\(\int \frac {\csc ^4(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [130]

Optimal result

Integrand size = 25, antiderivative size = 160 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {a b \tan (e+f x)}{3 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {(5 a-3 b) b \tan (e+f x)}{3 (a+b)^4 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {(3 a-5 b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 (a+b)^4 f}-\frac {\cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 (a+b)^3 f} \] Output:

-1/3*a*b*tan(f*x+e)/(a+b)^3/f/(a+b+b*tan(f*x+e)^2)^(3/2)-1/3*(5*a-3*b)*b*t 
an(f*x+e)/(a+b)^4/f/(a+b+b*tan(f*x+e)^2)^(1/2)-1/3*(3*a-5*b)*cot(f*x+e)*(a 
+b+b*tan(f*x+e)^2)^(1/2)/(a+b)^4/f-1/3*cot(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^( 
1/2)/(a+b)^3/f
 

Mathematica [A] (verified)

Time = 4.27 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.86 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {(a+2 b+a \cos (2 (e+f x)))^3 \left (\frac {4 b^2 (a+b)}{(a+2 b+a \cos (2 (e+f x)))^2}+\frac {4 b (-3 a+b)}{a+2 b+a \cos (2 (e+f x))}-2 (a-3 b) \csc ^2(e+f x)-(a+b) \csc ^4(e+f x)\right ) \sec ^4(e+f x) \tan (e+f x)}{24 (a+b)^4 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \] Input:

Integrate[Csc[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

Output:

((a + 2*b + a*Cos[2*(e + f*x)])^3*((4*b^2*(a + b))/(a + 2*b + a*Cos[2*(e + 
 f*x)])^2 + (4*b*(-3*a + b))/(a + 2*b + a*Cos[2*(e + f*x)]) - 2*(a - 3*b)* 
Csc[e + f*x]^2 - (a + b)*Csc[e + f*x]^4)*Sec[e + f*x]^4*Tan[e + f*x])/(24* 
(a + b)^4*f*(a + b*Sec[e + f*x]^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4620, 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.

\(\displaystyle \int \frac {\csc ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x)^4 \left (a+b \sec (e+f x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 4620

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

\(\Big \downarrow \) 359

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

\(\Big \downarrow \) 245

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

\(\Big \downarrow \) 209

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

\(\Big \downarrow \) 208

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

Input:

Int[Csc[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

Output:

(-1/3*Cot[e + f*x]^3/((a + b)*(a + b + b*Tan[e + f*x]^2)^(3/2)) + ((a - b) 
*(-(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))/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 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 = 11.16 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.29

method	result	size
default	\(\frac {\left (b +a \cos \left (f x +e \right )^{2}\right ) \left (\left (2 \cos \left (f x +e \right )^{6}-21 \cos \left (f x +e \right )^{4}+24 \cos \left (f x +e \right )^{2}\right ) a \,b^{2}+\left (-12 \cos \left (f x +e \right )^{6}+21 \cos \left (f x +e \right )^{4}-12 \cos \left (f x +e \right )^{2}\right ) b \,a^{2}-8 a \,b^{2}+\left (2 \cos \left (f x +e \right )^{6}-3 \cos \left (f x +e \right )^{4}\right ) a^{3}+\left (3 \cos \left (f x +e \right )^{4}-12 \cos \left (f x +e \right )^{2}+8\right ) b^{3}\right ) \sec \left (f x +e \right )^{5} \csc \left (f x +e \right )^{3}}{3 f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}\)	\(207\)

Input:

int(csc(f*x+e)^4/(a+b*sec(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/3/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)*((2*cos(f*x+e 
)^6-21*cos(f*x+e)^4+24*cos(f*x+e)^2)*a*b^2+(-12*cos(f*x+e)^6+21*cos(f*x+e) 
^4-12*cos(f*x+e)^2)*b*a^2-8*a*b^2+(2*cos(f*x+e)^6-3*cos(f*x+e)^4)*a^3+(3*c 
os(f*x+e)^4-12*cos(f*x+e)^2+8)*b^3)/(a+b*sec(f*x+e)^2)^(5/2)*sec(f*x+e)^5* 
csc(f*x+e)^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (144) = 288\).

Time = 2.07 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.00 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {{\left (2 \, {\left (a^{3} - 6 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{7} - 3 \, {\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{5} - 12 \, {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} - 8 \, {\left (a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left ({\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} f \cos \left (f x + e\right )^{6} - {\left (a^{6} + 2 \, a^{5} b - 2 \, a^{4} b^{2} - 8 \, a^{3} b^{3} - 7 \, a^{2} b^{4} - 2 \, a b^{5}\right )} f \cos \left (f x + e\right )^{4} - {\left (2 \, a^{5} b + 7 \, a^{4} b^{2} + 8 \, a^{3} b^{3} + 2 \, a^{2} b^{4} - 2 \, a b^{5} - b^{6}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{4} b^{2} + 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} + 4 \, a b^{5} + b^{6}\right )} f\right )} \sin \left (f x + e\right )} \] Input:

integrate(csc(f*x+e)^4/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")
 

Output:

-1/3*(2*(a^3 - 6*a^2*b + a*b^2)*cos(f*x + e)^7 - 3*(a^3 - 7*a^2*b + 7*a*b^ 
2 - b^3)*cos(f*x + e)^5 - 12*(a^2*b - 2*a*b^2 + b^3)*cos(f*x + e)^3 - 8*(a 
*b^2 - b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(((a 
^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*f*cos(f*x + e)^6 - (a^6 + 
2*a^5*b - 2*a^4*b^2 - 8*a^3*b^3 - 7*a^2*b^4 - 2*a*b^5)*f*cos(f*x + e)^4 - 
(2*a^5*b + 7*a^4*b^2 + 8*a^3*b^3 + 2*a^2*b^4 - 2*a*b^5 - b^6)*f*cos(f*x + 
e)^2 - (a^4*b^2 + 4*a^3*b^3 + 6*a^2*b^4 + 4*a*b^5 + b^6)*f)*sin(f*x + e))
 

Sympy [F]

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

integrate(csc(f*x+e)**4/(a+b*sec(f*x+e)**2)**(5/2),x)
 

Output:

Integral(csc(e + f*x)**4/(a + b*sec(e + f*x)**2)**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.35 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {8 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{3}} + \frac {4 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{2}} - \frac {16 \, b^{2} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{4}} - \frac {8 \, b^{2} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{3}} + \frac {3}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )} - \frac {6 \, b}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{2} \tan \left (f x + e\right )} + \frac {1}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )^{3}}}{3 \, f} \] Input:

integrate(csc(f*x+e)^4/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")
 

Output:

-1/3*(8*b*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^3) + 4*b*ta 
n(f*x + e)/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)^2) - 16*b^2*tan(f*x + 
 e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^4) - 8*b^2*tan(f*x + e)/((b*ta 
n(f*x + e)^2 + a + b)^(3/2)*(a + b)^3) + 3/((b*tan(f*x + e)^2 + a + b)^(3/ 
2)*(a + b)*tan(f*x + e)) - 6*b/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)^2 
*tan(f*x + e)) + 1/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)*tan(f*x + e)^ 
3))/f
 

Giac [F]

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

integrate(csc(f*x+e)^4/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")
 

Output:

integrate(csc(f*x + e)^4/(b*sec(f*x + e)^2 + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Hanged} \] Input:

int(1/(sin(e + f*x)^4*(a + b/cos(e + f*x)^2)^(5/2)),x)
 

Output:

\text{Hanged}
                                                                                    
                                                                                    
 

Reduce [F]

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

int(csc(f*x+e)^4/(a+b*sec(f*x+e)^2)^(5/2),x)
 

Output:

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