\(\int \cot (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\) [138]

Optimal result

Integrand size = 21, antiderivative size = 73 \[ \int \cot (c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {\sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d} \] Output:

2*a^(1/2)*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/d-2^(1/2)*a^(1/2)*arctan 
h(1/2*(a+a*sec(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99 \[ \int \cot (c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\frac {\left (2 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right )\right ) \sqrt {a (1+\sec (c+d x))}}{d \sqrt {1+\sec (c+d x)}} \] Input:

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

Output:

((2*ArcTanh[Sqrt[1 + Sec[c + d*x]]] - Sqrt[2]*ArcTanh[Sqrt[1 + Sec[c + d*x 
]]/Sqrt[2]])*Sqrt[a*(1 + Sec[c + d*x])])/(d*Sqrt[1 + Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 25, 4368, 25, 27, 97, 73, 219, 220}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4368

\(\displaystyle \frac {a^2 \int -\frac {\cos (c+d x)}{a (1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {a^2 \int \frac {\cos (c+d x)}{a (1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {a \int \frac {\cos (c+d x)}{(1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{d}\)

\(\Big \downarrow \) 97

\(\displaystyle -\frac {a \left (\int \frac {1}{(1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)+\int \frac {\cos (c+d x)}{\sqrt {\sec (c+d x) a+a}}d\sec (c+d x)\right )}{d}\)

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 219

\(\displaystyle -\frac {a \left (\frac {2 \int \frac {1}{\frac {\sec (c+d x) a+a}{a}-1}d\sqrt {\sec (c+d x) a+a}}{a}+\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a}}\right )}{d}\)

\(\Big \downarrow \) 220

\(\displaystyle -\frac {a \left (\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{\sqrt {a}}\right )}{d}\)

Input:

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

Output:

-((a*((-2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/Sqrt[a] + (Sqrt[2]*Ar 
cTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqrt[a])])/Sqrt[a]))/d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 97

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Simp[b/(b*c - a*d)   Int[(e + f*x)^p/(a + b*x), x], x] - Simp[d/(b*c 
 - a*d)   Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p}, 
 x] &&  !IntegerQ[p]
 

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 220

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

rule 3042

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

rule 4368

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n 
_), x_Symbol] :> Simp[-(d*b^(m - 1))^(-1)   Subst[Int[(-a + b*x)^((m - 1)/2 
)*((a + b*x)^((m - 1)/2 + n)/x), x], x, Csc[c + d*x]], x] /; FreeQ[{a, b, c 
, d, n}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]
 

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.30

method	result	size
default	\(-\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{2}\right )\right )}{d}\)	\(95\)

Input:

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

Output:

-1/d*(a*(1+sec(d*x+c)))^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(arctan 
(1/2*2^(1/2)/(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+2^(1/2)*arctan(1/2*2^(1/2 
)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 242, normalized size of antiderivative = 3.32 \[ \int \cot (c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\left [\frac {\sqrt {2} \sqrt {a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) - 1}\right ) + 2 \, \sqrt {a} \log \left (-2 \, a \cos \left (d x + c\right ) - 2 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - a\right )}{2 \, d}, \frac {\sqrt {2} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) - 2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right )}{d}\right ] \] Input:

integrate(cot(d*x+c)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

[1/2*(sqrt(2)*sqrt(a)*log(-(2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/co 
s(d*x + c))*cos(d*x + c) - 3*a*cos(d*x + c) - a)/(cos(d*x + c) - 1)) + 2*s 
qrt(a)*log(-2*a*cos(d*x + c) - 2*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x 
 + c))*cos(d*x + c) - a))/d, (sqrt(2)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqr 
t((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(a*cos(d*x + c) + a)) - 
2*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x 
 + c)/(a*cos(d*x + c) + a)))/d]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(cot(d*x+c)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.21 \[ \int \cot (c+d x) \sqrt {a+a \sec (c+d x)} \, dx=-\frac {\sqrt {2} {\left (\frac {\sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} - \frac {a \arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{d} \] Input:

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

Output:

-sqrt(2)*(sqrt(2)*a*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a) 
/sqrt(-a))/sqrt(-a) - a*arctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a 
))/sqrt(-a))*sgn(cos(d*x + c))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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