\(\int \cot (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\) [150]

Optimal result

Integrand size = 21, antiderivative size = 73 \[ \int \cot (c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d} \] Output:

2*a^(3/2)*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/d-2*2^(1/2)*a^(3/2)*arct 
anh(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 \[ \int \cot (c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\frac {\left (2 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )-2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right )\right ) (a (1+\sec (c+d x)))^{3/2}}{d (1+\sec (c+d x))^{3/2}} \] Input:

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

Output:

((2*ArcTanh[Sqrt[1 + Sec[c + d*x]]] - 2*Sqrt[2]*ArcTanh[Sqrt[1 + Sec[c + d 
*x]]/Sqrt[2]])*(a*(1 + Sec[c + d*x]))^(3/2))/(d*(1 + Sec[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, 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, 94, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^{3/2}}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )}dx\)

\(\Big \downarrow \) 4368

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 94

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 220

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

Input:

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

Output:

-((a*(-2*Sqrt[a]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]] + 2*Sqrt[2]*Sqr 
t[a]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*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 94

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

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 = 0.79 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.34

method	result	size
default	\(-\frac {a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{2}\right )+2 \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )\right )}{d}\)	\(98\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (58) = 116\).

Time = 0.09 (sec) , antiderivative size = 243, normalized size of antiderivative = 3.33 \[ \int \cot (c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\left [\frac {\sqrt {2} a^{\frac {3}{2}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) - 1}\right ) + a^{\frac {3}{2}} \log \left (-2 \, a \cos \left (d x + c\right ) - 2 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - a\right )}{d}, \frac {2 \, {\left (\sqrt {2} \sqrt {-a} a \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) - \sqrt {-a} a \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right )\right )}}{d}\right ] \] Input:

integrate(cot(d*x+c)*(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

[(sqrt(2)*a^(3/2)*log(-(2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d* 
x + c))*cos(d*x + c) - 3*a*cos(d*x + c) - a)/(cos(d*x + c) - 1)) + a^(3/2) 
*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, 2*(sqrt(2)*sqrt(-a)*a*arctan(sqrt(2)*sqrt(-a)*sqrt( 
(a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(a*cos(d*x + c) + a)) - sq 
rt(-a)*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]

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

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

Output:

Integral((a*(sec(c + d*x) + 1))**(3/2)*cot(c + d*x), x)
 

Maxima [F]

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

integrate(cot(d*x+c)*(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.22 \[ \int \cot (c+d x) (a+a \sec (c+d x))^{3/2} \, dx=-\frac {\sqrt {2} {\left (\frac {\sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} - \frac {2 \, a \arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}}\right )} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{d} \] Input:

integrate(cot(d*x+c)*(a+a*sec(d*x+c))^(3/2),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) - 2*a*arctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt( 
-a))/sqrt(-a))*a*sgn(cos(d*x + c))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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