\(\int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx\) [501]

Optimal result

Integrand size = 27, antiderivative size = 86 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {2 a \log (\sin (c+d x))}{d}-\frac {2 a \sin (c+d x)}{d}+\frac {a \sin ^2(c+d x)}{2 d}+\frac {a \sin ^3(c+d x)}{3 d} \]

[Out]

-a*csc(d*x+c)/d-1/2*a*csc(d*x+c)^2/d-2*a*ln(sin(d*x+c))/d-2*a*sin(d*x+c)/d+1/2*a*sin(d*x+c)^2/d+1/3*a*sin(d*x+
c)^3/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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2915, 12, 90} \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^2(c+d x)}{2 d}-\frac {2 a \sin (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {a \csc (c+d x)}{d}-\frac {2 a \log (\sin (c+d x))}{d} \]

[In]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}-\frac {2 a \sin (c+d x)}{d}+\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \left (\csc ^2(c+d x)+4 \log (\sin (c+d x))-\sin ^2(c+d x)\right )}{2 d} \]

[In]

Integrate[Cos[c + d*x]^2*Cot[c + d*x]^3*(a + a*Sin[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.22

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+a \left (-\frac {\cos ^{6}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{2}-\left (\cos ^{2}\left (d x +c \right )\right )-2 \ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(105\)
default	\(\frac {a \left (-\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+a \left (-\frac {\cos ^{6}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{2}-\left (\cos ^{2}\left (d x +c \right )\right )-2 \ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(105\)
parallelrisch	\(\frac {\left (192 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-192 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-12 \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+60\right ) \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\cos \left (\frac {9 d x}{2}+\frac {9 c}{2}\right )+\cos \left (\frac {7 d x}{2}+\frac {7 c}{2}\right )+20 \cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+20 \cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-90 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-96 \cos \left (d x +c \right )+24 \cos \left (2 d x +2 c \right )+12\right ) \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) a}{96 d}\)	\(167\)
risch	\(2 i a x +\frac {i a \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {7 i a \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {7 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i a \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {4 i a c}{d}-\frac {2 i a \left (i {\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(180\)
norman	\(\frac {-\frac {a}{8 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}-\frac {6 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {25 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {6 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {21 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {21 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {2 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(206\)

[In]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.19 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {6 \, a \cos \left (d x + c\right )^{4} - 9 \, a \cos \left (d x + c\right )^{2} + 24 \, {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} - 8 \, a\right )} \sin \left (d x + c\right ) - 3 \, a}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(6*a*cos(d*x + c)^4 - 9*a*cos(d*x + c)^2 + 24*(a*cos(d*x + c)^2 - a)*log(1/2*sin(d*x + c)) + 4*(a*cos(d*
x + c)^4 + 4*a*cos(d*x + c)^2 - 8*a)*sin(d*x + c) - 3*a)/(d*cos(d*x + c)^2 - d)

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*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 = 68, normalized size of antiderivative = 0.79 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {2 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - 12 \, a \log \left (\sin \left (d x + c\right )\right ) - 12 \, a \sin \left (d x + c\right ) - \frac {3 \, {\left (2 \, a \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )^{2}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/6*(2*a*sin(d*x + c)^3 + 3*a*sin(d*x + c)^2 - 12*a*log(sin(d*x + c)) - 12*a*sin(d*x + c) - 3*(2*a*sin(d*x + c
) + a)/sin(d*x + c)^2)/d

Giac [A] (verification not implemented)

none

Time = 0.76 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {2 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - 12 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 12 \, a \sin \left (d x + c\right ) + \frac {3 \, {\left (6 \, a \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right ) - a\right )}}{\sin \left (d x + c\right )^{2}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/6*(2*a*sin(d*x + c)^3 + 3*a*sin(d*x + c)^2 - 12*a*log(abs(sin(d*x + c))) - 12*a*sin(d*x + c) + 3*(6*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 = 10.05 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.66 \[ \int \cos ^2(c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {2\,a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {18\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {82\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-\frac {13\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+22\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {2\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

[In]

int((cos(c + d*x)^5*(a + a*sin(c + d*x)))/sin(c + d*x)^3,x)

[Out]

(2*a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (a/2 + 2*a*tan(c/2 + (d*x)/2) + (3*a*tan(c/2 + (d*x)/2)^2)/2 + 22*a*ta
n(c/2 + (d*x)/2)^3 - (13*a*tan(c/2 + (d*x)/2)^4)/2 + (82*a*tan(c/2 + (d*x)/2)^5)/3 - (15*a*tan(c/2 + (d*x)/2)^
6)/2 + 18*a*tan(c/2 + (d*x)/2)^7)/(d*(4*tan(c/2 + (d*x)/2)^2 + 12*tan(c/2 + (d*x)/2)^4 + 12*tan(c/2 + (d*x)/2)
^6 + 4*tan(c/2 + (d*x)/2)^8)) - (a*tan(c/2 + (d*x)/2))/(2*d) - (a*tan(c/2 + (d*x)/2)^2)/(8*d) - (2*a*log(tan(c
/2 + (d*x)/2)))/d