\(\int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx\) [499]

Optimal result

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

[Out]

a*ln(sin(d*x+c))/d+a*sin(d*x+c)/d-a*sin(d*x+c)^2/d-2/3*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 = 0.05 (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} \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin ^5(c+d x)}{5 d}+\frac {a \sin ^4(c+d x)}{4 d}-\frac {2 a \sin ^3(c+d x)}{3 d}-\frac {a \sin ^2(c+d x)}{d}+\frac {a \sin (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d} \]

[In]

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

[Out]

(a*Log[Sin[c + d*x]])/d + (a*Sin[c + d*x])/d - (a*Sin[c + d*x]^2)/d - (2*a*Sin[c + d*x]^3)/(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 (a-x)^2 (a+x)^3}{x} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \frac {(a-x)^2 (a+x)^3}{x} \, dx,x,a \sin (c+d x)\right )}{a^4 d} \\ & = \frac {\text {Subst}\left (\int \left (a^4+\frac {a^5}{x}-2 a^3 x-2 a^2 x^2+a x^3+x^4\right ) \, dx,x,a \sin (c+d x)\right )}{a^4 d} \\ & = \frac {a \log (\sin (c+d x))}{d}+\frac {a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{d}-\frac {2 a \sin ^3(c+d x)}{3 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 = 86, normalized size of antiderivative = 1.00 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \log (\sin (c+d x))}{d}+\frac {a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^4(c+d x)}{4 d}+\frac {a \sin ^5(c+d x)}{5 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a \left (\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(65\)
default	\(\frac {\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a \left (\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(65\)
parallelrisch	\(\frac {a \left (480 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-480 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-195+50 \sin \left (3 d x +3 c \right )+300 \sin \left (d x +c \right )+6 \sin \left (5 d x +5 c \right )+180 \cos \left (2 d x +2 c \right )+15 \cos \left (4 d x +4 c \right )\right )}{480 d}\)	\(87\)
risch	\(-i a x +\frac {3 a \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {3 a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}-\frac {2 i a c}{d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {5 a \sin \left (d x +c \right )}{8 d}+\frac {a \sin \left (5 d x +5 c \right )}{80 d}+\frac {a \cos \left (4 d x +4 c \right )}{32 d}+\frac {5 a \sin \left (3 d x +3 c \right )}{48 d}\)	\(119\)
norman	\(\frac {\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {8 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {116 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {8 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(205\)

[In]

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

[Out]

1/d*(1/5*a*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+a*(1/4*cos(d*x+c)^4+1/2*cos(d*x+c)^2+ln(sin(d*x+c)))
)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {15 \, a \cos \left (d x + c\right )^{4} + 30 \, a \cos \left (d x + c\right )^{2} + 60 \, a \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (3 \, a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right )}{60 \, d} \]

[In]

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

[Out]

1/60*(15*a*cos(d*x + c)^4 + 30*a*cos(d*x + c)^2 + 60*a*log(1/2*sin(d*x + c)) + 4*(3*a*cos(d*x + c)^4 + 4*a*cos
(d*x + c)^2 + 8*a)*sin(d*x + c))/d

Sympy [F]

\[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \cos ^{5}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.80 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {12 \, a \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 40 \, a \sin \left (d x + c\right )^{3} - 60 \, a \sin \left (d x + c\right )^{2} + 60 \, a \log \left (\sin \left (d x + c\right )\right ) + 60 \, a \sin \left (d x + c\right )}{60 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.81 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {12 \, a \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 40 \, a \sin \left (d x + c\right )^{3} - 60 \, a \sin \left (d x + c\right )^{2} + 60 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, a \sin \left (d x + c\right )}{60 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.86 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.47 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {a\,{\cos \left (c+d\,x\right )}^2}{2\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^4}{4\,d}+\frac {8\,a\,\sin \left (c+d\,x\right )}{15\,d}+\frac {4\,a\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{15\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{5\,d} \]

[In]

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

[Out]

(a*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (a*log(1/cos(c/2 + (d*x)/2)^2))/d + (a*cos(c + d*x)^2)/(2*d
) + (a*cos(c + d*x)^4)/(4*d) + (8*a*sin(c + d*x))/(15*d) + (4*a*cos(c + d*x)^2*sin(c + d*x))/(15*d) + (a*cos(c
+ d*x)^4*sin(c + d*x))/(5*d)