\(\int \frac {\csc ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\) [430]

Optimal result

Integrand size = 21, antiderivative size = 123 \[ \int \frac {\csc ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {3 \arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 b^{3/2} f}+\frac {3 \text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 b^{3/2} f}-\frac {3 \cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{16 b^3 f}-\frac {\cot ^4(e+f x) (b \sec (e+f x))^{3/2}}{4 b^3 f} \] Output:

-3/32*arctan((b*sec(f*x+e))^(1/2)/b^(1/2))/b^(3/2)/f+3/32*arctanh((b*sec(f 
*x+e))^(1/2)/b^(1/2))/b^(3/2)/f-3/16*cot(f*x+e)^2*(b*sec(f*x+e))^(3/2)/b^3 
/f-1/4*cot(f*x+e)^4*(b*sec(f*x+e))^(3/2)/b^3/f
 

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.89 \[ \int \frac {\csc ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {4 \csc ^2(e+f x)-16 \csc ^4(e+f x)-6 \arctan \left (\sqrt {\sec (e+f x)}\right ) \sqrt {\sec (e+f x)}+3 \left (-\log \left (1-\sqrt {\sec (e+f x)}\right )+\log \left (1+\sqrt {\sec (e+f x)}\right )\right ) \sqrt {\sec (e+f x)}}{64 b f \sqrt {b \sec (e+f x)}} \] Input:

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

Output:

(4*Csc[e + f*x]^2 - 16*Csc[e + f*x]^4 - 6*ArcTan[Sqrt[Sec[e + f*x]]]*Sqrt[ 
Sec[e + f*x]] + 3*(-Log[1 - Sqrt[Sec[e + f*x]]] + Log[1 + Sqrt[Sec[e + f*x 
]]])*Sqrt[Sec[e + f*x]])/(64*b*f*Sqrt[b*Sec[e + f*x]])
 

Rubi [A] (warning: unable to verify)

Time = 0.49 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3102, 25, 27, 252, 253, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3102

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 253

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-((b*((b*Sec[e + f*x])^(3/2)/(4*(b^2 - b^2*Sec[e + f*x]^2)^2) - (3*((-1/2* 
ArcTan[Sqrt[b]*Sec[e + f*x]]/Sqrt[b] + ArcTanh[Sqrt[b]*Sec[e + f*x]]/(2*Sq 
rt[b]))/(2*b^2) + (b*Sec[e + f*x])^(3/2)/(2*b^2*(b^2 - b^2*Sec[e + f*x]^2) 
)))/8))/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 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 253

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(213\) vs. \(2(99)=198\).

Time = 1.68 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.74

method	result	size
default	\(\frac {\left (-4 \cos \left (f x +e \right )^{2}-12\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \csc \left (f x +e \right )^{4}+\left (-3 \cos \left (f x +e \right )+3\right ) \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \csc \left (f x +e \right )^{2}+\left (3 \cos \left (f x +e \right )-3\right ) \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right ) \csc \left (f x +e \right )^{2}}{64 f \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {b \sec \left (f x +e \right )}\, b}\)	\(214\)

Input:

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

Output:

1/64/f/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(b*sec(f*x+e))^(1/2)/b*((-4*co 
s(f*x+e)^2-12)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*csc(f*x+e)^4+(-3*cos(f 
*x+e)+3)*ln((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))*csc(f*x+e)^2+(3* 
cos(f*x+e)-3)*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))*csc(f*x+e)^ 
2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 232 vs. \(2 (99) = 198\).

Time = 0.16 (sec) , antiderivative size = 473, normalized size of antiderivative = 3.85 \[ \int \frac {\csc ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\left [-\frac {6 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {-b} \arctan \left (\frac {2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) + b}\right ) + 3 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\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 (\cos \left (f x + e\right )^{3} + 3 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{128 \, {\left (b^{2} f \cos \left (f x + e\right )^{4} - 2 \, b^{2} f \cos \left (f x + e\right )^{2} + b^{2} f\right )}}, -\frac {6 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {b} \arctan \left (\frac {2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) - b}\right ) - 3 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\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 (\cos \left (f x + e\right )^{3} + 3 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{128 \, {\left (b^{2} f \cos \left (f x + e\right )^{4} - 2 \, b^{2} f \cos \left (f x + e\right )^{2} + b^{2} f\right )}}\right ] \] Input:

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

Output:

[-1/128*(6*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(-b)*arctan(2*sqrt( 
-b)*sqrt(b/cos(f*x + e))*cos(f*x + e)/(b*cos(f*x + e) + b)) + 3*(cos(f*x + 
 e)^4 - 2*cos(f*x + e)^2 + 1)*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*(cos(f*x + e)^3 + 3*cos(f*x 
+ e))*sqrt(b/cos(f*x + e)))/(b^2*f*cos(f*x + e)^4 - 2*b^2*f*cos(f*x + e)^2 
 + b^2*f), -1/128*(6*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(b)*arcta 
n(2*sqrt(b)*sqrt(b/cos(f*x + e))*cos(f*x + e)/(b*cos(f*x + e) - b)) - 3*(c 
os(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(b)*log((b*cos(f*x + e)^2 + 4*(c 
os(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*(cos(f*x + e)^3 + 3*co 
s(f*x + e))*sqrt(b/cos(f*x + e)))/(b^2*f*cos(f*x + e)^4 - 2*b^2*f*cos(f*x 
+ e)^2 + b^2*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.11 \[ \int \frac {\csc ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {b {\left (\frac {4 \, {\left (b^{2} \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}} + 3 \, \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {7}{2}}\right )}}{b^{6} - \frac {2 \, b^{6}}{\cos \left (f x + e\right )^{2}} + \frac {b^{6}}{\cos \left (f x + e\right )^{4}}} + \frac {6 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{b^{\frac {5}{2}}} + \frac {3 \, \log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{b^{\frac {5}{2}}}\right )}}{64 \, f} \] Input:

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

Output:

-1/64*b*(4*(b^2*(b/cos(f*x + e))^(3/2) + 3*(b/cos(f*x + e))^(7/2))/(b^6 - 
2*b^6/cos(f*x + e)^2 + b^6/cos(f*x + e)^4) + 6*arctan(sqrt(b/cos(f*x + e)) 
/sqrt(b))/b^(5/2) + 3*log(-(sqrt(b) - sqrt(b/cos(f*x + e)))/(sqrt(b) + sqr 
t(b/cos(f*x + e))))/b^(5/2))/f
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.04 \[ \int \frac {\csc ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {b^{2} {\left (\frac {3 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} - \frac {3 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {7}{2}}} + \frac {2 \, {\left (\sqrt {b \cos \left (f x + e\right )} b^{2} \cos \left (f x + e\right )^{2} + 3 \, \sqrt {b \cos \left (f x + e\right )} b^{2}\right )}}{{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )}^{2} b^{2}}\right )}}{32 \, f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \] Input:

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

Output:

-1/32*b^2*(3*arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/(sqrt(-b)*b^3) - 3*arct 
an(sqrt(b*cos(f*x + e))/sqrt(b))/b^(7/2) + 2*(sqrt(b*cos(f*x + e))*b^2*cos 
(f*x + e)^2 + 3*sqrt(b*cos(f*x + e))*b^2)/((b^2*cos(f*x + e)^2 - b^2)^2*b^ 
2))/(f*sgn(cos(f*x + e)))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^5(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^5\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \] Input:

int(1/(sin(e + f*x)^5*(b/cos(e + f*x))^(3/2)),x)
 

Output:

int(1/(sin(e + f*x)^5*(b/cos(e + f*x))^(3/2)), x)
 

Reduce [F]

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

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

Output:

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