\(\int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx\) [684]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 27, antiderivative size = 99 \[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\log (\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{a d}-\frac {\sin ^2(c+d x)}{a d}+\frac {2 \sin ^3(c+d x)}{3 a d}+\frac {\sin ^4(c+d x)}{4 a d}-\frac {\sin ^5(c+d x)}{5 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 99, 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} \[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sin ^5(c+d x)}{5 a d}+\frac {\sin ^4(c+d x)}{4 a d}+\frac {2 \sin ^3(c+d x)}{3 a d}-\frac {\sin ^2(c+d x)}{a d}-\frac {\sin (c+d x)}{a d}+\frac {\log (\sin (c+d x))}{a d} \]

[In]

Int[(Cos[c + d*x]^6*Cot[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {60 \log (\sin (c+d x))-60 \sin (c+d x)-60 \sin ^2(c+d x)+40 \sin ^3(c+d x)+15 \sin ^4(c+d x)-12 \sin ^5(c+d x)}{60 a d} \]

[In]

Integrate[(Cos[c + d*x]^6*Cot[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

(60*Log[Sin[c + d*x]] - 60*Sin[c + d*x] - 60*Sin[c + d*x]^2 + 40*Sin[c + d*x]^3 + 15*Sin[c + d*x]^4 - 12*Sin[c
 + d*x]^5)/(60*a*d)

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.65

method result size
derivativedivides \(\frac {-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\left (\sin ^{2}\left (d x +c \right )\right )-\sin \left (d x +c \right )+\ln \left (\sin \left (d x +c \right )\right )}{d a}\) \(64\)
default \(\frac {-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\left (\sin ^{2}\left (d x +c \right )\right )-\sin \left (d x +c \right )+\ln \left (\sin \left (d x +c \right )\right )}{d a}\) \(64\)
parallelrisch \(\frac {480 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-480 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-195-6 \sin \left (5 d x +5 c \right )-50 \sin \left (3 d x +3 c \right )-300 \sin \left (d x +c \right )+180 \cos \left (2 d x +2 c \right )+15 \cos \left (4 d x +4 c \right )}{480 a d}\) \(89\)
risch \(-\frac {i x}{a}+\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 a d}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a d}-\frac {2 i c}{a d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}-\frac {5 \sin \left (d x +c \right )}{8 a d}-\frac {\sin \left (5 d x +5 c \right )}{80 d a}+\frac {\cos \left (4 d x +4 c \right )}{32 a d}-\frac {5 \sin \left (3 d x +3 c \right )}{48 d a}\) \(137\)
norman \(\frac {\frac {2}{a d}+\frac {2 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {6 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {10 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {10 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {40 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {40 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {68 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {68 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {38 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {38 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}\) \(288\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {15 \, \cos \left (d x + c\right )^{4} + 30 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (3 \, \cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 60 \, \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right )}{60 \, a d} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} - 40 \, \sin \left (d x + c\right )^{3} + 60 \, \sin \left (d x + c\right )^{2} + 60 \, \sin \left (d x + c\right )}{a} - \frac {60 \, \log \left (\sin \left (d x + c\right )\right )}{a}}{60 \, d} \]

[In]

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

[Out]

-1/60*((12*sin(d*x + c)^5 - 15*sin(d*x + c)^4 - 40*sin(d*x + c)^3 + 60*sin(d*x + c)^2 + 60*sin(d*x + c))/a - 6
0*log(sin(d*x + c))/a)/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} - \frac {12 \, a^{4} \sin \left (d x + c\right )^{5} - 15 \, a^{4} \sin \left (d x + c\right )^{4} - 40 \, a^{4} \sin \left (d x + c\right )^{3} + 60 \, a^{4} \sin \left (d x + c\right )^{2} + 60 \, a^{4} \sin \left (d x + c\right )}{a^{5}}}{60 \, d} \]

[In]

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

[Out]

1/60*(60*log(abs(sin(d*x + c)))/a - (12*a^4*sin(d*x + c)^5 - 15*a^4*sin(d*x + c)^4 - 40*a^4*sin(d*x + c)^3 + 6
0*a^4*sin(d*x + c)^2 + 60*a^4*sin(d*x + c))/a^5)/d

Mupad [B] (verification not implemented)

Time = 10.28 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.41 \[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d}-\frac {\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{a\,d}-\frac {8\,\sin \left (c+d\,x\right )}{15\,a\,d}+\frac {{\cos \left (c+d\,x\right )}^2}{2\,a\,d}+\frac {{\cos \left (c+d\,x\right )}^4}{4\,a\,d}-\frac {4\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{15\,a\,d}-\frac {{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{5\,a\,d} \]

[In]

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

[Out]

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