\(\int \frac {\csc ^2(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [129]

Optimal result

Integrand size = 25, antiderivative size = 106 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\cot (e+f x)}{(a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {4 b \tan (e+f x)}{3 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {8 b \tan (e+f x)}{3 (a+b)^3 f \sqrt {a+b+b \tan ^2(e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.55 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.02 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (3 a^2-6 a b-b^2-6 \left (a^2-b^2\right ) \csc ^2(e+f x)+3 (a+b)^2 \csc ^4(e+f x)\right ) \sec ^2(e+f x) \tan ^3(e+f x)}{6 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 108, 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, 4620, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4620

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

\(\Big \downarrow \) 245

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

\(\Big \downarrow \) 209

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

\(\Big \downarrow \) 208

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

Input:

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

Output:

(-(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))/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 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 = 8.29 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.24

method	result	size
default	\(-\frac {\left (b +a \cos \left (f x +e \right )^{2}\right ) \left (\left (-6 \cos \left (f x +e \right )^{4}+12 \cos \left (f x +e \right )^{2}\right ) a b +3 \cos \left (f x +e \right )^{4} a^{2}+\left (-\cos \left (f x +e \right )^{4}-4 \cos \left (f x +e \right )^{2}+8\right ) b^{2}\right ) \sec \left (f x +e \right )^{5} \csc \left (f x +e \right )}{3 f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}\)	\(131\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.81 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {{\left ({\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{5} + 4 \, {\left (3 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} + 8 \, b^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left ({\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} f\right )} \sin \left (f x + e\right )} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.89 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {8 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{3}} + \frac {4 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{2}} + \frac {3}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )}}{3 \, f} \] Input:

integrate(csc(f*x+e)^2/(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) + 3/((b*tan(f*x + 
e)^2 + a + b)^(3/2)*(a + b)*tan(f*x + e)))/f
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 23.91 (sec) , antiderivative size = 336, normalized size of antiderivative = 3.17 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )\,\sqrt {a+\frac {b}{{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}\right )}^2}}\,\left (-a\,b\,6{}\mathrm {i}+a^2\,3{}\mathrm {i}-b^2\,1{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,12{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,18{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,12{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,3{}\mathrm {i}-b^2\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,20{}\mathrm {i}+b^2\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,90{}\mathrm {i}-b^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,20{}\mathrm {i}-b^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,1{}\mathrm {i}+a\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,24{}\mathrm {i}+a\,b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,60{}\mathrm {i}+a\,b\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,24{}\mathrm {i}-a\,b\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,6{}\mathrm {i}\right )}{3\,f\,{\left (a+b\right )}^3\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )\,{\left (a+2\,a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}+4\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\right )}^2} \] Input:

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

Output:

-((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*3i - a*b*6i - b^2*1i + a^2*exp(e*2i + f*x*2i)*12i + 
a^2*exp(e*4i + f*x*4i)*18i + a^2*exp(e*6i + f*x*6i)*12i + a^2*exp(e*8i + f 
*x*8i)*3i - b^2*exp(e*2i + f*x*2i)*20i + b^2*exp(e*4i + f*x*4i)*90i - b^2* 
exp(e*6i + f*x*6i)*20i - b^2*exp(e*8i + f*x*8i)*1i + a*b*exp(e*2i + f*x*2i 
)*24i + a*b*exp(e*4i + f*x*4i)*60i + a*b*exp(e*6i + f*x*6i)*24i - a*b*exp( 
e*8i + f*x*8i)*6i))/(3*f*(a + b)^3*(exp(e*2i + f*x*2i) - 1)*(a + 2*a*exp(e 
*2i + f*x*2i) + a*exp(e*4i + f*x*4i) + 4*b*exp(e*2i + f*x*2i))^2)
 

Reduce [F]

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

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

Output:

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