3.5.3 \(\int \csc ^5(e+f x) (b \sec (e+f x))^{5/2} \, dx\) [403]

3.5.3.1 Optimal result
3.5.3.2 Mathematica [A] (verified)
3.5.3.3 Rubi [A] (warning: unable to verify)
3.5.3.4 Maple [B] (verified)
3.5.3.5 Fricas [B] (verification not implemented)
3.5.3.6 Sympy [F(-1)]
3.5.3.7 Maxima [A] (verification not implemented)
3.5.3.8 Giac [A] (verification not implemented)
3.5.3.9 Mupad [F(-1)]

3.5.3.1 Optimal result

Integrand size = 21, antiderivative size = 143 \[ \int \csc ^5(e+f x) (b \sec (e+f x))^{5/2} \, dx=\frac {77 b^{5/2} \arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 f}-\frac {77 b^{5/2} \text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 f}+\frac {77 b (b \sec (e+f x))^{3/2}}{48 f}-\frac {11 \cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{16 b f}-\frac {\cot ^4(e+f x) (b \sec (e+f x))^{11/2}}{4 b^3 f} \]

output
77/32*b^(5/2)*arctan((b*sec(f*x+e))^(1/2)/b^(1/2))/f-77/32*b^(5/2)*arctanh 
((b*sec(f*x+e))^(1/2)/b^(1/2))/f+77/48*b*(b*sec(f*x+e))^(3/2)/f-11/16*cot( 
f*x+e)^2*(b*sec(f*x+e))^(7/2)/b/f-1/4*cot(f*x+e)^4*(b*sec(f*x+e))^(11/2)/b 
^3/f
 
3.5.3.2 Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.83 \[ \int \csc ^5(e+f x) (b \sec (e+f x))^{5/2} \, dx=\frac {b^3 \left (-180 \csc ^2(e+f x)-48 \csc ^4(e+f x)+462 \arctan \left (\sqrt {\sec (e+f x)}\right ) \sqrt {\sec (e+f x)}+231 \left (\log \left (1-\sqrt {\sec (e+f x)}\right )-\log \left (1+\sqrt {\sec (e+f x)}\right )\right ) \sqrt {\sec (e+f x)}+128 \sec ^2(e+f x)\right )}{192 f \sqrt {b \sec (e+f x)}} \]

input
Integrate[Csc[e + f*x]^5*(b*Sec[e + f*x])^(5/2),x]
 
output
(b^3*(-180*Csc[e + f*x]^2 - 48*Csc[e + f*x]^4 + 462*ArcTan[Sqrt[Sec[e + f* 
x]]]*Sqrt[Sec[e + f*x]] + 231*(Log[1 - Sqrt[Sec[e + f*x]]] - Log[1 + Sqrt[ 
Sec[e + f*x]]])*Sqrt[Sec[e + f*x]] + 128*Sec[e + f*x]^2))/(192*f*Sqrt[b*Se 
c[e + f*x]])
 
3.5.3.3 Rubi [A] (warning: unable to verify)

Time = 0.31 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 3102, 25, 27, 252, 252, 262, 266, 827, 216, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3102

\(\displaystyle \frac {\int -\frac {b^6 (b \sec (e+f x))^{13/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^3}d(b \sec (e+f x))}{b^5 f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {b^6 (b \sec (e+f x))^{13/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^3}d(b \sec (e+f x))}{b^5 f}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {b \int \frac {(b \sec (e+f x))^{13/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^3}d(b \sec (e+f x))}{f}\)

\(\Big \downarrow \) 252

\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{11/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {11}{8} \int \frac {(b \sec (e+f x))^{9/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^2}d(b \sec (e+f x))\right )}{f}\)

\(\Big \downarrow \) 252

\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{11/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {11}{8} \left (\frac {(b \sec (e+f x))^{7/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {7}{4} \int \frac {(b \sec (e+f x))^{5/2}}{b^2-b^2 \sec ^2(e+f x)}d(b \sec (e+f x))\right )\right )}{f}\)

\(\Big \downarrow \) 262

\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{11/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {11}{8} \left (\frac {(b \sec (e+f x))^{7/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {7}{4} \left (b^2 \int \frac {\sqrt {b \sec (e+f x)}}{b^2-b^2 \sec ^2(e+f x)}d(b \sec (e+f x))-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )\right )\right )}{f}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{11/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {11}{8} \left (\frac {(b \sec (e+f x))^{7/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {7}{4} \left (2 b^2 \int \frac {b^2 \sec ^2(e+f x)}{b^2-b^4 \sec ^4(e+f x)}d\sqrt {b \sec (e+f x)}-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )\right )\right )}{f}\)

\(\Big \downarrow \) 827

\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{11/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {11}{8} \left (\frac {(b \sec (e+f x))^{7/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {7}{4} \left (2 b^2 \left (\frac {1}{2} \int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}-\frac {1}{2} \int \frac {1}{b^2 \sec ^2(e+f x)+b}d\sqrt {b \sec (e+f x)}\right )-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )\right )\right )}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{11/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {11}{8} \left (\frac {(b \sec (e+f x))^{7/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {7}{4} \left (2 b^2 \left (\frac {1}{2} \int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}-\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}\right )-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )\right )\right )}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{11/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {11}{8} \left (\frac {(b \sec (e+f x))^{7/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {7}{4} \left (2 b^2 \left (\frac {\text {arctanh}\left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}-\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}\right )-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )\right )\right )}{f}\)

input
Int[Csc[e + f*x]^5*(b*Sec[e + f*x])^(5/2),x]
 
output
-((b*((b*Sec[e + f*x])^(11/2)/(4*(b^2 - b^2*Sec[e + f*x]^2)^2) - (11*((b*S 
ec[e + f*x])^(7/2)/(2*(b^2 - b^2*Sec[e + f*x]^2)) - (7*(2*b^2*(-1/2*ArcTan 
[Sqrt[b]*Sec[e + f*x]]/Sqrt[b] + ArcTanh[Sqrt[b]*Sec[e + f*x]]/(2*Sqrt[b]) 
) - (2*(b*Sec[e + f*x])^(3/2))/3))/4))/8))/f)
 

3.5.3.3.1 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 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 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 252
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 262
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 266
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 827
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3102
Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_S 
ymbol] :> Simp[1/(f*a^n)   Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/ 
2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1 
)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
 
3.5.3.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(555\) vs. \(2(115)=230\).

Time = 246.76 (sec) , antiderivative size = 556, normalized size of antiderivative = 3.89

method result size
default \(-\frac {\sqrt {b \sec \left (f x +e \right )}\, b^{2} \left (231 \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right ) \left (\cot ^{2}\left (f x +e \right )\right )+57 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \left (\cot ^{2}\left (f x +e \right )\right )-288 \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right ) \left (\cot ^{2}\left (f x +e \right )\right )-231 \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right ) \cot \left (f x +e \right ) \csc \left (f x +e \right )-57 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \cot \left (f x +e \right ) \csc \left (f x +e \right )+288 \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right ) \cot \left (f x +e \right ) \csc \left (f x +e \right )-308 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot ^{3}\left (f x +e \right )\right ) \csc \left (f x +e \right )+484 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cot \left (f x +e \right ) \left (\csc ^{3}\left (f x +e \right )\right )-128 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sec \left (f x +e \right ) \left (\csc ^{4}\left (f x +e \right )\right )\right )}{192 f \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\) \(556\)

input
int(csc(f*x+e)^5*(b*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
 
output
-1/192/f*(b*sec(f*x+e))^(1/2)*b^2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(23 
1*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))*cot(f*x+e)^2+57*ln((2*c 
os(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+ 
1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))*cot(f*x+e)^2-288*ln(2*(2*cos(f*x 
+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^ 
(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))*cot(f*x+e)^2-231*arctan(1/2/(-cos(f*x+ 
e)/(cos(f*x+e)+1)^2)^(1/2))*cot(f*x+e)*csc(f*x+e)-57*ln((2*cos(f*x+e)*(-co 
s(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-co 
s(f*x+e)+1)/(cos(f*x+e)+1))*cot(f*x+e)*csc(f*x+e)+288*ln(2*(2*cos(f*x+e)*( 
-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2) 
-cos(f*x+e)+1)/(cos(f*x+e)+1))*cot(f*x+e)*csc(f*x+e)-308*(-cos(f*x+e)/(cos 
(f*x+e)+1)^2)^(1/2)*cot(f*x+e)^3*csc(f*x+e)+484*(-cos(f*x+e)/(cos(f*x+e)+1 
)^2)^(1/2)*cot(f*x+e)*csc(f*x+e)^3-128*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2 
)*sec(f*x+e)*csc(f*x+e)^4)
 
3.5.3.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (115) = 230\).

Time = 0.36 (sec) , antiderivative size = 542, normalized size of antiderivative = 3.79 \[ \int \csc ^5(e+f x) (b \sec (e+f x))^{5/2} \, dx=\left [\frac {462 \, {\left (b^{2} \cos \left (f x + e\right )^{5} - 2 \, b^{2} \cos \left (f x + e\right )^{3} + b^{2} \cos \left (f x + e\right )\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) + 1\right )}}{2 \, b}\right ) + 231 \, {\left (b^{2} \cos \left (f x + e\right )^{5} - 2 \, b^{2} \cos \left (f x + e\right )^{3} + b^{2} \cos \left (f x + e\right )\right )} \sqrt {-b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, {\left (77 \, b^{2} \cos \left (f x + e\right )^{4} - 121 \, b^{2} \cos \left (f x + e\right )^{2} + 32 \, b^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{384 \, {\left (f \cos \left (f x + e\right )^{5} - 2 \, f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )\right )}}, -\frac {462 \, {\left (b^{2} \cos \left (f x + e\right )^{5} - 2 \, b^{2} \cos \left (f x + e\right )^{3} + b^{2} \cos \left (f x + e\right )\right )} \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {b}}\right ) - 231 \, {\left (b^{2} \cos \left (f x + e\right )^{5} - 2 \, b^{2} \cos \left (f x + e\right )^{3} + b^{2} \cos \left (f x + e\right )\right )} \sqrt {b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right ) - 8 \, {\left (77 \, b^{2} \cos \left (f x + e\right )^{4} - 121 \, b^{2} \cos \left (f x + e\right )^{2} + 32 \, b^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{384 \, {\left (f \cos \left (f x + e\right )^{5} - 2 \, f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )\right )}}\right ] \]

input
integrate(csc(f*x+e)^5*(b*sec(f*x+e))^(5/2),x, algorithm="fricas")
 
output
[1/384*(462*(b^2*cos(f*x + e)^5 - 2*b^2*cos(f*x + e)^3 + b^2*cos(f*x + e)) 
*sqrt(-b)*arctan(1/2*sqrt(-b)*sqrt(b/cos(f*x + e))*(cos(f*x + e) + 1)/b) + 
 231*(b^2*cos(f*x + e)^5 - 2*b^2*cos(f*x + e)^3 + b^2*cos(f*x + e))*sqrt(- 
b)*log((b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - cos(f*x + e))*sqrt(-b)*sqrt 
(b/cos(f*x + e)) - 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 + 2*cos(f*x + e) 
+ 1)) + 8*(77*b^2*cos(f*x + e)^4 - 121*b^2*cos(f*x + e)^2 + 32*b^2)*sqrt(b 
/cos(f*x + e)))/(f*cos(f*x + e)^5 - 2*f*cos(f*x + e)^3 + f*cos(f*x + e)), 
-1/384*(462*(b^2*cos(f*x + e)^5 - 2*b^2*cos(f*x + e)^3 + b^2*cos(f*x + e)) 
*sqrt(b)*arctan(1/2*sqrt(b/cos(f*x + e))*(cos(f*x + e) - 1)/sqrt(b)) - 231 
*(b^2*cos(f*x + e)^5 - 2*b^2*cos(f*x + e)^3 + b^2*cos(f*x + e))*sqrt(b)*lo 
g((b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(b)*sqrt(b/cos 
(f*x + e)) + 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 - 2*cos(f*x + e) + 1)) 
- 8*(77*b^2*cos(f*x + e)^4 - 121*b^2*cos(f*x + e)^2 + 32*b^2)*sqrt(b/cos(f 
*x + e)))/(f*cos(f*x + e)^5 - 2*f*cos(f*x + e)^3 + f*cos(f*x + e))]
 
3.5.3.6 Sympy [F(-1)]

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

input
integrate(csc(f*x+e)**5*(b*sec(f*x+e))**(5/2),x)
 
output
Timed out
 
3.5.3.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.08 \[ \int \csc ^5(e+f x) (b \sec (e+f x))^{5/2} \, dx=\frac {{\left (462 \, b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right ) + 231 \, b^{\frac {3}{2}} \log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right ) + 128 \, \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}} + \frac {12 \, {\left (15 \, b^{4} \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}} - 19 \, b^{2} \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {7}{2}}\right )}}{b^{4} - \frac {2 \, b^{4}}{\cos \left (f x + e\right )^{2}} + \frac {b^{4}}{\cos \left (f x + e\right )^{4}}}\right )} b}{192 \, f} \]

input
integrate(csc(f*x+e)^5*(b*sec(f*x+e))^(5/2),x, algorithm="maxima")
 
output
1/192*(462*b^(3/2)*arctan(sqrt(b/cos(f*x + e))/sqrt(b)) + 231*b^(3/2)*log( 
-(sqrt(b) - sqrt(b/cos(f*x + e)))/(sqrt(b) + sqrt(b/cos(f*x + e)))) + 128* 
(b/cos(f*x + e))^(3/2) + 12*(15*b^4*(b/cos(f*x + e))^(3/2) - 19*b^2*(b/cos 
(f*x + e))^(7/2))/(b^4 - 2*b^4/cos(f*x + e)^2 + b^4/cos(f*x + e)^4))*b/f
 
3.5.3.8 Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.05 \[ \int \csc ^5(e+f x) (b \sec (e+f x))^{5/2} \, dx=\frac {b^{10} {\left (\frac {231 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{7}} - \frac {231 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {15}{2}}} + \frac {6 \, {\left (15 \, \sqrt {b \cos \left (f x + e\right )} b^{2} \cos \left (f x + e\right )^{2} - 19 \, \sqrt {b \cos \left (f x + e\right )} b^{2}\right )}}{{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )}^{2} b^{6}} + \frac {64}{\sqrt {b \cos \left (f x + e\right )} b^{7} \cos \left (f x + e\right )}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{96 \, f} \]

input
integrate(csc(f*x+e)^5*(b*sec(f*x+e))^(5/2),x, algorithm="giac")
 
output
1/96*b^10*(231*arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/(sqrt(-b)*b^7) - 231* 
arctan(sqrt(b*cos(f*x + e))/sqrt(b))/b^(15/2) + 6*(15*sqrt(b*cos(f*x + e)) 
*b^2*cos(f*x + e)^2 - 19*sqrt(b*cos(f*x + e))*b^2)/((b^2*cos(f*x + e)^2 - 
b^2)^2*b^6) + 64/(sqrt(b*cos(f*x + e))*b^7*cos(f*x + e)))*sgn(cos(f*x + e) 
)/f
 
3.5.3.9 Mupad [F(-1)]

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

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