\(\int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx\) [1203]

Optimal result

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

[Out]

a*ln(sin(d*x+c))/d+b*sin(d*x+c)/d-a*sin(d*x+c)^2/d-2/3*b*sin(d*x+c)^3/d+1/4*a*sin(d*x+c)^4/d+1/5*b*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 = {2916, 12, 780} \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \sin ^4(c+d x)}{4 d}-\frac {a \sin ^2(c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d}+\frac {b \sin ^5(c+d x)}{5 d}-\frac {2 b \sin ^3(c+d x)}{3 d}+\frac {b \sin (c+d x)}{d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \log (\sin (c+d x))}{d}+\frac {b \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{d}-\frac {2 b \sin ^3(c+d x)}{3 d}+\frac {a \sin ^4(c+d x)}{4 d}+\frac {b \sin ^5(c+d x)}{5 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {a \left (\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+\frac {b \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(65\)
default	\(\frac {a \left (\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+\frac {b \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(65\)
parallelrisch	\(\frac {480 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-480 a \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \cos \left (4 d x +4 c \right ) a +180 a \cos \left (2 d x +2 c \right )+6 b \sin \left (5 d x +5 c \right )+50 b \sin \left (3 d x +3 c \right )+300 b \sin \left (d x +c \right )-195 a}{480 d}\)	\(95\)
risch	\(-i a x +\frac {3 a \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {3 a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}-\frac {2 i a c}{d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {5 b \sin \left (d x +c \right )}{8 d}+\frac {b \sin \left (5 d x +5 c \right )}{80 d}+\frac {a \cos \left (4 d x +4 c \right )}{32 d}+\frac {5 b \sin \left (3 d x +3 c \right )}{48 d}\)	\(119\)
norman	\(\frac {\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {8 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {116 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {8 b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(205\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {15 \, a \cos \left (d x + c\right )^{4} + 30 \, a \cos \left (d x + c\right )^{2} + 60 \, a \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (3 \, b \cos \left (d x + c\right )^{4} + 4 \, b \cos \left (d x + c\right )^{2} + 8 \, b\right )} \sin \left (d x + c\right )}{60 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)*(a+b*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*b*cos(d*x + c)^4 + 4*b*cos
(d*x + c)^2 + 8*b)*sin(d*x + c))/d

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.80 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {12 \, b \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 40 \, b \sin \left (d x + c\right )^{3} - 60 \, a \sin \left (d x + c\right )^{2} + 60 \, a \log \left (\sin \left (d x + c\right )\right ) + 60 \, b \sin \left (d x + c\right )}{60 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.81 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {12 \, b \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 40 \, b \sin \left (d x + c\right )^{3} - 60 \, a \sin \left (d x + c\right )^{2} + 60 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, b \sin \left (d x + c\right )}{60 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.79 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.47 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {a\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {a\,{\cos \left (c+d\,x\right )}^2}{2\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^4}{4\,d}+\frac {8\,b\,\sin \left (c+d\,x\right )}{15\,d}+\frac {4\,b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{15\,d}+\frac {b\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{5\,d} \]

[In]

int((cos(c + d*x)^5*(a + b*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*b*sin(c + d*x))/(15*d) + (4*b*cos(c + d*x)^2*sin(c + d*x))/(15*d) + (b*cos(c
+ d*x)^4*sin(c + d*x))/(5*d)