\(\int \frac {\cot (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\) [326]

Optimal result

Integrand size = 21, antiderivative size = 106 \[ \int \frac {\cot (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b} d} \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.94 \[ \int \frac {\cot (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}}{d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 25, 4373, 561, 1484, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4373

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

\(\Big \downarrow \) 561

\(\displaystyle -\frac {2 b^2 \int \frac {1}{\left (a-b^2 \sec ^2(c+d x)\right ) \left (b^4 \sec ^4(c+d x)-2 a b^2 \sec ^2(c+d x)+a^2-b^2\right )}d\sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 1484

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-2*b^2*(-(ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]]/(Sqrt[a]*b^2)) + ArcT 
anh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]]/(2*Sqrt[a - b]*b^2) + ArcTanh[Sq 
rt[a + b*Sec[c + d*x]]/Sqrt[a + b]]/(2*b^2*Sqrt[a + b])))/d
 

Defintions of rubi rules used

rule 25

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

rule 561

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{k = Denominator[n]}, Simp[k/d   Subst[Int[x^(k*(n + 1) - 1)*(-c 
/d + x^k/d)^m*Simp[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^k/d^2) + b*(x^(2*k)/d^2), 
 x]^p, x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, m, p}, x] && Frac 
tionQ[n] && IntegerQ[p] && IntegerQ[m]
 

rule 1484

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symb 
ol] :> Int[ExpandIntegrand[(d + e*x^2)^q/(a + b*x^2 + c*x^4), x], x] /; Fre 
eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 
 0] && IntegerQ[q]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4373

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(658\) vs. \(2(88)=176\).

Time = 0.81 (sec) , antiderivative size = 659, normalized size of antiderivative = 6.22

method	result	size
default	\(\frac {\left (2 a^{\frac {3}{2}} \ln \left (4 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sqrt {a}\, \cos \left (d x +c \right )+4 \sqrt {a}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}+4 a \cos \left (d x +c \right )+2 b \right ) \left (a -b \right )^{\frac {3}{2}}+2 \sqrt {a}\, \ln \left (4 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sqrt {a}\, \cos \left (d x +c \right )+4 \sqrt {a}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}+4 a \cos \left (d x +c \right )+2 b \right ) \left (a -b \right )^{\frac {3}{2}} b -\sqrt {a +b}\, \ln \left (-\frac {2 \left (2 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sqrt {a +b}\, \cos \left (d x +c \right )+2 \sqrt {a +b}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}+2 a \cos \left (d x +c \right )+b \cos \left (d x +c \right )+b \right )}{-1+\cos \left (d x +c \right )}\right ) \left (a -b \right )^{\frac {3}{2}} a +a^{3} \ln \left (\frac {2 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sqrt {a -b}\, \cos \left (d x +c \right )+2 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sqrt {a -b}-2 a \cos \left (d x +c \right )+b \cos \left (d x +c \right )-b}{\sqrt {a -b}\, \left (1+\cos \left (d x +c \right )\right )}\right )-\ln \left (\frac {2 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sqrt {a -b}\, \cos \left (d x +c \right )+2 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sqrt {a -b}-2 a \cos \left (d x +c \right )+b \cos \left (d x +c \right )-b}{\sqrt {a -b}\, \left (1+\cos \left (d x +c \right )\right )}\right ) a \,b^{2}\right ) \sqrt {a +b \sec \left (d x +c \right )}\, \cos \left (d x +c \right )}{2 d \left (a -b \right )^{\frac {3}{2}} \left (a +b \right ) a \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}}\)	\(659\)

Input:

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

Output:

1/2/d/(a-b)^(3/2)/(a+b)/a*(2*a^(3/2)*ln(4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+ 
cos(d*x+c))^2)^(1/2)*a^(1/2)*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d* 
x+c)/(1+cos(d*x+c))^2)^(1/2)+4*a*cos(d*x+c)+2*b)*(a-b)^(3/2)+2*a^(1/2)*ln( 
4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^(1/2)*cos(d*x+c)+ 
4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+4*a*cos(d*x 
+c)+2*b)*(a-b)^(3/2)*b-(a+b)^(1/2)*ln(-2*(2*((b+a*cos(d*x+c))*cos(d*x+c)/( 
1+cos(d*x+c))^2)^(1/2)*(a+b)^(1/2)*cos(d*x+c)+2*(a+b)^(1/2)*((b+a*cos(d*x+ 
c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+2*a*cos(d*x+c)+b*cos(d*x+c)+b)/(-1+ 
cos(d*x+c)))*(a-b)^(3/2)*a+a^3*ln(1/(a-b)^(1/2)*(2*((b+a*cos(d*x+c))*cos(d 
*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)*cos(d*x+c)+2*((b+a*cos(d*x+c))*c 
os(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)- 
b)/(1+cos(d*x+c)))-ln(1/(a-b)^(1/2)*(2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos 
(d*x+c))^2)^(1/2)*(a-b)^(1/2)*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1 
+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)-b)/(1+cos(d* 
x+c)))*a*b^2)*(a+b*sec(d*x+c))^(1/2)*cos(d*x+c)/(1+cos(d*x+c))/((b+a*cos(d 
*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (88) = 176\).

Time = 1.23 (sec) , antiderivative size = 2420, normalized size of antiderivative = 22.83 \[ \int \frac {\cot (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/4*(2*(a^2 - b^2)*sqrt(a)*log(-8*a^2*cos(d*x + c)^2 - 8*a*b*cos(d*x + c) 
 - b^2 - 4*(2*a*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + 
 c) + b)/cos(d*x + c))) + (a^2 + a*b)*sqrt(a - b)*log(-((8*a^2 - 8*a*b + b 
^2)*cos(d*x + c)^2 + b^2 - 4*((2*a - b)*cos(d*x + c)^2 + b*cos(d*x + c))*s 
qrt(a - b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)) + 2*(4*a*b - 3*b^2)*cos 
(d*x + c))/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + (a^2 - a*b)*sqrt(a + b 
)*log(-((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^2 + b^2 - 4*((2*a + b)*cos(d*x 
+ c)^2 + b*cos(d*x + c))*sqrt(a + b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c 
)) + 2*(4*a*b + 3*b^2)*cos(d*x + c))/(cos(d*x + c)^2 - 2*cos(d*x + c) + 1) 
))/((a^3 - a*b^2)*d), -1/4*(4*(a^2 - b^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt( 
(a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b)) - 
(a^2 + a*b)*sqrt(a - b)*log(-((8*a^2 - 8*a*b + b^2)*cos(d*x + c)^2 + b^2 - 
 4*((2*a - b)*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a - b)*sqrt((a*cos(d*x 
 + c) + b)/cos(d*x + c)) + 2*(4*a*b - 3*b^2)*cos(d*x + c))/(cos(d*x + c)^2 
 + 2*cos(d*x + c) + 1)) - (a^2 - a*b)*sqrt(a + b)*log(-((8*a^2 + 8*a*b + b 
^2)*cos(d*x + c)^2 + b^2 - 4*((2*a + b)*cos(d*x + c)^2 + b*cos(d*x + c))*s 
qrt(a + b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)) + 2*(4*a*b + 3*b^2)*cos 
(d*x + c))/(cos(d*x + c)^2 - 2*cos(d*x + c) + 1)))/((a^3 - a*b^2)*d), -1/4 
*(2*(a^2 + a*b)*sqrt(-a + b)*arctan(-2*sqrt(-a + b)*sqrt((a*cos(d*x + c) + 
 b)/cos(d*x + c))*cos(d*x + c)/((2*a - b)*cos(d*x + c) + b)) - 2*(a^2 -...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\cot (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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