\(\int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [436]

Optimal result

Integrand size = 29, antiderivative size = 54 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3 \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {\cot (c+d x)}{a^3 d}-\frac {4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))} \]

[Out]

3*arctanh(cos(d*x+c))/a^3/d-cot(d*x+c)/a^3/d-4*cos(d*x+c)/a^3/d/(1+sin(d*x+c))

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2954, 2951, 3855, 3852, 8, 2727} \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3 \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {\cot (c+d x)}{a^3 d}-\frac {4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)} \]

[In]

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

[Out]

(3*ArcTanh[Cos[c + d*x]])/(a^3*d) - Cot[c + d*x]/(a^3*d) - (4*Cos[c + d*x])/(a^3*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 2951

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

Rule 2954

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

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 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 \csc ^2(c+d x) \sec ^2(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6} \\ & = \frac {\int \left (-3 a \csc (c+d x)+a \csc ^2(c+d x)+\frac {4 a}{1+\sin (c+d x)}\right ) \, dx}{a^4} \\ & = \frac {\int \csc ^2(c+d x) \, dx}{a^3}-\frac {3 \int \csc (c+d x) \, dx}{a^3}+\frac {4 \int \frac {1}{1+\sin (c+d x)} \, dx}{a^3} \\ & = \frac {3 \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}-\frac {\text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d} \\ & = \frac {3 \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {\cot (c+d x)}{a^3 d}-\frac {4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(156\) vs. \(2(54)=108\).

Time = 1.18 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.89 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right ) \left (-17+\cot ^2\left (\frac {1}{2} (c+d x)\right )-6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+6 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\cot \left (\frac {1}{2} (c+d x)\right ) \left (1-6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+6 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \tan \left (\frac {1}{2} (c+d x)\right )}{2 a^3 d (1+\sin (c+d x))^3} \]

[In]

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

[Out]

-1/2*((Cos[(c + d*x)/2]*(-17 + Cot[(c + d*x)/2]^2 - 6*Log[Cos[(c + d*x)/2]] + 6*Log[Sin[(c + d*x)/2]] + Cot[(c
+ d*x)/2]*(1 - 6*Log[Cos[(c + d*x)/2]] + 6*Log[Sin[(c + d*x)/2]])) - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Si
n[(c + d*x)/2])^5*Tan[(c + d*x)/2])/(a^3*d*(1 + Sin[c + d*x])^3)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {16}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{2 d \,a^{3}}\)	\(59\)
default	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {16}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{2 d \,a^{3}}\)	\(59\)
parallelrisch	\(\frac {\left (-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-6\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\cot \left (\frac {d x}{2}+\frac {c}{2}\right )-18}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(70\)
risch	\(-\frac {2 \left (-5+i {\mathrm e}^{i \left (d x +c \right )}+4 \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) d \,a^{3}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{3}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}\)	\(102\)
norman	\(\frac {-\frac {13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {153 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {107 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {1}{2 a d}+\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}-\frac {50 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {15 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {180 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {158 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {203 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}-\frac {109 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}\)	\(260\)

[In]

int(cos(d*x+c)^4*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/2/d/a^3*(tan(1/2*d*x+1/2*c)-1/tan(1/2*d*x+1/2*c)-6*ln(tan(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. 165 vs. \(2 (54) = 108\).

Time = 0.29 (sec) , antiderivative size = 165, normalized size of antiderivative = 3.06 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {10 \, \cos \left (d x + c\right )^{2} + 3 \, {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (5 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right ) - 8}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d - {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (54) = 108\).

Time = 0.22 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.15 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {\frac {17 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1}{\frac {a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {\sin \left (d x + c\right )}{a^{3} {\left (\cos \left (d x + c\right ) + 1\right )}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.67 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {6 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} - \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 14 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a^{3}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(6*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - tan(1/2*d*x + 1/2*c)/a^3 - (3*tan(1/2*d*x + 1/2*c)^2 - 14*tan(1/2
*d*x + 1/2*c) - 1)/((tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c))*a^3))/d

Mupad [B] (verification not implemented)

Time = 10.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.61 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d}-\frac {17\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1}{d\,\left (2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d} \]

[In]

int(cos(c + d*x)^4/(sin(c + d*x)^2*(a + a*sin(c + d*x))^3),x)

[Out]

tan(c/2 + (d*x)/2)/(2*a^3*d) - (17*tan(c/2 + (d*x)/2) + 1)/(d*(2*a^3*tan(c/2 + (d*x)/2)^2 + 2*a^3*tan(c/2 + (d
*x)/2))) - (3*log(tan(c/2 + (d*x)/2)))/(a^3*d)