\(\int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [430]

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

Optimal result

Integrand size = 29, antiderivative size = 112 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \text {arctanh}(\cos (c+d x))}{4 a^2 d}-\frac {2 \cot (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{a^2 d}-\frac {\cot ^5(c+d x)}{5 a^2 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{4 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{2 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2948, 2836, 3852, 3853, 3855} \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \text {arctanh}(\cos (c+d x))}{4 a^2 d}-\frac {\cot ^5(c+d x)}{5 a^2 d}-\frac {\cot ^3(c+d x)}{a^2 d}-\frac {2 \cot (c+d x)}{a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{2 a^2 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{4 a^2 d} \]

[In]

Int[(Cot[c + d*x]^4*Csc[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 2836

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

Rule 2948

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

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

Mathematica [A] (verified)

Time = 1.40 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.69 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\csc ^5(c+d x) \left (-160 \cos (c+d x)+120 \cos (3 (c+d x))-24 \cos (5 (c+d x))+150 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-150 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+140 \sin (2 (c+d x))-75 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+75 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-30 \sin (4 (c+d x))+15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))-15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))\right )}{320 a^2 d} \]

[In]

Integrate[(Cot[c + d*x]^4*Csc[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]

[Out]

(Csc[c + d*x]^5*(-160*Cos[c + d*x] + 120*Cos[3*(c + d*x)] - 24*Cos[5*(c + d*x)] + 150*Log[Cos[(c + d*x)/2]]*Si
n[c + d*x] - 150*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] + 140*Sin[2*(c + d*x)] - 75*Log[Cos[(c + d*x)/2]]*Sin[3*(c
 + d*x)] + 75*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 30*Sin[4*(c + d*x)] + 15*Log[Cos[(c + d*x)/2]]*Sin[5*(c
 + d*x)] - 15*Log[Sin[(c + d*x)/2]]*Sin[5*(c + d*x)]))/(320*a^2*d)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.44 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.30

method result size
risch \(-\frac {15 \,{\mathrm e}^{9 i \left (d x +c \right )}-40 i {\mathrm e}^{6 i \left (d x +c \right )}-70 \,{\mathrm e}^{7 i \left (d x +c \right )}+200 i {\mathrm e}^{4 i \left (d x +c \right )}-120 i {\mathrm e}^{2 i \left (d x +c \right )}+70 \,{\mathrm e}^{3 i \left (d x +c \right )}+24 i-15 \,{\mathrm e}^{i \left (d x +c \right )}}{10 a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d \,a^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d \,a^{2}}\) \(146\)
parallelrisch \(\frac {-\left (\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )+5 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+40 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-110 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-120 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+110 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d \,a^{2}}\) \(146\)
derivativedivides \(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+22 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {3}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {22}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{32 d \,a^{2}}\) \(148\)
default \(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+22 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {3}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {22}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{32 d \,a^{2}}\) \(148\)
norman \(\frac {-\frac {1}{160 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{80 d a}-\frac {3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d a}+\frac {9 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d a}-\frac {3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {3 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}-\frac {9 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d a}+\frac {3 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d a}-\frac {\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )}{80 d a}+\frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d a}-\frac {45 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}-\frac {171 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}-\frac {99 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2}}\) \(283\)

[In]

int(cos(d*x+c)^4*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/10*(15*exp(9*I*(d*x+c))-40*I*exp(6*I*(d*x+c))-70*exp(7*I*(d*x+c))+200*I*exp(4*I*(d*x+c))-120*I*exp(2*I*(d*x
+c))+70*exp(3*I*(d*x+c))+24*I-15*exp(I*(d*x+c)))/a^2/d/(exp(2*I*(d*x+c))-1)^5-3/4/d/a^2*ln(exp(I*(d*x+c))-1)+3
/4/d/a^2*ln(exp(I*(d*x+c))+1)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.60 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {48 \, \cos \left (d x + c\right )^{5} - 120 \, \cos \left (d x + c\right )^{3} - 15 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 15 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 10 \, {\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 80 \, \cos \left (d x + c\right )}{40 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/40*(48*cos(d*x + c)^5 - 120*cos(d*x + c)^3 - 15*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x + c
) + 1/2)*sin(d*x + c) + 15*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) +
 10*(3*cos(d*x + c)^3 - 5*cos(d*x + c))*sin(d*x + c) + 80*cos(d*x + c))/((a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d
*x + c)^2 + a^2*d)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**6/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (104) = 208\).

Time = 0.33 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.08 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {\frac {110 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{2}} - \frac {120 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {{\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {15 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {110 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a^{2} \sin \left (d x + c\right )^{5}}}{160 \, d} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/160*((110*sin(d*x + c)/(cos(d*x + c) + 1) - 40*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 - 5*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^2 - 120*log(s
in(d*x + c)/(cos(d*x + c) + 1))/a^2 + (5*sin(d*x + c)/(cos(d*x + c) + 1) - 15*sin(d*x + c)^2/(cos(d*x + c) + 1
)^2 + 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 110*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 1)*(cos(d*x + c) + 1)
^5/(a^2*sin(d*x + c)^5))/d

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.66 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {274 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 110 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} - \frac {a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 15 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 110 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{10}}}{160 \, d} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/160*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - (274*tan(1/2*d*x + 1/2*c)^5 - 110*tan(1/2*d*x + 1/2*c)^4 + 40
*tan(1/2*d*x + 1/2*c)^3 - 15*tan(1/2*d*x + 1/2*c)^2 + 5*tan(1/2*d*x + 1/2*c) - 1)/(a^2*tan(1/2*d*x + 1/2*c)^5)
 - (a^8*tan(1/2*d*x + 1/2*c)^5 - 5*a^8*tan(1/2*d*x + 1/2*c)^4 + 15*a^8*tan(1/2*d*x + 1/2*c)^3 - 40*a^8*tan(1/2
*d*x + 1/2*c)^2 + 110*a^8*tan(1/2*d*x + 1/2*c))/a^10)/d

Mupad [B] (verification not implemented)

Time = 10.43 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.58 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-5\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+40\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-110\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+110\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-40\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+120\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \]

[In]

int(cos(c + d*x)^4/(sin(c + d*x)^6*(a + a*sin(c + d*x))^2),x)

[Out]

-(cos(c/2 + (d*x)/2)^10 - sin(c/2 + (d*x)/2)^10 + 5*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^9 - 5*cos(c/2 + (d*x
)/2)^9*sin(c/2 + (d*x)/2) - 15*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8 + 40*cos(c/2 + (d*x)/2)^3*sin(c/2 + (
d*x)/2)^7 - 110*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 + 110*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^4 - 40
*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^3 + 15*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)/(160*a^2*d*cos(c/2 + (d*x)/2)^5*sin(c/
2 + (d*x)/2)^5)