\(\int \frac {\csc ^6(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [118]

Optimal result

Integrand size = 25, antiderivative size = 167 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {a^2 b \tan (e+f x)}{(a+b)^4 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {\left (15 a^2-20 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^4 f}-\frac {(10 a+b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^3 f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 (a+b)^2 f} \] Output:

-a^2*b*tan(f*x+e)/(a+b)^4/f/(a+b+b*tan(f*x+e)^2)^(1/2)-1/15*(15*a^2-20*a*b 
-2*b^2)*cot(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/(a+b)^4/f-1/15*(10*a+b)*cot( 
f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(1/2)/(a+b)^3/f-1/5*cot(f*x+e)^5*(a+b+b*tan( 
f*x+e)^2)^(1/2)/(a+b)^2/f
 

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.75 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (-8 a^2 (a-5 b)+4 a \left (a^2-4 a b-5 b^2\right ) \csc ^2(e+f x)+(a-5 b) (a+b)^2 \csc ^4(e+f x)+3 (a+b)^3 \csc ^6(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{30 (a+b)^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \] Input:

Integrate[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

-1/30*((a + 2*b + a*Cos[2*(e + f*x)])*(-8*a^2*(a - 5*b) + 4*a*(a^2 - 4*a*b 
 - 5*b^2)*Csc[e + f*x]^2 + (a - 5*b)*(a + b)^2*Csc[e + f*x]^4 + 3*(a + b)^ 
3*Csc[e + f*x]^6)*Sec[e + f*x]^2*Tan[e + f*x])/((a + b)^4*f*(a + b*Sec[e + 
 f*x]^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4620, 365, 359, 245, 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.

\(\displaystyle \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x)^6 \left (a+b \sec (e+f x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 4620

\(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \left (\tan ^2(e+f x)+1\right )^2}{\left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 365

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

\(\Big \downarrow \) 359

\(\displaystyle \frac {\frac {\frac {\left (15 a^2-10 a b-b^2\right ) \int \frac {\cot ^2(e+f x)}{\left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 (a+b)}-\frac {2 (5 a+2 b) \cot ^3(e+f x)}{3 (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\)

\(\Big \downarrow \) 245

\(\displaystyle \frac {\frac {\frac {\left (15 a^2-10 a b-b^2\right ) \left (-\frac {2 b \int \frac {1}{\left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{a+b}-\frac {\cot (e+f x)}{(a+b) \sqrt {a+b \tan ^2(e+f x)+b}}\right )}{3 (a+b)}-\frac {2 (5 a+2 b) \cot ^3(e+f x)}{3 (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\)

\(\Big \downarrow \) 208

\(\displaystyle \frac {\frac {\frac {\left (15 a^2-10 a b-b^2\right ) \left (-\frac {2 b \tan (e+f x)}{(a+b)^2 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\cot (e+f x)}{(a+b) \sqrt {a+b \tan ^2(e+f x)+b}}\right )}{3 (a+b)}-\frac {2 (5 a+2 b) \cot ^3(e+f x)}{3 (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\)

Input:

Int[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

(-1/5*Cot[e + f*x]^5/((a + b)*Sqrt[a + b + b*Tan[e + f*x]^2]) + ((-2*(5*a 
+ 2*b)*Cot[e + f*x]^3)/(3*(a + b)*Sqrt[a + b + b*Tan[e + f*x]^2]) + ((15*a 
^2 - 10*a*b - b^2)*(-(Cot[e + f*x]/((a + b)*Sqrt[a + b + b*Tan[e + f*x]^2] 
)) - (2*b*Tan[e + f*x])/((a + b)^2*Sqrt[a + b + b*Tan[e + f*x]^2])))/(3*(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 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 = 9.13 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.22

method	result	size
default	\(-\frac {\left (b +a \cos \left (f x +e \right )^{2}\right ) \left (\left (-20 \cos \left (f x +e \right )^{4}+49 \cos \left (f x +e \right )^{2}\right ) a \,b^{2}+\left (-40 \cos \left (f x +e \right )^{6}+104 \cos \left (f x +e \right )^{4}-85 \cos \left (f x +e \right )^{2}\right ) b \,a^{2}-20 a \,b^{2}+30 a^{2} b +\left (8 \cos \left (f x +e \right )^{6}-20 \cos \left (f x +e \right )^{4}+15 \cos \left (f x +e \right )^{2}\right ) a^{3}+\left (5 \cos \left (f x +e \right )^{2}-2\right ) b^{3}\right ) \sec \left (f x +e \right )^{3} \csc \left (f x +e \right )^{5}}{15 f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\)	\(203\)

Input:

int(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/15/f/(a^4+4*a^3*b+6*a^2*b^2+4*a*b^3+b^4)*(b+a*cos(f*x+e)^2)*((-20*cos(f 
*x+e)^4+49*cos(f*x+e)^2)*a*b^2+(-40*cos(f*x+e)^6+104*cos(f*x+e)^4-85*cos(f 
*x+e)^2)*b*a^2-20*a*b^2+30*a^2*b+(8*cos(f*x+e)^6-20*cos(f*x+e)^4+15*cos(f* 
x+e)^2)*a^3+(5*cos(f*x+e)^2-2)*b^3)/(a+b*sec(f*x+e)^2)^(3/2)*sec(f*x+e)^3* 
csc(f*x+e)^5
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (153) = 306\).

Time = 2.01 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.88 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {{\left (8 \, {\left (a^{3} - 5 \, a^{2} b\right )} \cos \left (f x + e\right )^{7} - 4 \, {\left (5 \, a^{3} - 26 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{5} + {\left (15 \, a^{3} - 85 \, a^{2} b + 49 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (15 \, a^{2} b - 10 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left ({\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{6} - {\left (2 \, a^{5} + 7 \, a^{4} b + 8 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 2 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{5} + 2 \, a^{4} b - 2 \, a^{3} b^{2} - 8 \, a^{2} b^{3} - 7 \, a b^{4} - 2 \, b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} f\right )} \sin \left (f x + e\right )} \] Input:

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

-1/15*(8*(a^3 - 5*a^2*b)*cos(f*x + e)^7 - 4*(5*a^3 - 26*a^2*b + 5*a*b^2)*c 
os(f*x + e)^5 + (15*a^3 - 85*a^2*b + 49*a*b^2 + 5*b^3)*cos(f*x + e)^3 + 2* 
(15*a^2*b - 10*a*b^2 - b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos( 
f*x + e)^2)/(((a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*f*cos(f*x + 
e)^6 - (2*a^5 + 7*a^4*b + 8*a^3*b^2 + 2*a^2*b^3 - 2*a*b^4 - b^5)*f*cos(f*x 
 + e)^4 + (a^5 + 2*a^4*b - 2*a^3*b^2 - 8*a^2*b^3 - 7*a*b^4 - 2*b^5)*f*cos( 
f*x + e)^2 + (a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*f)*sin(f*x + 
e))
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.69 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {30 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2}} - \frac {80 \, b^{2} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{3}} + \frac {48 \, b^{3} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{4}} + \frac {15}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )} - \frac {40 \, b}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \tan \left (f x + e\right )} + \frac {24 \, b^{2}}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{3} \tan \left (f x + e\right )} + \frac {10}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )^{3}} - \frac {6 \, b}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \tan \left (f x + e\right )^{3}} + \frac {3}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \] Input:

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

-1/15*(30*b*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^2) - 80*b 
^2*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^3) + 48*b^3*tan(f* 
x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^4) + 15/(sqrt(b*tan(f*x + e 
)^2 + a + b)*(a + b)*tan(f*x + e)) - 40*b/(sqrt(b*tan(f*x + e)^2 + a + b)* 
(a + b)^2*tan(f*x + e)) + 24*b^2/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^3 
*tan(f*x + e)) + 10/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)*tan(f*x + e)^3 
) - 6*b/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^2*tan(f*x + e)^3) + 3/(sqr 
t(b*tan(f*x + e)^2 + a + b)*(a + b)*tan(f*x + e)^5))/f
 

Giac [F]

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

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

integrate(csc(f*x + e)^6/(b*sec(f*x + e)^2 + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Hanged} \] Input:

int(1/(sin(e + f*x)^6*(a + b/cos(e + f*x)^2)^(3/2)),x)
 

Output:

\text{Hanged}
                                                                                    
                                                                                    
 

Reduce [F]

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

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

Output:

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