\(\int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx\) [1058]

Optimal result

Integrand size = 25, antiderivative size = 90 \[ \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=a b x-\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {\left (2 a^2-b^2\right ) \cos (c+d x)}{3 d}+\frac {a b \cos (c+d x) \sin (c+d x)}{3 d}+\frac {\cos (c+d x) (a+b \sin (c+d x))^2}{3 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.01 \[ \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {3 \left (4 a^2-b^2\right ) \cos (c+d x)-b^2 \cos (3 (c+d x))+6 a \left (2 b c+2 b d x-2 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+b \sin (2 (c+d x))\right )}{12 d} \] Input:

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

Output:

(3*(4*a^2 - b^2)*Cos[c + d*x] - b^2*Cos[3*(c + d*x)] + 6*a*(2*b*c + 2*b*d* 
x - 2*a*Log[Cos[(c + d*x)/2]] + 2*a*Log[Sin[(c + d*x)/2]] + b*Sin[2*(c + d 
*x)]))/(12*d)
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 3368, 3042, 3529, 3042, 3512, 27, 3042, 3502, 27, 3042, 3214, 3042, 4257}

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.

\(\displaystyle \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3368

\(\displaystyle \int \left (1-\sin ^2(c+d x)\right ) \csc (c+d x) (a+b \sin (c+d x))^2dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3529

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3512

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

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

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 3042

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

rule 3214

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

rule 3368

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

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3512

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

rule 3529

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/(d*(m + n + 2))   Int[(a + b*Sin[e + f*x 
])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*( 
n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + C* 
(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f 
, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 
0])))
 

rule 4257

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

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {a^{2} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+2 a b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {b^{2} \cos \left (d x +c \right )^{3}}{3}}{d}\)	\(72\)
default	\(\frac {a^{2} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+2 a b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {b^{2} \cos \left (d x +c \right )^{3}}{3}}{d}\)	\(72\)
risch	\(a b x +\frac {a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {{\mathrm e}^{i \left (d x +c \right )} b^{2}}{8 d}+\frac {a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{8 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}-\frac {\cos \left (3 d x +3 c \right ) b^{2}}{12 d}+\frac {a b \sin \left (2 d x +2 c \right )}{2 d}\)	\(146\)

Input:

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

Output:

1/d*(a^2*(cos(d*x+c)+ln(csc(d*x+c)-cot(d*x+c)))+2*a*b*(1/2*sin(d*x+c)*cos( 
d*x+c)+1/2*d*x+1/2*c)-1/3*b^2*cos(d*x+c)^3)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.93 \[ \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {2 \, b^{2} \cos \left (d x + c\right )^{3} - 6 \, a b d x - 6 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, a^{2} \cos \left (d x + c\right ) + 3 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{6 \, d} \] Input:

integrate(cos(d*x+c)*cot(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="fricas")
 

Output:

-1/6*(2*b^2*cos(d*x + c)^3 - 6*a*b*d*x - 6*a*b*cos(d*x + c)*sin(d*x + c) - 
 6*a^2*cos(d*x + c) + 3*a^2*log(1/2*cos(d*x + c) + 1/2) - 3*a^2*log(-1/2*c 
os(d*x + c) + 1/2))/d
 

Sympy [F]

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

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

Output:

Integral((a + b*sin(c + d*x))**2*cos(c + d*x)*cot(c + d*x), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.82 \[ \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {2 \, b^{2} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b - 3 \, a^{2} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{6 \, d} \] Input:

integrate(cos(d*x+c)*cot(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.48 \[ \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {3 \, {\left (d x + c\right )} a b + 3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2} + b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \] Input:

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

Output:

1/3*(3*(d*x + c)*a*b + 3*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(3*a*b*tan 
(1/2*d*x + 1/2*c)^5 - 3*a^2*tan(1/2*d*x + 1/2*c)^4 + 3*b^2*tan(1/2*d*x + 1 
/2*c)^4 - 6*a^2*tan(1/2*d*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1/2*c) - 3*a^ 
2 + b^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d
 

Mupad [B] (verification not implemented)

Time = 34.39 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.50 \[ \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^2-2\,b^2\right )+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^2-\frac {2\,b^2}{3}-2\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2\,a\,b\,\mathrm {atan}\left (\frac {4\,a^2\,b^2}{4\,a^3\,b-4\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {4\,a^3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^3\,b-4\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \] Input:

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

Output:

(a^2*log(tan(c/2 + (d*x)/2)))/d + (tan(c/2 + (d*x)/2)^4*(2*a^2 - 2*b^2) + 
4*a^2*tan(c/2 + (d*x)/2)^2 + 2*a^2 - (2*b^2)/3 - 2*a*b*tan(c/2 + (d*x)/2)^ 
5 + 2*a*b*tan(c/2 + (d*x)/2))/(d*(3*tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d* 
x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) + (2*a*b*atan((4*a^2*b^2)/(4*a^3*b - 
4*a^2*b^2*tan(c/2 + (d*x)/2)) + (4*a^3*b*tan(c/2 + (d*x)/2))/(4*a^3*b - 4* 
a^2*b^2*tan(c/2 + (d*x)/2))))/d
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.07 \[ \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{2}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b +3 \cos \left (d x +c \right ) a^{2}-\cos \left (d x +c \right ) b^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}-3 a^{2}+3 a b c +3 a b d x +b^{2}}{3 d} \] Input:

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

Output:

(cos(c + d*x)*sin(c + d*x)**2*b**2 + 3*cos(c + d*x)*sin(c + d*x)*a*b + 3*c 
os(c + d*x)*a**2 - cos(c + d*x)*b**2 + 3*log(tan((c + d*x)/2))*a**2 - 3*a* 
*2 + 3*a*b*c + 3*a*b*d*x + b**2)/(3*d)