\(\int \frac {\csc ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [127]

Optimal result

Integrand size = 25, antiderivative size = 123 \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\left (15 a^2-20 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^3 f}-\frac {2 (5 a-2 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a f} \] Output:

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

Mathematica [A] (verified)

Time = 1.51 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.73 \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\cot (e+f x) \left (8 (a-b)^2+4 a (a-b) \csc ^2(e+f x)+3 a^2 \csc ^4(e+f x)\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{15 \sqrt {2} a^3 f} \] Input:

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

Output:

-1/15*(Cot[e + f*x]*(8*(a - b)^2 + 4*a*(a - b)*Csc[e + f*x]^2 + 3*a^2*Csc[ 
e + f*x]^4)*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(Sqrt 
[2]*a^3*f)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4146, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4146

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

\(\Big \downarrow \) 365

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

\(\Big \downarrow \) 359

\(\displaystyle \frac {\frac {\frac {\left (15 a^2-20 a b+8 b^2\right ) \int \frac {\cot ^2(e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{3 a}-\frac {2 (5 a-2 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}}{5 a}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a}}{f}\)

\(\Big \downarrow \) 242

\(\displaystyle \frac {\frac {-\frac {\left (15 a^2-20 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2}-\frac {2 (5 a-2 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}}{5 a}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a}}{f}\)

Input:

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

Output:

(-1/5*(Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/a + (-1/3*((15*a^2 - 20* 
a*b + 8*b^2)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/a^2 - (2*(5*a - 2*b) 
*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(3*a))/(5*a))/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 4146

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

Maple [A] (verified)

Time = 7.16 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.10

method	result	size
default	\(-\frac {\left (a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}\right ) \left (8 \sin \left (f x +e \right )^{4} b^{2}+16 a b \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2}+8 a^{2} \cos \left (f x +e \right )^{4}-20 a b \sin \left (f x +e \right )^{2}-20 a^{2} \cos \left (f x +e \right )^{2}+15 a^{2}\right ) \sec \left (f x +e \right ) \csc \left (f x +e \right )^{5}}{15 f \,a^{3} \sqrt {a +b \tan \left (f x +e \right )^{2}}}\)	\(135\)

Input:

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

Output:

-1/15/f/a^3*(a*cos(f*x+e)^2+b*sin(f*x+e)^2)*(8*sin(f*x+e)^4*b^2+16*a*b*cos 
(f*x+e)^2*sin(f*x+e)^2+8*a^2*cos(f*x+e)^4-20*a*b*sin(f*x+e)^2-20*a^2*cos(f 
*x+e)^2+15*a^2)/(a+b*tan(f*x+e)^2)^(1/2)*sec(f*x+e)*csc(f*x+e)^5
 

Fricas [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.15 \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {{\left (8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 4 \, {\left (5 \, a^{2} - 9 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} - 20 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )} \sin \left (f x + e\right )} \] Input:

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

Output:

-1/15*(8*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 4*(5*a^2 - 9*a*b + 4*b^2)*co 
s(f*x + e)^3 + (15*a^2 - 20*a*b + 8*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f 
*x + e)^2 + b)/cos(f*x + e)^2)/((a^3*f*cos(f*x + e)^4 - 2*a^3*f*cos(f*x + 
e)^2 + a^3*f)*sin(f*x + e))
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.41 \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\frac {15 \, \sqrt {b \tan \left (f x + e\right )^{2} + a}}{a \tan \left (f x + e\right )} - \frac {20 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b}{a^{2} \tan \left (f x + e\right )} + \frac {8 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b^{2}}{a^{3} \tan \left (f x + e\right )} + \frac {10 \, \sqrt {b \tan \left (f x + e\right )^{2} + a}}{a \tan \left (f x + e\right )^{3}} - \frac {4 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b}{a^{2} \tan \left (f x + e\right )^{3}} + \frac {3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a}}{a \tan \left (f x + e\right )^{5}}}{15 \, f} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 18.12 (sec) , antiderivative size = 761, normalized size of antiderivative = 6.19 \[ \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

((((a - b)*(32*a*b - 64*a^2 + 32*b^2))/(120*a^3*f*(a*1i - b*1i)) - ((a - b 
)*(64*a^2 - 96*a*b + 32*b^2))/(120*a^3*f*(a*1i - b*1i)))*(a + (b*(exp(e*2i 
 + f*x*2i)*1i - 1i)^2)/(exp(e*2i + f*x*2i) + 1)^2)^(1/2)*(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*2i + f*x*2i)*1i - 1i)^2)/(exp(e*2i + f*x*2i) 
 + 1)^2)^(1/2)*((2*(3*a - 3*b))/(3*a*f*(a*1i - b*1i)) + ((3*a - 3*b)*(96*a 
 - 64*b))/(240*a^2*f*(a*1i - b*1i)) + ((3*a - 3*b)*(256*a + 64*b))/(240*a^ 
2*f*(a*1i - b*1i)))*(2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 1))/((exp 
(e*2i + f*x*2i) - 1)^4*(exp(e*2i + f*x*2i) + 1)) + ((((a - b)*(32*a - 16*b 
))/(30*a^2*f*(a*1i - b*1i)) + ((a - b)*(32*a + 48*b))/(30*a^2*f*(a*1i - b* 
1i)))*(a + (b*(exp(e*2i + f*x*2i)*1i - 1i)^2)/(exp(e*2i + f*x*2i) + 1)^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)^2*(a + (b*(exp(e*2i + f*x*2i 
)*1i - 1i)^2)/(exp(e*2i + f*x*2i) + 1)^2)^(1/2)*(2*exp(e*2i + f*x*2i) + ex 
p(e*4i + f*x*4i) + 1)*8i)/(15*a^3*f*(exp(e*2i + f*x*2i) - 1)*(exp(e*2i + f 
*x*2i) + 1)) + (8*(2*a - 2*b)*(a + (b*(exp(e*2i + f*x*2i)*1i - 1i)^2)/(exp 
(e*2i + f*x*2i) + 1)^2)^(1/2)*(2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 
 1))/(5*a*f*(exp(e*2i + f*x*2i) - 1)^5*(exp(e*2i + f*x*2i) + 1)*(a*1i - b* 
1i))
 

Reduce [F]

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

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

Output:

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