\(\int \frac {\cot ^2(c+d x) \csc (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [341]

Optimal result

Integrand size = 29, antiderivative size = 100 \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 \sqrt {a} d}+\frac {\cot (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 d \sqrt {a+a \sin (c+d x)}} \] Output:

1/4*arctanh(a^(1/2)*cos(d*x+c)/(a+a*sin(d*x+c))^(1/2))/a^(1/2)/d+1/4*cot(d 
*x+c)/d/(a+a*sin(d*x+c))^(1/2)-1/2*cot(d*x+c)*csc(d*x+c)/d/(a+a*sin(d*x+c) 
)^(1/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(272\) vs. \(2(100)=200\).

Time = 2.54 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.72 \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (-8+4 \cot \left (\frac {1}{4} (c+d x)\right )-\csc ^2\left (\frac {1}{4} (c+d x)\right )+4 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\sec ^2\left (\frac {1}{4} (c+d x)\right )+\frac {2}{\left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}-\frac {8 \sin \left (\frac {1}{4} (c+d x)\right )}{\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )}-\frac {2}{\left (\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}+\frac {8 \sin \left (\frac {1}{4} (c+d x)\right )}{\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )}+4 \tan \left (\frac {1}{4} (c+d x)\right )\right )}{32 d \sqrt {a (1+\sin (c+d x))}} \] Input:

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

Output:

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-8 + 4*Cot[(c + d*x)/4] - Csc[(c + 
 d*x)/4]^2 + 4*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 4*Log[1 - Co 
s[(c + d*x)/2] + Sin[(c + d*x)/2]] + Sec[(c + d*x)/4]^2 + 2/(Cos[(c + d*x) 
/4] - Sin[(c + d*x)/4])^2 - (8*Sin[(c + d*x)/4])/(Cos[(c + d*x)/4] - Sin[( 
c + d*x)/4]) - 2/(Cos[(c + d*x)/4] + Sin[(c + d*x)/4])^2 + (8*Sin[(c + d*x 
)/4])/(Cos[(c + d*x)/4] + Sin[(c + d*x)/4]) + 4*Tan[(c + d*x)/4]))/(32*d*S 
qrt[a*(1 + Sin[c + d*x])])
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 3353, 3042, 3459, 3042, 3251, 3042, 3252, 219}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^2}{\sin (c+d x)^3 \sqrt {a \sin (c+d x)+a}}dx\)

\(\Big \downarrow \) 3353

\(\displaystyle \frac {\int \csc ^3(c+d x) (a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}}{\sin (c+d x)^3}dx}{a^2}\)

\(\Big \downarrow \) 3459

\(\displaystyle \frac {-\frac {1}{4} a \int \csc ^2(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{4} a \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)^2}dx-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}}{a^2}\)

\(\Big \downarrow \) 3251

\(\displaystyle \frac {-\frac {1}{4} a \left (\frac {1}{2} \int \csc (c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {a \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{4} a \left (\frac {1}{2} \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx-\frac {a \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}}{a^2}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {-\frac {1}{4} a \left (-\frac {a \int \frac {1}{a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}-\frac {a \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}}{a^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}-\frac {1}{4} a \left (-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {a \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{a^2}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3251

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e 
+ f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim 
p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2)))   Int[Sqrt[a + b*Sin[e 
+ f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x 
] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 
-1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
 

rule 3252

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

rule 3353

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

rule 3459

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) 
*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* 
c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d))   Int[Sqrt[a + b*Sin[e + f*x] 
]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x 
] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 
-1]
 

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.24

method	result	size
default	\(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (\left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} a^{\frac {3}{2}}-\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) a^{3} \sin \left (d x +c \right )^{2}+\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {5}{2}}\right )}{4 a^{\frac {7}{2}} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(124\)

Input:

int(cot(d*x+c)^2*csc(d*x+c)/(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE 
)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (84) = 168\).

Time = 0.09 (sec) , antiderivative size = 320, normalized size of antiderivative = 3.20 \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{16 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d + {\left (a d \cos \left (d x + c\right )^{2} - a d\right )} \sin \left (d x + c\right )\right )}} \] Input:

integrate(cot(d*x+c)^2*csc(d*x+c)/(a+a*sin(d*x+c))^(1/2),x, algorithm="fri 
cas")
 

Output:

1/16*((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)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 
4*(cos(d*x + c)^2 + (cos(d*x + c) + 3)*sin(d*x + c) - 2*cos(d*x + c) - 3)* 
sqrt(a*sin(d*x + c) + a)*sqrt(a) - 9*a*cos(d*x + c) + (a*cos(d*x + c)^2 + 
8*a*cos(d*x + c) - a)*sin(d*x + c) - a)/(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)) - 4*(cos(d*x + c)^ 
2 + (cos(d*x + c) + 3)*sin(d*x + c) - 2*cos(d*x + c) - 3)*sqrt(a*sin(d*x + 
 c) + a))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2 - a*d*cos(d*x + c) - a* 
d + (a*d*cos(d*x + c)^2 - a*d)*sin(d*x + c))
 

Sympy [F]

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

integrate(cot(d*x+c)**2*csc(d*x+c)/(a+a*sin(d*x+c))**(1/2),x)
 

Output:

Integral(cot(c + d*x)**2*csc(c + d*x)/sqrt(a*(sin(c + d*x) + 1)), x)
 

Maxima [F]

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

integrate(cot(d*x+c)^2*csc(d*x+c)/(a+a*sin(d*x+c))^(1/2),x, algorithm="max 
ima")
 

Output:

integrate(cot(d*x + c)^2*csc(d*x + c)/sqrt(a*sin(d*x + c) + a), x)
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.49 \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\sqrt {2} \sqrt {a} {\left (\frac {\sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {4 \, {\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{16 \, d} \] Input:

integrate(cot(d*x+c)^2*csc(d*x+c)/(a+a*sin(d*x+c))^(1/2),x, algorithm="gia 
c")
 

Output:

1/16*sqrt(2)*sqrt(a)*(sqrt(2)*log(abs(-2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x 
 + 1/2*c))/abs(2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x + 1/2*c)))/(a*sgn(cos(- 
1/4*pi + 1/2*d*x + 1/2*c))) + 4*(2*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 + sin( 
-1/4*pi + 1/2*d*x + 1/2*c))/((2*sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2*a* 
sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int -\frac {{\sin \left (c+d\,x\right )}^2-1}{{\sin \left (c+d\,x\right )}^3\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \] Input:

int(cot(c + d*x)^2/(sin(c + d*x)*(a + a*sin(c + d*x))^(1/2)),x)
 

Output:

int(-(sin(c + d*x)^2 - 1)/(sin(c + d*x)^3*(a + a*sin(c + d*x))^(1/2)), x)
 

Reduce [F]

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

int(cot(d*x+c)^2*csc(d*x+c)/(a+a*sin(d*x+c))^(1/2),x)
 

Output:

(sqrt(a)*int((sqrt(sin(c + d*x) + 1)*cot(c + d*x)**2*csc(c + d*x))/(sin(c 
+ d*x) + 1),x))/a