\(\int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [638]

Optimal result

Integrand size = 29, antiderivative size = 74 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {x}{2 a^2}+\frac {2 \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {2 \cos (c+d x)}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2954, 2788, 3855, 3852, 8, 2718, 2715} \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {2 \cos (c+d x)}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}+\frac {\sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {x}{2 a^2} \]

[In]

Int[(Cos[c + d*x]^4*Cot[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 8

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

Rule 2715

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

Rule 2718

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

Rule 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/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 \cot ^2(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4} \\ & = \frac {\int \left (-2 a^4 \csc (c+d x)+a^4 \csc ^2(c+d x)+2 a^4 \sin (c+d x)-a^4 \sin ^2(c+d x)\right ) \, dx}{a^6} \\ & = \frac {\int \csc ^2(c+d x) \, dx}{a^2}-\frac {\int \sin ^2(c+d x) \, dx}{a^2}-\frac {2 \int \csc (c+d x) \, dx}{a^2}+\frac {2 \int \sin (c+d x) \, dx}{a^2} \\ & = \frac {2 \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {2 \cos (c+d x)}{a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {\int 1 \, dx}{2 a^2}-\frac {\text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d} \\ & = -\frac {x}{2 a^2}+\frac {2 \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {2 \cos (c+d x)}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.07 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.57 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4 \left (-2 (c+d x)-8 \cos (c+d x)-2 \cot \left (\frac {1}{2} (c+d x)\right )+8 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-8 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\sin (2 (c+d x))+2 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{4 d (a+a \sin (c+d x))^2} \]

[In]

Integrate[(Cos[c + d*x]^4*Cot[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(-2*(c + d*x) - 8*Cos[c + d*x] - 2*Cot[(c + d*x)/2] + 8*Log[Cos[(c +
d*x)/2]] - 8*Log[Sin[(c + d*x)/2]] + Sin[2*(c + d*x)] + 2*Tan[(c + d*x)/2]))/(4*d*(a + a*Sin[c + d*x])^2)

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.16

method	result	size
parallelrisch	\(\frac {-8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 \cos \left (d x +c \right )-\cos \left (2 d x +2 c \right )-5\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-2 d x +2 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-8 \cos \left (d x +c \right )-8}{4 d \,a^{2}}\)	\(86\)
derivativedivides	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-8}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(112\)
default	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-8}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(112\)
risch	\(-\frac {x}{2 a^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 d \,a^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2}}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{d \,a^{2}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d \,a^{2}}-\frac {2 i}{a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{2}}\)	\(140\)
norman	\(\frac {-\frac {9 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {7 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {11 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {3 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {11 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {13 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {9 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {13 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {7 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {3 x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {1}{2 a d}-\frac {x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {5 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}-\frac {37 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {73 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {30 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {73 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}-\frac {101 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}-\frac {151 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}-\frac {81 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {61 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {25 \left (\tan ^{10}\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 )^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\)	\(500\)

[In]

int(cos(d*x+c)^6*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/4*(-8*ln(tan(1/2*d*x+1/2*c))+(2*cos(d*x+c)-cos(2*d*x+2*c)-5)*cot(1/2*d*x+1/2*c)-2*d*x+2*sec(1/2*d*x+1/2*c)*c
sc(1/2*d*x+1/2*c)-8*cos(d*x+c)-8)/d/a^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.19 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\cos \left (d x + c\right )^{3} + {\left (d x + 4 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 2 \, \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 2 \, \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + \cos \left (d x + c\right )}{2 \, a^{2} d \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/2*(cos(d*x + c)^3 + (d*x + 4*cos(d*x + c))*sin(d*x + c) - 2*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 2*lo
g(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + cos(d*x + c))/(a^2*d*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**6*csc(d*x+c)**2/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (70) = 140\).

Time = 0.31 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.73 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1}{\frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {4 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {\sin \left (d x + c\right )}{a^{2} {\left (\cos \left (d x + c\right ) + 1\right )}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.77 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {d x + c}{a^{2}} + \frac {4 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} - \frac {4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.87 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.36 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1}{d\,\left (2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}+\frac {\mathrm {atan}\left (\frac {1}{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-4}+\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-4}\right )}{a^2\,d} \]

[In]

int(cos(c + d*x)^6/(sin(c + d*x)^2*(a + a*sin(c + d*x))^2),x)

[Out]

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