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

Optimal result

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

-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
 

Mathematica [A] (verified)

Time = 1.86 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.57 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4 \left (-2 (c+d x)-8 \cos (c+d x)-2 \cot \left (\frac {1}{2} (c+d x)\right )+8 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-8 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\sin (2 (c+d x))+2 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{4 d (a+a \sin (c+d x))^2} \] Input:

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

Output:

((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]] + S 
in[2*(c + d*x)] + 2*Tan[(c + d*x)/2]))/(4*d*(a + a*Sin[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3354, 3042, 3188, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^6}{\sin (c+d x)^2 (a \sin (c+d x)+a)^2}dx\)

\(\Big \downarrow \) 3354

\(\displaystyle \frac {\int \cot ^2(c+d x) (a-a \sin (c+d x))^2dx}{a^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^2}{\tan (c+d x)^2}dx}{a^4}\)

\(\Big \downarrow \) 3188

\(\displaystyle \frac {\int \left (\csc ^2(c+d x) a^4-\sin ^2(c+d x) a^4-2 \csc (c+d x) a^4+2 \sin (c+d x) a^4\right )dx}{a^6}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {2 a^4 \text {arctanh}(\cos (c+d x))}{d}-\frac {2 a^4 \cos (c+d x)}{d}-\frac {a^4 \cot (c+d x)}{d}+\frac {a^4 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {a^4 x}{2}}{a^6}\)

Input:

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

Output:

(-1/2*(a^4*x) + (2*a^4*ArcTanh[Cos[c + d*x]])/d - (2*a^4*Cos[c + d*x])/d - 
 (a^4*Cot[c + d*x])/d + (a^4*Cos[c + d*x]*Sin[c + d*x])/(2*d))/a^6
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3188

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_ 
), x_Symbol] :> Simp[a^p   Int[ExpandIntegrand[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 3354

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(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]
 

Maple [A] (verified)

Time = 1.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.51

method	result	size
derivativedivides	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-8}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(112\)
default	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-8}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(112\)
risch	\(-\frac {x}{2 a^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 d \,a^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2}}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{d \,a^{2}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d \,a^{2}}-\frac {2 i}{a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{2}}\)	\(140\)

Input:

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

Output:

1/2/d/a^2*(tan(1/2*d*x+1/2*c)+16*(-1/8*tan(1/2*d*x+1/2*c)^3-1/2*tan(1/2*d* 
x+1/2*c)^2+1/8*tan(1/2*d*x+1/2*c)-1/2)/(1+tan(1/2*d*x+1/2*c)^2)^2-2*arctan 
(tan(1/2*d*x+1/2*c))-1/tan(1/2*d*x+1/2*c)-4*ln(tan(1/2*d*x+1/2*c)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.19 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\cos \left (d x + c\right )^{3} + {\left (d x + 4 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 2 \, \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 2 \, \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + \cos \left (d x + c\right )}{2 \, a^{2} d \sin \left (d x + c\right )} \] Input:

integrate(cos(d*x+c)^4*cot(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="frica 
s")
 

Output:

-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*log(-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]

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

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

Output:

Integral(cos(c + d*x)**4*cot(c + d*x)**2/(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. 202 vs. \(2 (70) = 140\).

Time = 0.12 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.73 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1}{\frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {4 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {\sin \left (d x + c\right )}{a^{2} {\left (\cos \left (d x + c\right ) + 1\right )}}}{2 \, d} \] Input:

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

Output:

-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)

Time = 0.17 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.77 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {d x + c}{a^{2}} + \frac {4 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} - \frac {4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}}}{2 \, d} \] Input:

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

Output:

-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 = 32.71 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.36 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1}{d\,\left (2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}+\frac {\mathrm {atan}\left (\frac {1}{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-4}+\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-4}\right )}{a^2\,d} \] Input:

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

Output:

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*t 
an(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))/(t 
an(c/2 + (d*x)/2) - 4))/(a^2*d)
 

Reduce [F]

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

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

Output:

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