\(\int \cot ^4(c+d x) (a+a \sec (c+d x))^{3/2} \, dx\) [157]

Optimal result

Integrand size = 23, antiderivative size = 144 \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\frac {2 a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}-\frac {a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} d}+\frac {3 a \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{2 d}-\frac {\cot ^3(c+d x) (a+a \sec (c+d x))^{3/2}}{3 d} \] Output:

2*a^(3/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d-1/4*a^(3/2)* 
arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)/d+3/ 
2*a*cot(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d-1/3*cot(d*x+c)^3*(a+a*sec(d*x+c))^ 
(3/2)/d
 

Mathematica [A] (warning: unable to verify)

Time = 6.56 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.57 \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=-\frac {\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )-4 \sqrt {2} \arctan \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}}}\right )\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {1+\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}}{8 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {3}{2}}(c+d x)}+\frac {\cos (c+d x) \sec ^3\left (\frac {1}{2} (c+d x)\right ) (a (1+\sec (c+d x)))^{3/2} \left (\frac {13}{24} \csc \left (\frac {1}{2} (c+d x)\right )-\frac {1}{24} \csc ^3\left (\frac {1}{2} (c+d x)\right )-\frac {11}{12} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{d} \] Input:

Integrate[Cot[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2),x]
 

Output:

-1/8*((ArcSin[Tan[(c + d*x)/2]] - 4*Sqrt[2]*ArcTan[Tan[(c + d*x)/2]/Sqrt[C 
os[c + d*x]/(1 + Cos[c + d*x])]])*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Se 
c[(c + d*x)/2]^4*Sqrt[1 + Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))/(d*S 
qrt[Sec[(c + d*x)/2]^2]*Sec[c + d*x]^(3/2)) + (Cos[c + d*x]*Sec[(c + d*x)/ 
2]^3*(a*(1 + Sec[c + d*x]))^(3/2)*((13*Csc[(c + d*x)/2])/24 - Csc[(c + d*x 
)/2]^3/24 - (11*Sin[(c + d*x)/2])/12))/d
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4375, 382, 27, 445, 27, 397, 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.

\(\displaystyle \int \cot ^4(c+d x) (a \sec (c+d x)+a)^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}{\cot \left (c+d x+\frac {\pi }{2}\right )^4}dx\)

\(\Big \downarrow \) 4375

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

\(\Big \downarrow \) 382

\(\displaystyle -\frac {2 \left (\frac {1}{6} \int -\frac {3 a \cot ^2(c+d x) (\sec (c+d x) a+a) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+3\right )}{\left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right ) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )+\frac {1}{6} \cot ^3(c+d x) (a \sec (c+d x)+a)^{3/2}\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2 \left (\frac {1}{6} \cot ^3(c+d x) (a \sec (c+d x)+a)^{3/2}-\frac {1}{2} a \int \frac {\cot ^2(c+d x) (\sec (c+d x) a+a) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+3\right )}{\left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right ) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )\right )}{d}\)

\(\Big \downarrow \) 445

\(\displaystyle -\frac {2 \left (\frac {1}{6} \cot ^3(c+d x) (a \sec (c+d x)+a)^{3/2}-\frac {1}{2} a \left (\frac {3}{2} \cot (c+d x) \sqrt {a \sec (c+d x)+a}-\frac {1}{2} \int \frac {a \left (\frac {3 a \tan ^2(c+d x)}{\sec (c+d x) a+a}+7\right )}{\left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right ) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )\right )\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2 \left (\frac {1}{6} \cot ^3(c+d x) (a \sec (c+d x)+a)^{3/2}-\frac {1}{2} a \left (\frac {3}{2} \cot (c+d x) \sqrt {a \sec (c+d x)+a}-\frac {1}{2} a \int \frac {\frac {3 a \tan ^2(c+d x)}{\sec (c+d x) a+a}+7}{\left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right ) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )\right )\right )}{d}\)

\(\Big \downarrow \) 397

\(\displaystyle -\frac {2 \left (\frac {1}{6} \cot ^3(c+d x) (a \sec (c+d x)+a)^{3/2}-\frac {1}{2} a \left (\frac {3}{2} \cot (c+d x) \sqrt {a \sec (c+d x)+a}-\frac {1}{2} a \left (4 \int \frac {1}{\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )-\int \frac {1}{\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )\right )\right )\right )}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {2 \left (\frac {1}{6} \cot ^3(c+d x) (a \sec (c+d x)+a)^{3/2}-\frac {1}{2} a \left (\frac {3}{2} \cot (c+d x) \sqrt {a \sec (c+d x)+a}-\frac {1}{2} a \left (\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {2} \sqrt {a}}-\frac {4 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a}}\right )\right )\right )}{d}\)

Input:

Int[Cot[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2),x]
 

Output:

(-2*((Cot[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2))/6 - (a*(-1/2*(a*((-4*ArcT 
an[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/Sqrt[a] + ArcTan[(Sqr 
t[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]/(Sqrt[2]*Sqrt[a]))) 
 + (3*Cot[c + d*x]*Sqrt[a + a*Sec[c + d*x]])/2))/2))/d
 

Defintions of rubi rules used

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 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 382

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) 
, x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/ 
(a*c*e*(m + 1))), x] - Simp[1/(a*c*e^2*(m + 1))   Int[(e*x)^(m + 2)*(a + b* 
x^2)^p*(c + d*x^2)^q*Simp[(b*c + a*d)*(m + 3) + 2*(b*c*p + a*d*q) + b*d*(m 
+ 2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[ 
b*c - a*d, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
 

rule 397

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

rule 445

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

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] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(315\) vs. \(2(119)=238\).

Time = 1.07 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.19

method	result	size
default	\(\frac {a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {2}\, \left (37 \cos \left (d x +c \right )^{3}+51 \cos \left (d x +c \right )^{2}-21 \cos \left (d x +c \right )-35\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cot \left (d x +c \right ) \csc \left (d x +c \right )^{2}+\left (30 \cos \left (d x +c \right )^{3}-24 \cos \left (d x +c \right )^{2}-42 \cos \left (d x +c \right )+36\right ) \cot \left (d x +c \right ) \csc \left (d x +c \right )^{2}-24 \left (1+\cos \left (d x +c \right )\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )-6 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{24 d \left (1+\cos \left (d x +c \right )\right )}\)	\(316\)

Input:

int(cot(d*x+c)^4*(a+a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/24/d*a*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))*(2^(1/2)*(37*cos(d*x+c)^3 
+51*cos(d*x+c)^2-21*cos(d*x+c)-35)*(-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+(30*cos(d*x+c)^3- 
24*cos(d*x+c)^2-42*cos(d*x+c)+36)*cot(d*x+c)*csc(d*x+c)^2-24*(1+cos(d*x+c) 
)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(2^(1/2)*(-csc(d*x+c 
)+cot(d*x+c))/(cot(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+csc(d*x+c)^2-1)^(1/2)) 
-6*(1+cos(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln((-2*cos(d*x+c)/( 
1+cos(d*x+c)))^(1/2)-cot(d*x+c)+csc(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 525, normalized size of antiderivative = 3.65 \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\left [\frac {3 \, \sqrt {\frac {1}{2}} {\left (a \cos \left (d x + c\right ) - a\right )} \sqrt {-a} \log \left (\frac {4 \, \sqrt {\frac {1}{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 ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) \sin \left (d x + c\right ) + 6 \, {\left (a \cos \left (d x + c\right ) - a\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 ) + 2 \, {\left (11 \, a \cos \left (d x + c\right )^{2} - 9 \, a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{12 \, {\left (d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right )}, \frac {3 \, \sqrt {\frac {1}{2}} {\left (a \cos \left (d x + c\right ) - a\right )} \sqrt {a} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 6 \, {\left (a \cos \left (d x + c\right ) - a\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 ) + {\left (11 \, a \cos \left (d x + c\right )^{2} - 9 \, a \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 )}\right ] \] Input:

integrate(cot(d*x+c)^4*(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

[1/12*(3*sqrt(1/2)*(a*cos(d*x + c) - a)*sqrt(-a)*log((4*sqrt(1/2)*sqrt(-a) 
*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + 3*a*c 
os(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1 
))*sin(d*x + c) + 6*(a*cos(d*x + c) - a)*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) + 2*(11*a*cos(d*x + c)^2 - 9*a*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)), 1/6*(3*sqrt(1/2 
)*(a*cos(d*x + c) - a)*sqrt(a)*arctan(2*sqrt(1/2)*sqrt((a*cos(d*x + c) + a 
)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))*sin(d*x + c) + 6*(a*c 
os(d*x + c) - a)*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 
))*sin(d*x + c) + (11*a*cos(d*x + c)^2 - 9*a*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. \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)**4*(a+a*sec(d*x+c))**(3/2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)^4*(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 369 vs. \(2 (119) = 238\).

Time = 0.81 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.56 \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=-\frac {3 \, \sqrt {2} \sqrt {-a} a \log \left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \frac {24 \, \sqrt {-a} a^{2} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{{\left | a \right |}} + \frac {8 \, \sqrt {2} {\left (6 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} \sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 9 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} \sqrt {-a} a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, \sqrt {-a} a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a\right )}^{3}}}{24 \, d} \] Input:

integrate(cot(d*x+c)^4*(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

-1/24*(3*sqrt(2)*sqrt(-a)*a*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*t 
an(1/2*d*x + 1/2*c)^2 + a))^2)*sgn(cos(d*x + c)) + 24*sqrt(-a)*a^2*log(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)/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) + 8*sqrt(2)*(6*(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^2*sgn(cos(d*x + c)) - 9*(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^3*sgn(co 
s(d*x + c)) + 5*sqrt(-a)*a^4*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. \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:

int(cot(c + d*x)^4*(a + a/cos(c + d*x))^(3/2),x)
                                                                                    
                                                                                    
 

Output:

int(cot(c + d*x)^4*(a + a/cos(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\sqrt {a}\, a \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{4} \sec \left (d x +c \right )d x +\int \sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{4}d x \right ) \] Input:

int(cot(d*x+c)^4*(a+a*sec(d*x+c))^(3/2),x)
 

Output:

sqrt(a)*a*(int(sqrt(sec(c + d*x) + 1)*cot(c + d*x)**4*sec(c + d*x),x) + in 
t(sqrt(sec(c + d*x) + 1)*cot(c + d*x)**4,x))