\(\int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [639]

Optimal result

Integrand size = 29, antiderivative size = 73 \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 x}{a^2}-\frac {\text {arctanh}(\cos (c+d x))}{2 a^2 d}+\frac {\cos (c+d x)}{a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2954, 2951, 3852, 8, 3853, 3855, 2718} \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{2 a^2 d}+\frac {\cos (c+d x)}{a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {2 x}{a^2} \]

[In]

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

[Out]

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

Rule 8

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

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2951

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))

Rule 2954

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e +
f*x])^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 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 \cot ^2(c+d x) \csc (c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4} \\ & = \frac {\int \left (2 a^4-2 a^4 \csc ^2(c+d x)+a^4 \csc ^3(c+d x)-a^4 \sin (c+d x)\right ) \, dx}{a^6} \\ & = \frac {2 x}{a^2}+\frac {\int \csc ^3(c+d x) \, dx}{a^2}-\frac {\int \sin (c+d x) \, dx}{a^2}-\frac {2 \int \csc ^2(c+d x) \, dx}{a^2} \\ & = \frac {2 x}{a^2}+\frac {\cos (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\int \csc (c+d x) \, dx}{2 a^2}+\frac {2 \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d} \\ & = \frac {2 x}{a^2}-\frac {\text {arctanh}(\cos (c+d x))}{2 a^2 d}+\frac {\cos (c+d x)}{a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.84 \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4 \left (16 (c+d x)+8 \cos (c+d x)+8 \cot \left (\frac {1}{2} (c+d x)\right )-\csc ^2\left (\frac {1}{2} (c+d x)\right )-4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\sec ^2\left (\frac {1}{2} (c+d x)\right )-8 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{8 d (a+a \sin (c+d x))^2} \]

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(16*(c + d*x) + 8*Cos[c + d*x] + 8*Cot[(c + d*x)/2] - Csc[(c + d*x)/2
]^2 - 4*Log[Cos[(c + d*x)/2]] + 4*Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)/2]^2 - 8*Tan[(c + d*x)/2]))/(8*d*(a +
a*Sin[c + d*x])^2)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.38

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {8}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+16 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2}}\)	\(101\)
default	\(\frac {\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {8}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+16 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2}}\)	\(101\)
parallelrisch	\(\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-8 \cos \left (d x +c \right )+23\right ) \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+16 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-17\right ) \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+16 d x}{8 d \,a^{2}}\)	\(104\)
risch	\(\frac {2 x}{a^{2}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{2}}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{2}}+\frac {{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}+4 i {\mathrm e}^{2 i \left (d x +c \right )}-4 i}{a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d \,a^{2}}\)	\(135\)
norman	\(\frac {-\frac {1}{8 a d}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}-\frac {5 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {2 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {6 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {12 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {20 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {24 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {24 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {20 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {12 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {6 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {2 x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {3 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {33 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {13 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {71 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {85 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {79 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {233 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {189 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {331 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(468\)

[In]

int(cos(d*x+c)^6*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/4/d/a^2*(1/2*tan(1/2*d*x+1/2*c)^2-4*tan(1/2*d*x+1/2*c)-1/2/tan(1/2*d*x+1/2*c)^2+4/tan(1/2*d*x+1/2*c)+2*ln(ta
n(1/2*d*x+1/2*c))+8/(1+tan(1/2*d*x+1/2*c)^2)+16*arctan(tan(1/2*d*x+1/2*c)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.62 \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {8 \, d x \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right )^{3} - 8 \, d x - {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 8 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}} \]

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**6*csc(d*x+c)**3/(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. 204 vs. \(2 (69) = 138\).

Time = 0.34 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.79 \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {15 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 1}{\frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}} + \frac {32 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {4 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{8 \, d} \]

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/8*((8*sin(d*x + c)/(cos(d*x + c) + 1) + 15*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 8*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 - 1)/(a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - (8*sin(d
*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a^2 + 32*arctan(sin(d*x + c)/(cos(d*x + c) +
1))/a^2 + 4*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.75 \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {16 \, {\left (d x + c\right )}}{a^{2}} + \frac {4 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{4}} + \frac {16}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} - \frac {6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*(16*(d*x + c)/a^2 + 4*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + (a^2*tan(1/2*d*x + 1/2*c)^2 - 8*a^2*tan(1/2*d*x
+ 1/2*c))/a^4 + 16/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^2) - (6*tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) +
1)/(a^2*tan(1/2*d*x + 1/2*c)^2))/d

Mupad [B] (verification not implemented)

Time = 10.15 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.55 \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}-\frac {4\,\mathrm {atan}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-4}+\frac {16}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-4}\right )}{a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^2\,d}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}}{d\,\left (4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2\,d} \]

[In]

int(cos(c + d*x)^6/(sin(c + d*x)^3*(a + a*sin(c + d*x))^2),x)

[Out]

tan(c/2 + (d*x)/2)^2/(8*a^2*d) - (4*atan((4*tan(c/2 + (d*x)/2))/(16*tan(c/2 + (d*x)/2) - 4) + 16/(16*tan(c/2 +
(d*x)/2) - 4)))/(a^2*d) + log(tan(c/2 + (d*x)/2))/(2*a^2*d) + (4*tan(c/2 + (d*x)/2) + (15*tan(c/2 + (d*x)/2)^
2)/2 + 4*tan(c/2 + (d*x)/2)^3 - 1/2)/(d*(4*a^2*tan(c/2 + (d*x)/2)^2 + 4*a^2*tan(c/2 + (d*x)/2)^4)) - tan(c/2 +
(d*x)/2)/(a^2*d)