\(\int \frac {\cot (c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx\) [231]

Optimal result

Integrand size = 27, antiderivative size = 82 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc (c+d x)}{a d}+\frac {\csc ^2(c+d x)}{2 a d}-\frac {\csc ^3(c+d x)}{3 a d}-\frac {\log (\sin (c+d x))}{a d}+\frac {\log (1+\sin (c+d x))}{a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 82, 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, 46} \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc ^3(c+d x)}{3 a d}+\frac {\csc ^2(c+d x)}{2 a d}-\frac {\csc (c+d x)}{a d}-\frac {\log (\sin (c+d x))}{a d}+\frac {\log (\sin (c+d x)+1)}{a d} \]

[In]

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

[Out]

-(Csc[c + d*x]/(a*d)) + Csc[c + d*x]^2/(2*a*d) - Csc[c + d*x]^3/(3*a*d) - Log[Sin[c + d*x]]/(a*d) + Log[1 + Si
n[c + d*x]]/(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 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[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^4}{x^4 (a+x)} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {a^3 \text {Subst}\left (\int \frac {1}{x^4 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^3 \text {Subst}\left (\int \left (\frac {1}{a x^4}-\frac {1}{a^2 x^3}+\frac {1}{a^3 x^2}-\frac {1}{a^4 x}+\frac {1}{a^4 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {\csc (c+d x)}{a d}+\frac {\csc ^2(c+d x)}{2 a d}-\frac {\csc ^3(c+d x)}{3 a d}-\frac {\log (\sin (c+d x))}{a d}+\frac {\log (1+\sin (c+d x))}{a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc (c+d x)}{a d}+\frac {\csc ^2(c+d x)}{2 a d}-\frac {\csc ^3(c+d x)}{3 a d}-\frac {\log (\sin (c+d x))}{a d}+\frac {\log (1+\sin (c+d x))}{a d} \]

[In]

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

[Out]

-(Csc[c + d*x]/(a*d)) + Csc[c + d*x]^2/(2*a*d) - Csc[c + d*x]^3/(3*a*d) - Log[Sin[c + d*x]]/(a*d) + Log[1 + Si
n[c + d*x]]/(a*d)

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.57

method	result	size
derivativedivides	\(-\frac {\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}+\csc \left (d x +c \right )-\ln \left (\csc \left (d x +c \right )+1\right )}{d a}\)	\(47\)
default	\(-\frac {\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}+\csc \left (d x +c \right )-\ln \left (\csc \left (d x +c \right )+1\right )}{d a}\)	\(47\)
parallelrisch	\(\frac {-\left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+48 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-15 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d a}\)	\(110\)
risch	\(-\frac {2 i \left (3 \,{\mathrm e}^{5 i \left (d x +c \right )}-10 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}+3 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\)	\(123\)
norman	\(\frac {-\frac {1}{24 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d a}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d a}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}\)	\(167\)

[In]

int(cos(d*x+c)*csc(d*x+c)^4/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.24 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {6 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 6 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 6 \, \cos \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 8}{6 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cos {\left (c + d x \right )} \csc ^{4}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)**4/(a+a*sin(d*x+c)),x)

[Out]

Integral(cos(c + d*x)*csc(c + d*x)**4/(sin(c + d*x) + 1), x)/a

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {6 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {6 \, \log \left (\sin \left (d x + c\right )\right )}{a} - \frac {6 \, \sin \left (d x + c\right )^{2} - 3 \, \sin \left (d x + c\right ) + 2}{a \sin \left (d x + c\right )^{3}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.82 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} - \frac {6 \, \sin \left (d x + c\right )^{2} - 3 \, \sin \left (d x + c\right ) + 2}{a \sin \left (d x + c\right )^{3}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.74 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.70 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}+\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a\,d}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {1}{3}\right )}{8\,a\,d} \]

[In]

int(cos(c + d*x)/(sin(c + d*x)^4*(a + a*sin(c + d*x))),x)

[Out]

tan(c/2 + (d*x)/2)^2/(8*a*d) - tan(c/2 + (d*x)/2)^3/(24*a*d) - log(tan(c/2 + (d*x)/2))/(a*d) + (2*log(tan(c/2
+ (d*x)/2) + 1))/(a*d) - (5*tan(c/2 + (d*x)/2))/(8*a*d) - (cot(c/2 + (d*x)/2)^3*(5*tan(c/2 + (d*x)/2)^2 - tan(
c/2 + (d*x)/2) + 1/3))/(8*a*d)