\(\int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [745]

Optimal result

Integrand size = 29, antiderivative size = 92 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {3 x}{a^3}-\frac {\text {arctanh}(\cos (c+d x))}{2 a^3 d}-\frac {\cos (c+d x)}{a^3 d}-\frac {3 \cot (c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a^3 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2954, 2951, 3855, 3852, 8, 3853, 2718} \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{2 a^3 d}-\frac {\cos (c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a^3 d}-\frac {3 \cot (c+d x)}{a^3 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac {3 x}{a^3} \]

[In]

Int[(Cos[c + d*x]^4*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]

[Out]

(-3*x)/a^3 - ArcTanh[Cos[c + d*x]]/(2*a^3*d) - Cos[c + d*x]/(a^3*d) - (3*Cot[c + d*x])/(a^3*d) - Cot[c + d*x]^
3/(3*a^3*d) + (3*Cot[c + d*x]*Csc[c + d*x])/(2*a^3*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 ^2(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6} \\ & = \frac {\int \left (-3 a^5+2 a^5 \csc (c+d x)+2 a^5 \csc ^2(c+d x)-3 a^5 \csc ^3(c+d x)+a^5 \csc ^4(c+d x)+a^5 \sin (c+d x)\right ) \, dx}{a^8} \\ & = -\frac {3 x}{a^3}+\frac {\int \csc ^4(c+d x) \, dx}{a^3}+\frac {\int \sin (c+d x) \, dx}{a^3}+\frac {2 \int \csc (c+d x) \, dx}{a^3}+\frac {2 \int \csc ^2(c+d x) \, dx}{a^3}-\frac {3 \int \csc ^3(c+d x) \, dx}{a^3} \\ & = -\frac {3 x}{a^3}-\frac {2 \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {\cos (c+d x)}{a^3 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac {3 \int \csc (c+d x) \, dx}{2 a^3}-\frac {\text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a^3 d}-\frac {2 \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d} \\ & = -\frac {3 x}{a^3}-\frac {\text {arctanh}(\cos (c+d x))}{2 a^3 d}-\frac {\cos (c+d x)}{a^3 d}-\frac {3 \cot (c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a^3 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.69 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.43 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\csc ^3(c+d x) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6 \left (-12 \left (6 (c+d x)+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin ^3(c+d x)+2 \cos (3 (c+d x)) (8+3 \sin (c+d x))+6 \cos (c+d x) (-4+5 \sin (c+d x))\right )}{24 a^3 d (1+\sin (c+d x))^3} \]

[In]

Integrate[(Cos[c + d*x]^4*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]

[Out]

(Csc[c + d*x]^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6*(-12*(6*(c + d*x) + Log[Cos[(c + d*x)/2]] - Log[Sin[(c
+ d*x)/2]])*Sin[c + d*x]^3 + 2*Cos[3*(c + d*x)]*(8 + 3*Sin[c + d*x]) + 6*Cos[c + d*x]*(-4 + 5*Sin[c + d*x])))
/(24*a^3*d*(1 + Sin[c + d*x])^3)

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.38

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {3}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {11}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {16}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-48 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{3}}\)	\(127\)
default	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {3}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {11}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {16}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-48 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{3}}\)	\(127\)
parallelrisch	\(\frac {12 \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )-9 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+34 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (-12 d x +5\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 d x +9 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-34 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(147\)
risch	\(-\frac {3 x}{a^{3}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{3}}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{3}}-\frac {12 i {\mathrm e}^{4 i \left (d x +c \right )}+9 \,{\mathrm e}^{5 i \left (d x +c \right )}-36 i {\mathrm e}^{2 i \left (d x +c \right )}+16 i-9 \,{\mathrm e}^{i \left (d x +c \right )}}{3 a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d \,a^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d \,a^{3}}\)	\(152\)

[In]

int(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/8/d/a^3*(1/3*tan(1/2*d*x+1/2*c)^3-3*tan(1/2*d*x+1/2*c)^2+11*tan(1/2*d*x+1/2*c)-1/3/tan(1/2*d*x+1/2*c)^3+3/ta
n(1/2*d*x+1/2*c)^2-11/tan(1/2*d*x+1/2*c)+4*ln(tan(1/2*d*x+1/2*c))-16/(1+tan(1/2*d*x+1/2*c)^2)-48*arctan(tan(1/
2*d*x+1/2*c)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.63 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {32 \, \cos \left (d x + c\right )^{3} + 3 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 6 \, {\left (6 \, d x \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right )^{3} - 6 \, d x + \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 36 \, \cos \left (d x + c\right )}{12 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/12*(32*cos(d*x + c)^3 + 3*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*(cos(d*x + c)^2
- 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 6*(6*d*x*cos(d*x + c)^2 + 2*cos(d*x + c)^3 - 6*d*x + cos(d*x
+ c))*sin(d*x + c) - 36*cos(d*x + c))/((a^3*d*cos(d*x + c)^2 - a^3*d)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**8*csc(d*x+c)**4/(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. 242 vs. \(2 (86) = 172\).

Time = 0.31 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.63 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {34 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {39 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {33 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 1}{\frac {a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {\frac {33 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{3}} - \frac {144 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{24 \, d} \]

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/24*((9*sin(d*x + c)/(cos(d*x + c) + 1) - 34*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 39*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 - 33*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 1)/(a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + a^3*sin(
d*x + c)^5/(cos(d*x + c) + 1)^5) + (33*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
+ sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^3 - 144*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 + 12*log(sin(d*x
+ c)/(cos(d*x + c) + 1))/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.71 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {72 \, {\left (d x + c\right )}}{a^{3}} - \frac {12 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {48}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}} + \frac {22 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - \frac {a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 33 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{9}}}{24 \, d} \]

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/24*(72*(d*x + c)/a^3 - 12*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 48/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3) + (22*
tan(1/2*d*x + 1/2*c)^3 + 33*tan(1/2*d*x + 1/2*c)^2 - 9*tan(1/2*d*x + 1/2*c) + 1)/(a^3*tan(1/2*d*x + 1/2*c)^3)
- (a^6*tan(1/2*d*x + 1/2*c)^3 - 9*a^6*tan(1/2*d*x + 1/2*c)^2 + 33*a^6*tan(1/2*d*x + 1/2*c))/a^9)/d

Mupad [B] (verification not implemented)

Time = 9.52 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.38 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^3\,d}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}+\frac {6\,\mathrm {atan}\left (\frac {36}{36\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6}-\frac {6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{36\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6}\right )}{a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^3\,d}+\frac {11\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^3\,d}-\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {34\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {1}{3}}{d\,\left (8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )} \]

[In]

int(cos(c + d*x)^8/(sin(c + d*x)^4*(a + a*sin(c + d*x))^3),x)

[Out]

tan(c/2 + (d*x)/2)^3/(24*a^3*d) - (3*tan(c/2 + (d*x)/2)^2)/(8*a^3*d) + (6*atan(36/(36*tan(c/2 + (d*x)/2) + 6)
- (6*tan(c/2 + (d*x)/2))/(36*tan(c/2 + (d*x)/2) + 6)))/(a^3*d) + log(tan(c/2 + (d*x)/2))/(2*a^3*d) + (11*tan(c
/2 + (d*x)/2))/(8*a^3*d) - ((34*tan(c/2 + (d*x)/2)^2)/3 - 3*tan(c/2 + (d*x)/2) + 13*tan(c/2 + (d*x)/2)^3 + 11*
tan(c/2 + (d*x)/2)^4 + 1/3)/(d*(8*a^3*tan(c/2 + (d*x)/2)^3 + 8*a^3*tan(c/2 + (d*x)/2)^5))