3.2.51 \(\int \frac {\csc ^6(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\) [151]

3.2.51.1 Optimal result

Integrand size = 25, antiderivative size = 219 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\left (5 a^2-20 a b+16 b^2\right ) \cot (e+f x)}{5 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {2 (5 a-4 b) \cot ^3(e+f x)}{15 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {4 b \left (5 a^2-20 a b+16 b^2\right ) \tan (e+f x)}{15 a^4 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {8 b \left (5 a^2-20 a b+16 b^2\right ) \tan (e+f x)}{15 a^5 f \sqrt {a+b \tan ^2(e+f x)}} \]

output

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

3.2.51.2 Mathematica [A] (verified)

Time = 2.92 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.79 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {\sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \left (-\cot (e+f x) \left (8 a^2-66 a b+73 b^2+2 a (2 a-7 b) \csc ^2(e+f x)+3 a^2 \csc ^4(e+f x)\right )+\frac {5 b (-a+b) \left (6 a^2-7 a b-11 b^2+\left (6 a^2-17 a b+11 b^2\right ) \cos (2 (e+f x))\right ) \sin (2 (e+f x))}{(a+b+(a-b) \cos (2 (e+f x)))^2}\right )}{15 \sqrt {2} a^5 f} \]

input

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

output

(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(-(Cot[e + f*x]*( 
8*a^2 - 66*a*b + 73*b^2 + 2*a*(2*a - 7*b)*Csc[e + f*x]^2 + 3*a^2*Csc[e + f 
*x]^4)) + (5*b*(-a + b)*(6*a^2 - 7*a*b - 11*b^2 + (6*a^2 - 17*a*b + 11*b^2 
)*Cos[2*(e + f*x)])*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[2*(e + f*x)])^2 
))/(15*Sqrt[2]*a^5*f)
 

3.2.51.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.89, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4146, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4146

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

\(\Big \downarrow \) 365

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

\(\Big \downarrow \) 359

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

\(\Big \downarrow \) 245

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

\(\Big \downarrow \) 209

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

\(\Big \downarrow \) 208

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

input

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

output

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

3.2.51.3.1 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 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]
 

3.2.51.4 Maple [A] (verified)

Time = 6.46 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.27

method	result	size
default	\(-\frac {\left (a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}\right ) \left (8 a^{4} \cos \left (f x +e \right )^{8}+112 \cos \left (f x +e \right )^{6} \sin \left (f x +e \right )^{2} a^{3} b +328 \cos \left (f x +e \right )^{4} \sin \left (f x +e \right )^{4} a^{2} b^{2}+352 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{6} a \,b^{3}+128 \sin \left (f x +e \right )^{8} b^{4}-20 a^{4} \cos \left (f x +e \right )^{6}-180 \cos \left (f x +e \right )^{4} \sin \left (f x +e \right )^{2} a^{3} b -320 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{4} a^{2} b^{2}-160 \sin \left (f x +e \right )^{6} a \,b^{3}+15 a^{4} \cos \left (f x +e \right )^{4}+60 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2} a^{3} b +40 \sin \left (f x +e \right )^{4} a^{2} b^{2}\right ) \sec \left (f x +e \right )^{5} \csc \left (f x +e \right )^{5}}{15 f \,a^{5} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}\)	\(279\)

input

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

output

-1/15/f/a^5*(a*cos(f*x+e)^2+b*sin(f*x+e)^2)*(8*a^4*cos(f*x+e)^8+112*cos(f* 
x+e)^6*sin(f*x+e)^2*a^3*b+328*cos(f*x+e)^4*sin(f*x+e)^4*a^2*b^2+352*cos(f* 
x+e)^2*sin(f*x+e)^6*a*b^3+128*sin(f*x+e)^8*b^4-20*a^4*cos(f*x+e)^6-180*cos 
(f*x+e)^4*sin(f*x+e)^2*a^3*b-320*cos(f*x+e)^2*sin(f*x+e)^4*a^2*b^2-160*sin 
(f*x+e)^6*a*b^3+15*a^4*cos(f*x+e)^4+60*cos(f*x+e)^2*sin(f*x+e)^2*a^3*b+40* 
sin(f*x+e)^4*a^2*b^2)/(a+b*tan(f*x+e)^2)^(5/2)*sec(f*x+e)^5*csc(f*x+e)^5
 

3.2.51.5 Fricas [F(-1)]

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

input

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

output

Timed out
 

3.2.51.6 Sympy [F]

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

input

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

output

Integral(csc(e + f*x)**6/(a + b*tan(e + f*x)**2)**(5/2), x)
 

3.2.51.7 Maxima [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.54 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {40 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{3}} + \frac {20 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {160 \, b^{2} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{4}} - \frac {80 \, b^{2} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{3}} + \frac {128 \, b^{3} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{5}} + \frac {64 \, b^{3} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{4}} + \frac {15}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \tan \left (f x + e\right )} - \frac {60 \, b}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2} \tan \left (f x + e\right )} + \frac {48 \, b^{2}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{3} \tan \left (f x + e\right )} + \frac {10}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \tan \left (f x + e\right )^{3}} - \frac {8 \, b}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2} \tan \left (f x + e\right )^{3}} + \frac {3}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \tan \left (f x + e\right )^{5}}}{15 \, f} \]

input

integrate(csc(f*x+e)^6/(a+b*tan(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)*a^3) + 20*b*tan(f*x + 
 e)/((b*tan(f*x + e)^2 + a)^(3/2)*a^2) - 160*b^2*tan(f*x + e)/(sqrt(b*tan( 
f*x + e)^2 + a)*a^4) - 80*b^2*tan(f*x + e)/((b*tan(f*x + e)^2 + a)^(3/2)*a 
^3) + 128*b^3*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a)*a^5) + 64*b^3*tan(f 
*x + e)/((b*tan(f*x + e)^2 + a)^(3/2)*a^4) + 15/((b*tan(f*x + e)^2 + a)^(3 
/2)*a*tan(f*x + e)) - 60*b/((b*tan(f*x + e)^2 + a)^(3/2)*a^2*tan(f*x + e)) 
 + 48*b^2/((b*tan(f*x + e)^2 + a)^(3/2)*a^3*tan(f*x + e)) + 10/((b*tan(f*x 
 + e)^2 + a)^(3/2)*a*tan(f*x + e)^3) - 8*b/((b*tan(f*x + e)^2 + a)^(3/2)*a 
^2*tan(f*x + e)^3) + 3/((b*tan(f*x + e)^2 + a)^(3/2)*a*tan(f*x + e)^5))/f
 

3.2.51.8 Giac [F]

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

input

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

output

sage0*x
 

3.2.51.9 Mupad [F(-1)]

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

input

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

output

\text{Hanged}