\(\int \frac {\sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [231]

Optimal result

Integrand size = 21, antiderivative size = 107 \[ \int \frac {\sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\left (2 a^2+b^2\right ) x}{2 b^3}-\frac {2 a^3 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d} \] Output:

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

Mathematica [A] (verified)

Time = 1.33 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.91 \[ \int \frac {\sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \left (2 a^2+b^2\right ) (c+d x)-\frac {8 a^3 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+4 a b \cos (c+d x)-b^2 \sin (2 (c+d x))}{4 b^3 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.16, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3272, 3042, 3502, 3042, 3214, 3042, 3139, 1083, 217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3272

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

\(\displaystyle \frac {\frac {\frac {x \left (2 a^2+b^2\right )}{b}-\frac {4 a^3 \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b}+\frac {2 a \cos (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {\frac {8 a^3 \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d}+\frac {x \left (2 a^2+b^2\right )}{b}}{b}+\frac {2 a \cos (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {\frac {x \left (2 a^2+b^2\right )}{b}-\frac {4 a^3 \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}}{b}+\frac {2 a \cos (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\)

Input:

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

Output:

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

Defintions of rubi rules used

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 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 3272

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* 
x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m 
 + n))   Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d 
*(m + n) + b^2*(b*c*(m - 2) + a*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 
 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m 
] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[a, 0] 
&& NeQ[c, 0])))
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.36

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{2}}{2}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a b -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}}{2}+a b \right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (2 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}-\frac {2 a^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{3} \sqrt {a^{2}-b^{2}}}}{d}\)	\(146\)
default	\(\frac {\frac {\frac {2 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{2}}{2}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a b -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}}{2}+a b \right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (2 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}-\frac {2 a^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{3} \sqrt {a^{2}-b^{2}}}}{d}\)	\(146\)
risch	\(\frac {x \,a^{2}}{b^{3}}+\frac {x}{2 b}+\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}+\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}+\frac {i a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}-\frac {i a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}-\frac {\sin \left (2 d x +2 c \right )}{4 b d}\)	\(215\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 359, normalized size of antiderivative = 3.36 \[ \int \frac {\sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} a^{3} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} d x + {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}, \frac {2 \, \sqrt {a^{2} - b^{2}} a^{3} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} d x - {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(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
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.41 \[ \int \frac {\sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {4 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{3}}{\sqrt {a^{2} - b^{2}} b^{3}} - \frac {{\left (2 \, a^{2} + b^{2}\right )} {\left (d x + c\right )}}{b^{3}} - \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \] Input:

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

Output:

-1/2*(4*(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/(sqrt(a^2 - b^2)*b^3) - (2*a^2 + b^2)*( 
d*x + c)/b^3 - 2*(b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x + 1/2*c)^2 - 
b*tan(1/2*d*x + 1/2*c) + 2*a)/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*b^2))/d
 

Mupad [B] (verification not implemented)

Time = 37.34 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.86 \[ \int \frac {\sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d}-\frac {\sin \left (2\,c+2\,d\,x\right )}{4\,b\,d}+\frac {2\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^3\,d}+\frac {a\,\cos \left (c+d\,x\right )}{b^2\,d}+\frac {a^3\,\mathrm {atan}\left (\frac {\left (-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2\right )\,1{}\mathrm {i}}{\sqrt {b^2-a^2}\,\left (a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )\,2{}\mathrm {i}}{b^3\,d\,\sqrt {b^2-a^2}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.56 \[ \int \frac {\sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{3}-\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{2}+\cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4}+2 \cos \left (d x +c \right ) a^{3} b -2 \cos \left (d x +c \right ) a \,b^{3}+2 a^{4} c +2 a^{4} d x -a^{2} b^{2} c -a^{2} b^{2} d x -b^{4} c -b^{4} d x}{2 b^{3} d \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

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