\(\int \frac {\cot ^2(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [313]

Optimal result

Integrand size = 27, antiderivative size = 78 \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {5 \text {arctanh}(\cos (c+d x))}{2 a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))} \]

[Out]

-5/2*arctanh(cos(d*x+c))/a^2/d+2*cot(d*x+c)/a^2/d-1/2*cot(d*x+c)*csc(d*x+c)/a^2/d+2*cos(d*x+c)/a^2/d/(1+sin(d*
x+c))

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2953, 3045, 3855, 3852, 8, 3853, 2727} \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {5 \text {arctanh}(\cos (c+d x))}{2 a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}+\frac {2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d} \]

[In]

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

[Out]

(-5*ArcTanh[Cos[c + d*x]])/(2*a^2*d) + (2*Cot[c + d*x])/(a^2*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a^2*d) + (2*C
os[c + d*x])/(a^2*d*(1 + Sin[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2953

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[n, 0])

Rule 3045

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\csc ^3(c+d x) (a-a \sin (c+d x))}{a+a \sin (c+d x)} \, dx}{a^2} \\ & = \frac {\int \left (2 \csc (c+d x)-2 \csc ^2(c+d x)+\csc ^3(c+d x)-\frac {2}{1+\sin (c+d x)}\right ) \, dx}{a^2} \\ & = \frac {\int \csc ^3(c+d x) \, dx}{a^2}+\frac {2 \int \csc (c+d x) \, dx}{a^2}-\frac {2 \int \csc ^2(c+d x) \, dx}{a^2}-\frac {2 \int \frac {1}{1+\sin (c+d x)} \, dx}{a^2} \\ & = -\frac {2 \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))}+\frac {\int \csc (c+d x) \, dx}{2 a^2}+\frac {2 \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d} \\ & = -\frac {5 \text {arctanh}(\cos (c+d x))}{2 a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(364\) vs. \(2(78)=156\).

Time = 1.21 (sec) , antiderivative size = 364, normalized size of antiderivative = 4.67 \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {4 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}{d (a+a \sin (c+d x))^2}+\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}{d (a+a \sin (c+d x))^2}-\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}{8 d (a+a \sin (c+d x))^2}-\frac {5 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}{2 d (a+a \sin (c+d x))^2}+\frac {5 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}{2 d (a+a \sin (c+d x))^2}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}{8 d (a+a \sin (c+d x))^2}-\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4 \tan \left (\frac {1}{2} (c+d x)\right )}{d (a+a \sin (c+d x))^2} \]

[In]

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

[Out]

(-4*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)/(d*(a + a*Sin[c + d*x])^2) + (Cot[(c + d*x)/2]*(
Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)/(d*(a + a*Sin[c + d*x])^2) - (Csc[(c + d*x)/2]^2*(Cos[(c + d*x)/2] + S
in[(c + d*x)/2])^4)/(8*d*(a + a*Sin[c + d*x])^2) - (5*Log[Cos[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/
2])^4)/(2*d*(a + a*Sin[c + d*x])^2) + (5*Log[Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)/(2*d*(
a + a*Sin[c + d*x])^2) + (Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)/(8*d*(a + a*Sin[c + d*x]
)^2) - ((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*Tan[(c + d*x)/2])/(d*(a + a*Sin[c + d*x])^2)

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+10 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {16}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{2}}\)	\(87\)
default	\(\frac {\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+10 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {16}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{2}}\)	\(87\)
parallelrisch	\(\frac {20 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )-\left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(105\)
risch	\(\frac {5 \,{\mathrm e}^{4 i \left (d x +c \right )}-11 \,{\mathrm e}^{2 i \left (d x +c \right )}+5 i {\mathrm e}^{3 i \left (d x +c \right )}+8-3 i {\mathrm e}^{i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) d \,a^{2}}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d \,a^{2}}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d \,a^{2}}\)	\(124\)
norman	\(\frac {\frac {10 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {1}{8 a d}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}-\frac {5 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {37 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {71 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(169\)

[In]

int(cos(d*x+c)^2*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/4/d/a^2*(1/2*tan(1/2*d*x+1/2*c)^2-4*tan(1/2*d*x+1/2*c)-1/2/tan(1/2*d*x+1/2*c)^2+4/tan(1/2*d*x+1/2*c)+10*ln(t
an(1/2*d*x+1/2*c))+16/(tan(1/2*d*x+1/2*c)+1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (74) = 148\).

Time = 0.28 (sec) , antiderivative size = 246, normalized size of antiderivative = 3.15 \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {16 \, \cos \left (d x + c\right )^{3} + 10 \, \cos \left (d x + c\right )^{2} - 5 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 5 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (8 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) - 14 \, \cos \left (d x + c\right ) - 8}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - a^{2} d + {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/4*(16*cos(d*x + c)^3 + 10*cos(d*x + c)^2 - 5*(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x
+ c) - cos(d*x + c) - 1)*log(1/2*cos(d*x + c) + 1/2) + 5*(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 -
1)*sin(d*x + c) - cos(d*x + c) - 1)*log(-1/2*cos(d*x + c) + 1/2) - 2*(8*cos(d*x + c)^2 + 3*cos(d*x + c) - 4)*
sin(d*x + c) - 14*cos(d*x + c) - 8)/(a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2 - a^2*d*cos(d*x + c) - a^2*d
+ (a^2*d*cos(d*x + c)^2 - a^2*d)*sin(d*x + c))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 161 vs. \(2 (74) = 148\).

Time = 0.25 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.06 \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1}{\frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}} - \frac {\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}} + \frac {20 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{8 \, d} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/8*((7*sin(d*x + c)/(cos(d*x + c) + 1) + 40*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)/(a^2*sin(d*x + c)^2/(cos
(d*x + c) + 1)^2 + a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3) - (8*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c
)^2/(cos(d*x + c) + 1)^2)/a^2 + 20*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.49 \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {20 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{4}} + \frac {32}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}} - \frac {30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*(20*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + (a^2*tan(1/2*d*x + 1/2*c)^2 - 8*a^2*tan(1/2*d*x + 1/2*c))/a^4 + 3
2/(a^2*(tan(1/2*d*x + 1/2*c) + 1)) - (30*tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 1)/(a^2*tan(1/2*d*x
+ 1/2*c)^2))/d

Mupad [B] (verification not implemented)

Time = 9.55 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.54 \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}+\frac {5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^2\,d}+\frac {20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-\frac {1}{2}}{d\,\left (4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2\,d} \]

[In]

int(cos(c + d*x)^2/(sin(c + d*x)^3*(a + a*sin(c + d*x))^2),x)

[Out]

tan(c/2 + (d*x)/2)^2/(8*a^2*d) + (5*log(tan(c/2 + (d*x)/2)))/(2*a^2*d) + ((7*tan(c/2 + (d*x)/2))/2 + 20*tan(c/
2 + (d*x)/2)^2 - 1/2)/(d*(4*a^2*tan(c/2 + (d*x)/2)^2 + 4*a^2*tan(c/2 + (d*x)/2)^3)) - tan(c/2 + (d*x)/2)/(a^2*
d)