\(\int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [561]

Optimal result

Integrand size = 21, antiderivative size = 96 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4 \csc (c+d x)}{a^3 d}-\frac {2 \csc ^2(c+d x)}{a^3 d}+\frac {\csc ^3(c+d x)}{a^3 d}-\frac {\csc ^4(c+d x)}{4 a^3 d}+\frac {4 \log (\sin (c+d x))}{a^3 d}-\frac {4 \log (1+\sin (c+d x))}{a^3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2786, 90} \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\csc ^4(c+d x)}{4 a^3 d}+\frac {\csc ^3(c+d x)}{a^3 d}-\frac {2 \csc ^2(c+d x)}{a^3 d}+\frac {4 \csc (c+d x)}{a^3 d}+\frac {4 \log (\sin (c+d x))}{a^3 d}-\frac {4 \log (\sin (c+d x)+1)}{a^3 d} \]

[In]

Int[Cot[c + d*x]^5/(a + a*Sin[c + d*x])^3,x]

[Out]

(4*Csc[c + d*x])/(a^3*d) - (2*Csc[c + d*x]^2)/(a^3*d) + Csc[c + d*x]^3/(a^3*d) - Csc[c + d*x]^4/(4*a^3*d) + (4
*Log[Sin[c + d*x]])/(a^3*d) - (4*Log[1 + Sin[c + d*x]])/(a^3*d)

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2786

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[x^p*((a + x)^(m - (p + 1)/2)/(a - x)^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a-x)^2}{x^5 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {a}{x^5}-\frac {3}{x^4}+\frac {4}{a x^3}-\frac {4}{a^2 x^2}+\frac {4}{a^3 x}-\frac {4}{a^3 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {4 \csc (c+d x)}{a^3 d}-\frac {2 \csc ^2(c+d x)}{a^3 d}+\frac {\csc ^3(c+d x)}{a^3 d}-\frac {\csc ^4(c+d x)}{4 a^3 d}+\frac {4 \log (\sin (c+d x))}{a^3 d}-\frac {4 \log (1+\sin (c+d x))}{a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.72 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {16 \csc (c+d x)-8 \csc ^2(c+d x)+4 \csc ^3(c+d x)-\csc ^4(c+d x)+16 \log (\sin (c+d x))-16 \log (1+\sin (c+d x))}{4 a^3 d} \]

[In]

Integrate[Cot[c + d*x]^5/(a + a*Sin[c + d*x])^3,x]

[Out]

(16*Csc[c + d*x] - 8*Csc[c + d*x]^2 + 4*Csc[c + d*x]^3 - Csc[c + d*x]^4 + 16*Log[Sin[c + d*x]] - 16*Log[1 + Si
n[c + d*x]])/(4*a^3*d)

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.70

method	result	size
derivativedivides	\(\frac {-\frac {1}{4 \sin \left (d x +c \right )^{4}}+\frac {1}{\sin \left (d x +c \right )^{3}}-\frac {2}{\sin \left (d x +c \right )^{2}}+\frac {4}{\sin \left (d x +c \right )}+4 \ln \left (\sin \left (d x +c \right )\right )-4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\)	\(67\)
default	\(\frac {-\frac {1}{4 \sin \left (d x +c \right )^{4}}+\frac {1}{\sin \left (d x +c \right )^{3}}-\frac {2}{\sin \left (d x +c \right )^{2}}+\frac {4}{\sin \left (d x +c \right )}+4 \ln \left (\sin \left (d x +c \right )\right )-4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\)	\(67\)
parallelrisch	\(\frac {-\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-36 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-36 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+152 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+152 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+256 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-512 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64 d \,a^{3}}\)	\(136\)
risch	\(\frac {4 i \left (-2 i {\mathrm e}^{6 i \left (d x +c \right )}+2 \,{\mathrm e}^{7 i \left (d x +c \right )}+5 i {\mathrm e}^{4 i \left (d x +c \right )}-8 \,{\mathrm e}^{5 i \left (d x +c \right )}-2 i {\mathrm e}^{2 i \left (d x +c \right )}+8 \,{\mathrm e}^{3 i \left (d x +c \right )}-2 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}-\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}+\frac {4 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{3}}\)	\(146\)
norman	\(\frac {-\frac {16 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {16 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {1}{64 a d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}-\frac {3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {21 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {21 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}-\frac {3 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {3 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d a}-\frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}-\frac {1339 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}-\frac {1339 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}-\frac {8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}\)	\(284\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^5/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.36 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {8 \, \cos \left (d x + c\right )^{2} + 16 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 16 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, {\left (4 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) - 9}{4 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(8*cos(d*x + c)^2 + 16*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*sin(d*x + c)) - 16*(cos(d*x + c)^4
- 2*cos(d*x + c)^2 + 1)*log(sin(d*x + c) + 1) - 4*(4*cos(d*x + c)^2 - 5)*sin(d*x + c) - 9)/(a^3*d*cos(d*x + c)
^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**5/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.78 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {16 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} - \frac {16 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}} - \frac {16 \, \sin \left (d x + c\right )^{3} - 8 \, \sin \left (d x + c\right )^{2} + 4 \, \sin \left (d x + c\right ) - 1}{a^{3} \sin \left (d x + c\right )^{4}}}{4 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.81 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {1536 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {768 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {1600 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 456 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 108 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}} + \frac {3 \, {\left (a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 152 \, a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{12}}}{192 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/192*(1536*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 768*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + (1600*tan(1/2*d
*x + 1/2*c)^4 - 456*tan(1/2*d*x + 1/2*c)^3 + 108*tan(1/2*d*x + 1/2*c)^2 - 24*tan(1/2*d*x + 1/2*c) + 3)/(a^3*ta
n(1/2*d*x + 1/2*c)^4) + 3*(a^9*tan(1/2*d*x + 1/2*c)^4 - 8*a^9*tan(1/2*d*x + 1/2*c)^3 + 36*a^9*tan(1/2*d*x + 1/
2*c)^2 - 152*a^9*tan(1/2*d*x + 1/2*c))/a^12)/d

Mupad [B] (verification not implemented)

Time = 10.00 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.78 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8\,a^3\,d}-\frac {9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^3\,d}+\frac {4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^3\,d}+\frac {19\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^3\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (38\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{4}\right )}{16\,a^3\,d} \]

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^5*(a + a*sin(c + d*x))^3),x)

[Out]

tan(c/2 + (d*x)/2)^3/(8*a^3*d) - (9*tan(c/2 + (d*x)/2)^2)/(16*a^3*d) - tan(c/2 + (d*x)/2)^4/(64*a^3*d) + (4*lo
g(tan(c/2 + (d*x)/2)))/(a^3*d) - (8*log(tan(c/2 + (d*x)/2) + 1))/(a^3*d) + (19*tan(c/2 + (d*x)/2))/(8*a^3*d) +
(cot(c/2 + (d*x)/2)^4*(2*tan(c/2 + (d*x)/2) - 9*tan(c/2 + (d*x)/2)^2 + 38*tan(c/2 + (d*x)/2)^3 - 1/4))/(16*a^
3*d)