\(\int \frac {\csc ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [105]

Optimal result

Integrand size = 25, antiderivative size = 132 \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\left (15 a^2+10 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^3 f}-\frac {2 (5 a+3 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 (a+b) f} \] Output:

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.76 \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (8 a^2+8 a b+3 b^2-2 a (3 a+b) \cos (2 (e+f x))+a^2 \cos (4 (e+f x))\right ) \csc ^5(e+f x) \sec (e+f x)}{30 (a+b)^3 f \sqrt {a+b \sec ^2(e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4620, 365, 359, 242}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4620

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

\(\Big \downarrow \) 365

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

\(\Big \downarrow \) 359

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

\(\Big \downarrow \) 242

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

Input:

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

Output:

(-1/5*(Cot[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])/(a + b) + (-1/3*((15 
*a^2 + 10*a*b + 3*b^2)*Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(a + b 
)^2 - (2*(5*a + 3*b)*Cot[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(3*(a 
+ b)))/(5*(a + b)))/f
 

Defintions of rubi rules used

rule 242

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x 
] && EqQ[m + 2*p + 3, 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 = 6.23 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {\left (b +a \cos \left (f x +e \right )^{2}\right ) \left (8 \cos \left (f x +e \right )^{4} a^{2}-20 a^{2} \cos \left (f x +e \right )^{2}-4 \cos \left (f x +e \right )^{2} a b +15 a^{2}+10 a b +3 b^{2}\right ) \sec \left (f x +e \right ) \csc \left (f x +e \right )^{5}}{15 f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \sqrt {a +b \sec \left (f x +e \right )^{2}}}\)	\(120\)

Input:

int(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/15/f/(a^3+3*a^2*b+3*a*b^2+b^3)*(b+a*cos(f*x+e)^2)*(8*cos(f*x+e)^4*a^2-2 
0*a^2*cos(f*x+e)^2-4*cos(f*x+e)^2*a*b+15*a^2+10*a*b+3*b^2)/(a+b*sec(f*x+e) 
^2)^(1/2)*sec(f*x+e)*csc(f*x+e)^5
 

Fricas [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.30 \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 4 \, {\left (5 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 10 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )} \sin \left (f x + e\right )} \] Input:

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

-1/15*(8*a^2*cos(f*x + e)^5 - 4*(5*a^2 + a*b)*cos(f*x + e)^3 + (15*a^2 + 1 
0*a*b + 3*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/( 
((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^4 - 2*(a^3 + 3*a^2*b + 3*a 
*b^2 + b^3)*f*cos(f*x + e)^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f)*sin(f*x 
+ e))
 

Sympy [F]

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

integrate(csc(f*x+e)**6/(a+b*sec(f*x+e)**2)**(1/2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.45 \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\frac {15 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b}}{{\left (a + b\right )} \tan \left (f x + e\right )} - \frac {20 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} b}{{\left (a + b\right )}^{2} \tan \left (f x + e\right )} + \frac {8 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} b^{2}}{{\left (a + b\right )}^{3} \tan \left (f x + e\right )} + \frac {10 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b}}{{\left (a + b\right )} \tan \left (f x + e\right )^{3}} - \frac {4 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} b}{{\left (a + b\right )}^{2} \tan \left (f x + e\right )^{3}} + \frac {3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b}}{{\left (a + b\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \] Input:

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

-1/15*(15*sqrt(b*tan(f*x + e)^2 + a + b)/((a + b)*tan(f*x + e)) - 20*sqrt( 
b*tan(f*x + e)^2 + a + b)*b/((a + b)^2*tan(f*x + e)) + 8*sqrt(b*tan(f*x + 
e)^2 + a + b)*b^2/((a + b)^3*tan(f*x + e)) + 10*sqrt(b*tan(f*x + e)^2 + a 
+ b)/((a + b)*tan(f*x + e)^3) - 4*sqrt(b*tan(f*x + e)^2 + a + b)*b/((a + b 
)^2*tan(f*x + e)^3) + 3*sqrt(b*tan(f*x + e)^2 + a + b)/((a + b)*tan(f*x + 
e)^5))/f
 

Giac [F]

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

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 24.05 (sec) , antiderivative size = 723, normalized size of antiderivative = 5.48 \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

(((32*a + 16*b)/(5*f*(6*a + 6*b)*(a*1i + b*1i)) + (32*a + 80*b)/(5*f*(6*a 
+ 6*b)*(a*1i + b*1i)))*(a + b/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i) 
/2)^2)^(1/2)*(2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 1))/((exp(e*2i + 
 f*x*2i) - 1)^3*(exp(e*2i + f*x*2i) + 1)) - ((a + b/(exp(- e*1i - f*x*1i)/ 
2 + exp(e*1i + f*x*1i)/2)^2)^(1/2)*((a*(2*a + b)*32i)/(15*f*(a*1i + b*1i)^ 
2*(8*a + 8*b)) + (a*(2*a + 3*b)*32i)/(15*f*(a*1i + b*1i)^2*(8*a + 8*b)))*( 
2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 1))/((exp(e*2i + f*x*2i) - 1)^ 
2*(exp(e*2i + f*x*2i) + 1)) + ((a + b/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + 
 f*x*1i)/2)^2)^(1/2)*(2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 1)*((96* 
a + 32*b)/(5*f*(a*1i + b*1i)*(16*a + 16*b)) + (160*a + 160*b)/(5*f*(a*1i + 
 b*1i)*(16*a + 16*b)) + (256*a + 320*b)/(5*f*(a*1i + b*1i)*(16*a + 16*b))) 
)/((exp(e*2i + f*x*2i) - 1)^4*(exp(e*2i + f*x*2i) + 1)) - ((a + b/(exp(- e 
*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2)^2)^(1/2)*(2*exp(e*2i + f*x*2i) + e 
xp(e*4i + f*x*4i) + 1)*32i)/(f*(exp(e*2i + f*x*2i) - 1)^5*(exp(e*2i + f*x* 
2i) + 1)*(10*a + 10*b)) - (8*a^2*(a + b/(exp(- e*1i - f*x*1i)/2 + exp(e*1i 
 + f*x*1i)/2)^2)^(1/2)*(2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 1))/(1 
5*f*(exp(e*2i + f*x*2i) - 1)*(exp(e*2i + f*x*2i) + 1)*(a*1i + b*1i)^3)
 

Reduce [F]

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

int(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(1/2),x)
 

Output:

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