\(\int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [1230]

Optimal result

Integrand size = 27, antiderivative size = 226 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \csc (c+d x)}{a^6 d}-\frac {2 b \left (a^2-b^2\right ) \csc ^2(c+d x)}{a^5 d}+\frac {\left (2 a^2-3 b^2\right ) \csc ^3(c+d x)}{3 a^4 d}+\frac {b \csc ^4(c+d x)}{2 a^3 d}-\frac {\csc ^5(c+d x)}{5 a^2 d}-\frac {2 b \left (a^4-4 a^2 b^2+3 b^4\right ) \log (\sin (c+d x))}{a^7 d}+\frac {2 b \left (a^4-4 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))}{a^7 d}-\frac {b \left (a^2-b^2\right )^2}{a^6 d (a+b \sin (c+d x))} \] Output:

-(a^4-6*a^2*b^2+5*b^4)*csc(d*x+c)/a^6/d-2*b*(a^2-b^2)*csc(d*x+c)^2/a^5/d+1 
/3*(2*a^2-3*b^2)*csc(d*x+c)^3/a^4/d+1/2*b*csc(d*x+c)^4/a^3/d-1/5*csc(d*x+c 
)^5/a^2/d-2*b*(a^4-4*a^2*b^2+3*b^4)*ln(sin(d*x+c))/a^7/d+2*b*(a^4-4*a^2*b^ 
2+3*b^4)*ln(a+b*sin(d*x+c))/a^7/d-b*(a^2-b^2)^2/a^6/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 3.18 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.97 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-30 a^2 \left (a^4-4 a^2 b^2+3 b^4\right ) \csc ^2(c+d x)+\left (-40 a^5 b+30 a^3 b^3\right ) \csc ^3(c+d x)+5 a^4 \left (4 a^2-3 b^2\right ) \csc ^4(c+d x)+9 a^5 b \csc ^5(c+d x)-6 a^6 \csc ^6(c+d x)-60 b^2 \left (a^4-4 a^2 b^2+3 b^4\right ) (\log (\sin (c+d x))-\log (a+b \sin (c+d x)))-60 a b \left (a^4-4 a^2 b^2+3 b^4\right ) \csc (c+d x) (1+\log (\sin (c+d x))-\log (a+b \sin (c+d x)))}{30 a^7 d (b+a \csc (c+d x))} \] Input:

Integrate[(Cot[c + d*x]^5*Csc[c + d*x])/(a + b*Sin[c + d*x])^2,x]
 

Output:

(-30*a^2*(a^4 - 4*a^2*b^2 + 3*b^4)*Csc[c + d*x]^2 + (-40*a^5*b + 30*a^3*b^ 
3)*Csc[c + d*x]^3 + 5*a^4*(4*a^2 - 3*b^2)*Csc[c + d*x]^4 + 9*a^5*b*Csc[c + 
 d*x]^5 - 6*a^6*Csc[c + d*x]^6 - 60*b^2*(a^4 - 4*a^2*b^2 + 3*b^4)*(Log[Sin 
[c + d*x]] - Log[a + b*Sin[c + d*x]]) - 60*a*b*(a^4 - 4*a^2*b^2 + 3*b^4)*C 
sc[c + d*x]*(1 + Log[Sin[c + d*x]] - Log[a + b*Sin[c + d*x]]))/(30*a^7*d*( 
b + a*Csc[c + d*x]))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3316, 27, 522, 2009}

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.

\(\displaystyle \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^5}{\sin (c+d x)^6 (a+b \sin (c+d x))^2}dx\)

\(\Big \downarrow \) 3316

\(\displaystyle \frac {\int \frac {\csc ^6(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{(a+b \sin (c+d x))^2}d(b \sin (c+d x))}{b^5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b \int \frac {\csc ^6(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{b^6 (a+b \sin (c+d x))^2}d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 522

\(\displaystyle \frac {b \int \left (\frac {\csc ^6(c+d x)}{a^2 b^2}-\frac {2 \csc ^5(c+d x)}{a^3 b}+\frac {\left (3 b^4-2 a^2 b^2\right ) \csc ^4(c+d x)}{a^4 b^4}+\frac {4 \left (a^2-b^2\right ) \csc ^3(c+d x)}{a^5 b}+\frac {\left (a^4-6 b^2 a^2+5 b^4\right ) \csc ^2(c+d x)}{a^6 b^2}-\frac {2 \left (a^4-4 b^2 a^2+3 b^4\right ) \csc (c+d x)}{a^7 b}+\frac {2 \left (a^4-4 b^2 a^2+3 b^4\right )}{a^7 (a+b \sin (c+d x))}+\frac {\left (a^2-b^2\right )^2}{a^6 (a+b \sin (c+d x))^2}\right )d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b \left (\frac {\csc ^4(c+d x)}{2 a^3}-\frac {\csc ^5(c+d x)}{5 a^2 b}-\frac {\left (a^2-b^2\right )^2}{a^6 (a+b \sin (c+d x))}-\frac {2 \left (a^2-b^2\right ) \csc ^2(c+d x)}{a^5}+\frac {\left (2 a^2-3 b^2\right ) \csc ^3(c+d x)}{3 a^4 b}-\frac {2 \left (a^4-4 a^2 b^2+3 b^4\right ) \log (b \sin (c+d x))}{a^7}+\frac {2 \left (a^4-4 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))}{a^7}-\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \csc (c+d x)}{a^6 b}\right )}{d}\)

Input:

Int[(Cot[c + d*x]^5*Csc[c + d*x])/(a + b*Sin[c + d*x])^2,x]
 

Output:

(b*(-(((a^4 - 6*a^2*b^2 + 5*b^4)*Csc[c + d*x])/(a^6*b)) - (2*(a^2 - b^2)*C 
sc[c + d*x]^2)/a^5 + ((2*a^2 - 3*b^2)*Csc[c + d*x]^3)/(3*a^4*b) + Csc[c + 
d*x]^4/(2*a^3) - Csc[c + d*x]^5/(5*a^2*b) - (2*(a^4 - 4*a^2*b^2 + 3*b^4)*L 
og[b*Sin[c + d*x]])/a^7 + (2*(a^4 - 4*a^2*b^2 + 3*b^4)*Log[a + b*Sin[c + d 
*x]])/a^7 - (a^2 - b^2)^2/(a^6*(a + b*Sin[c + d*x]))))/d
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 522

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3316

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* 
f)   Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* 
Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) 
/2] && NeQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 1.54 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {-\frac {\frac {\csc \left (d x +c \right )^{5} a^{4}}{5}-\frac {b \csc \left (d x +c \right )^{4} a^{3}}{2}-\frac {2 a^{4} \csc \left (d x +c \right )^{3}}{3}+a^{2} b^{2} \csc \left (d x +c \right )^{3}+2 a^{3} b \csc \left (d x +c \right )^{2}-2 a \,b^{3} \csc \left (d x +c \right )^{2}+\csc \left (d x +c \right ) a^{4}-6 \csc \left (d x +c \right ) a^{2} b^{2}+5 \csc \left (d x +c \right ) b^{4}}{a^{6}}+\frac {2 b \left (a^{4}-4 a^{2} b^{2}+3 b^{4}\right ) \ln \left (\csc \left (d x +c \right ) a +b \right )}{a^{7}}+\frac {b^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{a^{7} \left (\csc \left (d x +c \right ) a +b \right )}}{d}\)	\(198\)
default	\(\frac {-\frac {\frac {\csc \left (d x +c \right )^{5} a^{4}}{5}-\frac {b \csc \left (d x +c \right )^{4} a^{3}}{2}-\frac {2 a^{4} \csc \left (d x +c \right )^{3}}{3}+a^{2} b^{2} \csc \left (d x +c \right )^{3}+2 a^{3} b \csc \left (d x +c \right )^{2}-2 a \,b^{3} \csc \left (d x +c \right )^{2}+\csc \left (d x +c \right ) a^{4}-6 \csc \left (d x +c \right ) a^{2} b^{2}+5 \csc \left (d x +c \right ) b^{4}}{a^{6}}+\frac {2 b \left (a^{4}-4 a^{2} b^{2}+3 b^{4}\right ) \ln \left (\csc \left (d x +c \right ) a +b \right )}{a^{7}}+\frac {b^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{a^{7} \left (\csc \left (d x +c \right ) a +b \right )}}{d}\)	\(198\)
risch	\(-\frac {4 i \left (234 i a^{4} b \,{\mathrm e}^{5 i \left (d x +c \right )}-15 i a^{4} b \,{\mathrm e}^{11 i \left (d x +c \right )}-115 i a^{4} b \,{\mathrm e}^{3 i \left (d x +c \right )}-690 i a^{2} b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+15 i a^{4} b \,{\mathrm e}^{i \left (d x +c \right )}-60 i a^{2} b^{3} {\mathrm e}^{i \left (d x +c \right )}+330 i a^{2} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+60 i a^{2} b^{3} {\mathrm e}^{11 i \left (d x +c \right )}+115 i a^{4} b \,{\mathrm e}^{9 i \left (d x +c \right )}-330 i a^{2} b^{3} {\mathrm e}^{9 i \left (d x +c \right )}-234 i a^{4} b \,{\mathrm e}^{7 i \left (d x +c \right )}+690 i a^{2} b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-60 a^{3} b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+58 a^{5} {\mathrm e}^{6 i \left (d x +c \right )}+15 a^{5} {\mathrm e}^{10 i \left (d x +c \right )}+15 a^{5} {\mathrm e}^{2 i \left (d x +c \right )}-20 a^{5} {\mathrm e}^{4 i \left (d x +c \right )}-20 a^{5} {\mathrm e}^{8 i \left (d x +c \right )}-450 i b^{5} {\mathrm e}^{7 i \left (d x +c \right )}+270 a \,b^{4} {\mathrm e}^{6 i \left (d x +c \right )}+210 a^{3} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-60 a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-180 a \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+45 a \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-300 a^{3} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-180 a \,b^{4} {\mathrm e}^{8 i \left (d x +c \right )}+210 a^{3} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+45 a \,b^{4} {\mathrm e}^{10 i \left (d x +c \right )}+450 i b^{5} {\mathrm e}^{5 i \left (d x +c \right )}-225 i b^{5} {\mathrm e}^{3 i \left (d x +c \right )}+45 i b^{5} {\mathrm e}^{i \left (d x +c \right )}-45 i b^{5} {\mathrm e}^{11 i \left (d x +c \right )}+225 i b^{5} {\mathrm e}^{9 i \left (d x +c \right )}\right )}{15 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right ) d \,a^{6}}-\frac {2 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{3}}+\frac {8 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{5} d}-\frac {6 b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{7} d}+\frac {2 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{3} d}-\frac {8 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{5} d}+\frac {6 b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{7} d}\)	\(763\)

Input:

int(cot(d*x+c)^5*csc(d*x+c)/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-1/a^6*(1/5*csc(d*x+c)^5*a^4-1/2*b*csc(d*x+c)^4*a^3-2/3*a^4*csc(d*x+c 
)^3+a^2*b^2*csc(d*x+c)^3+2*a^3*b*csc(d*x+c)^2-2*a*b^3*csc(d*x+c)^2+csc(d*x 
+c)*a^4-6*csc(d*x+c)*a^2*b^2+5*csc(d*x+c)*b^4)+2/a^7*b*(a^4-4*a^2*b^2+3*b^ 
4)*ln(csc(d*x+c)*a+b)+b^2*(a^4-2*a^2*b^2+b^4)/a^7/(csc(d*x+c)*a+b))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 696 vs. \(2 (220) = 440\).

Time = 0.13 (sec) , antiderivative size = 696, normalized size of antiderivative = 3.08 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:

integrate(cot(d*x+c)^5*csc(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="fricas" 
)
 

Output:

1/30*(16*a^6 - 105*a^4*b^2 + 90*a^2*b^4 + 30*(a^6 - 4*a^4*b^2 + 3*a^2*b^4) 
*cos(d*x + c)^4 - 5*(8*a^6 - 45*a^4*b^2 + 36*a^2*b^4)*cos(d*x + c)^2 + 60* 
((a^4*b^2 - 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^6 - a^4*b^2 + 4*a^2*b^4 - 3*b^ 
6 - 3*(a^4*b^2 - 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^4 + 3*(a^4*b^2 - 4*a^2*b^ 
4 + 3*b^6)*cos(d*x + c)^2 - (a^5*b - 4*a^3*b^3 + 3*a*b^5 + (a^5*b - 4*a^3* 
b^3 + 3*a*b^5)*cos(d*x + c)^4 - 2*(a^5*b - 4*a^3*b^3 + 3*a*b^5)*cos(d*x + 
c)^2)*sin(d*x + c))*log(b*sin(d*x + c) + a) - 60*((a^4*b^2 - 4*a^2*b^4 + 3 
*b^6)*cos(d*x + c)^6 - a^4*b^2 + 4*a^2*b^4 - 3*b^6 - 3*(a^4*b^2 - 4*a^2*b^ 
4 + 3*b^6)*cos(d*x + c)^4 + 3*(a^4*b^2 - 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^2 
 - (a^5*b - 4*a^3*b^3 + 3*a*b^5 + (a^5*b - 4*a^3*b^3 + 3*a*b^5)*cos(d*x + 
c)^4 - 2*(a^5*b - 4*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(1 
/2*sin(d*x + c)) + (91*a^5*b - 270*a^3*b^3 + 180*a*b^5 + 60*(a^5*b - 4*a^3 
*b^3 + 3*a*b^5)*cos(d*x + c)^4 - 10*(16*a^5*b - 51*a^3*b^3 + 36*a*b^5)*cos 
(d*x + c)^2)*sin(d*x + c))/(a^7*b*d*cos(d*x + c)^6 - 3*a^7*b*d*cos(d*x + c 
)^4 + 3*a^7*b*d*cos(d*x + c)^2 - a^7*b*d - (a^8*d*cos(d*x + c)^4 - 2*a^8*d 
*cos(d*x + c)^2 + a^8*d)*sin(d*x + c))
 

Sympy [F]

\[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\cot ^{5}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \] Input:

integrate(cot(d*x+c)**5*csc(d*x+c)/(a+b*sin(d*x+c))**2,x)
 

Output:

Integral(cot(c + d*x)**5*csc(c + d*x)/(a + b*sin(c + d*x))**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.00 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {9 \, a^{4} b \sin \left (d x + c\right ) - 60 \, {\left (a^{4} b - 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \sin \left (d x + c\right )^{5} - 6 \, a^{5} - 30 \, {\left (a^{5} - 4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \sin \left (d x + c\right )^{4} - 10 \, {\left (4 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )^{3} + 5 \, {\left (4 \, a^{5} - 3 \, a^{3} b^{2}\right )} \sin \left (d x + c\right )^{2}}{a^{6} b \sin \left (d x + c\right )^{6} + a^{7} \sin \left (d x + c\right )^{5}} + \frac {60 \, {\left (a^{4} b - 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{7}} - \frac {60 \, {\left (a^{4} b - 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{7}}}{30 \, d} \] Input:

integrate(cot(d*x+c)^5*csc(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="maxima" 
)
 

Output:

1/30*((9*a^4*b*sin(d*x + c) - 60*(a^4*b - 4*a^2*b^3 + 3*b^5)*sin(d*x + c)^ 
5 - 6*a^5 - 30*(a^5 - 4*a^3*b^2 + 3*a*b^4)*sin(d*x + c)^4 - 10*(4*a^4*b - 
3*a^2*b^3)*sin(d*x + c)^3 + 5*(4*a^5 - 3*a^3*b^2)*sin(d*x + c)^2)/(a^6*b*s 
in(d*x + c)^6 + a^7*sin(d*x + c)^5) + 60*(a^4*b - 4*a^2*b^3 + 3*b^5)*log(b 
*sin(d*x + c) + a)/a^7 - 60*(a^4*b - 4*a^2*b^3 + 3*b^5)*log(sin(d*x + c))/ 
a^7)/d
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.04 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 \, {\left (a^{4} b - 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{7} d} + \frac {2 \, {\left (a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{7} b d} + \frac {9 \, a^{5} b \sin \left (d x + c\right ) - 6 \, a^{6} - 60 \, {\left (a^{5} b - 4 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \sin \left (d x + c\right )^{5} - 30 \, {\left (a^{6} - 4 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \sin \left (d x + c\right )^{4} - 10 \, {\left (4 \, a^{5} b - 3 \, a^{3} b^{3}\right )} \sin \left (d x + c\right )^{3} + 5 \, {\left (4 \, a^{6} - 3 \, a^{4} b^{2}\right )} \sin \left (d x + c\right )^{2}}{30 \, {\left (b \sin \left (d x + c\right ) + a\right )} a^{7} d \sin \left (d x + c\right )^{5}} \] Input:

integrate(cot(d*x+c)^5*csc(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="giac")
 

Output:

-2*(a^4*b - 4*a^2*b^3 + 3*b^5)*log(abs(sin(d*x + c)))/(a^7*d) + 2*(a^4*b^2 
 - 4*a^2*b^4 + 3*b^6)*log(abs(b*sin(d*x + c) + a))/(a^7*b*d) + 1/30*(9*a^5 
*b*sin(d*x + c) - 6*a^6 - 60*(a^5*b - 4*a^3*b^3 + 3*a*b^5)*sin(d*x + c)^5 
- 30*(a^6 - 4*a^4*b^2 + 3*a^2*b^4)*sin(d*x + c)^4 - 10*(4*a^5*b - 3*a^3*b^ 
3)*sin(d*x + c)^3 + 5*(4*a^6 - 3*a^4*b^2)*sin(d*x + c)^2)/((b*sin(d*x + c) 
 + a)*a^7*d*sin(d*x + c)^5)
 

Mupad [B] (verification not implemented)

Time = 23.38 (sec) , antiderivative size = 628, normalized size of antiderivative = 2.78 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:

int(cot(c + d*x)^5/(sin(c + d*x)*(a + b*sin(c + d*x))^2),x)
 

Output:

(tan(c/2 + (d*x)/2)^3*((64*a^2 + 128*b^2)/(3072*a^4) + 1/(32*a^2) - b^2/(6 
*a^4)))/d - (tan(c/2 + (d*x)/2)*(b^2/(2*a^4) - (4*b*((b*(64*a^2 + 128*b^2) 
)/(256*a^5) - b/(8*a^3) + (4*b*((64*a^2 + 128*b^2)/(1024*a^4) + 3/(32*a^2) 
 - b^2/(2*a^4)))/a))/a + ((64*a^2 + 128*b^2)*((64*a^2 + 128*b^2)/(1024*a^4 
) + 3/(32*a^2) - b^2/(2*a^4)))/(32*a^2)))/d - tan(c/2 + (d*x)/2)^5/(160*a^ 
2*d) - (tan(c/2 + (d*x)/2)^3*((23*a^4*b)/3 - 8*a^2*b^3) + tan(c/2 + (d*x)/ 
2)^4*(48*a*b^4 + (25*a^5)/3 - 56*a^3*b^2) + tan(c/2 + (d*x)/2)^5*(32*a^4*b 
 + 160*b^5 - 184*a^2*b^3) + a^5/5 - tan(c/2 + (d*x)/2)^2*((22*a^5)/15 - 2* 
a^3*b^2) - (3*a^4*b*tan(c/2 + (d*x)/2))/5 + (2*tan(c/2 + (d*x)/2)^6*(5*a^6 
 - 32*b^6 + 104*a^2*b^4 - 74*a^4*b^2))/a)/(d*(32*a^7*tan(c/2 + (d*x)/2)^5 
+ 32*a^7*tan(c/2 + (d*x)/2)^7 + 64*a^6*b*tan(c/2 + (d*x)/2)^6)) - (tan(c/2 
 + (d*x)/2)^2*((b*(64*a^2 + 128*b^2))/(512*a^5) - b/(16*a^3) + (2*b*((64*a 
^2 + 128*b^2)/(1024*a^4) + 3/(32*a^2) - b^2/(2*a^4)))/a))/d - (log(tan(c/2 
 + (d*x)/2))*(2*a^4*b + 6*b^5 - 8*a^2*b^3))/(a^7*d) + (b*tan(c/2 + (d*x)/2 
)^4)/(32*a^3*d) + (log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2 
)*(2*a^4*b + 6*b^5 - 8*a^2*b^3))/(a^7*d)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 675, normalized size of antiderivative = 2.99 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:

int(cot(d*x+c)^5*csc(d*x+c)/(a+b*sin(d*x+c))^2,x)
 

Output:

(480*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**6 
*a**4*b**3 - 1920*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*si 
n(c + d*x)**6*a**2*b**5 + 1440*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x) 
/2)*b + a)*sin(c + d*x)**6*b**7 + 480*log(tan((c + d*x)/2)**2*a + 2*tan((c 
 + d*x)/2)*b + a)*sin(c + d*x)**5*a**5*b**2 - 1920*log(tan((c + d*x)/2)**2 
*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**5*a**3*b**4 + 1440*log(tan((c 
 + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**5*a*b**6 - 480*l 
og(tan((c + d*x)/2))*sin(c + d*x)**6*a**4*b**3 + 1920*log(tan((c + d*x)/2) 
)*sin(c + d*x)**6*a**2*b**5 - 1440*log(tan((c + d*x)/2))*sin(c + d*x)**6*b 
**7 - 480*log(tan((c + d*x)/2))*sin(c + d*x)**5*a**5*b**2 + 1920*log(tan(( 
c + d*x)/2))*sin(c + d*x)**5*a**3*b**4 - 1440*log(tan((c + d*x)/2))*sin(c 
+ d*x)**5*a*b**6 + 75*sin(c + d*x)**6*a**6*b - 435*sin(c + d*x)**6*a**4*b* 
*3 + 360*sin(c + d*x)**6*a**2*b**5 + 75*sin(c + d*x)**5*a**7 - 915*sin(c + 
 d*x)**5*a**5*b**2 + 2280*sin(c + d*x)**5*a**3*b**4 - 1440*sin(c + d*x)**5 
*a*b**6 - 240*sin(c + d*x)**4*a**6*b + 960*sin(c + d*x)**4*a**4*b**3 - 720 
*sin(c + d*x)**4*a**2*b**5 - 320*sin(c + d*x)**3*a**5*b**2 + 240*sin(c + d 
*x)**3*a**3*b**4 + 160*sin(c + d*x)**2*a**6*b - 120*sin(c + d*x)**2*a**4*b 
**3 + 72*sin(c + d*x)*a**5*b**2 - 48*a**6*b)/(240*sin(c + d*x)**5*a**7*b*d 
*(sin(c + d*x)*b + a))