\(\int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx\) [230]

Optimal result

Integrand size = 27, antiderivative size = 63 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc (c+d x)}{a d}-\frac {\csc ^2(c+d x)}{2 a d}+\frac {\log (\sin (c+d x))}{a d}-\frac {\log (1+\sin (c+d x))}{a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 63, 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} \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc ^2(c+d x)}{2 a d}+\frac {\csc (c+d x)}{a d}+\frac {\log (\sin (c+d x))}{a d}-\frac {\log (\sin (c+d x)+1)}{a d} \]

[In]

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

[Out]

Csc[c + d*x]/(a*d) - Csc[c + d*x]^2/(2*a*d) + Log[Sin[c + d*x]]/(a*d) - Log[1 + Sin[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^3}{x^3 (a+x)} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {a^2 \text {Subst}\left (\int \frac {1}{x^3 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^2 \text {Subst}\left (\int \left (\frac {1}{a x^3}-\frac {1}{a^2 x^2}+\frac {1}{a^3 x}-\frac {1}{a^3 (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 {\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 = 63, normalized size of antiderivative = 1.00 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc (c+d x)}{a d}-\frac {\csc ^2(c+d x)}{2 a d}+\frac {\log (\sin (c+d x))}{a d}-\frac {\log (1+\sin (c+d x))}{a d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.59

method	result	size
derivativedivides	\(-\frac {\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}-\csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )+1\right )}{d a}\)	\(37\)
default	\(-\frac {\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}-\csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )+1\right )}{d a}\)	\(37\)
parallelrisch	\(\frac {-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-16 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d a}\)	\(84\)
risch	\(\frac {2 i \left (-i {\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\)	\(97\)
norman	\(\frac {-\frac {1}{8 a d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}\)	\(128\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.14 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, \sin \left (d x + c\right ) + 1}{2 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cos {\left (c + d x \right )} \csc ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

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

[Out]

Integral(cos(c + d*x)*csc(c + d*x)**3/(sin(c + d*x) + 1), x)/a

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.87 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {2 \, \log \left (\sin \left (d x + c\right )\right )}{a} - \frac {2 \, \sin \left (d x + c\right ) - 1}{a \sin \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} - \frac {2 \, \sin \left (d x + c\right ) - 1}{a \sin \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.12 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.68 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}\right )}{4\,a\,d} \]

[In]

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

[Out]

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