\(\int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx\) [354]

Optimal result

Integrand size = 25, antiderivative size = 56 \[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \log (\sin (c+d x))}{d}+\frac {a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{2 d}-\frac {a \sin ^3(c+d x)}{3 d} \]

[Out]

a*ln(sin(d*x+c))/d+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 = 0.04 (sec) , antiderivative size = 56, 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, 76} \[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \sin ^2(c+d x)}{2 d}+\frac {a \sin (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d} \]

[In]

Int[Cos[c + d*x]^2*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)/(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 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

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) (a+x)^2}{x} \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {\text {Subst}\left (\int \frac {(a-x) (a+x)^2}{x} \, dx,x,a \sin (c+d x)\right )}{a^2 d} \\ & = \frac {\text {Subst}\left (\int \left (a^2+\frac {a^3}{x}-a x-x^2\right ) \, dx,x,a \sin (c+d x)\right )}{a^2 d} \\ & = \frac {a \log (\sin (c+d x))}{d}+\frac {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 = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \log (\sin (c+d x))}{d}+\frac {a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{2 d}-\frac {a \sin ^3(c+d x)}{3 d} \]

[In]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {\frac {a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a \left (\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(45\)
default	\(\frac {\frac {a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a \left (\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(45\)
parallelrisch	\(\frac {a \left (12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3+\sin \left (3 d x +3 c \right )+9 \sin \left (d x +c \right )+3 \cos \left (2 d x +2 c \right )\right )}{12 d}\)	\(63\)
risch	\(-i a x +\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 i a c}{d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {3 a \sin \left (d x +c \right )}{4 d}+\frac {a \sin \left (3 d x +3 c \right )}{12 d}\)	\(89\)
norman	\(\frac {\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(137\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.91 \[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {3 \, a \cos \left (d x + c\right )^{2} + 6 \, a \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 2 \, {\left (a \cos \left (d x + c\right )^{2} + 2 \, a\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \cos ^{3}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.84 \[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {2 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - 6 \, a \log \left (\sin \left (d x + c\right )\right ) - 6 \, a \sin \left (d x + c\right )}{6 \, d} \]

[In]

integrate(cos(d*x+c)^3*csc(d*x+c)*(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 - 6*a*log(sin(d*x + c)) - 6*a*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.86 \[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {2 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - 6 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 6 \, a \sin \left (d x + c\right )}{6 \, d} \]

[In]

integrate(cos(d*x+c)^3*csc(d*x+c)*(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 - 6*a*log(abs(sin(d*x + c))) - 6*a*sin(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 10.31 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.64 \[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {a\,{\cos \left (c+d\,x\right )}^2}{2\,d}+\frac {2\,a\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \]

[In]

int((cos(c + d*x)^3*(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
) + (2*a*sin(c + d*x))/(3*d) + (a*cos(c + d*x)^2*sin(c + d*x))/(3*d)