\(\int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx\) [418]

Optimal result

Integrand size = 29, antiderivative size = 100 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{8 a d}-\frac {\cot ^3(c+d x)}{3 a d}-\frac {\cot ^5(c+d x)}{5 a d}-\frac {\cot (c+d x) \csc (c+d x)}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2918, 2687, 14, 2691, 3853, 3855} \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{8 a d}-\frac {\cot ^5(c+d x)}{5 a d}-\frac {\cot ^3(c+d x)}{3 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {\cot (c+d x) \csc (c+d x)}{8 a d} \]

[In]

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

[Out]

-1/8*ArcTanh[Cos[c + d*x]]/(a*d) - Cot[c + d*x]^3/(3*a*d) - Cot[c + d*x]^5/(5*a*d) - (Cot[c + d*x]*Csc[c + d*x
])/(8*a*d) + (Cot[c + d*x]*Csc[c + d*x]^3)/(4*a*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2918

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^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 \cot ^2(c+d x) \csc ^3(c+d x) \, dx}{a}+\frac {\int \cot ^2(c+d x) \csc ^4(c+d x) \, dx}{a} \\ & = \frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {\int \csc ^3(c+d x) \, dx}{4 a}+\frac {\text {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{a d} \\ & = -\frac {\cot (c+d x) \csc (c+d x)}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {\int \csc (c+d x) \, dx}{8 a}+\frac {\text {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,-\cot (c+d x)\right )}{a d} \\ & = -\frac {\text {arctanh}(\cos (c+d x))}{8 a d}-\frac {\cot ^3(c+d x)}{3 a d}-\frac {\cot ^5(c+d x)}{5 a d}-\frac {\cot (c+d x) \csc (c+d x)}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.89 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc ^5(c+d x) \left (320 \cos (c+d x)+80 \cos (3 (c+d x))-16 \cos (5 (c+d x))+150 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-150 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-180 \sin (2 (c+d x))-75 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+75 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-30 \sin (4 (c+d x))+15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))-15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))\right )}{1920 a d} \]

[In]

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

[Out]

-1/1920*(Csc[c + d*x]^5*(320*Cos[c + d*x] + 80*Cos[3*(c + d*x)] - 16*Cos[5*(c + d*x)] + 150*Log[Cos[(c + d*x)/
2]]*Sin[c + d*x] - 150*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] - 180*Sin[2*(c + d*x)] - 75*Log[Cos[(c + d*x)/2]]*Si
n[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]]*Si
n[5*(c + d*x)] - 15*Log[Sin[(c + d*x)/2]]*Sin[5*(c + d*x)]))/(a*d)

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.22

method	result	size
parallelrisch	\(\frac {6 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \left (\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+120 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-60 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+60 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{960 d a}\)	\(122\)
derivativedivides	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{32 d a}\)	\(124\)
default	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{32 d a}\)	\(124\)
risch	\(\frac {15 \,{\mathrm e}^{9 i \left (d x +c \right )}-240 i {\mathrm e}^{6 i \left (d x +c \right )}+90 \,{\mathrm e}^{7 i \left (d x +c \right )}-80 i {\mathrm e}^{4 i \left (d x +c \right )}-80 i {\mathrm e}^{2 i \left (d x +c \right )}-90 \,{\mathrm e}^{3 i \left (d x +c \right )}+16 i-15 \,{\mathrm e}^{i \left (d x +c \right )}}{60 a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d a}\)	\(146\)
norman	\(\frac {-\frac {1}{160 a d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{320 d a}+\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{192 d a}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{96 d a}+\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d a}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d a}+\frac {\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )}{96 d a}-\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{192 d a}-\frac {3 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{320 d a}+\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d a}+\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}\)	\(242\)

[In]

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

[Out]

1/960*(6*tan(1/2*d*x+1/2*c)^5-6*cot(1/2*d*x+1/2*c)^5-15*tan(1/2*d*x+1/2*c)^4+15*cot(1/2*d*x+1/2*c)^4+10*tan(1/
2*d*x+1/2*c)^3-10*cot(1/2*d*x+1/2*c)^3+120*ln(tan(1/2*d*x+1/2*c))-60*tan(1/2*d*x+1/2*c)+60*cot(1/2*d*x+1/2*c))
/d/a

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.61 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {32 \, \cos \left (d x + c\right )^{5} - 80 \, \cos \left (d x + c\right )^{3} - 15 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 15 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 30 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )} \sin \left (d x + c\right )} \]

[In]

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

[Out]

1/240*(32*cos(d*x + c)^5 - 80*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) +
30*(cos(d*x + c)^3 + cos(d*x + c))*sin(d*x + c))/((a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d)*sin(d*x +
c))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (90) = 180\).

Time = 0.22 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.95 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {\frac {60 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {6 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a} - \frac {120 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {{\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {60 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 6\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a \sin \left (d x + c\right )^{5}}}{960 \, d} \]

[In]

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

[Out]

-1/960*((60*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/(cos(
d*x + c) + 1)^4 - 6*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a - 120*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - (15*
sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 60*sin(d*x + c)^4/(cos(d*x + c) + 1
)^4 - 6)*(cos(d*x + c) + 1)^5/(a*sin(d*x + c)^5))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.57 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {6 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{5}} - \frac {274 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 60 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]

[In]

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

[Out]

1/960*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a + (6*a^4*tan(1/2*d*x + 1/2*c)^5 - 15*a^4*tan(1/2*d*x + 1/2*c)^4 +
10*a^4*tan(1/2*d*x + 1/2*c)^3 - 60*a^4*tan(1/2*d*x + 1/2*c))/a^5 - (274*tan(1/2*d*x + 1/2*c)^5 - 60*tan(1/2*d*
x + 1/2*c)^4 + 10*tan(1/2*d*x + 1/2*c)^2 - 15*tan(1/2*d*x + 1/2*c) + 6)/(a*tan(1/2*d*x + 1/2*c)^5))/d

Mupad [B] (verification not implemented)

Time = 9.56 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.51 \[ \int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-\frac {1}{5}\right )}{32\,a\,d} \]

[In]

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

[Out]

tan(c/2 + (d*x)/2)^3/(96*a*d) - tan(c/2 + (d*x)/2)^4/(64*a*d) + tan(c/2 + (d*x)/2)^5/(160*a*d) + log(tan(c/2 +
(d*x)/2))/(8*a*d) - tan(c/2 + (d*x)/2)/(16*a*d) + (cot(c/2 + (d*x)/2)^5*(tan(c/2 + (d*x)/2)/2 - tan(c/2 + (d*
x)/2)^2/3 + 2*tan(c/2 + (d*x)/2)^4 - 1/5))/(32*a*d)