\(\int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx\) [233]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 32, antiderivative size = 105 \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^3 A \text {arctanh}(\cos (c+d x))}{4 d}-\frac {2 a^3 A \cot ^3(c+d x)}{3 d}-\frac {a^3 A \cot ^5(c+d x)}{5 d}+\frac {a^3 A \cot (c+d x) \csc (c+d x)}{4 d}-\frac {a^3 A \cot (c+d x) \csc ^3(c+d x)}{2 d} \]

[Out]

1/4*a^3*A*arctanh(cos(d*x+c))/d-2/3*a^3*A*cot(d*x+c)^3/d-1/5*a^3*A*cot(d*x+c)^5/d+1/4*a^3*A*cot(d*x+c)*csc(d*x
+c)/d-1/2*a^3*A*cot(d*x+c)*csc(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3029, 2788, 3852, 8, 3853, 3855} \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^3 A \text {arctanh}(\cos (c+d x))}{4 d}-\frac {a^3 A \cot ^5(c+d x)}{5 d}-\frac {2 a^3 A \cot ^3(c+d x)}{3 d}-\frac {a^3 A \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {a^3 A \cot (c+d x) \csc (c+d x)}{4 d} \]

[In]

Int[Csc[c + d*x]^6*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

(a^3*A*ArcTanh[Cos[c + d*x]])/(4*d) - (2*a^3*A*Cot[c + d*x]^3)/(3*d) - (a^3*A*Cot[c + d*x]^5)/(5*d) + (a^3*A*C
ot[c + d*x]*Csc[c + d*x])/(4*d) - (a^3*A*Cot[c + d*x]*Csc[c + d*x]^3)/(2*d)

Rule 8

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

Rule 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a,
 b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 3029

Int[sin[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[a^n*c^n, Int[Tan[e + f*x]^p*(a + b*Sin[e + f*x])^(m - n), x], x] /; FreeQ[{a,
b, c, d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[p + 2*n, 0] && IntegerQ[n]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

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

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \left (a^3 A^3\right ) \int \frac {\cot ^6(c+d x)}{(A-A \sin (c+d x))^2} \, dx \\ & = \frac {a^3 \int \left (-A^4 \csc ^2(c+d x)-2 A^4 \csc ^3(c+d x)+2 A^4 \csc ^5(c+d x)+A^4 \csc ^6(c+d x)\right ) \, dx}{A^3} \\ & = -\left (\left (a^3 A\right ) \int \csc ^2(c+d x) \, dx\right )+\left (a^3 A\right ) \int \csc ^6(c+d x) \, dx-\left (2 a^3 A\right ) \int \csc ^3(c+d x) \, dx+\left (2 a^3 A\right ) \int \csc ^5(c+d x) \, dx \\ & = \frac {a^3 A \cot (c+d x) \csc (c+d x)}{d}-\frac {a^3 A \cot (c+d x) \csc ^3(c+d x)}{2 d}-\left (a^3 A\right ) \int \csc (c+d x) \, dx+\frac {1}{2} \left (3 a^3 A\right ) \int \csc ^3(c+d x) \, dx+\frac {\left (a^3 A\right ) \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}-\frac {\left (a^3 A\right ) \text {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,\cot (c+d x)\right )}{d} \\ & = \frac {a^3 A \text {arctanh}(\cos (c+d x))}{d}-\frac {2 a^3 A \cot ^3(c+d x)}{3 d}-\frac {a^3 A \cot ^5(c+d x)}{5 d}+\frac {a^3 A \cot (c+d x) \csc (c+d x)}{4 d}-\frac {a^3 A \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {1}{4} \left (3 a^3 A\right ) \int \csc (c+d x) \, dx \\ & = \frac {a^3 A \text {arctanh}(\cos (c+d x))}{4 d}-\frac {2 a^3 A \cot ^3(c+d x)}{3 d}-\frac {a^3 A \cot ^5(c+d x)}{5 d}+\frac {a^3 A \cot (c+d x) \csc (c+d x)}{4 d}-\frac {a^3 A \cot (c+d x) \csc ^3(c+d x)}{2 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(268\) vs. \(2(105)=210\).

Time = 0.27 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.55 \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=a^3 A \left (\frac {7 \cot \left (\frac {1}{2} (c+d x)\right )}{30 d}+\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right )}{16 d}-\frac {19 \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{480 d}-\frac {\csc ^4\left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^4\left (\frac {1}{2} (c+d x)\right )}{160 d}+\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}-\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}-\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right )}{16 d}+\frac {\sec ^4\left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {7 \tan \left (\frac {1}{2} (c+d x)\right )}{30 d}+\frac {19 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{480 d}+\frac {\sec ^4\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{160 d}\right ) \]

[In]

Integrate[Csc[c + d*x]^6*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

a^3*A*((7*Cot[(c + d*x)/2])/(30*d) + Csc[(c + d*x)/2]^2/(16*d) - (19*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(480
*d) - Csc[(c + d*x)/2]^4/(32*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^4)/(160*d) + Log[Cos[(c + d*x)/2]]/(4*d)
- Log[Sin[(c + d*x)/2]]/(4*d) - Sec[(c + d*x)/2]^2/(16*d) + Sec[(c + d*x)/2]^4/(32*d) - (7*Tan[(c + d*x)/2])/(
30*d) + (19*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(480*d) + (Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2])/(160*d))

Maple [A] (verified)

Time = 1.85 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.15

method result size
parallelrisch \(-\frac {a^{3} \left (\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )-\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {25 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {25 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-30 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+30 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+40 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) A}{160 d}\) \(121\)
derivativedivides \(\frac {A \,a^{3} \cot \left (d x +c \right )-2 A \,a^{3} \left (-\frac {\csc \left (d x +c \right ) \cot \left (d x +c \right )}{2}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 A \,a^{3} \left (\left (-\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \csc \left (d x +c \right )}{8}\right ) \cot \left (d x +c \right )+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\csc ^{2}\left (d x +c \right )\right )}{15}\right ) \cot \left (d x +c \right )}{d}\) \(140\)
default \(\frac {A \,a^{3} \cot \left (d x +c \right )-2 A \,a^{3} \left (-\frac {\csc \left (d x +c \right ) \cot \left (d x +c \right )}{2}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 A \,a^{3} \left (\left (-\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \csc \left (d x +c \right )}{8}\right ) \cot \left (d x +c \right )+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\csc ^{2}\left (d x +c \right )\right )}{15}\right ) \cot \left (d x +c \right )}{d}\) \(140\)
risch \(-\frac {A \,a^{3} \left (-60 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{9 i \left (d x +c \right )}+240 i {\mathrm e}^{6 i \left (d x +c \right )}+90 \,{\mathrm e}^{7 i \left (d x +c \right )}-40 i {\mathrm e}^{4 i \left (d x +c \right )}+80 i {\mathrm e}^{2 i \left (d x +c \right )}-90 \,{\mathrm e}^{3 i \left (d x +c \right )}-28 i-15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d}+\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d}\) \(161\)

[In]

int(csc(d*x+c)^6*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/160*a^3*(cot(1/2*d*x+1/2*c)^5-tan(1/2*d*x+1/2*c)^5+5*cot(1/2*d*x+1/2*c)^4-5*tan(1/2*d*x+1/2*c)^4+25/3*cot(1
/2*d*x+1/2*c)^3-25/3*tan(1/2*d*x+1/2*c)^3-30*cot(1/2*d*x+1/2*c)+30*tan(1/2*d*x+1/2*c)+40*ln(tan(1/2*d*x+1/2*c)
))*A/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (95) = 190\).

Time = 0.26 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.91 \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {56 \, A a^{3} \cos \left (d x + c\right )^{5} - 80 \, A a^{3} \cos \left (d x + c\right )^{3} + 15 \, {\left (A a^{3} \cos \left (d x + c\right )^{4} - 2 \, A a^{3} \cos \left (d x + c\right )^{2} + A a^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 15 \, {\left (A a^{3} \cos \left (d x + c\right )^{4} - 2 \, A a^{3} \cos \left (d x + c\right )^{2} + A a^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left (A a^{3} \cos \left (d x + c\right )^{3} + A a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(csc(d*x+c)^6*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/120*(56*A*a^3*cos(d*x + c)^5 - 80*A*a^3*cos(d*x + c)^3 + 15*(A*a^3*cos(d*x + c)^4 - 2*A*a^3*cos(d*x + c)^2 +
 A*a^3)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 15*(A*a^3*cos(d*x + c)^4 - 2*A*a^3*cos(d*x + c)^2 + A*a^3)*
log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 30*(A*a^3*cos(d*x + c)^3 + A*a^3*cos(d*x + c))*sin(d*x + c))/((d*c
os(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)**6*(a+a*sin(d*x+c))**3*(A-A*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.67 \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {15 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{3} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {120 \, A a^{3}}{\tan \left (d x + c\right )} - \frac {8 \, {\left (15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} + 3\right )} A a^{3}}{\tan \left (d x + c\right )^{5}}}{120 \, d} \]

[In]

integrate(csc(d*x+c)^6*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/120*(15*A*a^3*(2*(3*cos(d*x + c)^3 - 5*cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - 3*log(cos(d*x
 + c) + 1) + 3*log(cos(d*x + c) - 1)) - 60*A*a^3*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) - log(cos(d*x + c) + 1)
+ log(cos(d*x + c) - 1)) + 120*A*a^3/tan(d*x + c) - 8*(15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 + 3)*A*a^3/tan(d*
x + c)^5)/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.66 \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 25 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 90 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {274 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 90 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 25 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \]

[In]

integrate(csc(d*x+c)^6*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="giac")

[Out]

1/480*(3*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*A*a^3*tan(1/2*d*x + 1/2*c)^4 + 25*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 12
0*A*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 90*A*a^3*tan(1/2*d*x + 1/2*c) + (274*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 9
0*A*a^3*tan(1/2*d*x + 1/2*c)^4 - 25*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 15*A*a^3*tan(1/2*d*x + 1/2*c) - 3*A*a^3)/ta
n(1/2*d*x + 1/2*c)^5)/d

Mupad [B] (verification not implemented)

Time = 12.87 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.32 \[ \int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {A\,a^3\,\left (3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-15\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-25\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+90\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-90\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+25\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+120\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}{480\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \]

[In]

int(((A - A*sin(c + d*x))*(a + a*sin(c + d*x))^3)/sin(c + d*x)^6,x)

[Out]

-(A*a^3*(3*cos(c/2 + (d*x)/2)^10 - 3*sin(c/2 + (d*x)/2)^10 - 15*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^9 + 15*c
os(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2) - 25*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8 + 90*cos(c/2 + (d*x)/2)^
4*sin(c/2 + (d*x)/2)^6 - 90*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4 + 25*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x
)/2)^2 + 120*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5))/(480*d*cos
(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5)