\(\int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [475]

Optimal result

Integrand size = 29, antiderivative size = 98 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}+\frac {10 \cos (c+d x)}{3 a d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 a^2 d} \] Output:

-2*arctanh(a^(1/2)*cos(d*x+c)/(a+a*sin(d*x+c))^(1/2))/a^(3/2)/d+10/3*cos(d 
*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)-2/3*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/a^2 
/d
 

Mathematica [A] (verified)

Time = 1.09 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.50 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (9 \cos \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {3}{2} (c+d x)\right )-3 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-9 \sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )\right )}{3 d (a (1+\sin (c+d x)))^{3/2}} \] Input:

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

Output:

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(9*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]] - 9*Sin[(c + d*x)/2] - Sin[(3*(c + d*x)) 
/2]))/(3*d*(a*(1 + Sin[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.37, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {3042, 3359, 3042, 3125, 3525, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^4}{\sin (c+d x) (a \sin (c+d x)+a)^{3/2}}dx\)

\(\Big \downarrow \) 3359

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3125

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

\(\Big \downarrow \) 3525

\(\displaystyle \frac {\frac {2 \int \frac {1}{2} \csc (c+d x) \sqrt {\sin (c+d x) a+a} (\sin (c+d x) a+3 a)dx}{3 a}-\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{a^2}+\frac {4 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \csc (c+d x) \sqrt {\sin (c+d x) a+a} (\sin (c+d x) a+3 a)dx}{3 a}-\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{a^2}+\frac {4 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\sqrt {\sin (c+d x) a+a} (\sin (c+d x) a+3 a)}{\sin (c+d x)}dx}{3 a}-\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{a^2}+\frac {4 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3460

\(\displaystyle \frac {\frac {3 a \int \csc (c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{3 a}-\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{a^2}+\frac {4 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 a \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{3 a}-\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{a^2}+\frac {4 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {-\frac {6 a^2 \int \frac {1}{a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{3 a}-\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{a^2}+\frac {4 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {-\frac {6 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{3 a}-\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{a^2}+\frac {4 \cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}}\)

Input:

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

Output:

(4*Cos[c + d*x])/(a*d*Sqrt[a + a*Sin[c + d*x]]) + ((-2*Cos[c + d*x]*Sqrt[a 
 + a*Sin[c + d*x]])/(3*d) + ((-6*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sq 
rt[a + a*Sin[c + d*x]]])/d - (2*a^2*Cos[c + d*x])/(d*Sqrt[a + a*Sin[c + d* 
x]]))/(3*a))/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 3125

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos 
[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq 
Q[a^2 - b^2, 0]
 

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 3359

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + 
(b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[-2/(a*b*d)   Int[(d* 
Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 2), x], x] + Simp[1/a^2   I 
nt[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 2)*(1 + Sin[e + f*x]^2), x] 
, x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1]
 

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]
 

rule 3525

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

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05

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

Input:

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

Output:

2/3/a^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)/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. 260 vs. \(2 (84) = 168\).

Time = 0.10 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.65 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {3 \, \sqrt {a} {\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right ) - 4 \, \cos \left (d x + c\right ) - 5\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{6 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \] Input:

integrate(cos(d*x+c)^3*cot(d*x+c)/(a+a*sin(d*x+c))^(3/2),x, algorithm="fri 
cas")
 

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) + 5)*sin(d*x + c) - 4*cos(d*x + c) - 
 5)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) + a 
^2*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

integrate(cos(d*x+c)^3*cot(d*x+c)/(a+a*sin(d*x+c))^(3/2),x, algorithm="max 
ima")
 

Output:

integrate(cos(d*x + c)^3*cot(d*x + c)/(a*sin(d*x + c) + a)^(3/2), x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.41 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {\sqrt {2} \sqrt {a} {\left (\frac {3 \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {4 \, {\left (2 \, a^{4} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{4} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{6 \, d} \] Input:

integrate(cos(d*x+c)^3*cot(d*x+c)/(a+a*sin(d*x+c))^(3/2),x, algorithm="gia 
c")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(a)*int((sqrt(sin(c + d*x) + 1)*cos(c + d*x)**3*cot(c + d*x))/(sin(c 
+ d*x)**2 + 2*sin(c + d*x) + 1),x))/a**2