\(\int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [245]

Optimal result

Integrand size = 19, antiderivative size = 74 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\log (\sin (c+d x))}{a^3 d}-\frac {\log (1+\sin (c+d x))}{a^3 d}+\frac {1}{2 a d (a+a \sin (c+d x))^2}+\frac {1}{d \left (a^3+a^3 \sin (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2786, 46} \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {1}{d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {\log (\sin (c+d x))}{a^3 d}-\frac {\log (\sin (c+d x)+1)}{a^3 d}+\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]

[In]

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

[Out]

Log[Sin[c + d*x]]/(a^3*d) - Log[1 + Sin[c + d*x]]/(a^3*d) + 1/(2*a*d*(a + a*Sin[c + d*x])^2) + 1/(d*(a^3 + a^3
*Sin[c + d*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 2786

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[x^p*((a + x)^(m - (p + 1)/2)/(a - x)^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.70 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {2 \log (\sin (c+d x))-2 \log (1+\sin (c+d x))+\frac {3+2 \sin (c+d x)}{(1+\sin (c+d x))^2}}{2 a^3 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.58

method	result	size
derivativedivides	\(-\frac {\frac {2}{\csc \left (d x +c \right )+1}-\frac {1}{2 \left (\csc \left (d x +c \right )+1\right )^{2}}+\ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{3}}\)	\(43\)
default	\(-\frac {\frac {2}{\csc \left (d x +c \right )+1}-\frac {1}{2 \left (\csc \left (d x +c \right )+1\right )^{2}}+\ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{3}}\)	\(43\)
risch	\(\frac {2 i \left (-{\mathrm e}^{i \left (d x +c \right )}+3 i {\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{3 i \left (d x +c \right )}\right )}{d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{3}}\)	\(98\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.41 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {2 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 2 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, \sin \left (d x + c\right ) - 3}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(cos(c + d*x)*csc(c + d*x)/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x)/a**3

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.97 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {2 \, \sin \left (d x + c\right ) + 3}{a^{3} \sin \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) + a^{3}} - \frac {2 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {2 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.80 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac {2 \, \sin \left (d x + c\right ) + 3}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.79 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.00 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^3\right )}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^3\,d} \]

[In]

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

[Out]

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