\(\int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx\) [169]

Optimal result

Integrand size = 23, antiderivative size = 96 \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^2 \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 a \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2}}{3 d} \] Output:

2*a^(5/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2*a^2*cot(d* 
x+c)*(a+a*sec(d*x+c))^(1/2)/d-2/3*a*cot(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)/d
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.84 \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=-\frac {2 \left (\frac {1}{1+\cos (c+d x)}\right )^{3/2} \cot ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) (a (1+\sec (c+d x)))^{5/2}}{3 d \sqrt {\frac {1}{1+\sec (c+d x)}}} \] Input:

Integrate[Cot[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2),x]
 

Output:

(-2*((1 + Cos[c + d*x])^(-1))^(3/2)*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 
 -3/2, -1/2, 2*Sin[(c + d*x)/2]^2]*(a*(1 + Sec[c + d*x]))^(5/2))/(3*d*Sqrt 
[(1 + Sec[c + d*x])^(-1)])
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4375, 264, 264, 216}

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.

\(\displaystyle \int \cot ^4(c+d x) (a \sec (c+d x)+a)^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\cot \left (c+d x+\frac {\pi }{2}\right )^4}dx\)

\(\Big \downarrow \) 4375

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 216

\(\displaystyle -\frac {2 a \left (\frac {1}{3} \cot ^3(c+d x) (a \sec (c+d x)+a)^{3/2}-a \left (\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )+\cot (c+d x) \sqrt {a \sec (c+d x)+a}\right )\right )}{d}\)

Input:

Int[Cot[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2),x]
 

Output:

(-2*a*((Cot[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2))/3 - a*(Sqrt[a]*ArcTan[( 
Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]] + Cot[c + d*x]*Sqrt[a + a* 
Sec[c + d*x]])))/d
 

Defintions of rubi rules used

rule 216

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

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 3042

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

rule 4375

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(213\) vs. \(2(84)=168\).

Time = 91.42 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.23

method	result	size
default	\(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (3 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+\sqrt {2}\, \left (7 \cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )-5\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cot \left (d x +c \right ) \csc \left (d x +c \right )^{2}+\left (6 \cos \left (d x +c \right )^{2}-12 \cos \left (d x +c \right )+6\right ) \cot \left (d x +c \right ) \csc \left (d x +c \right )^{2}\right )}{3 d}\)	\(214\)

Input:

int(cot(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/3/d*a^2*(a*(1+sec(d*x+c)))^(1/2)*(3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2) 
*2^(1/2)*arctanh(2^(1/2)/(cot(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+csc(d*x+c)^ 
2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c)))+2^(1/2)*(7*cos(d*x+c)^2+2*cos(d*x+c)-5 
)*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)* 
cot(d*x+c)*csc(d*x+c)^2+(6*cos(d*x+c)^2-12*cos(d*x+c)+6)*cot(d*x+c)*csc(d* 
x+c)^2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 355, normalized size of antiderivative = 3.70 \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\left [\frac {3 \, {\left (a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \sqrt {-a} \log \left (-\frac {8 \, a \cos \left (d x + c\right )^{3} - 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right ) + 1}\right ) \sin \left (d x + c\right ) + 4 \, {\left (4 \, a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{6 \, {\left (d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right )}, \frac {3 \, {\left (a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \sqrt {a} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) - a}\right ) \sin \left (d x + c\right ) + 2 \, {\left (4 \, a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{3 \, {\left (d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right )}\right ] \] Input:

integrate(cot(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

[1/6*(3*(a^2*cos(d*x + c) - a^2)*sqrt(-a)*log(-(8*a*cos(d*x + c)^3 - 4*(2* 
cos(d*x + c)^2 - cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x 
+ c))*sin(d*x + c) - 7*a*cos(d*x + c) + a)/(cos(d*x + c) + 1))*sin(d*x + c 
) + 4*(4*a^2*cos(d*x + c)^2 - 3*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a 
)/cos(d*x + c)))/((d*cos(d*x + c) - d)*sin(d*x + c)), 1/3*(3*(a^2*cos(d*x 
+ c) - a^2)*sqrt(a)*arctan(2*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c 
))*cos(d*x + c)*sin(d*x + c)/(2*a*cos(d*x + c)^2 + a*cos(d*x + c) - a))*si 
n(d*x + c) + 2*(4*a^2*cos(d*x + c)^2 - 3*a^2*cos(d*x + c))*sqrt((a*cos(d*x 
 + c) + a)/cos(d*x + c)))/((d*cos(d*x + c) - d)*sin(d*x + c))]
 

Sympy [F(-1)]

Timed out. \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)**4*(a+a*sec(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (84) = 168\).

Time = 1.01 (sec) , antiderivative size = 311, normalized size of antiderivative = 3.24 \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=-\frac {\frac {3 \, \sqrt {-a} a^{3} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{{\left | a \right |}} + \frac {\sqrt {2} {\left (9 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} \sqrt {-a} a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 12 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} \sqrt {-a} a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 7 \, \sqrt {-a} a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a\right )}^{3}}}{3 \, d} \] Input:

integrate(cot(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

-1/3*(3*sqrt(-a)*a^3*log(abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*ta 
n(1/2*d*x + 1/2*c)^2 + a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(-a)*tan 
(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*sqrt(2)*abs 
(a) - 6*a))*sgn(cos(d*x + c))/abs(a) + sqrt(2)*(9*(sqrt(-a)*tan(1/2*d*x + 
1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*sqrt(-a)*a^3*sgn(cos(d*x + 
 c)) - 12*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 
+ a))^2*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 7*sqrt(-a)*a^5*sgn(cos(d*x + c))) 
/((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 
- a)^3)/d
 

Mupad [F(-1)]

Timed out. \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \] Input:

int(cot(c + d*x)^4*(a + a/cos(c + d*x))^(5/2),x)
 

Output:

int(cot(c + d*x)^4*(a + a/cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}d x +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{4} \sec \left (d x +c \right )d x \right )+\int \sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{4}d x \right ) \] Input:

int(cot(d*x+c)^4*(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)**4*sec(c + d*x)**2,x 
) + 2*int(sqrt(sec(c + d*x) + 1)*cot(c + d*x)**4*sec(c + d*x),x) + int(sqr 
t(sec(c + d*x) + 1)*cot(c + d*x)**4,x))