\(\int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [743]

Optimal result

Integrand size = 29, antiderivative size = 92 \[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {x}{2 a^3}+\frac {3 \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {3 \cos (c+d x)}{a^3 d}+\frac {\cos ^3(c+d x)}{3 a^3 d}-\frac {\cot (c+d x)}{a^3 d}+\frac {3 \cos (c+d x) \sin (c+d x)}{2 a^3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2954, 2788, 3855, 3852, 8, 2718, 2715, 2713} \[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3 \text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {\cos ^3(c+d x)}{3 a^3 d}-\frac {3 \cos (c+d x)}{a^3 d}-\frac {\cot (c+d x)}{a^3 d}+\frac {3 \sin (c+d x) \cos (c+d x)}{2 a^3 d}+\frac {x}{2 a^3} \]

[In]

Int[(Cos[c + d*x]^6*Cot[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]

[Out]

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

Rule 8

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

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2718

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

Rule 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/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 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) (a-a \sin (c+d x))^3 \, dx}{a^6} \\ & = \frac {\int \left (2 a^5-3 a^5 \csc (c+d x)+a^5 \csc ^2(c+d x)+2 a^5 \sin (c+d x)-3 a^5 \sin ^2(c+d x)+a^5 \sin ^3(c+d x)\right ) \, dx}{a^8} \\ & = \frac {2 x}{a^3}+\frac {\int \csc ^2(c+d x) \, dx}{a^3}+\frac {\int \sin ^3(c+d x) \, dx}{a^3}+\frac {2 \int \sin (c+d x) \, dx}{a^3}-\frac {3 \int \csc (c+d x) \, dx}{a^3}-\frac {3 \int \sin ^2(c+d x) \, dx}{a^3} \\ & = \frac {2 x}{a^3}+\frac {3 \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {2 \cos (c+d x)}{a^3 d}+\frac {3 \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {3 \int 1 \, dx}{2 a^3}-\frac {\text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}-\frac {\text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^3 d} \\ & = \frac {x}{2 a^3}+\frac {3 \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {3 \cos (c+d x)}{a^3 d}+\frac {\cos ^3(c+d x)}{3 a^3 d}-\frac {\cot (c+d x)}{a^3 d}+\frac {3 \cos (c+d x) \sin (c+d x)}{2 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.54 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6 \left (6 (c+d x)-33 \cos (c+d x)+\cos (3 (c+d x))-6 \cot \left (\frac {1}{2} (c+d x)\right )+36 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-36 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 \sin (2 (c+d x))+6 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{12 d (a+a \sin (c+d x))^3} \]

[In]

Integrate[(Cos[c + d*x]^6*Cot[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6*(6*(c + d*x) - 33*Cos[c + d*x] + Cos[3*(c + d*x)] - 6*Cot[(c + d*x)/2
] + 36*Log[Cos[(c + d*x)/2]] - 36*Log[Sin[(c + d*x)/2]] + 9*Sin[2*(c + d*x)] + 6*Tan[(c + d*x)/2]))/(12*d*(a +
a*Sin[c + d*x])^3)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.03

method	result	size
parallelrisch	\(\frac {-36 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (18 \cos \left (d x +c \right )-9 \cos \left (2 d x +2 c \right )-21\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+6 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+6 d x -33 \cos \left (d x +c \right )+\cos \left (3 d x +3 c \right )-32}{12 d \,a^{3}}\)	\(95\)
derivativedivides	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {-6 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {32}{3}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{3}}\)	\(125\)
default	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {-6 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {32}{3}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{3}}\)	\(125\)
risch	\(\frac {x}{2 a^{3}}-\frac {3 i {\mathrm e}^{2 i \left (d x +c \right )}}{8 d \,a^{3}}-\frac {11 \,{\mathrm e}^{i \left (d x +c \right )}}{8 d \,a^{3}}-\frac {11 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d \,a^{3}}+\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d \,a^{3}}-\frac {2 i}{a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{3}}+\frac {\cos \left (3 d x +3 c \right )}{12 d \,a^{3}}\)	\(157\)

[In]

int(cos(d*x+c)^8*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/12*(-36*ln(tan(1/2*d*x+1/2*c))+(18*cos(d*x+c)-9*cos(2*d*x+2*c)-21)*cot(1/2*d*x+1/2*c)+6*sec(1/2*d*x+1/2*c)*c
sc(1/2*d*x+1/2*c)+6*d*x-33*cos(d*x+c)+cos(3*d*x+3*c)-32)/d/a^3

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.13 \[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {9 \, \cos \left (d x + c\right )^{3} - {\left (2 \, \cos \left (d x + c\right )^{3} + 3 \, d x - 18 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 9 \, \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 9 \, \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )}{6 \, a^{3} d \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/6*(9*cos(d*x + c)^3 - (2*cos(d*x + c)^3 + 3*d*x - 18*cos(d*x + c))*sin(d*x + c) - 9*log(1/2*cos(d*x + c) +
1/2)*sin(d*x + c) + 9*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*cos(d*x + c))/(a^3*d*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**8*csc(d*x+c)**2/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (86) = 172\).

Time = 0.33 (sec) , antiderivative size = 285, normalized size of antiderivative = 3.10 \[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {\frac {32 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {72 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {9 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {24 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {21 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 3}{\frac {a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} - \frac {6 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac {18 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {3 \, \sin \left (d x + c\right )}{a^{3} {\left (\cos \left (d x + c\right ) + 1\right )}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/6*((32*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 72*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 9*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 24*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 21*sin(d*x + c
)^6/(cos(d*x + c) + 1)^6 + 3)/(a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 3*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3
+ 3*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7) - 6*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^3 + 18*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - 3*sin(d*x + c)/(a^3*(cos(d*x + c) +
1)))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.60 \[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {3 \, {\left (d x + c\right )}}{a^{3}} - \frac {18 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} + \frac {3 \, {\left (6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 36 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 16\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/6*(3*(d*x + c)/a^3 - 18*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 3*tan(1/2*d*x + 1/2*c)/a^3 + 3*(6*tan(1/2*d*x +
1/2*c) - 1)/(a^3*tan(1/2*d*x + 1/2*c)) - 2*(9*tan(1/2*d*x + 1/2*c)^5 + 12*tan(1/2*d*x + 1/2*c)^4 + 36*tan(1/2
*d*x + 1/2*c)^2 - 9*tan(1/2*d*x + 1/2*c) + 16)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^3))/d

Mupad [B] (verification not implemented)

Time = 10.53 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.51 \[ \int \frac {\cos ^6(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d}-\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {32\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+1}{d\,\left (2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {\mathrm {atan}\left (\frac {1}{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6}-\frac {6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6}\right )}{a^3\,d} \]

[In]

int(cos(c + d*x)^8/(sin(c + d*x)^2*(a + a*sin(c + d*x))^3),x)

[Out]

tan(c/2 + (d*x)/2)/(2*a^3*d) - ((32*tan(c/2 + (d*x)/2))/3 - 3*tan(c/2 + (d*x)/2)^2 + 24*tan(c/2 + (d*x)/2)^3 +
3*tan(c/2 + (d*x)/2)^4 + 8*tan(c/2 + (d*x)/2)^5 + 7*tan(c/2 + (d*x)/2)^6 + 1)/(d*(6*a^3*tan(c/2 + (d*x)/2)^3
+ 6*a^3*tan(c/2 + (d*x)/2)^5 + 2*a^3*tan(c/2 + (d*x)/2)^7 + 2*a^3*tan(c/2 + (d*x)/2))) - (3*log(tan(c/2 + (d*x
)/2)))/(a^3*d) - atan(1/(tan(c/2 + (d*x)/2) + 6) - (6*tan(c/2 + (d*x)/2))/(tan(c/2 + (d*x)/2) + 6))/(a^3*d)