\(\int \frac {\cot ^2(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx\) [305]

Optimal result

Integrand size = 29, antiderivative size = 95 \[ \int \frac {\cot ^2(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \text {arctanh}(\cos (c+d x))}{8 a d}+\frac {\cot (c+d x)}{a d}+\frac {\cot ^3(c+d x)}{3 a d}-\frac {3 \cot (c+d x) \csc (c+d x)}{8 a d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d} \]

[Out]

-3/8*arctanh(cos(d*x+c))/a/d+cot(d*x+c)/a/d+1/3*cot(d*x+c)^3/a/d-3/8*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 = 0.10 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2918, 3853, 3855, 3852} \[ \int \frac {\cot ^2(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \text {arctanh}(\cos (c+d x))}{8 a d}+\frac {\cot ^3(c+d x)}{3 a d}+\frac {\cot (c+d x)}{a d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {3 \cot (c+d x) \csc (c+d x)}{8 a d} \]

[In]

Int[(Cot[c + d*x]^2*Csc[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

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

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 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 ^4(c+d x) \, dx}{a}+\frac {\int \csc ^5(c+d x) \, dx}{a} \\ & = -\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {3 \int \csc ^3(c+d x) \, dx}{4 a}+\frac {\text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a d} \\ & = \frac {\cot (c+d x)}{a d}+\frac {\cot ^3(c+d x)}{3 a d}-\frac {3 \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 {3 \int \csc (c+d x) \, dx}{8 a} \\ & = -\frac {3 \text {arctanh}(\cos (c+d x))}{8 a d}+\frac {\cot (c+d x)}{a d}+\frac {\cot ^3(c+d x)}{3 a d}-\frac {3 \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 = 1.42 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.32 \[ \int \frac {\cot ^2(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc ^4(c+d x) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (66 \cos (c+d x)+72 \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin ^4(c+d x)+2 \cos (3 (c+d x)) (-9+16 \sin (c+d x))-48 \sin (2 (c+d x))\right )}{192 a d (1+\sin (c+d x))} \]

[In]

Integrate[(Cot[c + d*x]^2*Csc[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

-1/192*(Csc[c + d*x]^4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(66*Cos[c + d*x] + 72*(Log[Cos[(c + d*x)/2]] -
Log[Sin[(c + d*x)/2]])*Sin[c + d*x]^4 + 2*Cos[3*(c + d*x)]*(-9 + 16*Sin[c + d*x]) - 48*Sin[2*(c + d*x)]))/(a*d
*(1 + Sin[c + d*x]))

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.28

method	result	size
parallelrisch	\(\frac {-3 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+72 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+72 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{192 d a}\)	\(122\)
derivativedivides	\(\frac {\frac {\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {6}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {2}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{16 d a}\)	\(124\)
default	\(\frac {\frac {\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {6}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {2}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{16 d a}\)	\(124\)
risch	\(\frac {9 \,{\mathrm e}^{7 i \left (d x +c \right )}-48 i {\mathrm e}^{4 i \left (d x +c \right )}-33 \,{\mathrm e}^{5 i \left (d x +c \right )}+64 i {\mathrm e}^{2 i \left (d x +c \right )}-33 \,{\mathrm e}^{3 i \left (d x +c \right )}-16 i+9 \,{\mathrm e}^{i \left (d x +c \right )}}{12 a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d a}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d a}\)	\(134\)
norman	\(\frac {-\frac {1}{64 a d}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{192 d a}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d a}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}-\frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d a}-\frac {5 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}+\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}+\frac {3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}\)	\(204\)

[In]

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

[Out]

1/192*(-3*cot(1/2*d*x+1/2*c)^4+3*tan(1/2*d*x+1/2*c)^4+8*cot(1/2*d*x+1/2*c)^3-8*tan(1/2*d*x+1/2*c)^3-24*cot(1/2
*d*x+1/2*c)^2+24*tan(1/2*d*x+1/2*c)^2+72*cot(1/2*d*x+1/2*c)+72*ln(tan(1/2*d*x+1/2*c))-72*tan(1/2*d*x+1/2*c))/d
/a

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.51 \[ \int \frac {\cot ^2(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {18 \, \cos \left (d x + c\right )^{3} - 9 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 9 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 16 \, {\left (2 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 30 \, \cos \left (d x + c\right )}{48 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )}} \]

[In]

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

[Out]

1/48*(18*cos(d*x + c)^3 - 9*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2) + 9*(cos(d*x +
c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2) - 16*(2*cos(d*x + c)^3 - 3*cos(d*x + c))*sin(d*x +
c) - 30*cos(d*x + c))/(a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d)

Sympy [F]

\[ \int \frac {\cot ^2(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cos ^{2}{\left (c + d x \right )} \csc ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

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

[Out]

Integral(cos(c + d*x)**2*csc(c + d*x)**5/(sin(c + d*x) + 1), x)/a

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (87) = 174\).

Time = 0.21 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.05 \[ \int \frac {\cot ^2(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {\frac {72 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a} - \frac {72 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {{\left (\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {72 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 3\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{4}}{a \sin \left (d x + c\right )^{4}}}{192 \, d} \]

[In]

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

[Out]

-1/192*((72*sin(d*x + c)/(cos(d*x + c) + 1) - 24*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 8*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 - 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/a - 72*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - (8*sin
(d*x + c)/(cos(d*x + c) + 1) - 24*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 72*sin(d*x + c)^3/(cos(d*x + c) + 1)^3
- 3)*(cos(d*x + c) + 1)^4/(a*sin(d*x + c)^4))/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.65 \[ \int \frac {\cot ^2(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {72 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{4}} - \frac {150 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 72 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

[In]

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

[Out]

1/192*(72*log(abs(tan(1/2*d*x + 1/2*c)))/a + (3*a^3*tan(1/2*d*x + 1/2*c)^4 - 8*a^3*tan(1/2*d*x + 1/2*c)^3 + 24
*a^3*tan(1/2*d*x + 1/2*c)^2 - 72*a^3*tan(1/2*d*x + 1/2*c))/a^4 - (150*tan(1/2*d*x + 1/2*c)^4 - 72*tan(1/2*d*x
+ 1/2*c)^3 + 24*tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 3)/(a*tan(1/2*d*x + 1/2*c)^4))/d

Mupad [B] (verification not implemented)

Time = 9.77 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.59 \[ \int \frac {\cot ^2(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a\,d}+\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a\,d}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-\frac {1}{4}\right )}{16\,a\,d} \]

[In]

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

[Out]

tan(c/2 + (d*x)/2)^2/(8*a*d) - tan(c/2 + (d*x)/2)^3/(24*a*d) + tan(c/2 + (d*x)/2)^4/(64*a*d) + (3*log(tan(c/2
+ (d*x)/2)))/(8*a*d) - (3*tan(c/2 + (d*x)/2))/(8*a*d) + (cot(c/2 + (d*x)/2)^4*((2*tan(c/2 + (d*x)/2))/3 - 2*ta
n(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^3 - 1/4))/(16*a*d)