - 2*a^2*b^2)) - (b^2*(2*a^3*cos(c + d*x) + 2*a^3))/(d*cos(c 
 + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) + (b^3*(2*a^2*sin(c + d*x) - 
6*a^2*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))))/(d*cos(c 
+ d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (b^5*(sin(c + d*x) - 2*cos(c 
 + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))))/(d*cos(c + d*x)*(a^2 
 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (a^3*atan(((2*b^14*sin(c/2 + (d*x)/2)*( 
b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 2*a^14*sin(c/2 + (d*x)/2)*(b^6 
- a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 2*a^8*sin(c/2 + (d*x)/2)*(b^6 - a^6 
 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) - 6*a^3*b^11*cos(c/2 + (d*x)/2)*(b^6 - a^6 
 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 15*a^5*b^9*cos(c/2 + (d*x)/2)*(b^6 - a^6 
 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 20*a^7*b^7*cos(c/2 + (d*x)/2)*(b^6 - a^6 
 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 15*a^9*b^5*cos(c/2 + (d*x)/2)*(b^6 - a^6 
 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 6*a^11*b^3*cos(c/2 + (d*x)/2)*(b^6 - a^6 
 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 3*a^6*b^2*sin(c/2 + (d*x)/2)*(b^6 - a^6 
- 3*a^2*b^4 + 3*a^4*b^2)^(3/2) - 13*a^2*b^12*sin(c/2 + (d*x)/2)*(b^6 - ...