\(\int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x)) \, dx\) [504]

Optimal result

Integrand size = 25, antiderivative size = 86 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}+\frac {a \csc ^2(c+d x)}{d}+\frac {2 a \csc ^3(c+d x)}{3 d}-\frac {a \csc ^4(c+d x)}{4 d}-\frac {a \csc ^5(c+d x)}{5 d}+\frac {a \log (\sin (c+d x))}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2915, 12, 90} \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc ^5(c+d x)}{5 d}-\frac {a \csc ^4(c+d x)}{4 d}+\frac {2 a \csc ^3(c+d x)}{3 d}+\frac {a \csc ^2(c+d x)}{d}-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.17 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \cot ^2(c+d x)}{2 d}-\frac {a \cot ^4(c+d x)}{4 d}-\frac {a \csc (c+d x)}{d}+\frac {2 a \csc ^3(c+d x)}{3 d}-\frac {a \csc ^5(c+d x)}{5 d}+\frac {a \log (\cos (c+d x))}{d}+\frac {a \log (\tan (c+d x))}{d} \]

[In]

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

[Out]

(a*Cot[c + d*x]^2)/(2*d) - (a*Cot[c + d*x]^4)/(4*d) - (a*Csc[c + d*x])/d + (2*a*Csc[c + d*x]^3)/(3*d) - (a*Csc
[c + d*x]^5)/(5*d) + (a*Log[Cos[c + d*x]])/d + (a*Log[Tan[c + d*x]])/d

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(-\frac {a \left (\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{4}-\frac {2 \left (\csc ^{3}\left (d x +c \right )\right )}{3}-\left (\csc ^{2}\left (d x +c \right )\right )+\csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )\right )\right )}{d}\)	\(61\)
default	\(-\frac {a \left (\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{4}-\frac {2 \left (\csc ^{3}\left (d x +c \right )\right )}{3}-\left (\csc ^{2}\left (d x +c \right )\right )+\csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )\right )\right )}{d}\)	\(61\)
risch	\(-i a x -\frac {2 i a c}{d}-\frac {2 i a \left (15 \,{\mathrm e}^{9 i \left (d x +c \right )}-20 \,{\mathrm e}^{7 i \left (d x +c \right )}-30 i {\mathrm e}^{8 i \left (d x +c \right )}+58 \,{\mathrm e}^{5 i \left (d x +c \right )}+60 i {\mathrm e}^{6 i \left (d x +c \right )}-20 \,{\mathrm e}^{3 i \left (d x +c \right )}-60 i {\mathrm e}^{4 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}+30 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(156\)
parallelrisch	\(-\frac {\left (\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )+\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {5 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {25 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {25 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-30 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-30 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+50 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+50 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-160 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+160 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) a}{160 d}\)	\(156\)
norman	\(\frac {-\frac {a}{160 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d}+\frac {11 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{240 d}+\frac {11 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}-\frac {25 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}-\frac {5 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {25 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}+\frac {11 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}+\frac {11 a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{240 d}-\frac {a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}-\frac {a \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(239\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.44 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {60 \, a \cos \left (d x + c\right )^{4} - 80 \, a \cos \left (d x + c\right )^{2} - 60 \, {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 15 \, {\left (4 \, a \cos \left (d x + c\right )^{2} - 3 \, a\right )} \sin \left (d x + c\right ) + 32 \, a}{60 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

[In]

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

[Out]

-1/60*(60*a*cos(d*x + c)^4 - 80*a*cos(d*x + c)^2 - 60*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a)*log(1/2*sin(
d*x + c))*sin(d*x + c) + 15*(4*a*cos(d*x + c)^2 - 3*a)*sin(d*x + c) + 32*a)/((d*cos(d*x + c)^4 - 2*d*cos(d*x +
c)^2 + d)*sin(d*x + c))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x)) \, dx=\frac {60 \, a \log \left (\sin \left (d x + c\right )\right ) - \frac {60 \, a \sin \left (d x + c\right )^{4} - 60 \, a \sin \left (d x + c\right )^{3} - 40 \, a \sin \left (d x + c\right )^{2} + 15 \, a \sin \left (d x + c\right ) + 12 \, a}{\sin \left (d x + c\right )^{5}}}{60 \, d} \]

[In]

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

[Out]

1/60*(60*a*log(sin(d*x + c)) - (60*a*sin(d*x + c)^4 - 60*a*sin(d*x + c)^3 - 40*a*sin(d*x + c)^2 + 15*a*sin(d*x
+ c) + 12*a)/sin(d*x + c)^5)/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.98 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x)) \, dx=\frac {60 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {137 \, a \sin \left (d x + c\right )^{5} + 60 \, a \sin \left (d x + c\right )^{4} - 60 \, a \sin \left (d x + c\right )^{3} - 40 \, a \sin \left (d x + c\right )^{2} + 15 \, a \sin \left (d x + c\right ) + 12 \, a}{\sin \left (d x + c\right )^{5}}}{60 \, d} \]

[In]

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

[Out]

1/60*(60*a*log(abs(sin(d*x + c))) - (137*a*sin(d*x + c)^5 + 60*a*sin(d*x + c)^4 - 60*a*sin(d*x + c)^3 - 40*a*s
in(d*x + c)^2 + 15*a*sin(d*x + c) + 12*a)/sin(d*x + c)^5)/d

Mupad [B] (verification not implemented)

Time = 9.93 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.24 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x)) \, dx=\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {5\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}+\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {a}{5}\right )}{32\,d} \]

[In]

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

[Out]

(3*a*tan(c/2 + (d*x)/2)^2)/(16*d) - (a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (5*a*tan(c/2 + (d*x)/2))/(16*d) + (5
*a*tan(c/2 + (d*x)/2)^3)/(96*d) - (a*tan(c/2 + (d*x)/2)^4)/(64*d) - (a*tan(c/2 + (d*x)/2)^5)/(160*d) + (a*log(
tan(c/2 + (d*x)/2)))/d - (cot(c/2 + (d*x)/2)^5*(a/5 + (a*tan(c/2 + (d*x)/2))/2 - (5*a*tan(c/2 + (d*x)/2)^2)/3
- 6*a*tan(c/2 + (d*x)/2)^3 + 10*a*tan(c/2 + (d*x)/2)^4))/(32*d)