\(\int \cos (c+d x) \cot (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [325]

Optimal result

Integrand size = 27, antiderivative size = 93 \[ \int \cos (c+d x) \cot (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}+\frac {2 a \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.54 \[ \int \cos (c+d x) \cot (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {a (1+\sin (c+d x))} \left (3 \cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )-3 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:

Integrate[Cos[c + d*x]*Cot[c + d*x]*Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

(Sqrt[a*(1 + Sin[c + d*x])]*(3*Cos[(c + d*x)/2] + Cos[(3*(c + d*x))/2] - 3 
*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 3*Log[1 - Cos[(c + d*x)/2] 
 + Sin[(c + d*x)/2]] - 3*Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2]))/(3*d*(C 
os[(c + d*x)/2] + Sin[(c + d*x)/2]))
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 3353, 3042, 3455, 27, 3042, 3460, 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.

\(\displaystyle \int \cos (c+d x) \cot (c+d x) \sqrt {a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3353

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3455

\(\displaystyle \frac {\frac {2}{3} \int \frac {1}{2} \csc (c+d x) \sqrt {\sin (c+d x) a+a} \left (3 a^2-a^2 \sin (c+d x)\right )dx+\frac {2 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \int \csc (c+d x) \sqrt {\sin (c+d x) a+a} \left (3 a^2-a^2 \sin (c+d x)\right )dx+\frac {2 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \frac {\sqrt {\sin (c+d x) a+a} \left (3 a^2-a^2 \sin (c+d x)\right )}{\sin (c+d x)}dx+\frac {2 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{a^2}\)

\(\Big \downarrow \) 3460

\(\displaystyle \frac {\frac {1}{3} \left (3 a^2 \int \csc (c+d x) \sqrt {\sin (c+d x) a+a}dx+\frac {2 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (3 a^2 \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx+\frac {2 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{a^2}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {6 a^3 \int \frac {1}{a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}\right )+\frac {2 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{a^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {2 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}+\frac {1}{3} \left (\frac {2 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {6 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}\right )}{a^2}\)

Input:

Int[Cos[c + d*x]*Cot[c + d*x]*Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

((2*a^2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d) + ((-6*a^(5/2)*ArcTan 
h[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d + (2*a^3*Cos[c + d*x 
])/(d*Sqrt[a + a*Sin[c + d*x]]))/3)/a^2
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 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 3455

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

rule 3460

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + 
b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b 
*d*(2*n + 3))   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] 
 /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 = 0.37 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.11

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (3 a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )+\left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-3 \sqrt {a -a \sin \left (d x +c \right )}\, a \right )}{3 a \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(103\)

Input:

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

Output:

-2/3*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(3*a^(3/2)*arctanh((a-a*sin( 
d*x+c))^(1/2)/a^(1/2))+(a-a*sin(d*x+c))^(3/2)-3*(a-a*sin(d*x+c))^(1/2)*a)/ 
a/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. 250 vs. \(2 (79) = 158\).

Time = 0.09 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.69 \[ \int \cos (c+d x) \cot (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {3 \, \sqrt {a} {\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{6 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:

integrate(cos(d*x+c)*cot(d*x+c)*(a+a*sin(d*x+c))^(1/2),x, algorithm="frica 
s")
 

Output:

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

Sympy [F]

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

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

Output:

Integral(sqrt(a*(sin(c + d*x) + 1))*cos(c + d*x)*cot(c + d*x), x)
 

Maxima [F]

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

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

Output:

integrate(sqrt(a*sin(d*x + c) + a)*cos(d*x + c)*cot(d*x + c), x)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.42 \[ \int \cos (c+d x) \cot (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} {\left (8 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 12 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{6 \, d} \] Input:

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

Output:

1/6*sqrt(2)*(8*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 
 1/2*c)^3 - 3*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)))*sgn(cos(-1/4*pi + 1/ 
2*d*x + 1/2*c)) - 12*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2 
*d*x + 1/2*c))*sqrt(a)/d
 

Mupad [F(-1)]

Timed out. \[ \int \cos (c+d x) \cot (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \cos \left (c+d\,x\right )\,\mathrm {cot}\left (c+d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)*cot(c + d*x),x)