\(\int \csc ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) [72]

Optimal result

Integrand size = 25, antiderivative size = 183 \[ \int \csc ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f}-\frac {\left (3 a^2+12 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{8 (a+b)^{3/2} f}-\frac {(3 a+4 b) \cot (e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 (a+b) f}-\frac {\cot (e+f x) \csc ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 f} \] Output:

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

Mathematica [A] (verified)

Time = 1.07 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.14 \[ \int \csc ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\cos (e+f x) \left (8 \sqrt {b} (a+b)^2 \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {b}}\right )-\sqrt {a+b} \left (3 a^2+12 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {a+b}}\right )+\frac {(a+b) \sqrt {a+2 b+a \cos (2 (e+f x))} (-7 a-8 b+(3 a+4 b) \cos (2 (e+f x))) \csc ^4(e+f x)}{2 \sqrt {2}}\right ) \sqrt {a+b \sec ^2(e+f x)}}{4 \sqrt {2} (a+b)^2 f \sqrt {a+2 b+a \cos (2 (e+f x))}} \] Input:

Integrate[Csc[e + f*x]^5*Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

(Cos[e + f*x]*(8*Sqrt[b]*(a + b)^2*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/ 
Sqrt[b]] - Sqrt[a + b]*(3*a^2 + 12*a*b + 8*b^2)*ArcTanh[Sqrt[a + b - a*Sin 
[e + f*x]^2]/Sqrt[a + b]] + ((a + b)*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(- 
7*a - 8*b + (3*a + 4*b)*Cos[2*(e + f*x)])*Csc[e + f*x]^4)/(2*Sqrt[2]))*Sqr 
t[a + b*Sec[e + f*x]^2])/(4*Sqrt[2]*(a + b)^2*f*Sqrt[a + 2*b + a*Cos[2*(e 
+ f*x)]])
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4622, 25, 369, 440, 25, 398, 224, 219, 291, 219}

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.

\(\displaystyle \int \csc ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a+b \sec (e+f x)^2}}{\sin (e+f x)^5}dx\)

\(\Big \downarrow \) 4622

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 369

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

\(\Big \downarrow \) 440

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 398

\(\displaystyle \frac {\frac {1}{4} \left (\frac {(3 a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )}-\frac {\left (3 a^2+12 a b+8 b^2\right ) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)-8 b (a+b) \int \frac {1}{\sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{2 (a+b)}\right )-\frac {\sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{4} \left (\frac {(3 a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )}-\frac {\left (3 a^2+12 a b+8 b^2\right ) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)-8 b (a+b) \int \frac {1}{1-\frac {b \sec ^2(e+f x)}{b \sec ^2(e+f x)+a}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}}{2 (a+b)}\right )-\frac {\sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{4} \left (\frac {(3 a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )}-\frac {\left (3 a^2+12 a b+8 b^2\right ) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)-8 \sqrt {b} (a+b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 (a+b)}\right )-\frac {\sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {1}{4} \left (\frac {(3 a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )}-\frac {\left (3 a^2+12 a b+8 b^2\right ) \int \frac {1}{1-\frac {(a+b) \sec ^2(e+f x)}{b \sec ^2(e+f x)+a}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}-8 \sqrt {b} (a+b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 (a+b)}\right )-\frac {\sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{4} \left (\frac {(3 a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) \left (1-\sec ^2(e+f x)\right )}-\frac {\frac {\left (3 a^2+12 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{\sqrt {a+b}}-8 \sqrt {b} (a+b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 (a+b)}\right )-\frac {\sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)}}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\)

Input:

Int[Csc[e + f*x]^5*Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

(-1/4*(Sec[e + f*x]^3*Sqrt[a + b*Sec[e + f*x]^2])/(1 - Sec[e + f*x]^2)^2 + 
 (-1/2*(-8*Sqrt[b]*(a + b)*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e 
 + f*x]^2]] + ((3*a^2 + 12*a*b + 8*b^2)*ArcTanh[(Sqrt[a + b]*Sec[e + f*x]) 
/Sqrt[a + b*Sec[e + f*x]^2]])/Sqrt[a + b])/(a + b) + ((3*a + 4*b)*Sec[e + 
f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(2*(a + b)*(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 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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 369

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

rule 398

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

rule 440

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

rule 3042

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

rule 4622

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si 
mp[1/(f*ff^m)   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p 
/x^(m + 1)), 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])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2006\) vs. \(2(161)=322\).

Time = 4.12 (sec) , antiderivative size = 2007, normalized size of antiderivative = 10.97

method	result	size
default	\(\text {Expression too large to display}\)	\(2007\)

Input:

int(csc(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/16/f/(a+b)^(9/2)*((-16*cos(f*x+e)+16)*sin(f*x+e)^2*(a+b)^(5/2)*b^(5/2)*l 
n(4*(b*cot(f*x+e)^2-2*b*cot(f*x+e)*csc(f*x+e)+b*csc(f*x+e)^2+2*b^(1/2)*((b 
+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+b)/(cot(f*x+e)^2-2*csc(f*x+e)*cot 
(f*x+e)+csc(f*x+e)^2-1))+(-32*cos(f*x+e)+32)*sin(f*x+e)^2*(a+b)^(5/2)*b^(3 
/2)*ln(4*(b*cot(f*x+e)^2-2*b*cot(f*x+e)*csc(f*x+e)+b*csc(f*x+e)^2+2*b^(1/2 
)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+b)/(cot(f*x+e)^2-2*csc(f*x+e 
)*cot(f*x+e)+csc(f*x+e)^2-1))*a+(-16*cos(f*x+e)+16)*sin(f*x+e)^2*(a+b)^(5/ 
2)*b^(1/2)*ln(4*(b*cot(f*x+e)^2-2*b*cot(f*x+e)*csc(f*x+e)+b*csc(f*x+e)^2+2 
*b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+b)/(cot(f*x+e)^2-2*cs 
c(f*x+e)*cot(f*x+e)+csc(f*x+e)^2-1))*a^2+(6*cos(f*x+e)^2-10)*((b+a*cos(f*x 
+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*(a+b)^(5/2)+(14*cos(f*x+e)^2-22)*(a+b)^ 
(5/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b+(8*cos(f*x+e)^2-12)* 
((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2*(a+b)^(5/2)+(3*cos(f*x+e)- 
3)*sin(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+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^5+(21*cos(f*x+e)-21)*sin(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+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*b+(53*cos(f*x+e)-53)*sin(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+e))^2)^(1/2)*cos(f...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (161) = 322\).

Time = 0.45 (sec) , antiderivative size = 1516, normalized size of antiderivative = 8.28 \[ \int \csc ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\text {Too large to display} \] Input:

integrate(csc(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[1/16*(((3*a^2 + 12*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a^2 + 12*a*b + 8*b^ 
2)*cos(f*x + e)^2 + 3*a^2 + 12*a*b + 8*b^2)*sqrt(a + b)*log(2*(a*cos(f*x + 
 e)^2 - 2*sqrt(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x 
+ e) + a + 2*b)/(cos(f*x + e)^2 - 1)) + 8*((a^2 + 2*a*b + b^2)*cos(f*x + e 
)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(b)*lo 
g((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2 
)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) + 2*((3*a^2 + 7*a*b + 4*b^2)*cos(f*x 
 + e)^3 - (5*a^2 + 11*a*b + 6*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + 
b)/cos(f*x + e)^2))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^4 - 2*(a^2 + 2*a*b 
 + b^2)*f*cos(f*x + e)^2 + (a^2 + 2*a*b + b^2)*f), 1/8*(((3*a^2 + 12*a*b + 
 8*b^2)*cos(f*x + e)^4 - 2*(3*a^2 + 12*a*b + 8*b^2)*cos(f*x + e)^2 + 3*a^2 
 + 12*a*b + 8*b^2)*sqrt(-a - b)*arctan(sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 
 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a*cos(f*x + e)^2 + b)) + 4*((a^2 + 2*a 
*b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x + e)^2 + a^2 + 2* 
a*b + b^2)*sqrt(b)*log((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*cos(f*x + e)^ 
2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) + ((3*a^2 + 7*a 
*b + 4*b^2)*cos(f*x + e)^3 - (5*a^2 + 11*a*b + 6*b^2)*cos(f*x + e))*sqrt(( 
a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e) 
^4 - 2*(a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 + (a^2 + 2*a*b + b^2)*f), -1/1 
6*(16*((a^2 + 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f...
 

Sympy [F]

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

integrate(csc(f*x+e)**5*(a+b*sec(f*x+e)**2)**(1/2),x)
 

Output:

Integral(sqrt(a + b*sec(e + f*x)**2)*csc(e + f*x)**5, x)
                                                                                    
                                                                                    
 

Maxima [F]

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

integrate(csc(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*sec(f*x + e)^2 + a)*csc(f*x + e)^5, x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 867 vs. \(2 (161) = 322\).

Time = 0.96 (sec) , antiderivative size = 867, normalized size of antiderivative = 4.74 \[ \int \csc ^5(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\text {Too large to display} \] Input:

integrate(csc(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

-1/64*(sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan( 
1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b)*(tan(1/2*f*x + 1/ 
2*e)^2 + (9*a + 11*b)/(a + b)) + 128*b*arctan(-1/2*(sqrt(a + b)*tan(1/2*f* 
x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 
2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b) - sqrt(a 
+ b))/sqrt(-b))/sqrt(-b) - 8*(3*a^2 + 12*a*b + 8*b^2)*arctan(-(sqrt(a + b) 
*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 
1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b 
))/sqrt(-a - b))/((a + b)*sqrt(-a - b)) - 4*(3*a^2 + 12*a*b + 8*b^2)*log(a 
bs(-(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 
b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 
1/2*e)^2 + a + b))*(a + b) + sqrt(a + b)*(a - b)))/(a + b)^(3/2) + 4*(2*(s 
qrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan( 
1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) 
^2 + a + b))^3*(2*a^2 - 3*b^2) - (sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqr 
t(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 
1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^2*(3*a^2 + 10*a*b + 7*b^2) 
*sqrt(a + b) - 2*(3*a^3 + 3*a^2*b - 2*a*b^2 - 2*b^3)*(sqrt(a + b)*tan(1/2* 
f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 
- 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b)) + 5...
 

Mupad [F(-1)]

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

int((a + b/cos(e + f*x)^2)^(1/2)/sin(e + f*x)^5,x)
 

Output:

int((a + b/cos(e + f*x)^2)^(1/2)/sin(e + f*x)^5, x)
 

Reduce [F]

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

int(csc(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x)
 

Output:

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