\(\int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [239]

Optimal result

Integrand size = 27, antiderivative size = 101 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {3 \csc (c+d x)}{a^2 d}+\frac {\csc ^2(c+d x)}{a^2 d}-\frac {\csc ^3(c+d x)}{3 a^2 d}-\frac {4 \log (\sin (c+d x))}{a^2 d}+\frac {4 \log (1+\sin (c+d x))}{a^2 d}-\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 101, 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} \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )}-\frac {\csc ^3(c+d x)}{3 a^2 d}+\frac {\csc ^2(c+d x)}{a^2 d}-\frac {3 \csc (c+d x)}{a^2 d}-\frac {4 \log (\sin (c+d x))}{a^2 d}+\frac {4 \log (\sin (c+d x)+1)}{a^2 d} \]

[In]

Int[(Cot[c + d*x]*Csc[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

(-3*Csc[c + d*x])/(a^2*d) + Csc[c + d*x]^2/(a^2*d) - Csc[c + d*x]^3/(3*a^2*d) - (4*Log[Sin[c + d*x]])/(a^2*d)
+ (4*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^4}{x^4 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {a^3 \text {Subst}\left (\int \frac {1}{x^4 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^3 \text {Subst}\left (\int \left (\frac {1}{a^2 x^4}-\frac {2}{a^3 x^3}+\frac {3}{a^4 x^2}-\frac {4}{a^5 x}+\frac {1}{a^4 (a+x)^2}+\frac {4}{a^5 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {3 \csc (c+d x)}{a^2 d}+\frac {\csc ^2(c+d x)}{a^2 d}-\frac {\csc ^3(c+d x)}{3 a^2 d}-\frac {4 \log (\sin (c+d x))}{a^2 d}+\frac {4 \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 = 0.66 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.97 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {3 \csc (c+d x)}{a^2 d}+\frac {\csc ^2(c+d x)}{a^2 d}-\frac {\csc ^3(c+d x)}{3 a^2 d}-\frac {4 \log (\sin (c+d x))}{a^2 d}+\frac {4 \log (1+\sin (c+d x))}{a^2 d}-\frac {1}{a^2 d (1+\sin (c+d x))} \]

[In]

Integrate[(Cot[c + d*x]*Csc[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.60

method	result	size
derivativedivides	\(-\frac {\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\left (\csc ^{2}\left (d x +c \right )\right )+3 \csc \left (d x +c \right )-\frac {1}{\csc \left (d x +c \right )+1}-4 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{2}}\)	\(61\)
default	\(-\frac {\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\left (\csc ^{2}\left (d x +c \right )\right )+3 \csc \left (d x +c \right )-\frac {1}{\csc \left (d x +c \right )+1}-4 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{2}}\)	\(61\)
risch	\(-\frac {8 i \left (3 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}-8 i {\mathrm e}^{4 i \left (d x +c \right )}-7 \,{\mathrm e}^{5 i \left (d x +c \right )}+3 i {\mathrm e}^{2 i \left (d x +c \right )}+7 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d \,a^{2}}+\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}-\frac {4 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{2}}\)	\(160\)
parallelrisch	\(\frac {192 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-96 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-28 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-28 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+114 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}\)	\(162\)
norman	\(\frac {-\frac {1}{24 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}-\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d a}+\frac {33 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {33 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}+\frac {8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}\)	\(208\)

[In]

int(cos(d*x+c)*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.93 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {6 \, \cos \left (d x + c\right )^{2} - 12 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 12 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) - 7}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d - {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*(6*cos(d*x + c)^2 - 12*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^2 - 1)*sin(d*x + c) + 1)*log(1/2
*sin(d*x + c)) + 12*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^2 - 1)*sin(d*x + c) + 1)*log(sin(d*x +
c) + 1) + 2*(6*cos(d*x + c)^2 - 5)*sin(d*x + c) - 7)/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d -
(a^2*d*cos(d*x + c)^2 - a^2*d)*sin(d*x + c))

Sympy [F]

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

[In]

integrate(cos(d*x+c)*csc(d*x+c)**4/(a+a*sin(d*x+c))**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.89 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {12 \, \sin \left (d x + c\right )^{3} + 6 \, \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}{a^{2} \sin \left (d x + c\right )^{4} + a^{2} \sin \left (d x + c\right )^{3}} - \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {12 \, \log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{3 \, d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/3*((12*sin(d*x + c)^3 + 6*sin(d*x + c)^2 - 2*sin(d*x + c) + 1)/(a^2*sin(d*x + c)^4 + a^2*sin(d*x + c)^3) -
12*log(sin(d*x + c) + 1)/a^2 + 12*log(sin(d*x + c))/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.02 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {12 \, \log \left ({\left | -\frac {a}{a \sin \left (d x + c\right ) + a} + 1 \right |}\right )}{a^{2}} + \frac {3}{{\left (a \sin \left (d x + c\right ) + a\right )} a} + \frac {\frac {30 \, a}{a \sin \left (d x + c\right ) + a} - \frac {18 \, a^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} - 13}{a^{2} {\left (\frac {a}{a \sin \left (d x + c\right ) + a} - 1\right )}^{3}}}{3 \, d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/3*(12*log(abs(-a/(a*sin(d*x + c) + a) + 1))/a^2 + 3/((a*sin(d*x + c) + a)*a) + (30*a/(a*sin(d*x + c) + a) -
18*a^2/(a*sin(d*x + c) + a)^2 - 13)/(a^2*(a/(a*sin(d*x + c) + a) - 1)^3))/d

Mupad [B] (verification not implemented)

Time = 9.72 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.00 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^2\,d}-\frac {-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {1}{3}}{d\,\left (8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}+\frac {8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^2\,d}-\frac {13\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^2\,d} \]

[In]

int(cos(c + d*x)/(sin(c + d*x)^4*(a + a*sin(c + d*x))^2),x)

[Out]

tan(c/2 + (d*x)/2)^2/(4*a^2*d) - tan(c/2 + (d*x)/2)^3/(24*a^2*d) - ((28*tan(c/2 + (d*x)/2)^2)/3 - (4*tan(c/2 +
(d*x)/2))/3 + 24*tan(c/2 + (d*x)/2)^3 - 3*tan(c/2 + (d*x)/2)^4 + 1/3)/(d*(8*a^2*tan(c/2 + (d*x)/2)^3 + 16*a^2
*tan(c/2 + (d*x)/2)^4 + 8*a^2*tan(c/2 + (d*x)/2)^5)) - (4*log(tan(c/2 + (d*x)/2)))/(a^2*d) + (8*log(tan(c/2 +
(d*x)/2) + 1))/(a^2*d) - (13*tan(c/2 + (d*x)/2))/(8*a^2*d)