\(\int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx\) [519]

Optimal result

Integrand size = 27, antiderivative size = 112 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \csc (c+d x)}{d}+\frac {2 a^2 \csc ^2(c+d x)}{d}+\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {a^2 \csc ^4(c+d x)}{2 d}-\frac {a^2 \csc ^5(c+d x)}{5 d}+\frac {2 a^2 \log (\sin (c+d x))}{d}+\frac {a^2 \sin (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 112, 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} \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \sin (c+d x)}{d}-\frac {a^2 \csc ^5(c+d x)}{5 d}-\frac {a^2 \csc ^4(c+d x)}{2 d}+\frac {a^2 \csc ^3(c+d x)}{3 d}+\frac {2 a^2 \csc ^2(c+d x)}{d}+\frac {a^2 \csc (c+d x)}{d}+\frac {2 a^2 \log (\sin (c+d x))}{d} \]

[In]

Int[Cot[c + d*x]^5*Csc[c + d*x]*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*Csc[c + d*x])/d + (2*a^2*Csc[c + d*x]^2)/d + (a^2*Csc[c + d*x]^3)/(3*d) - (a^2*Csc[c + d*x]^4)/(2*d) - (a
^2*Csc[c + d*x]^5)/(5*d) + (2*a^2*Log[Sin[c + d*x]])/d + (a^2*Sin[c + d*x])/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)^4}{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)^4}{x^6} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a \text {Subst}\left (\int \left (1+\frac {a^6}{x^6}+\frac {2 a^5}{x^5}-\frac {a^4}{x^4}-\frac {4 a^3}{x^3}-\frac {a^2}{x^2}+\frac {2 a}{x}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^2 \csc (c+d x)}{d}+\frac {2 a^2 \csc ^2(c+d x)}{d}+\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {a^2 \csc ^4(c+d x)}{2 d}-\frac {a^2 \csc ^5(c+d x)}{5 d}+\frac {2 a^2 \log (\sin (c+d x))}{d}+\frac {a^2 \sin (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.68 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (30 \csc (c+d x)+60 \csc ^2(c+d x)+10 \csc ^3(c+d x)-15 \csc ^4(c+d x)-6 \csc ^5(c+d x)+60 \log (\sin (c+d x))+30 \sin (c+d x)\right )}{30 d} \]

[In]

Integrate[Cot[c + d*x]^5*Csc[c + d*x]*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(30*Csc[c + d*x] + 60*Csc[c + d*x]^2 + 10*Csc[c + d*x]^3 - 15*Csc[c + d*x]^4 - 6*Csc[c + d*x]^5 + 60*Log[
Sin[c + d*x]] + 30*Sin[c + d*x]))/(30*d)

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.69

method	result	size
derivativedivides	\(-\frac {a^{2} \left (\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{2}-\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-2 \left (\csc ^{2}\left (d x +c \right )\right )-\csc \left (d x +c \right )+2 \ln \left (\csc \left (d x +c \right )\right )-\frac {1}{\csc \left (d x +c \right )}\right )}{d}\)	\(77\)
default	\(-\frac {a^{2} \left (\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{2}-\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-2 \left (\csc ^{2}\left (d x +c \right )\right )-\csc \left (d x +c \right )+2 \ln \left (\csc \left (d x +c \right )\right )-\frac {1}{\csc \left (d x +c \right )}\right )}{d}\)	\(77\)
parallelrisch	\(\frac {\left (\csc ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\left (-\sin \left (5 d x +5 c \right )+5 \sin \left (3 d x +3 c \right )-10 \sin \left (d x +c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\sin \left (5 d x +5 c \right )-5 \sin \left (3 d x +3 c \right )+10 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {19 \sin \left (5 d x +5 c \right )}{32}-\frac {109 \cos \left (2 d x +2 c \right )}{12}+\frac {5 \cos \left (4 d x +4 c \right )}{2}-\frac {\cos \left (6 d x +6 c \right )}{4}+\frac {33 \sin \left (d x +c \right )}{16}-\frac {33 \sin \left (3 d x +3 c \right )}{32}+\frac {157}{30}\right ) \left (\sec ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{256 d}\)	\(180\)
risch	\(-2 i a^{2} x -\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {4 i a^{2} c}{d}+\frac {2 i a^{2} \left (15 \,{\mathrm e}^{9 i \left (d x +c \right )}-80 \,{\mathrm e}^{7 i \left (d x +c \right )}+60 i {\mathrm e}^{8 i \left (d x +c \right )}+82 \,{\mathrm e}^{5 i \left (d x +c \right )}-120 i {\mathrm e}^{6 i \left (d x +c \right )}-80 \,{\mathrm e}^{3 i \left (d x +c \right )}+120 i {\mathrm e}^{4 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}-60 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(201\)
norman	\(\frac {-\frac {a^{2}}{160 d}-\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32 d}-\frac {a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{480 d}+\frac {5 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}+\frac {277 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{480 d}+\frac {355 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}+\frac {355 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}+\frac {277 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{480 d}+\frac {5 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}-\frac {a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{480 d}-\frac {a^{2} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d}-\frac {a^{2} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}-\frac {11 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(304\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.37 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {30 \, a^{2} \cos \left (d x + c\right )^{6} - 120 \, a^{2} \cos \left (d x + c\right )^{4} + 160 \, a^{2} \cos \left (d x + c\right )^{2} - 60 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 64 \, a^{2} + 15 \, {\left (4 \, a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2}\right )} \sin \left (d x + c\right )}{30 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/30*(30*a^2*cos(d*x + c)^6 - 120*a^2*cos(d*x + c)^4 + 160*a^2*cos(d*x + c)^2 - 60*(a^2*cos(d*x + c)^4 - 2*a^
2*cos(d*x + c)^2 + a^2)*log(1/2*sin(d*x + c))*sin(d*x + c) - 64*a^2 + 15*(4*a^2*cos(d*x + c)^2 - 3*a^2)*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. \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**6*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.84 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {60 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + 30 \, a^{2} \sin \left (d x + c\right ) + \frac {30 \, a^{2} \sin \left (d x + c\right )^{4} + 60 \, a^{2} \sin \left (d x + c\right )^{3} + 10 \, a^{2} \sin \left (d x + c\right )^{2} - 15 \, a^{2} \sin \left (d x + c\right ) - 6 \, a^{2}}{\sin \left (d x + c\right )^{5}}}{30 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/30*(60*a^2*log(sin(d*x + c)) + 30*a^2*sin(d*x + c) + (30*a^2*sin(d*x + c)^4 + 60*a^2*sin(d*x + c)^3 + 10*a^2
*sin(d*x + c)^2 - 15*a^2*sin(d*x + c) - 6*a^2)/sin(d*x + c)^5)/d

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.97 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 30 \, a^{2} \sin \left (d x + c\right ) - \frac {137 \, a^{2} \sin \left (d x + c\right )^{5} - 30 \, a^{2} \sin \left (d x + c\right )^{4} - 60 \, a^{2} \sin \left (d x + c\right )^{3} - 10 \, a^{2} \sin \left (d x + c\right )^{2} + 15 \, a^{2} \sin \left (d x + c\right ) + 6 \, a^{2}}{\sin \left (d x + c\right )^{5}}}{30 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/30*(60*a^2*log(abs(sin(d*x + c))) + 30*a^2*sin(d*x + c) - (137*a^2*sin(d*x + c)^5 - 30*a^2*sin(d*x + c)^4 -
60*a^2*sin(d*x + c)^3 - 10*a^2*sin(d*x + c)^2 + 15*a^2*sin(d*x + c) + 6*a^2)/sin(d*x + c)^5)/d

Mupad [B] (verification not implemented)

Time = 9.76 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.38 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {82\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {55\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a^2}{5}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {2\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {9\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}-\frac {2\,a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

[In]

int((cos(c + d*x)^5*(a + a*sin(c + d*x))^2)/sin(c + d*x)^6,x)

[Out]

((2*a^2*tan(c/2 + (d*x)/2)^2)/15 + 11*a^2*tan(c/2 + (d*x)/2)^3 + (55*a^2*tan(c/2 + (d*x)/2)^4)/3 + 12*a^2*tan(
c/2 + (d*x)/2)^5 + 82*a^2*tan(c/2 + (d*x)/2)^6 - a^2/5 - a^2*tan(c/2 + (d*x)/2))/(d*(32*tan(c/2 + (d*x)/2)^5 +
32*tan(c/2 + (d*x)/2)^7)) + (3*a^2*tan(c/2 + (d*x)/2)^2)/(8*d) + (a^2*tan(c/2 + (d*x)/2)^3)/(96*d) - (a^2*tan
(c/2 + (d*x)/2)^4)/(32*d) - (a^2*tan(c/2 + (d*x)/2)^5)/(160*d) + (2*a^2*log(tan(c/2 + (d*x)/2)))/d + (9*a^2*ta
n(c/2 + (d*x)/2))/(16*d) - (2*a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d