\(\int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\) [211]

Optimal result

Integrand size = 27, antiderivative size = 61 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {3 a^3 \csc (c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}+\frac {3 a^3 \log (\sin (c+d x))}{d}+\frac {a^3 \sin (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 61, 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 = {2912, 12, 45} \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \sin (c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}-\frac {3 a^3 \csc (c+d x)}{d}+\frac {3 a^3 \log (\sin (c+d x))}{d} \]

[In]

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

[Out]

(-3*a^3*Csc[c + d*x])/d - (a^3*Csc[c + d*x]^2)/(2*d) + (3*a^3*Log[Sin[c + d*x]])/d + (a^3*Sin[c + d*x])/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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {a^3 (a+x)^3}{x^3} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {a^2 \text {Subst}\left (\int \frac {(a+x)^3}{x^3} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^2 \text {Subst}\left (\int \left (1+\frac {a^3}{x^3}+\frac {3 a^2}{x^2}+\frac {3 a}{x}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {3 a^3 \csc (c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}+\frac {3 a^3 \log (\sin (c+d x))}{d}+\frac {a^3 \sin (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=a^3 \left (-\frac {3 \csc (c+d x)}{d}-\frac {\csc ^2(c+d x)}{2 d}+\frac {3 \log (\sin (c+d x))}{d}+\frac {\sin (c+d x)}{d}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(-\frac {a^{3} \left (\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}+3 \csc \left (d x +c \right )+3 \ln \left (\csc \left (d x +c \right )\right )-\frac {1}{\csc \left (d x +c \right )}\right )}{d}\)	\(47\)
default	\(-\frac {a^{3} \left (\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}+3 \csc \left (d x +c \right )+3 \ln \left (\csc \left (d x +c \right )\right )-\frac {1}{\csc \left (d x +c \right )}\right )}{d}\)	\(47\)
risch	\(-3 i a^{3} x -\frac {i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {6 i a^{3} c}{d}-\frac {2 i a^{3} \left (i {\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {3 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(132\)
parallelrisch	\(-\frac {a^{3} \left (-24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \cot \left (\frac {d x}{2}+\frac {c}{2}\right ) \cos \left (d x +c \right )+3 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}\)	\(133\)
norman	\(\frac {-\frac {a^{3}}{8 d}-\frac {3 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}-\frac {4 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {5 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {5 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {5 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {3 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a^{3} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(228\)

[In]

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

[Out]

-1/d*a^3*(1/2*csc(d*x+c)^2+3*csc(d*x+c)+3*ln(csc(d*x+c))-1/csc(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.26 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} + 6 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 2 \, {\left (a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

a**3*(Integral(cos(c + d*x)*csc(c + d*x)**3, x) + Integral(3*sin(c + d*x)*cos(c + d*x)*csc(c + d*x)**3, x) + I
ntegral(3*sin(c + d*x)**2*cos(c + d*x)*csc(c + d*x)**3, x) + Integral(sin(c + d*x)**3*cos(c + d*x)*csc(c + d*x
)**3, x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {6 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 2 \, a^{3} \sin \left (d x + c\right ) - \frac {6 \, a^{3} \sin \left (d x + c\right ) + a^{3}}{\sin \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {6 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 2 \, a^{3} \sin \left (d x + c\right ) - \frac {6 \, a^{3} \sin \left (d x + c\right ) + a^{3}}{\sin \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.37 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.67 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {3\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+6\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^3}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {3\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

[In]

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

[Out]

(3*a^3*log(tan(c/2 + (d*x)/2)))/d - ((a^3*tan(c/2 + (d*x)/2)^2)/2 - 2*a^3*tan(c/2 + (d*x)/2)^3 + a^3/2 + 6*a^3
*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 + 4*tan(c/2 + (d*x)/2)^4)) - (a^3*tan(c/2 + (d*x)/2)^2)/(8*d)
- (3*a^3*tan(c/2 + (d*x)/2))/(2*d) - (3*a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d