\(\int \cos ^4(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx\) [664]

Optimal result

Integrand size = 27, antiderivative size = 115 \[ \int \cos ^4(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {3 a \log (\sin (c+d x))}{d}-\frac {3 a \sin (c+d x)}{d}+\frac {3 a \sin ^2(c+d x)}{2 d}+\frac {a \sin ^3(c+d x)}{d}-\frac {a \sin ^4(c+d x)}{4 d}-\frac {a \sin ^5(c+d x)}{5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 115, 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} \[ \int \cos ^4(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \sin ^4(c+d x)}{4 d}+\frac {a \sin ^3(c+d x)}{d}+\frac {3 a \sin ^2(c+d x)}{2 d}-\frac {3 a \sin (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {a \csc (c+d x)}{d}-\frac {3 a \log (\sin (c+d x))}{d} \]

[In]

Int[Cos[c + d*x]^4*Cot[c + d*x]^3*(a + a*Sin[c + d*x]),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00 \[ \int \cos ^4(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {3 a \log (\sin (c+d x))}{d}-\frac {3 a \sin (c+d x)}{d}+\frac {3 a \sin ^2(c+d x)}{2 d}+\frac {a \sin ^3(c+d x)}{d}-\frac {a \sin ^4(c+d x)}{4 d}-\frac {a \sin ^5(c+d x)}{5 d} \]

[In]

Integrate[Cos[c + d*x]^4*Cot[c + d*x]^3*(a + a*Sin[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\cos ^{8}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )\right )+a \left (-\frac {\cos ^{8}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\cos ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\cos ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(125\)
default	\(\frac {a \left (-\frac {\cos ^{8}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )\right )+a \left (-\frac {\cos ^{8}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\cos ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\cos ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(125\)
parallelrisch	\(\frac {a \left (960 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (2 d x +2 c \right )-960 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (2 d x +2 c \right )-960 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+960 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1750 \sin \left (d x +c \right )+2 \sin \left (7 d x +7 c \right )+26 \sin \left (5 d x +5 c \right )+322 \sin \left (3 d x +3 c \right )+5 \cos \left (6 d x +6 c \right )+90 \cos \left (4 d x +4 c \right )-685 \cos \left (2 d x +2 c \right )+270\right ) \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2560 d}\)	\(175\)
risch	\(3 i a x +\frac {3 i a \,{\mathrm e}^{3 i \left (d x +c \right )}}{32 d}-\frac {5 a \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {19 i a \,{\mathrm e}^{i \left (d x +c \right )}}{16 d}-\frac {19 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{16 d}-\frac {5 a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}-\frac {3 i a \,{\mathrm e}^{-3 i \left (d x +c \right )}}{32 d}+\frac {6 i a c}{d}-\frac {2 i a \left (i {\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {a \sin \left (5 d x +5 c \right )}{80 d}-\frac {a \cos \left (4 d x +4 c \right )}{32 d}\)	\(210\)
norman	\(\frac {-\frac {a}{8 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}-\frac {9 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {47 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {182 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}-\frac {47 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {9 a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {a \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {31 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {31 a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {147 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {147 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(274\)

[In]

int(cos(d*x+c)^7*csc(d*x+c)^3*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(a*(-1/sin(d*x+c)*cos(d*x+c)^8-(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))+a*(-1/2/s
in(d*x+c)^2*cos(d*x+c)^8-1/2*cos(d*x+c)^6-3/4*cos(d*x+c)^4-3/2*cos(d*x+c)^2-3*ln(sin(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.08 \[ \int \cos ^4(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {40 \, a \cos \left (d x + c\right )^{6} + 120 \, a \cos \left (d x + c\right )^{4} - 255 \, a \cos \left (d x + c\right )^{2} + 480 \, {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 32 \, {\left (a \cos \left (d x + c\right )^{6} + 2 \, a \cos \left (d x + c\right )^{4} + 8 \, a \cos \left (d x + c\right )^{2} - 16 \, a\right )} \sin \left (d x + c\right ) + 15 \, a}{160 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/160*(40*a*cos(d*x + c)^6 + 120*a*cos(d*x + c)^4 - 255*a*cos(d*x + c)^2 + 480*(a*cos(d*x + c)^2 - a)*log(1/2
*sin(d*x + c)) + 32*(a*cos(d*x + c)^6 + 2*a*cos(d*x + c)^4 + 8*a*cos(d*x + c)^2 - 16*a)*sin(d*x + c) + 15*a)/(
d*cos(d*x + c)^2 - d)

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**7*csc(d*x+c)**3*(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.78 \[ \int \cos ^4(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {4 \, a \sin \left (d x + c\right )^{5} + 5 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3} - 30 \, a \sin \left (d x + c\right )^{2} + 60 \, a \log \left (\sin \left (d x + c\right )\right ) + 60 \, a \sin \left (d x + c\right ) + \frac {10 \, {\left (2 \, a \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )^{2}}}{20 \, d} \]

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/20*(4*a*sin(d*x + c)^5 + 5*a*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3 - 30*a*sin(d*x + c)^2 + 60*a*log(sin(d*x
+ c)) + 60*a*sin(d*x + c) + 10*(2*a*sin(d*x + c) + a)/sin(d*x + c)^2)/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90 \[ \int \cos ^4(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {4 \, a \sin \left (d x + c\right )^{5} + 5 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3} - 30 \, a \sin \left (d x + c\right )^{2} + 60 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, a \sin \left (d x + c\right ) - \frac {10 \, {\left (9 \, a \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right ) - a\right )}}{\sin \left (d x + c\right )^{2}}}{20 \, d} \]

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/20*(4*a*sin(d*x + c)^5 + 5*a*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3 - 30*a*sin(d*x + c)^2 + 60*a*log(abs(sin(
d*x + c))) + 60*a*sin(d*x + c) - 10*(9*a*sin(d*x + c)^2 - 2*a*sin(d*x + c) - a)/sin(d*x + c)^2)/d

Mupad [B] (verification not implemented)

Time = 10.72 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.70 \[ \int \cos ^4(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3\,a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {26\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-\frac {47\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{2}+74\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-\frac {107\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+\frac {628\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{5}-51\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+84\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-19\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+34\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {3\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

[In]

int((cos(c + d*x)^7*(a + a*sin(c + d*x)))/sin(c + d*x)^3,x)

[Out]

(3*a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (a/2 + 2*a*tan(c/2 + (d*x)/2) + (5*a*tan(c/2 + (d*x)/2)^2)/2 + 34*a*ta
n(c/2 + (d*x)/2)^3 - 19*a*tan(c/2 + (d*x)/2)^4 + 84*a*tan(c/2 + (d*x)/2)^5 - 51*a*tan(c/2 + (d*x)/2)^6 + (628*
a*tan(c/2 + (d*x)/2)^7)/5 - (107*a*tan(c/2 + (d*x)/2)^8)/2 + 74*a*tan(c/2 + (d*x)/2)^9 - (47*a*tan(c/2 + (d*x)
/2)^10)/2 + 26*a*tan(c/2 + (d*x)/2)^11)/(d*(4*tan(c/2 + (d*x)/2)^2 + 20*tan(c/2 + (d*x)/2)^4 + 40*tan(c/2 + (d
*x)/2)^6 + 40*tan(c/2 + (d*x)/2)^8 + 20*tan(c/2 + (d*x)/2)^10 + 4*tan(c/2 + (d*x)/2)^12)) - (a*tan(c/2 + (d*x)
/2))/(2*d) - (a*tan(c/2 + (d*x)/2)^2)/(8*d) - (3*a*log(tan(c/2 + (d*x)/2)))/d