\(\int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [237]

Optimal result

Integrand size = 25, antiderivative size = 68 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc (c+d x)}{a^2 d}-\frac {2 \log (\sin (c+d x))}{a^2 d}+\frac {2 \log (1+\sin (c+d x))}{a^2 d}-\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 68, 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 = {2912, 12, 46} \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )}-\frac {\csc (c+d x)}{a^2 d}-\frac {2 \log (\sin (c+d x))}{a^2 d}+\frac {2 \log (\sin (c+d x)+1)}{a^2 d} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.66 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc (c+d x)+2 \log (\sin (c+d x))-2 \log (1+\sin (c+d x))+\frac {1}{1+\sin (c+d x)}}{a^2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.57

method	result	size
derivativedivides	\(-\frac {\csc \left (d x +c \right )-\frac {1}{\csc \left (d x +c \right )+1}-2 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{2}}\)	\(39\)
default	\(-\frac {\csc \left (d x +c \right )-\frac {1}{\csc \left (d x +c \right )+1}-2 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{2}}\)	\(39\)
parallelrisch	\(\frac {\left (8 \sin \left (d x +c \right )+8\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-4 \sin \left (d x +c \right )-4\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-4 \cos \left (d x +c \right )+4\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )}\)	\(104\)
risch	\(-\frac {4 i \left (i {\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d \,a^{2}}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{2}}\)	\(112\)
norman	\(\frac {-\frac {1}{2 a d}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {9 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {9 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}}{a \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}+\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}\)	\(134\)

[In]

int(cos(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.53 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 \, {\left (\cos \left (d x + c\right )^{2} - \sin \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 2 \, {\left (\cos \left (d x + c\right )^{2} - \sin \left (d x + c\right ) - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, \sin \left (d x + c\right ) - 1}{a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \sin \left (d x + c\right ) - a^{2} d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\cos {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)**2/(a+a*sin(d*x+c))**2,x)

[Out]

Integral(cos(c + d*x)*csc(c + d*x)**2/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, \sin \left (d x + c\right ) + 1}{a^{2} \sin \left (d x + c\right )^{2} + a^{2} \sin \left (d x + c\right )} - \frac {2 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {2 \, \log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, \log \left ({\left | -\frac {a}{a \sin \left (d x + c\right ) + a} + 1 \right |}\right )}{a^{2}} + \frac {1}{{\left (a \sin \left (d x + c\right ) + a\right )} a} - \frac {1}{a^{2} {\left (\frac {a}{a \sin \left (d x + c\right ) + a} - 1\right )}}}{d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.26 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.00 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^2\,d}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1}{d\,\left (2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d} \]

[In]

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

[Out]

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