\(\int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx\) [532]

Optimal result

Integrand size = 27, antiderivative size = 146 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {10 a^4 \csc (c+d x)}{d}+\frac {2 a^4 \csc ^2(c+d x)}{d}-\frac {4 a^4 \csc ^3(c+d x)}{3 d}-\frac {a^4 \csc ^4(c+d x)}{d}-\frac {a^4 \csc ^5(c+d x)}{5 d}-\frac {4 a^4 \log (\sin (c+d x))}{d}+\frac {4 a^4 \sin (c+d x)}{d}+\frac {2 a^4 \sin ^2(c+d x)}{d}+\frac {a^4 \sin ^3(c+d x)}{3 d} \]

[Out]

10*a^4*csc(d*x+c)/d+2*a^4*csc(d*x+c)^2/d-4/3*a^4*csc(d*x+c)^3/d-a^4*csc(d*x+c)^4/d-1/5*a^4*csc(d*x+c)^5/d-4*a^
4*ln(sin(d*x+c))/d+4*a^4*sin(d*x+c)/d+2*a^4*sin(d*x+c)^2/d+1/3*a^4*sin(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2915, 12, 90} \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \sin ^3(c+d x)}{3 d}+\frac {2 a^4 \sin ^2(c+d x)}{d}+\frac {4 a^4 \sin (c+d x)}{d}-\frac {a^4 \csc ^5(c+d x)}{5 d}-\frac {a^4 \csc ^4(c+d x)}{d}-\frac {4 a^4 \csc ^3(c+d x)}{3 d}+\frac {2 a^4 \csc ^2(c+d x)}{d}+\frac {10 a^4 \csc (c+d x)}{d}-\frac {4 a^4 \log (\sin (c+d x))}{d} \]

[In]

Int[Cot[c + d*x]^5*Csc[c + d*x]*(a + a*Sin[c + d*x])^4,x]

[Out]

(10*a^4*Csc[c + d*x])/d + (2*a^4*Csc[c + d*x]^2)/d - (4*a^4*Csc[c + d*x]^3)/(3*d) - (a^4*Csc[c + d*x]^4)/d - (
a^4*Csc[c + d*x]^5)/(5*d) - (4*a^4*Log[Sin[c + d*x]])/d + (4*a^4*Sin[c + d*x])/d + (2*a^4*Sin[c + d*x]^2)/d +
(a^4*Sin[c + d*x]^3)/(3*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2915

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {a^6 (a-x)^2 (a+x)^6}{x^6} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {a \text {Subst}\left (\int \frac {(a-x)^2 (a+x)^6}{x^6} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a \text {Subst}\left (\int \left (4 a^2+\frac {a^8}{x^6}+\frac {4 a^7}{x^5}+\frac {4 a^6}{x^4}-\frac {4 a^5}{x^3}-\frac {10 a^4}{x^2}-\frac {4 a^3}{x}+4 a x+x^2\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {10 a^4 \csc (c+d x)}{d}+\frac {2 a^4 \csc ^2(c+d x)}{d}-\frac {4 a^4 \csc ^3(c+d x)}{3 d}-\frac {a^4 \csc ^4(c+d x)}{d}-\frac {a^4 \csc ^5(c+d x)}{5 d}-\frac {4 a^4 \log (\sin (c+d x))}{d}+\frac {4 a^4 \sin (c+d x)}{d}+\frac {2 a^4 \sin ^2(c+d x)}{d}+\frac {a^4 \sin ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.66 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \left (150 \csc (c+d x)+30 \csc ^2(c+d x)-20 \csc ^3(c+d x)-15 \csc ^4(c+d x)-3 \csc ^5(c+d x)-60 \log (\sin (c+d x))+60 \sin (c+d x)+30 \sin ^2(c+d x)+5 \sin ^3(c+d x)\right )}{15 d} \]

[In]

Integrate[Cot[c + d*x]^5*Csc[c + d*x]*(a + a*Sin[c + d*x])^4,x]

[Out]

(a^4*(150*Csc[c + d*x] + 30*Csc[c + d*x]^2 - 20*Csc[c + d*x]^3 - 15*Csc[c + d*x]^4 - 3*Csc[c + d*x]^5 - 60*Log
[Sin[c + d*x]] + 60*Sin[c + d*x] + 30*Sin[c + d*x]^2 + 5*Sin[c + d*x]^3))/(15*d)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(-\frac {a^{4} \left (\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\csc ^{4}\left (d x +c \right )+\frac {4 \left (\csc ^{3}\left (d x +c \right )\right )}{3}-2 \left (\csc ^{2}\left (d x +c \right )\right )-10 \csc \left (d x +c \right )-4 \ln \left (\csc \left (d x +c \right )\right )-\frac {4}{\csc \left (d x +c \right )}-\frac {2}{\csc \left (d x +c \right )^{2}}-\frac {1}{3 \csc \left (d x +c \right )^{3}}\right )}{d}\)	\(95\)
default	\(-\frac {a^{4} \left (\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\csc ^{4}\left (d x +c \right )+\frac {4 \left (\csc ^{3}\left (d x +c \right )\right )}{3}-2 \left (\csc ^{2}\left (d x +c \right )\right )-10 \csc \left (d x +c \right )-4 \ln \left (\csc \left (d x +c \right )\right )-\frac {4}{\csc \left (d x +c \right )}-\frac {2}{\csc \left (d x +c \right )^{2}}-\frac {1}{3 \csc \left (d x +c \right )^{3}}\right )}{d}\)	\(95\)
parallelrisch	\(\frac {5 \left (\sec ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\left (-\sin \left (3 d x +3 c \right )+\frac {\sin \left (5 d x +5 c \right )}{5}+2 \sin \left (d x +c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\sin \left (3 d x +3 c \right )-\frac {\sin \left (5 d x +5 c \right )}{5}-2 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {201 \sin \left (3 d x +3 c \right )}{320}+\frac {37 \sin \left (5 d x +5 c \right )}{320}-\frac {\sin \left (7 d x +7 c \right )}{40}-\frac {61 \cos \left (2 d x +2 c \right )}{12}+\frac {199 \cos \left (4 d x +4 c \right )}{120}-\frac {7 \cos \left (6 d x +6 c \right )}{60}+\frac {\cos \left (8 d x +8 c \right )}{480}+\frac {109 \sin \left (d x +c \right )}{160}+\frac {8111}{2400}\right ) \left (\csc ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}{128 d}\)	\(202\)
risch	\(4 i a^{4} x +\frac {i a^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {a^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{2 d}-\frac {17 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {17 i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {a^{4} {\mathrm e}^{-2 i \left (d x +c \right )}}{2 d}-\frac {i a^{4} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {8 i a^{4} c}{d}+\frac {4 i a^{4} \left (75 \,{\mathrm e}^{9 i \left (d x +c \right )}-260 \,{\mathrm e}^{7 i \left (d x +c \right )}+30 i {\mathrm e}^{8 i \left (d x +c \right )}+346 \,{\mathrm e}^{5 i \left (d x +c \right )}-30 i {\mathrm e}^{6 i \left (d x +c \right )}-260 \,{\mathrm e}^{3 i \left (d x +c \right )}+30 i {\mathrm e}^{4 i \left (d x +c \right )}+75 \,{\mathrm e}^{i \left (d x +c \right )}-30 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {4 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(271\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

-a^4/d*(1/5*csc(d*x+c)^5+csc(d*x+c)^4+4/3*csc(d*x+c)^3-2*csc(d*x+c)^2-10*csc(d*x+c)-4*ln(csc(d*x+c))-4/csc(d*x
+c)-2/csc(d*x+c)^2-1/3/csc(d*x+c)^3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.32 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {5 \, a^{4} \cos \left (d x + c\right )^{8} - 80 \, a^{4} \cos \left (d x + c\right )^{6} + 360 \, a^{4} \cos \left (d x + c\right )^{4} - 480 \, a^{4} \cos \left (d x + c\right )^{2} + 192 \, a^{4} - 60 \, {\left (a^{4} \cos \left (d x + c\right )^{4} - 2 \, a^{4} \cos \left (d x + c\right )^{2} + a^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 15 \, {\left (2 \, a^{4} \cos \left (d x + c\right )^{6} - 5 \, a^{4} \cos \left (d x + c\right )^{4} + 6 \, a^{4} \cos \left (d x + c\right )^{2} - 2 \, a^{4}\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/15*(5*a^4*cos(d*x + c)^8 - 80*a^4*cos(d*x + c)^6 + 360*a^4*cos(d*x + c)^4 - 480*a^4*cos(d*x + c)^2 + 192*a^4
- 60*(a^4*cos(d*x + c)^4 - 2*a^4*cos(d*x + c)^2 + a^4)*log(1/2*sin(d*x + c))*sin(d*x + c) - 15*(2*a^4*cos(d*x
+ c)^6 - 5*a^4*cos(d*x + c)^4 + 6*a^4*cos(d*x + c)^2 - 2*a^4)*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x
+ c)^2 + d)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**6*(a+a*sin(d*x+c))**4,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.82 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {5 \, a^{4} \sin \left (d x + c\right )^{3} + 30 \, a^{4} \sin \left (d x + c\right )^{2} - 60 \, a^{4} \log \left (\sin \left (d x + c\right )\right ) + 60 \, a^{4} \sin \left (d x + c\right ) + \frac {150 \, a^{4} \sin \left (d x + c\right )^{4} + 30 \, a^{4} \sin \left (d x + c\right )^{3} - 20 \, a^{4} \sin \left (d x + c\right )^{2} - 15 \, a^{4} \sin \left (d x + c\right ) - 3 \, a^{4}}{\sin \left (d x + c\right )^{5}}}{15 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

1/15*(5*a^4*sin(d*x + c)^3 + 30*a^4*sin(d*x + c)^2 - 60*a^4*log(sin(d*x + c)) + 60*a^4*sin(d*x + c) + (150*a^4
*sin(d*x + c)^4 + 30*a^4*sin(d*x + c)^3 - 20*a^4*sin(d*x + c)^2 - 15*a^4*sin(d*x + c) - 3*a^4)/sin(d*x + c)^5)
/d

Giac [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.92 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {5 \, a^{4} \sin \left (d x + c\right )^{3} + 30 \, a^{4} \sin \left (d x + c\right )^{2} - 60 \, a^{4} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, a^{4} \sin \left (d x + c\right ) + \frac {137 \, a^{4} \sin \left (d x + c\right )^{5} + 150 \, a^{4} \sin \left (d x + c\right )^{4} + 30 \, a^{4} \sin \left (d x + c\right )^{3} - 20 \, a^{4} \sin \left (d x + c\right )^{2} - 15 \, a^{4} \sin \left (d x + c\right ) - 3 \, a^{4}}{\sin \left (d x + c\right )^{5}}}{15 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/15*(5*a^4*sin(d*x + c)^3 + 30*a^4*sin(d*x + c)^2 - 60*a^4*log(abs(sin(d*x + c))) + 60*a^4*sin(d*x + c) + (13
7*a^4*sin(d*x + c)^5 + 150*a^4*sin(d*x + c)^4 + 30*a^4*sin(d*x + c)^3 - 20*a^4*sin(d*x + c)^2 - 15*a^4*sin(d*x
+ c) - 3*a^4)/sin(d*x + c)^5)/d

Mupad [B] (verification not implemented)

Time = 9.35 (sec) , antiderivative size = 357, normalized size of antiderivative = 2.45 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}-\frac {19\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{16\,d}-\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}-\frac {4\,a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {398\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+264\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+1017\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+278\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {3314\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{5}+18\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {612\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {104\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a^4}{5}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+96\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+96\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {71\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}+\frac {4\,a^4\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

[In]

int((cos(c + d*x)^5*(a + a*sin(c + d*x))^4)/sin(c + d*x)^6,x)

[Out]

(a^4*tan(c/2 + (d*x)/2)^2)/(4*d) - (19*a^4*tan(c/2 + (d*x)/2)^3)/(96*d) - (a^4*tan(c/2 + (d*x)/2)^4)/(16*d) -
(a^4*tan(c/2 + (d*x)/2)^5)/(160*d) - (4*a^4*log(tan(c/2 + (d*x)/2)))/d + (2*a^4*tan(c/2 + (d*x)/2)^3 - (104*a^
4*tan(c/2 + (d*x)/2)^2)/15 + (612*a^4*tan(c/2 + (d*x)/2)^4)/5 + 18*a^4*tan(c/2 + (d*x)/2)^5 + (3314*a^4*tan(c/
2 + (d*x)/2)^6)/5 + 278*a^4*tan(c/2 + (d*x)/2)^7 + 1017*a^4*tan(c/2 + (d*x)/2)^8 + 264*a^4*tan(c/2 + (d*x)/2)^
9 + 398*a^4*tan(c/2 + (d*x)/2)^10 - a^4/5 - 2*a^4*tan(c/2 + (d*x)/2))/(d*(32*tan(c/2 + (d*x)/2)^5 + 96*tan(c/2
+ (d*x)/2)^7 + 96*tan(c/2 + (d*x)/2)^9 + 32*tan(c/2 + (d*x)/2)^11)) + (71*a^4*tan(c/2 + (d*x)/2))/(16*d) + (4
*a^4*log(tan(c/2 + (d*x)/2)^2 + 1))/d