\(\int \csc ^5(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx\) [232]

Optimal result

Integrand size = 32, antiderivative size = 86 \[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {5 a^3 A \text {arctanh}(\cos (c+d x))}{8 d}-\frac {2 a^3 A \cot ^3(c+d x)}{3 d}-\frac {3 a^3 A \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a^3 A \cot (c+d x) \csc ^3(c+d x)}{4 d} \]

[Out]

5/8*a^3*A*arctanh(cos(d*x+c))/d-2/3*a^3*A*cot(d*x+c)^3/d-3/8*a^3*A*cot(d*x+c)*csc(d*x+c)/d-1/4*a^3*A*cot(d*x+c
)*csc(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3045, 3855, 3852, 8, 3853} \[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {5 a^3 A \text {arctanh}(\cos (c+d x))}{8 d}-\frac {2 a^3 A \cot ^3(c+d x)}{3 d}-\frac {a^3 A \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3 a^3 A \cot (c+d x) \csc (c+d x)}{8 d} \]

[In]

Int[Csc[c + d*x]^5*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

(5*a^3*A*ArcTanh[Cos[c + d*x]])/(8*d) - (2*a^3*A*Cot[c + d*x]^3)/(3*d) - (3*a^3*A*Cot[c + d*x]*Csc[c + d*x])/(
8*d) - (a^3*A*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d)

Rule 8

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

Rule 3045

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && 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}& = \int \left (-a^3 A \csc (c+d x)-2 a^3 A \csc ^2(c+d x)+2 a^3 A \csc ^4(c+d x)+a^3 A \csc ^5(c+d x)\right ) \, dx \\ & = -\left (\left (a^3 A\right ) \int \csc (c+d x) \, dx\right )+\left (a^3 A\right ) \int \csc ^5(c+d x) \, dx-\left (2 a^3 A\right ) \int \csc ^2(c+d x) \, dx+\left (2 a^3 A\right ) \int \csc ^4(c+d x) \, dx \\ & = \frac {a^3 A \text {arctanh}(\cos (c+d x))}{d}-\frac {a^3 A \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {1}{4} \left (3 a^3 A\right ) \int \csc ^3(c+d x) \, dx+\frac {\left (2 a^3 A\right ) \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}-\frac {\left (2 a^3 A\right ) \text {Subst}\left (\int \left (1+x^2\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 {3 a^3 A \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a^3 A \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {1}{8} \left (3 a^3 A\right ) \int \csc (c+d x) \, dx \\ & = \frac {5 a^3 A \text {arctanh}(\cos (c+d x))}{8 d}-\frac {2 a^3 A \cot ^3(c+d x)}{3 d}-\frac {3 a^3 A \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a^3 A \cot (c+d x) \csc ^3(c+d x)}{4 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(210\) vs. \(2(86)=172\).

Time = 0.23 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.44 \[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=a^3 A \left (\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{3 d}-\frac {3 \csc ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{12 d}-\frac {\csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}+\frac {5 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {5 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {3 \sec ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {\sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}-\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{3 d}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{12 d}\right ) \]

[In]

Integrate[Csc[c + d*x]^5*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

a^3*A*(Cot[(c + d*x)/2]/(3*d) - (3*Csc[(c + d*x)/2]^2)/(32*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(12*d) -
Csc[(c + d*x)/2]^4/(64*d) + (5*Log[Cos[(c + d*x)/2]])/(8*d) - (5*Log[Sin[(c + d*x)/2]])/(8*d) + (3*Sec[(c + d
*x)/2]^2)/(32*d) + Sec[(c + d*x)/2]^4/(64*d) - Tan[(c + d*x)/2]/(3*d) + (Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/
(12*d))

Maple [A] (verified)

Time = 1.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.33

method	result	size
derivativedivides	\(\frac {-A \,a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+2 A \,a^{3} \cot \left (d x +c \right )+2 A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{3}\right ) \cot \left (d x +c \right )+A \,a^{3} \left (\left (-\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \csc \left (d x +c \right )}{8}\right ) \cot \left (d x +c \right )+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\)	\(114\)
default	\(\frac {-A \,a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+2 A \,a^{3} \cot \left (d x +c \right )+2 A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{3}\right ) \cot \left (d x +c \right )+A \,a^{3} \left (\left (-\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \csc \left (d x +c \right )}{8}\right ) \cot \left (d x +c \right )+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\)	\(114\)
parallelrisch	\(-\frac {a^{3} \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )-\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {16 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {16 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+8 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-16 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+40 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) A}{64 d}\)	\(121\)
risch	\(\frac {A \,a^{3} \left (9 \,{\mathrm e}^{7 i \left (d x +c \right )}-33 \,{\mathrm e}^{5 i \left (d x +c \right )}+48 i {\mathrm e}^{6 i \left (d x +c \right )}-33 \,{\mathrm e}^{3 i \left (d x +c \right )}-48 i {\mathrm e}^{4 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}+16 i {\mathrm e}^{2 i \left (d x +c \right )}-16 i\right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}-\frac {5 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}+\frac {5 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}\)	\(149\)
norman	\(\frac {-\frac {A \,a^{3}}{64 d}-\frac {57 A \,a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}-\frac {19 A \,a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}-\frac {27 A \,a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}-\frac {49 A \,a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}-\frac {A \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d}-\frac {3 A \,a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}-\frac {A \,a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {A \,a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {A \,a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {A \,a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {A \,a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {A \,a^{3} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {3 A \,a^{3} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}+\frac {A \,a^{3} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {A \,a^{3} \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {5 A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}\)	\(356\)

[In]

int(csc(d*x+c)^5*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(-A*a^3*ln(csc(d*x+c)-cot(d*x+c))+2*A*a^3*cot(d*x+c)+2*A*a^3*(-2/3-1/3*csc(d*x+c)^2)*cot(d*x+c)+A*a^3*((-1
/4*csc(d*x+c)^3-3/8*csc(d*x+c))*cot(d*x+c)+3/8*ln(csc(d*x+c)-cot(d*x+c))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (78) = 156\).

Time = 0.29 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.93 \[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {32 \, A a^{3} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 18 \, A a^{3} \cos \left (d x + c\right )^{3} + 30 \, A a^{3} \cos \left (d x + c\right ) - 15 \, {\left (A a^{3} \cos \left (d x + c\right )^{4} - 2 \, A a^{3} \cos \left (d x + c\right )^{2} + A a^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (A a^{3} \cos \left (d x + c\right )^{4} - 2 \, A a^{3} \cos \left (d x + c\right )^{2} + A a^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

[In]

integrate(csc(d*x+c)^5*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/48*(32*A*a^3*cos(d*x + c)^3*sin(d*x + c) - 18*A*a^3*cos(d*x + c)^3 + 30*A*a^3*cos(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) + 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))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

Sympy [F(-1)]

Timed out. \[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)**5*(a+a*sin(d*x+c))**3*(A-A*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.69 \[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {3 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 24 \, A a^{3} {\left (\log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {96 \, A a^{3}}{\tan \left (d x + c\right )} - \frac {32 \, {\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} A a^{3}}{\tan \left (d x + c\right )^{3}}}{48 \, d} \]

[In]

integrate(csc(d*x+c)^5*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/48*(3*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)) + 24*A*a^3*(log(cos(d*x + c) + 1) - log(cos(d*x + c) - 1)) + 96*A*a^3/tan(
d*x + c) - 32*(3*tan(d*x + c)^2 + 1)*A*a^3/tan(d*x + c)^3)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 174 vs. \(2 (78) = 156\).

Time = 0.32 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.02 \[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 16 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 120 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 48 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {250 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 48 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 16 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

[In]

integrate(csc(d*x+c)^5*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="giac")

[Out]

1/192*(3*A*a^3*tan(1/2*d*x + 1/2*c)^4 + 16*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 24*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 12
0*A*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 48*A*a^3*tan(1/2*d*x + 1/2*c) + (250*A*a^3*tan(1/2*d*x + 1/2*c)^4 + 4
8*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 24*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 16*A*a^3*tan(1/2*d*x + 1/2*c) - 3*A*a^3)/ta
n(1/2*d*x + 1/2*c)^4)/d

Mupad [B] (verification not implemented)

Time = 13.20 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.84 \[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {A\,a^3\,\left (3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-16\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-24\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+48\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-48\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+24\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+120\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}{192\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4} \]

[In]

int(((A - A*sin(c + d*x))*(a + a*sin(c + d*x))^3)/sin(c + d*x)^5,x)

[Out]

-(A*a^3*(3*cos(c/2 + (d*x)/2)^8 - 3*sin(c/2 + (d*x)/2)^8 - 16*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^7 + 16*cos
(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2) - 24*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^6 + 48*cos(c/2 + (d*x)/2)^3*
sin(c/2 + (d*x)/2)^5 - 48*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^3 + 24*cos(c/2 + (d*x)/2)^6*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)^4*sin(c/2 + (d*x)/2)^4))/(192*d*cos(c
/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^4)