\(\int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx\) [528]

Optimal result

Integrand size = 27, antiderivative size = 133 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {5 a^3 \csc (c+d x)}{d}+\frac {5 a^3 \csc ^2(c+d x)}{2 d}-\frac {a^3 \csc ^3(c+d x)}{3 d}-\frac {3 a^3 \csc ^4(c+d x)}{4 d}-\frac {a^3 \csc ^5(c+d x)}{5 d}+\frac {a^3 \log (\sin (c+d x))}{d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {a^3 \sin ^2(c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 133, 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} \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \sin ^2(c+d x)}{2 d}+\frac {3 a^3 \sin (c+d x)}{d}-\frac {a^3 \csc ^5(c+d x)}{5 d}-\frac {3 a^3 \csc ^4(c+d x)}{4 d}-\frac {a^3 \csc ^3(c+d x)}{3 d}+\frac {5 a^3 \csc ^2(c+d x)}{2 d}+\frac {5 a^3 \csc (c+d x)}{d}+\frac {a^3 \log (\sin (c+d x))}{d} \]

[In]

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

[Out]

(5*a^3*Csc[c + d*x])/d + (5*a^3*Csc[c + d*x]^2)/(2*d) - (a^3*Csc[c + d*x]^3)/(3*d) - (3*a^3*Csc[c + d*x]^4)/(4
*d) - (a^3*Csc[c + d*x]^5)/(5*d) + (a^3*Log[Sin[c + d*x]])/d + (3*a^3*Sin[c + d*x])/d + (a^3*Sin[c + d*x]^2)/(
2*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)^5}{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)^5}{x^6} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a \text {Subst}\left (\int \left (3 a+\frac {a^7}{x^6}+\frac {3 a^6}{x^5}+\frac {a^5}{x^4}-\frac {5 a^4}{x^3}-\frac {5 a^3}{x^2}+\frac {a^2}{x}+x\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {5 a^3 \csc (c+d x)}{d}+\frac {5 a^3 \csc ^2(c+d x)}{2 d}-\frac {a^3 \csc ^3(c+d x)}{3 d}-\frac {3 a^3 \csc ^4(c+d x)}{4 d}-\frac {a^3 \csc ^5(c+d x)}{5 d}+\frac {a^3 \log (\sin (c+d x))}{d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {a^3 \sin ^2(c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.65 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \left (300 \csc (c+d x)+150 \csc ^2(c+d x)-20 \csc ^3(c+d x)-45 \csc ^4(c+d x)-12 \csc ^5(c+d x)+60 \log (\sin (c+d x))+180 \sin (c+d x)+30 \sin ^2(c+d x)\right )}{60 d} \]

[In]

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

[Out]

(a^3*(300*Csc[c + d*x] + 150*Csc[c + d*x]^2 - 20*Csc[c + d*x]^3 - 45*Csc[c + d*x]^4 - 12*Csc[c + d*x]^5 + 60*L
og[Sin[c + d*x]] + 180*Sin[c + d*x] + 30*Sin[c + d*x]^2))/(60*d)

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.64

method	result	size
derivativedivides	\(-\frac {a^{3} \left (\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {3 \left (\csc ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\frac {5 \left (\csc ^{2}\left (d x +c \right )\right )}{2}-5 \csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )\right )-\frac {3}{\csc \left (d x +c \right )}-\frac {1}{2 \csc \left (d x +c \right )^{2}}\right )}{d}\)	\(85\)
default	\(-\frac {a^{3} \left (\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {3 \left (\csc ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\frac {5 \left (\csc ^{2}\left (d x +c \right )\right )}{2}-5 \csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )\right )-\frac {3}{\csc \left (d x +c \right )}-\frac {1}{2 \csc \left (d x +c \right )^{2}}\right )}{d}\)	\(85\)
parallelrisch	\(\frac {\left (\left (-\sin \left (5 d x +5 c \right )+5 \sin \left (3 d x +3 c \right )-10 \sin \left (d x +c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\sin \left (5 d x +5 c \right )-5 \sin \left (3 d x +3 c \right )+10 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {59 \sin \left (5 d x +5 c \right )}{64}-\frac {\sin \left (7 d x +7 c \right )}{8}-\frac {359 \cos \left (2 d x +2 c \right )}{6}+19 \cos \left (4 d x +4 c \right )-\frac {3 \cos \left (6 d x +6 c \right )}{2}+\frac {141 \sin \left (d x +c \right )}{32}-\frac {233 \sin \left (3 d x +3 c \right )}{64}+\frac {587}{15}\right ) \left (\sec ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{512 d}\)	\(191\)
risch	\(-i a^{3} x -\frac {a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {3 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {3 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{3} c}{d}+\frac {2 i a^{3} \left (75 \,{\mathrm e}^{9 i \left (d x +c \right )}-280 \,{\mathrm e}^{7 i \left (d x +c \right )}+75 i {\mathrm e}^{8 i \left (d x +c \right )}+362 \,{\mathrm e}^{5 i \left (d x +c \right )}-135 i {\mathrm e}^{6 i \left (d x +c \right )}-280 \,{\mathrm e}^{3 i \left (d x +c \right )}+135 i {\mathrm e}^{4 i \left (d x +c \right )}+75 \,{\mathrm e}^{i \left (d x +c \right )}-75 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(234\)
norman	\(\frac {-\frac {a^{3}}{160 d}-\frac {3 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d}-\frac {11 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{120 d}+\frac {19 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}+\frac {83 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d}+\frac {601 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d}+\frac {1235 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 d}+\frac {601 a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d}+\frac {83 a^{3} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d}+\frac {19 a^{3} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}-\frac {11 a^{3} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{120 d}-\frac {3 a^{3} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}-\frac {a^{3} \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}+\frac {3 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}+\frac {3 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(341\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.35 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {180 \, a^{3} \cos \left (d x + c\right )^{6} - 840 \, a^{3} \cos \left (d x + c\right )^{4} + 1120 \, a^{3} \cos \left (d x + c\right )^{2} - 448 \, a^{3} - 60 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 15 \, {\left (2 \, a^{3} \cos \left (d x + c\right )^{6} - 5 \, a^{3} \cos \left (d x + c\right )^{4} + 14 \, a^{3} \cos \left (d x + c\right )^{2} - 8 \, a^{3}\right )} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(180*a^3*cos(d*x + c)^6 - 840*a^3*cos(d*x + c)^4 + 1120*a^3*cos(d*x + c)^2 - 448*a^3 - 60*(a^3*cos(d*x +
c)^4 - 2*a^3*cos(d*x + c)^2 + a^3)*log(1/2*sin(d*x + c))*sin(d*x + c) + 15*(2*a^3*cos(d*x + c)^6 - 5*a^3*cos(
d*x + c)^4 + 14*a^3*cos(d*x + c)^2 - 8*a^3)*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. \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**6*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.80 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {30 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 180 \, a^{3} \sin \left (d x + c\right ) + \frac {300 \, a^{3} \sin \left (d x + c\right )^{4} + 150 \, a^{3} \sin \left (d x + c\right )^{3} - 20 \, a^{3} \sin \left (d x + c\right )^{2} - 45 \, a^{3} \sin \left (d x + c\right ) - 12 \, a^{3}}{\sin \left (d x + c\right )^{5}}}{60 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(30*a^3*sin(d*x + c)^2 + 60*a^3*log(sin(d*x + c)) + 180*a^3*sin(d*x + c) + (300*a^3*sin(d*x + c)^4 + 150*
a^3*sin(d*x + c)^3 - 20*a^3*sin(d*x + c)^2 - 45*a^3*sin(d*x + c) - 12*a^3)/sin(d*x + c)^5)/d

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.92 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {30 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 180 \, a^{3} \sin \left (d x + c\right ) - \frac {137 \, a^{3} \sin \left (d x + c\right )^{5} - 300 \, a^{3} \sin \left (d x + c\right )^{4} - 150 \, a^{3} \sin \left (d x + c\right )^{3} + 20 \, a^{3} \sin \left (d x + c\right )^{2} + 45 \, a^{3} \sin \left (d x + c\right ) + 12 \, a^{3}}{\sin \left (d x + c\right )^{5}}}{60 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(30*a^3*sin(d*x + c)^2 + 60*a^3*log(abs(sin(d*x + c))) + 180*a^3*sin(d*x + c) - (137*a^3*sin(d*x + c)^5 -
300*a^3*sin(d*x + c)^4 - 150*a^3*sin(d*x + c)^3 + 20*a^3*sin(d*x + c)^2 + 45*a^3*sin(d*x + c) + 12*a^3)/sin(d
*x + c)^5)/d

Mupad [B] (verification not implemented)

Time = 9.85 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.34 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {7\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {7\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {266\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+78\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {1013\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {53\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}+\frac {1037\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}+11\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {41\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-\frac {a^3}{5}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {37\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}-\frac {a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

[In]

int((cos(c + d*x)^5*(a + a*sin(c + d*x))^3)/sin(c + d*x)^6,x)

[Out]

(7*a^3*tan(c/2 + (d*x)/2)^2)/(16*d) - (7*a^3*tan(c/2 + (d*x)/2)^3)/(96*d) - (3*a^3*tan(c/2 + (d*x)/2)^4)/(64*d
) - (a^3*tan(c/2 + (d*x)/2)^5)/(160*d) + (a^3*log(tan(c/2 + (d*x)/2)))/d + (11*a^3*tan(c/2 + (d*x)/2)^3 - (41*
a^3*tan(c/2 + (d*x)/2)^2)/15 + (1037*a^3*tan(c/2 + (d*x)/2)^4)/15 + (53*a^3*tan(c/2 + (d*x)/2)^5)/2 + (1013*a^
3*tan(c/2 + (d*x)/2)^6)/3 + 78*a^3*tan(c/2 + (d*x)/2)^7 + 266*a^3*tan(c/2 + (d*x)/2)^8 - a^3/5 - (3*a^3*tan(c/
2 + (d*x)/2))/2)/(d*(32*tan(c/2 + (d*x)/2)^5 + 64*tan(c/2 + (d*x)/2)^7 + 32*tan(c/2 + (d*x)/2)^9)) + (37*a^3*t
an(c/2 + (d*x)/2))/(16*d) - (a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d