\(\int \frac {\csc ^6(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [131]

Optimal result

Integrand size = 25, antiderivative size = 215 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {a^2 b \tan (e+f x)}{3 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (5 a-6 b) b \tan (e+f x)}{3 (a+b)^5 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\left (15 a^2-50 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^5 f}-\frac {2 (5 a-2 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^4 f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 (a+b)^3 f} \] Output:

-1/3*a^2*b*tan(f*x+e)/(a+b)^4/f/(a+b+b*tan(f*x+e)^2)^(3/2)-1/3*a*(5*a-6*b) 
*b*tan(f*x+e)/(a+b)^5/f/(a+b+b*tan(f*x+e)^2)^(1/2)-1/15*(15*a^2-50*a*b+8*b 
^2)*cot(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/(a+b)^5/f-2/15*(5*a-2*b)*cot(f*x 
+e)^3*(a+b+b*tan(f*x+e)^2)^(1/2)/(a+b)^4/f-1/5*cot(f*x+e)^5*(a+b+b*tan(f*x 
+e)^2)^(1/2)/(a+b)^3/f
 

Mathematica [A] (verified)

Time = 4.78 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.80 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {(a+2 b+a \cos (2 (e+f x)))^3 \left (\frac {20 a b^2 (a+b)}{(a+2 b+a \cos (2 (e+f x)))^2}+\frac {10 a b (-6 a+5 b)}{a+2 b+a \cos (2 (e+f x))}+\left (-8 a^2+50 a b-15 b^2\right ) \csc ^2(e+f x)+2 (a+b) (-2 a+5 b) \csc ^4(e+f x)-3 (a+b)^2 \csc ^6(e+f x)\right ) \sec ^4(e+f x) \tan (e+f x)}{120 (a+b)^5 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \] Input:

Integrate[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

Output:

((a + 2*b + a*Cos[2*(e + f*x)])^3*((20*a*b^2*(a + b))/(a + 2*b + a*Cos[2*( 
e + f*x)])^2 + (10*a*b*(-6*a + 5*b))/(a + 2*b + a*Cos[2*(e + f*x)]) + (-8* 
a^2 + 50*a*b - 15*b^2)*Csc[e + f*x]^2 + 2*(a + b)*(-2*a + 5*b)*Csc[e + f*x 
]^4 - 3*(a + b)^2*Csc[e + f*x]^6)*Sec[e + f*x]^4*Tan[e + f*x])/(120*(a + b 
)^5*f*(a + b*Sec[e + f*x]^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4620, 365, 359, 245, 209, 208}

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.

\(\displaystyle \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x)^6 \left (a+b \sec (e+f x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 4620

\(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \left (\tan ^2(e+f x)+1\right )^2}{\left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 365

\(\displaystyle \frac {\frac {\int \frac {\cot ^4(e+f x) \left (5 (a+b) \tan ^2(e+f x)+2 (5 a+b)\right )}{\left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 359

\(\displaystyle \frac {\frac {\frac {\left (5 a^2-10 a b+b^2\right ) \int \frac {\cot ^2(e+f x)}{\left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{a+b}-\frac {2 (5 a+b) \cot ^3(e+f x)}{3 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 245

\(\displaystyle \frac {\frac {\frac {\left (5 a^2-10 a b+b^2\right ) \left (-\frac {4 b \int \frac {1}{\left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{a+b}-\frac {\cot (e+f x)}{(a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}\right )}{a+b}-\frac {2 (5 a+b) \cot ^3(e+f x)}{3 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 209

\(\displaystyle \frac {\frac {\frac {\left (5 a^2-10 a b+b^2\right ) \left (-\frac {4 b \left (\frac {2 \int \frac {1}{\left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 (a+b)}+\frac {\tan (e+f x)}{3 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}\right )}{a+b}-\frac {\cot (e+f x)}{(a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}\right )}{a+b}-\frac {2 (5 a+b) \cot ^3(e+f x)}{3 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 208

\(\displaystyle \frac {\frac {\frac {\left (5 a^2-10 a b+b^2\right ) \left (-\frac {4 b \left (\frac {2 \tan (e+f x)}{3 (a+b)^2 \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {\tan (e+f x)}{3 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}\right )}{a+b}-\frac {\cot (e+f x)}{(a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}\right )}{a+b}-\frac {2 (5 a+b) \cot ^3(e+f x)}{3 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

Input:

Int[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

Output:

(-1/5*Cot[e + f*x]^5/((a + b)*(a + b + b*Tan[e + f*x]^2)^(3/2)) + ((-2*(5* 
a + b)*Cot[e + f*x]^3)/(3*(a + b)*(a + b + b*Tan[e + f*x]^2)^(3/2)) + ((5* 
a^2 - 10*a*b + b^2)*(-(Cot[e + f*x]/((a + b)*(a + b + b*Tan[e + f*x]^2)^(3 
/2))) - (4*b*(Tan[e + f*x]/(3*(a + b)*(a + b + b*Tan[e + f*x]^2)^(3/2)) + 
(2*Tan[e + f*x])/(3*(a + b)^2*Sqrt[a + b + b*Tan[e + f*x]^2])))/(a + b)))/ 
(a + b))/(5*(a + b)))/f
 

Defintions of rubi rules used

rule 208

Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), 
x] /; FreeQ[{a, b}, x]
 

rule 209

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
 

rule 245

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + 
 b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) 
   Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si 
mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
 

rule 359

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + 
Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] 
&& LtQ[m, -1] &&  !ILtQ[p, -1]
 

rule 365

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

rule 3042

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

rule 4620

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ 
)]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m 
+ 1)/f   Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f 
f^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, 
 x] && IntegerQ[m/2] && IntegerQ[n/2]
 

Maple [A] (verified)

Time = 9.23 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.35

method	result	size
default	\(-\frac {\left (b +a \cos \left (f x +e \right )^{2}\right ) \left (\left (60 \cos \left (f x +e \right )^{6}-180 \cos \left (f x +e \right )^{4}+212 \cos \left (f x +e \right )^{2}\right ) a \,b^{3}+\left (-80 \cos \left (f x +e \right )^{8}+212 \cos \left (f x +e \right )^{6}-180 \cos \left (f x +e \right )^{4}+60 \cos \left (f x +e \right )^{2}\right ) b \,a^{3}-80 a \,b^{3}+40 a^{2} b^{2}+\left (40 \cos \left (f x +e \right )^{8}-220 \cos \left (f x +e \right )^{6}+378 \cos \left (f x +e \right )^{4}-220 \cos \left (f x +e \right )^{2}\right ) a^{2} b^{2}+\left (15 \cos \left (f x +e \right )^{4}-20 \cos \left (f x +e \right )^{2}+8\right ) b^{4}+\left (8 \cos \left (f x +e \right )^{8}-20 \cos \left (f x +e \right )^{6}+15 \cos \left (f x +e \right )^{4}\right ) a^{4}\right ) \sec \left (f x +e \right )^{5} \csc \left (f x +e \right )^{5}}{15 f \left (a^{5}+5 b \,a^{4}+10 a^{3} b^{2}+10 a^{2} b^{3}+5 a \,b^{4}+b^{5}\right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}\)	\(291\)

Input:

int(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/15/f/(a^5+5*a^4*b+10*a^3*b^2+10*a^2*b^3+5*a*b^4+b^5)*(b+a*cos(f*x+e)^2) 
*((60*cos(f*x+e)^6-180*cos(f*x+e)^4+212*cos(f*x+e)^2)*a*b^3+(-80*cos(f*x+e 
)^8+212*cos(f*x+e)^6-180*cos(f*x+e)^4+60*cos(f*x+e)^2)*b*a^3-80*a*b^3+40*a 
^2*b^2+(40*cos(f*x+e)^8-220*cos(f*x+e)^6+378*cos(f*x+e)^4-220*cos(f*x+e)^2 
)*a^2*b^2+(15*cos(f*x+e)^4-20*cos(f*x+e)^2+8)*b^4+(8*cos(f*x+e)^8-20*cos(f 
*x+e)^6+15*cos(f*x+e)^4)*a^4)/(a+b*sec(f*x+e)^2)^(5/2)*sec(f*x+e)^5*csc(f* 
x+e)^5
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 460 vs. \(2 (195) = 390\).

Time = 8.38 (sec) , antiderivative size = 460, normalized size of antiderivative = 2.14 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {{\left (8 \, {\left (a^{4} - 10 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{9} - 4 \, {\left (5 \, a^{4} - 53 \, a^{3} b + 55 \, a^{2} b^{2} - 15 \, a b^{3}\right )} \cos \left (f x + e\right )^{7} + 3 \, {\left (5 \, a^{4} - 60 \, a^{3} b + 126 \, a^{2} b^{2} - 60 \, a b^{3} + 5 \, b^{4}\right )} \cos \left (f x + e\right )^{5} + 4 \, {\left (15 \, a^{3} b - 55 \, a^{2} b^{2} + 53 \, a b^{3} - 5 \, b^{4}\right )} \cos \left (f x + e\right )^{3} + 8 \, {\left (5 \, a^{2} b^{2} - 10 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left ({\left (a^{7} + 5 \, a^{6} b + 10 \, a^{5} b^{2} + 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{8} - 2 \, {\left (a^{7} + 4 \, a^{6} b + 5 \, a^{5} b^{2} - 5 \, a^{3} b^{4} - 4 \, a^{2} b^{5} - a b^{6}\right )} f \cos \left (f x + e\right )^{6} + {\left (a^{7} + a^{6} b - 9 \, a^{5} b^{2} - 25 \, a^{4} b^{3} - 25 \, a^{3} b^{4} - 9 \, a^{2} b^{5} + a b^{6} + b^{7}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + 4 \, a^{5} b^{2} + 5 \, a^{4} b^{3} - 5 \, a^{2} b^{5} - 4 \, a b^{6} - b^{7}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} + 5 \, a^{4} b^{3} + 10 \, a^{3} b^{4} + 10 \, a^{2} b^{5} + 5 \, a b^{6} + b^{7}\right )} f\right )} \sin \left (f x + e\right )} \] Input:

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")
 

Output:

-1/15*(8*(a^4 - 10*a^3*b + 5*a^2*b^2)*cos(f*x + e)^9 - 4*(5*a^4 - 53*a^3*b 
 + 55*a^2*b^2 - 15*a*b^3)*cos(f*x + e)^7 + 3*(5*a^4 - 60*a^3*b + 126*a^2*b 
^2 - 60*a*b^3 + 5*b^4)*cos(f*x + e)^5 + 4*(15*a^3*b - 55*a^2*b^2 + 53*a*b^ 
3 - 5*b^4)*cos(f*x + e)^3 + 8*(5*a^2*b^2 - 10*a*b^3 + b^4)*cos(f*x + e))*s 
qrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(((a^7 + 5*a^6*b + 10*a^5*b^2 + 
 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^8 - 2*(a^7 + 4*a^6*b + 5 
*a^5*b^2 - 5*a^3*b^4 - 4*a^2*b^5 - a*b^6)*f*cos(f*x + e)^6 + (a^7 + a^6*b 
- 9*a^5*b^2 - 25*a^4*b^3 - 25*a^3*b^4 - 9*a^2*b^5 + a*b^6 + b^7)*f*cos(f*x 
 + e)^4 + 2*(a^6*b + 4*a^5*b^2 + 5*a^4*b^3 - 5*a^2*b^5 - 4*a*b^6 - b^7)*f* 
cos(f*x + e)^2 + (a^5*b^2 + 5*a^4*b^3 + 10*a^3*b^4 + 10*a^2*b^5 + 5*a*b^6 
+ b^7)*f)*sin(f*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(csc(f*x+e)**6/(a+b*sec(f*x+e)**2)**(5/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.73 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {40 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{3}} + \frac {20 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{2}} - \frac {160 \, b^{2} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{4}} - \frac {80 \, b^{2} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{3}} + \frac {128 \, b^{3} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{5}} + \frac {64 \, b^{3} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{4}} + \frac {15}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )} - \frac {60 \, b}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{2} \tan \left (f x + e\right )} + \frac {48 \, b^{2}}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{3} \tan \left (f x + e\right )} + \frac {10}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )^{3}} - \frac {8 \, b}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{2} \tan \left (f x + e\right )^{3}} + \frac {3}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \] Input:

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")
 

Output:

-1/15*(40*b*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^3) + 20*b 
*tan(f*x + e)/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)^2) - 160*b^2*tan(f 
*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^4) - 80*b^2*tan(f*x + e)/( 
(b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)^3) + 128*b^3*tan(f*x + e)/(sqrt(b 
*tan(f*x + e)^2 + a + b)*(a + b)^5) + 64*b^3*tan(f*x + e)/((b*tan(f*x + e) 
^2 + a + b)^(3/2)*(a + b)^4) + 15/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b 
)*tan(f*x + e)) - 60*b/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)^2*tan(f*x 
 + e)) + 48*b^2/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)^3*tan(f*x + e)) 
+ 10/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)*tan(f*x + e)^3) - 8*b/((b*t 
an(f*x + e)^2 + a + b)^(3/2)*(a + b)^2*tan(f*x + e)^3) + 3/((b*tan(f*x + e 
)^2 + a + b)^(3/2)*(a + b)*tan(f*x + e)^5))/f
 

Giac [F]

\[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\csc \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")
 

Output:

integrate(csc(f*x + e)^6/(b*sec(f*x + e)^2 + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Hanged} \] Input:

int(1/(sin(e + f*x)^6*(a + b/cos(e + f*x)^2)^(5/2)),x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \csc \left (f x +e \right )^{6}}{\sec \left (f x +e \right )^{6} b^{3}+3 \sec \left (f x +e \right )^{4} a \,b^{2}+3 \sec \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:

int(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x)
 

Output:

int((sqrt(sec(e + f*x)**2*b + a)*csc(e + f*x)**6)/(sec(e + f*x)**6*b**3 + 
3*sec(e + f*x)**4*a*b**2 + 3*sec(e + f*x)**2*a**2*b + a**3),x)