\(\int \csc ^5(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [85]

Optimal result

Integrand size = 25, antiderivative size = 218 \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {3 \sqrt {b} (a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f}-\frac {3 \left (a^2+8 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{8 \sqrt {a+b} f}+\frac {3 (a+4 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {3 (a+2 b) \csc ^2(e+f x) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 f}-\frac {\cot (e+f x) \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{4 f} \] Output:

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

Mathematica [A] (verified)

Time = 2.55 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.20 \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {\left (b+a \cos ^2(e+f x)\right ) \left (-12 b^{3/2} \left (a^2+3 a b+2 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {b}}\right ) \cos ^2(e+f x)+3 b \sqrt {a+b} \left (a^2+8 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {a+b}}\right ) \cos ^2(e+f x)+\frac {b (a+b) \sqrt {a+2 b+a \cos (2 (e+f x))} (11 a+4 b+8 (a+3 b) \cos (2 (e+f x))-3 (a+4 b) \cos (4 (e+f x))) \csc ^4(e+f x)}{8 \sqrt {2}}\right ) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 \sqrt {2} b (a+b) f (a+2 b+a \cos (2 (e+f x)))^{3/2}} \] Input:

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

Output:

-1/2*((b + a*Cos[e + f*x]^2)*(-12*b^(3/2)*(a^2 + 3*a*b + 2*b^2)*ArcTanh[Sq 
rt[a + b - a*Sin[e + f*x]^2]/Sqrt[b]]*Cos[e + f*x]^2 + 3*b*Sqrt[a + b]*(a^ 
2 + 8*a*b + 8*b^2)*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[a + b]]*Cos 
[e + f*x]^2 + (b*(a + b)*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(11*a + 4*b + 
8*(a + 3*b)*Cos[2*(e + f*x)] - 3*(a + 4*b)*Cos[4*(e + f*x)])*Csc[e + f*x]^ 
4)/(8*Sqrt[2]))*Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(Sqrt[2]*b*(a + b 
)*f*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4622, 25, 369, 27, 439, 25, 444, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4622

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 369

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 439

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 444

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(-1/4*(Sec[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2))/(1 - Sec[e + f*x]^2)^2 
 + (3*(((a + 2*b)*Sec[e + f*x]^3*Sqrt[a + b*Sec[e + f*x]^2])/(2*(1 - Sec[e 
 + f*x]^2)) + (4*Sqrt[b]*(a + 2*b)*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + 
 b*Sec[e + f*x]^2]] - ((a^2 + 8*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 + 4*b)*Sec[e + f*x]*Sq 
rt[a + b*Sec[e + f*x]^2])/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 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 439

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a 
 + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*b*g*(p + 1))), x] + Simp[1/(2*a*b*(p 
+ 1))   Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(2*b*e*( 
p + 1) + (b*e - a*f)*(m + 1)) + d*(2*b*e*(p + 1) + (b*e - a*f)*(m + 2*q + 1 
))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && G 
tQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])
 

rule 444

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q 
_.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[f*g*(g*x)^(m - 1)*(a + b*x^2)^ 
(p + 1)*((c + d*x^2)^(q + 1)/(b*d*(m + 2*(p + q + 1) + 1))), x] - Simp[g^2/ 
(b*d*(m + 2*(p + q + 1) + 1))   Int[(g*x)^(m - 2)*(a + b*x^2)^p*(c + d*x^2) 
^q*Simp[a*f*c*(m - 1) + (a*f*d*(m + 2*q + 1) + b*(f*c*(m + 2*p + 1) - e*d*( 
m + 2*(p + q + 1) + 1)))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, 
q}, x] && 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. \(2889\) vs. \(2(190)=380\).

Time = 9.27 (sec) , antiderivative size = 2890, normalized size of antiderivative = 13.26

method	result	size
default	\(\text {Expression too large to display}\)	\(2890\)

Input:

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

Output:

1/16/f/b/(a+b)^(11/2)*(cos(f*x+e)^2*(-48*cos(f*x+e)+48)*sin(f*x+e)^2*b^(9/ 
2)*(a+b)^(7/2)*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))+cos(f*x+e)^2*(-120*cos(f*x+e)+120 
)*sin(f*x+e)^2*b^(7/2)*(a+b)^(7/2)*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+cos(f*x+e)^ 
2*(-96*cos(f*x+e)+96)*sin(f*x+e)^2*b^(5/2)*(a+b)^(7/2)*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^2+cos(f*x+e)^2*(-24*cos(f*x+e)+24)*sin(f*x+e)^2*b^(3/2)*(a+b)^(7/ 
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^3+cos(f*x+e)^2*(6*cos(f*x+e)^2-10)*((b+a*co 
s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(7/2)*a^3*b+(36*cos(f*x+e)^4-56* 
cos(f*x+e)^2+8)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(7/2)*a^ 
2*b^2+(54*cos(f*x+e)^4-82*cos(f*x+e)^2+16)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ 
e))^2)^(1/2)*(a+b)^(7/2)*a*b^3+(24*cos(f*x+e)^4-36*cos(f*x+e)^2+8)*((b+a*c 
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(7/2)*b^4+cos(f*x+e)^2*(3*cos(f 
*x+e)-3)*sin(f*x+e)^2*ln(-4*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x...
 

Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 1551, normalized size of antiderivative = 7.11 \[ \int \csc ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/16*(3*((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^5 - 2*(a^2 + 8*a*b + 8*b^2)*c 
os(f*x + e)^3 + (a^2 + 8*a*b + 8*b^2)*cos(f*x + e))*sqrt(a + b)*log(2*(a*c 
os(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)) + 12*((a^2 + 3*a*b + 2*b^2)* 
cos(f*x + e)^5 - 2*(a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 + (a^2 + 3*a*b + 2 
*b^2)*cos(f*x + e))*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) + 2*(3 
*(a^2 + 5*a*b + 4*b^2)*cos(f*x + e)^4 - (5*a^2 + 23*a*b + 18*b^2)*cos(f*x 
+ e)^2 + 4*a*b + 4*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a + 
 b)*f*cos(f*x + e)^5 - 2*(a + b)*f*cos(f*x + e)^3 + (a + b)*f*cos(f*x + e) 
), 1/8*(3*((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^5 - 2*(a^2 + 8*a*b + 8*b^2)* 
cos(f*x + e)^3 + (a^2 + 8*a*b + 8*b^2)*cos(f*x + e))*sqrt(-a - b)*arctan(s 
qrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a*co 
s(f*x + e)^2 + b)) + 6*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^5 - 2*(a^2 + 3* 
a*b + 2*b^2)*cos(f*x + e)^3 + (a^2 + 3*a*b + 2*b^2)*cos(f*x + e))*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 + 5*a*b + 4*b^2)*cos(f*x 
 + e)^4 - (5*a^2 + 23*a*b + 18*b^2)*cos(f*x + e)^2 + 4*a*b + 4*b^2)*sqrt(( 
a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a + b)*f*cos(f*x + e)^5 - 2*(a + 
b)*f*cos(f*x + e)^3 + (a + b)*f*cos(f*x + e)), -1/16*(24*((a^2 + 3*a*b ...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(csc(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)*csc(f*x + e)^5, x)
 

Giac [F(-2)]

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

integrate(csc(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:Unable to divide, perhaps due to ro 
unding error%%%{%%%{262144,[6,6]%%%},[12]%%%}+%%%{%%{[%%%{1572864,[6,6]%%% 
},0]:[1,0
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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