\(\int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx\) [254]

Optimal result

Integrand size = 19, antiderivative size = 97 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\log (\sin (c+d x))}{a^4 d}-\frac {\log (1+\sin (c+d x))}{a^4 d}+\frac {1}{3 a d (a+a \sin (c+d x))^3}+\frac {1}{2 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {1}{d \left (a^4+a^4 \sin (c+d x)\right )} \]

[Out]

ln(sin(d*x+c))/a^4/d-ln(1+sin(d*x+c))/a^4/d+1/3/a/d/(a+a*sin(d*x+c))^3+1/2/d/(a^2+a^2*sin(d*x+c))^2+1/d/(a^4+a
^4*sin(d*x+c))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 97, 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} \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {1}{d \left (a^4 \sin (c+d x)+a^4\right )}+\frac {\log (\sin (c+d x))}{a^4 d}-\frac {\log (\sin (c+d x)+1)}{a^4 d}+\frac {1}{2 d \left (a^2 \sin (c+d x)+a^2\right )^2}+\frac {1}{3 a d (a \sin (c+d x)+a)^3} \]

[In]

Int[Cot[c + d*x]/(a + a*Sin[c + d*x])^4,x]

[Out]

Log[Sin[c + d*x]]/(a^4*d) - Log[1 + Sin[c + d*x]]/(a^4*d) + 1/(3*a*d*(a + a*Sin[c + d*x])^3) + 1/(2*d*(a^2 + a
^2*Sin[c + d*x])^2) + 1/(d*(a^4 + a^4*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)^4} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{a^4 x}-\frac {1}{a (a+x)^4}-\frac {1}{a^2 (a+x)^3}-\frac {1}{a^3 (a+x)^2}-\frac {1}{a^4 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {\log (\sin (c+d x))}{a^4 d}-\frac {\log (1+\sin (c+d x))}{a^4 d}+\frac {1}{3 a d (a+a \sin (c+d x))^3}+\frac {1}{2 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {1}{d \left (a^4+a^4 \sin (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.64 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {6 \log (\sin (c+d x))-6 \log (1+\sin (c+d x))+\frac {11+15 \sin (c+d x)+6 \sin ^2(c+d x)}{(1+\sin (c+d x))^3}}{6 a^4 d} \]

[In]

Integrate[Cot[c + d*x]/(a + a*Sin[c + d*x])^4,x]

[Out]

(6*Log[Sin[c + d*x]] - 6*Log[1 + Sin[c + d*x]] + (11 + 15*Sin[c + d*x] + 6*Sin[c + d*x]^2)/(1 + Sin[c + d*x])^
3)/(6*a^4*d)

Maple [A] (verified)

Time = 3.50 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.57

method	result	size
derivativedivides	\(-\frac {\frac {3}{\csc \left (d x +c \right )+1}-\frac {3}{2 \left (\csc \left (d x +c \right )+1\right )^{2}}+\frac {1}{3 \left (\csc \left (d x +c \right )+1\right )^{3}}+\ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{4}}\)	\(55\)
default	\(-\frac {\frac {3}{\csc \left (d x +c \right )+1}-\frac {3}{2 \left (\csc \left (d x +c \right )+1\right )^{2}}+\frac {1}{3 \left (\csc \left (d x +c \right )+1\right )^{3}}+\ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{4}}\)	\(55\)
risch	\(\frac {2 i \left (-28 \,{\mathrm e}^{3 i \left (d x +c \right )}-15 i {\mathrm e}^{2 i \left (d x +c \right )}+15 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{4}}\)	\(123\)

[In]

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

[Out]

-1/d/a^4*(3/(csc(d*x+c)+1)-3/2/(csc(d*x+c)+1)^2+1/3/(csc(d*x+c)+1)^3+ln(csc(d*x+c)+1))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.57 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {6 \, \cos \left (d x + c\right )^{2} + 6 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 6 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, \sin \left (d x + c\right ) - 17}{6 \, {\left (3 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

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

[Out]

1/6*(6*cos(d*x + c)^2 + 6*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)*log(1/2*sin(d*x + c)) - 6
*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)*log(sin(d*x + c) + 1) - 15*sin(d*x + c) - 17)/(3*a
^4*d*cos(d*x + c)^2 - 4*a^4*d + (a^4*d*cos(d*x + c)^2 - 4*a^4*d)*sin(d*x + c))

Sympy [F]

\[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\int \frac {\cos {\left (c + d x \right )} \csc {\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

[In]

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

[Out]

Integral(cos(c + d*x)*csc(c + d*x)/(sin(c + d*x)**4 + 4*sin(c + d*x)**3 + 6*sin(c + d*x)**2 + 4*sin(c + d*x) +
1), x)/a**4

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\frac {6 \, \sin \left (d x + c\right )^{2} + 15 \, \sin \left (d x + c\right ) + 11}{a^{4} \sin \left (d x + c\right )^{3} + 3 \, a^{4} \sin \left (d x + c\right )^{2} + 3 \, a^{4} \sin \left (d x + c\right ) + a^{4}} - \frac {6 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {6 \, \log \left (\sin \left (d x + c\right )\right )}{a^{4}}}{6 \, d} \]

[In]

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

[Out]

1/6*((6*sin(d*x + c)^2 + 15*sin(d*x + c) + 11)/(a^4*sin(d*x + c)^3 + 3*a^4*sin(d*x + c)^2 + 3*a^4*sin(d*x + c)
+ a^4) - 6*log(sin(d*x + c) + 1)/a^4 + 6*log(sin(d*x + c))/a^4)/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.71 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4}} - \frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{4}} - \frac {6 \, \sin \left (d x + c\right )^{2} + 15 \, \sin \left (d x + c\right ) + 11}{a^{4} {\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{6 \, d} \]

[In]

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

[Out]

-1/6*(6*log(abs(sin(d*x + c) + 1))/a^4 - 6*log(abs(sin(d*x + c)))/a^4 - (6*sin(d*x + c)^2 + 15*sin(d*x + c) +
11)/(a^4*(sin(d*x + c) + 1)^3))/d

Mupad [B] (verification not implemented)

Time = 11.19 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.12 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^4\,d}-\frac {6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {80\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+15\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+20\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+15\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^4\right )} \]

[In]

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

[Out]

log(tan(c/2 + (d*x)/2))/(a^4*d) - (2*log(tan(c/2 + (d*x)/2) + 1))/(a^4*d) - (6*tan(c/2 + (d*x)/2) + 18*tan(c/2
+ (d*x)/2)^2 + (80*tan(c/2 + (d*x)/2)^3)/3 + 18*tan(c/2 + (d*x)/2)^4 + 6*tan(c/2 + (d*x)/2)^5)/(d*(15*a^4*tan
(c/2 + (d*x)/2)^2 + 20*a^4*tan(c/2 + (d*x)/2)^3 + 15*a^4*tan(c/2 + (d*x)/2)^4 + 6*a^4*tan(c/2 + (d*x)/2)^5 + a
^4*tan(c/2 + (d*x)/2)^6 + a^4 + 6*a^4*tan(c/2 + (d*x)/2)))