\(\int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx\) [162]

Optimal result

Integrand size = 21, antiderivative size = 95 \[ \int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {4 \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 a^2 \sqrt {a+a \sec (c+d x)}}{d} \] Output:

2*a^(5/2)*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/d-4*2^(1/2)*a^(5/2)*arct 
anh(1/2*(a+a*sec(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d+2*a^2*(a+a*sec(d*x+c))^( 
1/2)/d
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.87 \[ \int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {2 (a (1+\sec (c+d x)))^{5/2} \left (\text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )-2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right )+\sqrt {1+\sec (c+d x)}\right )}{d (1+\sec (c+d x))^{5/2}} \] Input:

Integrate[Cot[c + d*x]*(a + a*Sec[c + d*x])^(5/2),x]
 

Output:

(2*(a*(1 + Sec[c + d*x]))^(5/2)*(ArcTanh[Sqrt[1 + Sec[c + d*x]]] - 2*Sqrt[ 
2]*ArcTanh[Sqrt[1 + Sec[c + d*x]]/Sqrt[2]] + Sqrt[1 + Sec[c + d*x]]))/(d*( 
1 + Sec[c + d*x])^(5/2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 25, 4368, 25, 27, 95, 25, 27, 174, 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.

\(\displaystyle \int \cot (c+d x) (a \sec (c+d x)+a)^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\cot \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^{5/2}}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )}dx\)

\(\Big \downarrow \) 4368

\(\displaystyle \frac {a^2 \int -\frac {\cos (c+d x) (\sec (c+d x) a+a)^{3/2}}{a (1-\sec (c+d x))}d\sec (c+d x)}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {a^2 \int \frac {\cos (c+d x) (\sec (c+d x) a+a)^{3/2}}{a (1-\sec (c+d x))}d\sec (c+d x)}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {a \int \frac {\cos (c+d x) (\sec (c+d x) a+a)^{3/2}}{1-\sec (c+d x)}d\sec (c+d x)}{d}\)

\(\Big \downarrow \) 95

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

\(\displaystyle -\frac {a \left (a^2 \left (\frac {8 \int \frac {1}{2-\frac {\sec (c+d x) a+a}{a}}d\sqrt {\sec (c+d x) a+a}}{a}+\frac {2 \int \frac {1}{\frac {\sec (c+d x) a+a}{a}-1}d\sqrt {\sec (c+d x) a+a}}{a}\right )-2 a \sqrt {a \sec (c+d x)+a}\right )}{d}\)

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 220

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

Input:

Int[Cot[c + d*x]*(a + a*Sec[c + d*x])^(5/2),x]
 

Output:

-((a*(a^2*((-2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/Sqrt[a] + (4*Sqr 
t[2]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqrt[a])])/Sqrt[a]) - 2*a*S 
qrt[a + a*Sec[c + d*x]]))/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 95

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Simp[f*((e + f*x)^(p - 1)/(b*d*(p - 1))), x] + Simp[1/(b*d)   Int[(b 
*d*e^2 - a*c*f^2 + f*(2*b*d*e - b*c*f - a*d*f)*x)*((e + f*x)^(p - 2)/((a + 
b*x)*(c + d*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 1]
 

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 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 = 2.53 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.27

method	result	size
default	\(\frac {\left (-\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-4 \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+2\right ) a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d}\)	\(121\)

Input:

int(cot(d*x+c)*(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/d*(-2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*(-2 
*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-4*arctan(1/2*2^(1/2)/(-cos(d*x+c)/(1+cos 
(d*x+c)))^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+2)*a^2*(a*(1+sec(d*x 
+c)))^(1/2)
                                                                                    
                                                                                    
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.16 \[ \int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\left [\frac {2 \, \sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) - 1}\right ) + a^{\frac {5}{2}} \log \left (-2 \, a \cos \left (d x + c\right ) - 2 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - a\right ) + 2 \, a^{2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{d}, \frac {2 \, {\left (2 \, \sqrt {2} \sqrt {-a} a^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) - \sqrt {-a} a^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) + a^{2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}\right )}}{d}\right ] \] Input:

integrate(cot(d*x+c)*(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

[(2*sqrt(2)*a^(5/2)*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^(5/ 
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) + 2*a^2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/d, 2 
*(2*sqrt(2)*sqrt(-a)*a^2*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)) - sqrt(-a)*a^2*arctan(sq 
rt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(a*cos(d*x + c 
) + a)) + a^2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/d]
 

Sympy [F(-1)]

Timed out. \[ \int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)*(a+a*sec(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right ) \,d x } \] Input:

integrate(cot(d*x+c)*(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate((a*sec(d*x + c) + a)^(5/2)*cot(d*x + c), x)
 

Giac [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.18 \[ \int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=-\frac {\sqrt {2} {\left (\frac {\sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} - \frac {4 \, a \arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {2 \, a}{\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{d} \] Input:

integrate(cot(d*x+c)*(a+a*sec(d*x+c))^(5/2),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) - 4*a*arctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt( 
-a))/sqrt(-a) - 2*a/sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))*a^2*sgn(cos(d*x + 
 c))/d
 

Mupad [F(-1)]

Timed out. \[ \int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int \mathrm {cot}\left (c+d\,x\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \] Input:

int(cot(c + d*x)*(a + a/cos(c + d*x))^(5/2),x)
 

Output:

int(cot(c + d*x)*(a + a/cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right ) \sec \left (d x +c \right )^{2}d x +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right ) \sec \left (d x +c \right )d x \right )+\int \sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right )d x \right ) \] Input:

int(cot(d*x+c)*(a+a*sec(d*x+c))^(5/2),x)
 

Output:

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