\(\int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx\) [1208]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.06 (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 = {2916, 12, 780} \[ \int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \csc ^5(c+d x)}{5 d}+\frac {2 a \csc ^3(c+d x)}{3 d}-\frac {a \csc (c+d x)}{d}-\frac {b \csc ^4(c+d x)}{4 d}+\frac {b \csc ^2(c+d x)}{d}+\frac {b \log (\sin (c+d x))}{d} \]

[In]

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

[Out]

-((a*Csc[c + d*x])/d) + (b*Csc[c + d*x]^2)/d + (2*a*Csc[c + d*x]^3)/(3*d) - (b*Csc[c + d*x]^4)/(4*d) - (a*Csc[
c + d*x]^5)/(5*d) + (b*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 780

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(e*x
)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, m}, x] && IGtQ[p, 0]

Rule 2916

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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {b^6 (a+x) \left (b^2-x^2\right )^2}{x^6} \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = \frac {b \text {Subst}\left (\int \frac {(a+x) \left (b^2-x^2\right )^2}{x^6} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b \text {Subst}\left (\int \left (\frac {a b^4}{x^6}+\frac {b^4}{x^5}-\frac {2 a b^2}{x^4}-\frac {2 b^2}{x^3}+\frac {a}{x^2}+\frac {1}{x}\right ) \, dx,x,b \sin (c+d x)\right )}{d} \\ & = -\frac {a \csc (c+d x)}{d}+\frac {b \csc ^2(c+d x)}{d}+\frac {2 a \csc ^3(c+d x)}{3 d}-\frac {b \csc ^4(c+d x)}{4 d}-\frac {a \csc ^5(c+d x)}{5 d}+\frac {b \log (\sin (c+d x))}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.17 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx=\frac {b \cot ^2(c+d x)}{2 d}-\frac {b \cot ^4(c+d x)}{4 d}-\frac {a \csc (c+d x)}{d}+\frac {2 a \csc ^3(c+d x)}{3 d}-\frac {a \csc ^5(c+d x)}{5 d}+\frac {b \log (\cos (c+d x))}{d}+\frac {b \log (\tan (c+d x))}{d} \]

[In]

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

[Out]

(b*Cot[c + d*x]^2)/(2*d) - (b*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) + (b*Log[Cos[c + d*x]])/d + (b*Log[Tan[c + d*x]])/d

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(-\frac {\frac {\left (\csc ^{5}\left (d x +c \right )\right ) a}{5}+\frac {b \left (\csc ^{4}\left (d x +c \right )\right )}{4}-\frac {2 \left (\csc ^{3}\left (d x +c \right )\right ) a}{3}-b \left (\csc ^{2}\left (d x +c \right )\right )+\csc \left (d x +c \right ) a +b \ln \left (\csc \left (d x +c \right )\right )}{d}\)	\(68\)
default	\(-\frac {\frac {\left (\csc ^{5}\left (d x +c \right )\right ) a}{5}+\frac {b \left (\csc ^{4}\left (d x +c \right )\right )}{4}-\frac {2 \left (\csc ^{3}\left (d x +c \right )\right ) a}{3}-b \left (\csc ^{2}\left (d x +c \right )\right )+\csc \left (d x +c \right ) a +b \ln \left (\csc \left (d x +c \right )\right )}{d}\)	\(68\)
risch	\(-i x b -\frac {2 i b c}{d}-\frac {2 i \left (15 a \,{\mathrm e}^{9 i \left (d x +c \right )}-20 a \,{\mathrm e}^{7 i \left (d x +c \right )}-30 i b \,{\mathrm e}^{8 i \left (d x +c \right )}+58 a \,{\mathrm e}^{5 i \left (d x +c \right )}+60 i b \,{\mathrm e}^{6 i \left (d x +c \right )}-20 a \,{\mathrm e}^{3 i \left (d x +c \right )}-60 i b \,{\mathrm e}^{4 i \left (d x +c \right )}+15 a \,{\mathrm e}^{i \left (d x +c \right )}+30 i b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(164\)
parallelrisch	\(\frac {-6 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -6 a \left (\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -15 b \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+50 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+50 a \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+180 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +180 b \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-300 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+960 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -960 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -300 a \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{960 d}\)	\(171\)
norman	\(\frac {-\frac {a}{160 d}+\frac {11 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{240 d}-\frac {25 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}-\frac {5 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {25 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}+\frac {11 a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{240 d}-\frac {a \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d}+\frac {11 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}+\frac {11 b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}-\frac {b \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {b \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(239\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.44 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx=-\frac {60 \, a \cos \left (d x + c\right )^{4} - 80 \, a \cos \left (d x + c\right )^{2} - 60 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 15 \, {\left (4 \, b \cos \left (d x + c\right )^{2} - 3 \, b\right )} \sin \left (d x + c\right ) + 32 \, a}{60 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

[In]

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

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**6*(a+b*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 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx=\frac {60 \, b \log \left (\sin \left (d x + c\right )\right ) - \frac {60 \, a \sin \left (d x + c\right )^{4} - 60 \, b \sin \left (d x + c\right )^{3} - 40 \, a \sin \left (d x + c\right )^{2} + 15 \, b \sin \left (d x + c\right ) + 12 \, a}{\sin \left (d x + c\right )^{5}}}{60 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.98 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx=\frac {60 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {137 \, b \sin \left (d x + c\right )^{5} + 60 \, a \sin \left (d x + c\right )^{4} - 60 \, b \sin \left (d x + c\right )^{3} - 40 \, a \sin \left (d x + c\right )^{2} + 15 \, b \sin \left (d x + c\right ) + 12 \, a}{\sin \left (d x + c\right )^{5}}}{60 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.61 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.24 \[ \int \cot ^5(c+d x) \csc (c+d x) (a+b \sin (c+d x)) \, dx=\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {5\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {a}{5}\right )}{32\,d} \]

[In]

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

[Out]

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