\(\int \cot ^5(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [394]

Optimal result

Integrand size = 25, antiderivative size = 159 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{f}-\frac {\left (8 a^2+4 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 \sqrt {a+b} f}+\frac {(4 a-b) \cot ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {(a+b) \cot ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 f} \] Output:

a^(3/2)*arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2))/f-1/8*(8*a^2+4*a*b-b^2)* 
arctanh((a+b*sec(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(1/2)/f+1/8*(4*a-b)*co 
t(f*x+e)^2*(a+b*sec(f*x+e)^2)^(1/2)/f-1/4*(a+b)*cot(f*x+e)^4*(a+b*sec(f*x+ 
e)^2)^(1/2)/f
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 4.00 (sec) , antiderivative size = 684, normalized size of antiderivative = 4.30 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \cos ^3(e+f x) \left (-\frac {\left (1+e^{2 i (e+f x)}\right ) \left (b \left (1+6 e^{2 i (e+f x)}+e^{4 i (e+f x)}\right )+a \left (6-4 e^{2 i (e+f x)}+6 e^{4 i (e+f x)}\right )\right )}{\left (-1+e^{2 i (e+f x)}\right )^4}+\frac {-8 i a^{3/2} \sqrt {a+b} f x+\left (8 a^2+4 a b-b^2\right ) \log \left (1-e^{2 i (e+f x)}\right )+4 a^{3/2} \sqrt {a+b} \log \left (a+2 b+a e^{2 i (e+f x)}+\sqrt {a} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )+4 a^{3/2} \sqrt {a+b} \log \left (a+a e^{2 i (e+f x)}+2 b e^{2 i (e+f x)}+\sqrt {a} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )-8 a^2 \log \left (a+b+a e^{2 i (e+f x)}+b e^{2 i (e+f x)}+\sqrt {a+b} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )-4 a b \log \left (a+b+a e^{2 i (e+f x)}+b e^{2 i (e+f x)}+\sqrt {a+b} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )+b^2 \log \left (a+b+a e^{2 i (e+f x)}+b e^{2 i (e+f x)}+\sqrt {a+b} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}\right )}{\sqrt {a+b} \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 \sqrt {2} f (a+2 b+a \cos (2 e+2 f x))^{3/2}} \] Input:

Integrate[Cot[e + f*x]^5*(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

(E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I)*(e + 
f*x))]*Cos[e + f*x]^3*(-(((1 + E^((2*I)*(e + f*x)))*(b*(1 + 6*E^((2*I)*(e 
+ f*x)) + E^((4*I)*(e + f*x))) + a*(6 - 4*E^((2*I)*(e + f*x)) + 6*E^((4*I) 
*(e + f*x)))))/(-1 + E^((2*I)*(e + f*x)))^4) + ((-8*I)*a^(3/2)*Sqrt[a + b] 
*f*x + (8*a^2 + 4*a*b - b^2)*Log[1 - E^((2*I)*(e + f*x))] + 4*a^(3/2)*Sqrt 
[a + b]*Log[a + 2*b + a*E^((2*I)*(e + f*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(e 
 + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]] + 4*a^(3/2)*Sqrt[a + b]*Log[a + 
 a*E^((2*I)*(e + f*x)) + 2*b*E^((2*I)*(e + f*x)) + Sqrt[a]*Sqrt[4*b*E^((2* 
I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]] - 8*a^2*Log[a + b + a*E^(( 
2*I)*(e + f*x)) + b*E^((2*I)*(e + f*x)) + Sqrt[a + b]*Sqrt[4*b*E^((2*I)*(e 
 + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]] - 4*a*b*Log[a + b + a*E^((2*I)* 
(e + f*x)) + b*E^((2*I)*(e + f*x)) + Sqrt[a + b]*Sqrt[4*b*E^((2*I)*(e + f* 
x)) + a*(1 + E^((2*I)*(e + f*x)))^2]] + b^2*Log[a + b + a*E^((2*I)*(e + f* 
x)) + b*E^((2*I)*(e + f*x)) + Sqrt[a + b]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a 
*(1 + E^((2*I)*(e + f*x)))^2]])/(Sqrt[a + b]*Sqrt[4*b*E^((2*I)*(e + f*x)) 
+ a*(1 + E^((2*I)*(e + f*x)))^2]))*(a + b*Sec[e + f*x]^2)^(3/2))/(2*Sqrt[2 
]*f*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4627, 25, 354, 109, 27, 168, 27, 174, 73, 221}

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.

\(\displaystyle \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^{3/2}}{\tan (e+f x)^5}dx\)

\(\Big \downarrow \) 4627

\(\displaystyle \frac {\int -\frac {\cos (e+f x) \left (b \sec ^2(e+f x)+a\right )^{3/2}}{\left (1-\sec ^2(e+f x)\right )^3}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {\cos (e+f x) \left (b \sec ^2(e+f x)+a\right )^{3/2}}{\left (1-\sec ^2(e+f x)\right )^3}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 354

\(\displaystyle -\frac {\int \frac {\cos (e+f x) \left (b \sec ^2(e+f x)+a\right )^{3/2}}{\left (1-\sec ^2(e+f x)\right )^3}d\sec ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 109

\(\displaystyle -\frac {\frac {(a+b) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )^2}-\frac {1}{2} \int -\frac {\cos (e+f x) \left (4 a^2+(3 a-b) b \sec ^2(e+f x)\right )}{2 \left (1-\sec ^2(e+f x)\right )^2 \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {1}{4} \int \frac {\cos (e+f x) \left (4 a^2+(3 a-b) b \sec ^2(e+f x)\right )}{\left (1-\sec ^2(e+f x)\right )^2 \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)+\frac {(a+b) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\)

\(\Big \downarrow \) 168

\(\displaystyle -\frac {\frac {1}{4} \left (\frac {(4 a-b) \sqrt {a+b \sec ^2(e+f x)}}{1-\sec ^2(e+f x)}-\frac {\int -\frac {(a+b) \cos (e+f x) \left (8 a^2+(4 a-b) b \sec ^2(e+f x)\right )}{2 \left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{a+b}\right )+\frac {(a+b) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {\cos (e+f x) \left (8 a^2+(4 a-b) b \sec ^2(e+f x)\right )}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)+\frac {(4 a-b) \sqrt {a+b \sec ^2(e+f x)}}{1-\sec ^2(e+f x)}\right )+\frac {(a+b) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\)

\(\Big \downarrow \) 174

\(\displaystyle -\frac {\frac {1}{4} \left (\frac {1}{2} \left (\left (8 a^2+4 a b-b^2\right ) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)+8 a^2 \int \frac {\cos (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)\right )+\frac {(4 a-b) \sqrt {a+b \sec ^2(e+f x)}}{1-\sec ^2(e+f x)}\right )+\frac {(a+b) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {\frac {1}{4} \left (\frac {1}{2} \left (\frac {2 \left (8 a^2+4 a b-b^2\right ) \int \frac {1}{\frac {a+b}{b}-\frac {\sec ^4(e+f x)}{b}}d\sqrt {b \sec ^2(e+f x)+a}}{b}+\frac {16 a^2 \int \frac {1}{\frac {\sec ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sec ^2(e+f x)+a}}{b}\right )+\frac {(4 a-b) \sqrt {a+b \sec ^2(e+f x)}}{1-\sec ^2(e+f x)}\right )+\frac {(a+b) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {\frac {1}{4} \left (\frac {1}{2} \left (\frac {2 \left (8 a^2+4 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-16 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )\right )+\frac {(4 a-b) \sqrt {a+b \sec ^2(e+f x)}}{1-\sec ^2(e+f x)}\right )+\frac {(a+b) \sqrt {a+b \sec ^2(e+f x)}}{2 \left (1-\sec ^2(e+f x)\right )^2}}{2 f}\)

Input:

Int[Cot[e + f*x]^5*(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

-1/2*(((a + b)*Sqrt[a + b*Sec[e + f*x]^2])/(2*(1 - Sec[e + f*x]^2)^2) + (( 
-16*a^(3/2)*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]] + (2*(8*a^2 + 4*a* 
b - b^2)*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a + b]])/Sqrt[a + b])/2 + 
 ((4*a - b)*Sqrt[a + b*Sec[e + f*x]^2])/(1 - Sec[e + f*x]^2))/4)/f
 

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 109

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

rule 168

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 3042

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

rule 4627

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si 
mp[1/f   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), x] 
, x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[( 
m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4] || IGtQ[p, 0] || Integers 
Q[2*n, p])
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1450\) vs. \(2(137)=274\).

Time = 1.61 (sec) , antiderivative size = 1451, normalized size of antiderivative = 9.13

method	result	size
default	\(\text {Expression too large to display}\)	\(1451\)

Input:

int(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/16/f/(a+b)^(5/2)*((-16*cos(f*x+e)+16)*sin(f*x+e)^2*(a+b)^(5/2)*a^(3/2)*l 
n(4*a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*a^(1/ 
2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+4*cos(f*x+e)*a)+cos(f*x+e)* 
(-12*cos(f*x+e)^2+8)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(5/ 
2)*a+cos(f*x+e)*(-2*cos(f*x+e)^2-2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^ 
(1/2)*(a+b)^(5/2)*b+(8*cos(f*x+e)-8)*sin(f*x+e)^2*ln(-4*((a+b)^(1/2)*((b+a 
*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f* 
x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+cos(f*x+e)*a+b)/(-1+cos(f*x+e)))*a^4+(20*c 
os(f*x+e)-20)*sin(f*x+e)^2*ln(-4*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f 
*x+e))^2)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^ 
2)^(1/2)+cos(f*x+e)*a+b)/(-1+cos(f*x+e)))*a^3*b+(15*cos(f*x+e)-15)*sin(f*x 
+e)^2*ln(-4*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f 
*x+e)+(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+cos(f*x+e)*a 
+b)/(-1+cos(f*x+e)))*a^2*b^2+(-2+2*cos(f*x+e))*sin(f*x+e)^2*ln(-4*((a+b)^( 
1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*(( 
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+cos(f*x+e)*a+b)/(-1+cos(f*x+e))) 
*a*b^3+(1-cos(f*x+e))*sin(f*x+e)^2*ln(-4*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/ 
(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos( 
f*x+e))^2)^(1/2)+cos(f*x+e)*a+b)/(-1+cos(f*x+e)))*b^4+(-8*cos(f*x+e)+8)*si 
n(f*x+e)^2*ln(2/(a+b)^(1/2)*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 404 vs. \(2 (137) = 274\).

Time = 1.38 (sec) , antiderivative size = 1801, normalized size of antiderivative = 11.33 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \] Input:

integrate(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[1/32*(4*((a^2 + a*b)*cos(f*x + e)^4 - 2*(a^2 + a*b)*cos(f*x + e)^2 + a^2 
+ a*b)*sqrt(a)*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b*cos(f*x + e)^6 + 160 
*a^2*b^2*cos(f*x + e)^4 + 32*a*b^3*cos(f*x + e)^2 + b^4 + 8*(16*a^3*cos(f* 
x + e)^8 + 24*a^2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f*x + e)^4 + b^3*cos(f*x 
 + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) - ((8*a^2 + 
4*a*b - b^2)*cos(f*x + e)^4 - 2*(8*a^2 + 4*a*b - b^2)*cos(f*x + e)^2 + 8*a 
^2 + 4*a*b - b^2)*sqrt(a + b)*log(2*((8*a^2 + 8*a*b + b^2)*cos(f*x + e)^4 
+ 2*(4*a*b + 3*b^2)*cos(f*x + e)^2 + b^2 + 4*((2*a + b)*cos(f*x + e)^4 + b 
*cos(f*x + e)^2)*sqrt(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/ 
(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)) - 4*((6*a^2 + 7*a*b + b^2)*cos(f* 
x + e)^4 - (4*a^2 + 3*a*b - b^2)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + 
b)/cos(f*x + e)^2))/((a + b)*f*cos(f*x + e)^4 - 2*(a + b)*f*cos(f*x + e)^2 
 + (a + b)*f), 1/16*(((8*a^2 + 4*a*b - b^2)*cos(f*x + e)^4 - 2*(8*a^2 + 4* 
a*b - b^2)*cos(f*x + e)^2 + 8*a^2 + 4*a*b - b^2)*sqrt(-a - b)*arctan(1/2*( 
(2*a + b)*cos(f*x + e)^2 + b)*sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos 
(f*x + e)^2)/((a^2 + a*b)*cos(f*x + e)^2 + a*b + b^2)) + 2*((a^2 + a*b)*co 
s(f*x + e)^4 - 2*(a^2 + a*b)*cos(f*x + e)^2 + a^2 + a*b)*sqrt(a)*log(128*a 
^4*cos(f*x + e)^8 + 256*a^3*b*cos(f*x + e)^6 + 160*a^2*b^2*cos(f*x + e)^4 
+ 32*a*b^3*cos(f*x + e)^2 + b^4 + 8*(16*a^3*cos(f*x + e)^8 + 24*a^2*b*cos( 
f*x + e)^6 + 10*a*b^2*cos(f*x + e)^4 + b^3*cos(f*x + e)^2)*sqrt(a)*sqrt...
 

Sympy [F(-1)]

Timed out. \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \] Input:

integrate(cot(f*x+e)**5*(a+b*sec(f*x+e)**2)**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cot \left (f x + e\right )^{5} \,d x } \] Input:

integrate(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^(3/2)*cot(f*x + e)^5, x)
 

Giac [F(-2)]

Exception generated. \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^5\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \] Input:

int(cot(e + f*x)^5*(a + b/cos(e + f*x)^2)^(3/2),x)
 

Output:

int(cot(e + f*x)^5*(a + b/cos(e + f*x)^2)^(3/2), x)
 

Reduce [F]

\[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \cot \left (f x +e \right )^{5} \sec \left (f x +e \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \cot \left (f x +e \right )^{5}d x \right ) a \] Input:

int(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^(3/2),x)
 

Output:

int(sqrt(sec(e + f*x)**2*b + a)*cot(e + f*x)**5*sec(e + f*x)**2,x)*b + int 
(sqrt(sec(e + f*x)**2*b + a)*cot(e + f*x)**5,x)*a