\(\int \frac {\csc ^2(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\) [149]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [F]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 25, antiderivative size = 97 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\cot (e+f x)}{a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {4 b \tan (e+f x)}{3 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {8 b \tan (e+f x)}{3 a^3 f \sqrt {a+b \tan ^2(e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 2.41 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.37 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\left (3 \left (3 a^2+4 a b+8 b^2\right )+4 \left (3 a^2-8 b^2\right ) \cos (2 (e+f x))+\left (3 a^2-12 a b+8 b^2\right ) \cos (4 (e+f x))\right ) \cot (e+f x) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{6 \sqrt {2} a^3 f (a+b+(a-b) \cos (2 (e+f x)))^2} \] Input:

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

Output:

-1/6*((3*(3*a^2 + 4*a*b + 8*b^2) + 4*(3*a^2 - 8*b^2)*Cos[2*(e + f*x)] + (3 
*a^2 - 12*a*b + 8*b^2)*Cos[4*(e + f*x)])*Cot[e + f*x]*Sqrt[(a + b + (a - b 
)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(Sqrt[2]*a^3*f*(a + b + (a - b)*Cos[2 
*(e + f*x)])^2)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4146, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4146

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

\(\Big \downarrow \) 245

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

\(\Big \downarrow \) 209

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

\(\Big \downarrow \) 208

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

Input:

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

Output:

(-(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)/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 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 = 0.78 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.93

method result size
derivativedivides \(\frac {-\frac {1}{a \tan \left (f x +e \right ) \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {4 b \left (\frac {\tan \left (f x +e \right )}{3 a \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \tan \left (f x +e \right )}{3 a^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{a}}{f}\) \(90\)
default \(\frac {-\frac {1}{a \tan \left (f x +e \right ) \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {4 b \left (\frac {\tan \left (f x +e \right )}{3 a \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \tan \left (f x +e \right )}{3 a^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{a}}{f}\) \(90\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 4.85 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.61 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {{\left ({\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{5} + 4 \, {\left (3 \, a b - 4 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 8 \, b^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left (a^{3} b^{2} f + {\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b - a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.88 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {8 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{3}} + \frac {4 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2}} + \frac {3}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \tan \left (f x + e\right )}}{3 \, f} \] Input:

integrate(csc(f*x+e)^2/(a+b*tan(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)*a^3) + 4*b*tan(f*x + e) 
/((b*tan(f*x + e)^2 + a)^(3/2)*a^2) + 3/((b*tan(f*x + e)^2 + a)^(3/2)*a*ta 
n(f*x + e)))/f
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 22.15 (sec) , antiderivative size = 324, normalized size of antiderivative = 3.34 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )\,\sqrt {a+\frac {b\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^2}{{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2}}\,\left (-a\,b\,12{}\mathrm {i}+a^2\,3{}\mathrm {i}+b^2\,8{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,12{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,18{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,12{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,3{}\mathrm {i}-b^2\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,32{}\mathrm {i}+b^2\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,48{}\mathrm {i}-b^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,32{}\mathrm {i}+b^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,8{}\mathrm {i}+a\,b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,24{}\mathrm {i}-a\,b\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,12{}\mathrm {i}\right )}{3\,a^3\,f\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )\,{\left (a-b+2\,a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}+2\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\right )}^2} \] Input:

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

Output:

-((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)*(a^2*3i - a*b*12i + b^2*8i + 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*e 
xp(e*8i + f*x*8i)*3i - b^2*exp(e*2i + f*x*2i)*32i + b^2*exp(e*4i + f*x*4i) 
*48i - b^2*exp(e*6i + f*x*6i)*32i + b^2*exp(e*8i + f*x*8i)*8i + a*b*exp(e* 
4i + f*x*4i)*24i - a*b*exp(e*8i + f*x*8i)*12i))/(3*a^3*f*(exp(e*2i + f*x*2 
i) - 1)*(a - b + 2*a*exp(e*2i + f*x*2i) + a*exp(e*4i + f*x*4i) + 2*b*exp(e 
*2i + f*x*2i) - b*exp(e*4i + f*x*4i))^2)
 

Reduce [F]

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

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

Output:

int((sqrt(tan(e + f*x)**2*b + a)*csc(e + f*x)**2)/(tan(e + f*x)**6*b**3 + 
3*tan(e + f*x)**4*a*b**2 + 3*tan(e + f*x)**2*a**2*b + a**3),x)