\(\int \frac {\csc ^5(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\) [46]

Optimal result

Integrand size = 23, antiderivative size = 197 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {3 \sqrt {a} (a-b) \sqrt {b} \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{2 (a+b)^4 f}-\frac {3 \left (a^2-6 a b+b^2\right ) \text {arctanh}(\cos (e+f x))}{8 (a+b)^4 f}+\frac {3 a (a-3 b) \cos (e+f x)}{8 (a+b)^3 f \left (b+a \cos ^2(e+f x)\right )}-\frac {(a-5 b) \cot (e+f x) \csc (e+f x)}{8 (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )}-\frac {\cot (e+f x) \csc ^3(e+f x)}{4 (a+b) f \left (b+a \cos ^2(e+f x)\right )} \] Output:

3/2*a^(1/2)*(a-b)*b^(1/2)*arctan(a^(1/2)*cos(f*x+e)/b^(1/2))/(a+b)^4/f-3/8 
*(a^2-6*a*b+b^2)*arctanh(cos(f*x+e))/(a+b)^4/f+3/8*a*(a-3*b)*cos(f*x+e)/(a 
+b)^3/f/(b+a*cos(f*x+e)^2)-1/8*(a-5*b)*cot(f*x+e)*csc(f*x+e)/(a+b)^2/f/(b+ 
a*cos(f*x+e)^2)-1/4*cot(f*x+e)*csc(f*x+e)^3/(a+b)/f/(b+a*cos(f*x+e)^2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.66 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.28 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \left (96 \sqrt {a} (a-b) \sqrt {b} \arctan \left (\frac {\left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right ) (a+2 b+a \cos (2 (e+f x)))+96 \sqrt {a} (a-b) \sqrt {b} \arctan \left (\frac {\left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right ) (a+2 b+a \cos (2 (e+f x)))-2 (a+b) \left (11 a^2+43 a b-4 b^2+4 \left (2 a^2-5 a b+5 b^2\right ) \cos (2 (e+f x))-3 a (a-3 b) \cos (4 (e+f x))\right ) \cot (e+f x) \csc ^3(e+f x)-24 \left (a^2-6 a b+b^2\right ) (a+2 b+a \cos (2 (e+f x))) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )+24 \left (a^2-6 a b+b^2\right ) (a+2 b+a \cos (2 (e+f x))) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sec ^4(e+f x)}{256 (a+b)^4 f \left (a+b \sec ^2(e+f x)\right )^2} \] Input:

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

Output:

((a + 2*b + a*Cos[2*(e + f*x)])*(96*Sqrt[a]*(a - b)*Sqrt[b]*ArcTan[((-Sqrt 
[a] - I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos 
[e]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt 
[b]]*(a + 2*b + a*Cos[2*(e + f*x)]) + 96*Sqrt[a]*(a - b)*Sqrt[b]*ArcTan[(( 
-Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)/2] 
+ Cos[e]*(Sqrt[a] + Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2])) 
/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)]) - 2*(a + b)*(11*a^2 + 43*a*b - 4* 
b^2 + 4*(2*a^2 - 5*a*b + 5*b^2)*Cos[2*(e + f*x)] - 3*a*(a - 3*b)*Cos[4*(e 
+ f*x)])*Cot[e + f*x]*Csc[e + f*x]^3 - 24*(a^2 - 6*a*b + b^2)*(a + 2*b + a 
*Cos[2*(e + f*x)])*Log[Cos[(e + f*x)/2]] + 24*(a^2 - 6*a*b + b^2)*(a + 2*b 
 + a*Cos[2*(e + f*x)])*Log[Sin[(e + f*x)/2]])*Sec[e + f*x]^4)/(256*(a + b) 
^4*f*(a + b*Sec[e + f*x]^2)^2)
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 4621, 372, 402, 27, 402, 27, 397, 218, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4621

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 397

\(\displaystyle -\frac {\frac {\cos (e+f x)}{4 (a+b) \left (1-\cos ^2(e+f x)\right )^2 \left (a \cos ^2(e+f x)+b\right )}-\frac {\frac {3 \left (\frac {\frac {4 a b (a-b) \int \frac {1}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{a+b}-\frac {\left (a^2-6 a b+b^2\right ) \int \frac {1}{1-\cos ^2(e+f x)}d\cos (e+f x)}{a+b}}{a+b}+\frac {a (a-3 b) \cos (e+f x)}{(a+b) \left (a \cos ^2(e+f x)+b\right )}\right )}{2 (a+b)}-\frac {(a-5 b) \cos (e+f x)}{2 (a+b) \left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )}}{4 (a+b)}}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {\frac {\cos (e+f x)}{4 (a+b) \left (1-\cos ^2(e+f x)\right )^2 \left (a \cos ^2(e+f x)+b\right )}-\frac {\frac {3 \left (\frac {\frac {4 \sqrt {a} \sqrt {b} (a-b) \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{a+b}-\frac {\left (a^2-6 a b+b^2\right ) \int \frac {1}{1-\cos ^2(e+f x)}d\cos (e+f x)}{a+b}}{a+b}+\frac {a (a-3 b) \cos (e+f x)}{(a+b) \left (a \cos ^2(e+f x)+b\right )}\right )}{2 (a+b)}-\frac {(a-5 b) \cos (e+f x)}{2 (a+b) \left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )}}{4 (a+b)}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\frac {\cos (e+f x)}{4 (a+b) \left (1-\cos ^2(e+f x)\right )^2 \left (a \cos ^2(e+f x)+b\right )}-\frac {\frac {3 \left (\frac {\frac {4 \sqrt {a} \sqrt {b} (a-b) \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{a+b}-\frac {\left (a^2-6 a b+b^2\right ) \text {arctanh}(\cos (e+f x))}{a+b}}{a+b}+\frac {a (a-3 b) \cos (e+f x)}{(a+b) \left (a \cos ^2(e+f x)+b\right )}\right )}{2 (a+b)}-\frac {(a-5 b) \cos (e+f x)}{2 (a+b) \left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )}}{4 (a+b)}}{f}\)

Input:

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

Output:

-((Cos[e + f*x]/(4*(a + b)*(1 - Cos[e + f*x]^2)^2*(b + a*Cos[e + f*x]^2)) 
- (-1/2*((a - 5*b)*Cos[e + f*x])/((a + b)*(1 - Cos[e + f*x]^2)*(b + a*Cos[ 
e + f*x]^2)) + (3*(((4*Sqrt[a]*(a - b)*Sqrt[b]*ArcTan[(Sqrt[a]*Cos[e + f*x 
])/Sqrt[b]])/(a + b) - ((a^2 - 6*a*b + b^2)*ArcTanh[Cos[e + f*x]])/(a + b) 
)/(a + b) + (a*(a - 3*b)*Cos[e + f*x])/((a + b)*(b + a*Cos[e + f*x]^2))))/ 
(2*(a + b)))/(4*(a + b)))/f)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 372

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 
)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 
))   Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + 
 (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, 
e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a 
, b, c, d, e, m, 2, p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4621

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

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {\frac {a b \left (\frac {\left (-\frac {a}{2}-\frac {b}{2}\right ) \cos \left (f x +e \right )}{b +a \cos \left (f x +e \right )^{2}}+\frac {3 \left (a -b \right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{\left (a +b \right )^{4}}+\frac {1}{16 \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )^{2}}-\frac {-3 a +5 b}{16 \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-3 a^{2}+18 a b -3 b^{2}\right ) \ln \left (1+\cos \left (f x +e \right )\right )}{16 \left (a +b \right )^{4}}-\frac {1}{16 \left (a +b \right )^{2} \left (-1+\cos \left (f x +e \right )\right )^{2}}-\frac {-3 a +5 b}{16 \left (a +b \right )^{3} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (3 a^{2}-18 a b +3 b^{2}\right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{16 \left (a +b \right )^{4}}}{f}\)	\(213\)
default	\(\frac {\frac {a b \left (\frac {\left (-\frac {a}{2}-\frac {b}{2}\right ) \cos \left (f x +e \right )}{b +a \cos \left (f x +e \right )^{2}}+\frac {3 \left (a -b \right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{\left (a +b \right )^{4}}+\frac {1}{16 \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )^{2}}-\frac {-3 a +5 b}{16 \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-3 a^{2}+18 a b -3 b^{2}\right ) \ln \left (1+\cos \left (f x +e \right )\right )}{16 \left (a +b \right )^{4}}-\frac {1}{16 \left (a +b \right )^{2} \left (-1+\cos \left (f x +e \right )\right )^{2}}-\frac {-3 a +5 b}{16 \left (a +b \right )^{3} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (3 a^{2}-18 a b +3 b^{2}\right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{16 \left (a +b \right )^{4}}}{f}\)	\(213\)
risch	\(-\frac {-3 a^{2} {\mathrm e}^{11 i \left (f x +e \right )}+9 a b \,{\mathrm e}^{11 i \left (f x +e \right )}+5 a^{2} {\mathrm e}^{9 i \left (f x +e \right )}-11 a b \,{\mathrm e}^{9 i \left (f x +e \right )}+20 b^{2} {\mathrm e}^{9 i \left (f x +e \right )}+30 a^{2} {\mathrm e}^{7 i \left (f x +e \right )}+66 a b \,{\mathrm e}^{7 i \left (f x +e \right )}+12 b^{2} {\mathrm e}^{7 i \left (f x +e \right )}+30 a^{2} {\mathrm e}^{5 i \left (f x +e \right )}+66 a b \,{\mathrm e}^{5 i \left (f x +e \right )}+12 b^{2} {\mathrm e}^{5 i \left (f x +e \right )}+5 a^{2} {\mathrm e}^{3 i \left (f x +e \right )}-11 a b \,{\mathrm e}^{3 i \left (f x +e \right )}+20 b^{2} {\mathrm e}^{3 i \left (f x +e \right )}-3 a^{2} {\mathrm e}^{i \left (f x +e \right )}+9 a b \,{\mathrm e}^{i \left (f x +e \right )}}{4 f \left (a +b \right )^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) a^{2}}{8 f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {9 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) a b}{4 f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b^{2}}{8 f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) a^{2}}{8 f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}+\frac {9 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) a b}{4 f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b^{2}}{8 f \left (a^{4}+4 b \,a^{3}+6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}+\frac {3 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) a}{4 \left (a +b \right )^{4} f}-\frac {3 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b}{4 \left (a +b \right )^{4} f}-\frac {3 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) a}{4 \left (a +b \right )^{4} f}+\frac {3 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b}{4 \left (a +b \right )^{4} f}\)	\(771\)

Input:

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

Output:

1/f*(a*b/(a+b)^4*((-1/2*a-1/2*b)*cos(f*x+e)/(b+a*cos(f*x+e)^2)+3/2*(a-b)/( 
a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2)))+1/16/(a+b)^2/(1+cos(f*x+e))^2 
-1/16*(-3*a+5*b)/(a+b)^3/(1+cos(f*x+e))+1/16/(a+b)^4*(-3*a^2+18*a*b-3*b^2) 
*ln(1+cos(f*x+e))-1/16/(a+b)^2/(-1+cos(f*x+e))^2-1/16*(-3*a+5*b)/(a+b)^3/( 
-1+cos(f*x+e))+1/16/(a+b)^4*(3*a^2-18*a*b+3*b^2)*ln(-1+cos(f*x+e)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 586 vs. \(2 (179) = 358\).

Time = 0.20 (sec) , antiderivative size = 1202, normalized size of antiderivative = 6.10 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/16*(6*(a^3 - 2*a^2*b - 3*a*b^2)*cos(f*x + e)^5 - 2*(5*a^3 - 9*a^2*b - 9 
*a*b^2 + 5*b^3)*cos(f*x + e)^3 - 12*((a^2 - a*b)*cos(f*x + e)^6 - (2*a^2 - 
 3*a*b + b^2)*cos(f*x + e)^4 + (a^2 - 3*a*b + 2*b^2)*cos(f*x + e)^2 + a*b 
- b^2)*sqrt(-a*b)*log((a*cos(f*x + e)^2 - 2*sqrt(-a*b)*cos(f*x + e) - b)/( 
a*cos(f*x + e)^2 + b)) - 6*(3*a^2*b + 2*a*b^2 - b^3)*cos(f*x + e) - 3*((a^ 
3 - 6*a^2*b + a*b^2)*cos(f*x + e)^6 - (2*a^3 - 13*a^2*b + 8*a*b^2 - b^3)*c 
os(f*x + e)^4 + a^2*b - 6*a*b^2 + b^3 + (a^3 - 8*a^2*b + 13*a*b^2 - 2*b^3) 
*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2) + 3*((a^3 - 6*a^2*b + a*b^2)* 
cos(f*x + e)^6 - (2*a^3 - 13*a^2*b + 8*a*b^2 - b^3)*cos(f*x + e)^4 + a^2*b 
 - 6*a*b^2 + b^3 + (a^3 - 8*a^2*b + 13*a*b^2 - 2*b^3)*cos(f*x + e)^2)*log( 
-1/2*cos(f*x + e) + 1/2))/((a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4) 
*f*cos(f*x + e)^6 - (2*a^5 + 7*a^4*b + 8*a^3*b^2 + 2*a^2*b^3 - 2*a*b^4 - b 
^5)*f*cos(f*x + e)^4 + (a^5 + 2*a^4*b - 2*a^3*b^2 - 8*a^2*b^3 - 7*a*b^4 - 
2*b^5)*f*cos(f*x + e)^2 + (a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)* 
f), 1/16*(6*(a^3 - 2*a^2*b - 3*a*b^2)*cos(f*x + e)^5 - 2*(5*a^3 - 9*a^2*b 
- 9*a*b^2 + 5*b^3)*cos(f*x + e)^3 + 24*((a^2 - a*b)*cos(f*x + e)^6 - (2*a^ 
2 - 3*a*b + b^2)*cos(f*x + e)^4 + (a^2 - 3*a*b + 2*b^2)*cos(f*x + e)^2 + a 
*b - b^2)*sqrt(a*b)*arctan(sqrt(a*b)*cos(f*x + e)/b) - 6*(3*a^2*b + 2*a*b^ 
2 - b^3)*cos(f*x + e) - 3*((a^3 - 6*a^2*b + a*b^2)*cos(f*x + e)^6 - (2*a^3 
 - 13*a^2*b + 8*a*b^2 - b^3)*cos(f*x + e)^4 + a^2*b - 6*a*b^2 + b^3 + (...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 369 vs. \(2 (179) = 358\).

Time = 0.12 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.87 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {3 \, {\left (a^{2} - 6 \, a b + b^{2}\right )} \log \left (\cos \left (f x + e\right ) + 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac {3 \, {\left (a^{2} - 6 \, a b + b^{2}\right )} \log \left (\cos \left (f x + e\right ) - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac {24 \, {\left (a^{2} b - a b^{2}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sqrt {a b}} - \frac {2 \, {\left (3 \, {\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{5} - {\left (5 \, a^{2} - 14 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (3 \, a b - b^{2}\right )} \cos \left (f x + e\right )\right )}}{{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (2 \, a^{4} + 5 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4} + {\left (a^{4} + a^{3} b - 3 \, a^{2} b^{2} - 5 \, a b^{3} - 2 \, b^{4}\right )} \cos \left (f x + e\right )^{2}}}{16 \, f} \] Input:

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

Output:

-1/16*(3*(a^2 - 6*a*b + b^2)*log(cos(f*x + e) + 1)/(a^4 + 4*a^3*b + 6*a^2* 
b^2 + 4*a*b^3 + b^4) - 3*(a^2 - 6*a*b + b^2)*log(cos(f*x + e) - 1)/(a^4 + 
4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - 24*(a^2*b - a*b^2)*arctan(a*cos(f*x 
 + e)/sqrt(a*b))/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*sqrt(a*b)) - 
 2*(3*(a^2 - 3*a*b)*cos(f*x + e)^5 - (5*a^2 - 14*a*b + 5*b^2)*cos(f*x + e) 
^3 - 3*(3*a*b - b^2)*cos(f*x + e))/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*co 
s(f*x + e)^6 - (2*a^4 + 5*a^3*b + 3*a^2*b^2 - a*b^3 - b^4)*cos(f*x + e)^4 
+ a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4 + (a^4 + a^3*b - 3*a^2*b^2 - 5*a*b^3 - 
 2*b^4)*cos(f*x + e)^2))/f
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.65 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {a b \cos \left (f x + e\right )}{2 \, {\left (a^{3} f + 3 \, a^{2} b f + 3 \, a b^{2} f + b^{3} f\right )} {\left (a \cos \left (f x + e\right )^{2} + b\right )}} + \frac {3 \, {\left (a^{2} - 6 \, a b + b^{2}\right )} \log \left ({\left | -\cos \left (f x + e\right ) + 1 \right |}\right )}{16 \, {\left (a^{4} f + 4 \, a^{3} b f + 6 \, a^{2} b^{2} f + 4 \, a b^{3} f + b^{4} f\right )}} - \frac {3 \, {\left (a^{2} - 6 \, a b + b^{2}\right )} \log \left ({\left | -\cos \left (f x + e\right ) - 1 \right |}\right )}{16 \, {\left (a^{4} f + 4 \, a^{3} b f + 6 \, a^{2} b^{2} f + 4 \, a b^{3} f + b^{4} f\right )}} + \frac {3 \, {\left (a^{2} b - a b^{2}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{2 \, {\left (a^{4} f + 4 \, a^{3} b f + 6 \, a^{2} b^{2} f + 4 \, a b^{3} f + b^{4} f\right )} \sqrt {a b}} + \frac {3 \, a \cos \left (f x + e\right )^{3} - 5 \, b \cos \left (f x + e\right )^{3} - 5 \, a \cos \left (f x + e\right ) + 3 \, b \cos \left (f x + e\right )}{8 \, {\left (a^{3} f + 3 \, a^{2} b f + 3 \, a b^{2} f + b^{3} f\right )} {\left (\cos \left (f x + e\right )^{2} - 1\right )}^{2}} \] Input:

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

Output:

-1/2*a*b*cos(f*x + e)/((a^3*f + 3*a^2*b*f + 3*a*b^2*f + b^3*f)*(a*cos(f*x 
+ e)^2 + b)) + 3/16*(a^2 - 6*a*b + b^2)*log(abs(-cos(f*x + e) + 1))/(a^4*f 
 + 4*a^3*b*f + 6*a^2*b^2*f + 4*a*b^3*f + b^4*f) - 3/16*(a^2 - 6*a*b + b^2) 
*log(abs(-cos(f*x + e) - 1))/(a^4*f + 4*a^3*b*f + 6*a^2*b^2*f + 4*a*b^3*f 
+ b^4*f) + 3/2*(a^2*b - a*b^2)*arctan(a*cos(f*x + e)/sqrt(a*b))/((a^4*f + 
4*a^3*b*f + 6*a^2*b^2*f + 4*a*b^3*f + b^4*f)*sqrt(a*b)) + 1/8*(3*a*cos(f*x 
 + e)^3 - 5*b*cos(f*x + e)^3 - 5*a*cos(f*x + e) + 3*b*cos(f*x + e))/((a^3* 
f + 3*a^2*b*f + 3*a*b^2*f + b^3*f)*(cos(f*x + e)^2 - 1)^2)
 

Mupad [B] (verification not implemented)

Time = 17.14 (sec) , antiderivative size = 4338, normalized size of antiderivative = 22.02 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan((((-a*b)^(1/2)*((cos(e + f*x)*(9*a^7 - 108*a^6*b + 153*a^3*b^4 - 396 
*a^4*b^3 + 486*a^5*b^2))/(32*(6*a*b^5 + 6*a^5*b + a^6 + b^6 + 15*a^2*b^4 + 
 20*a^3*b^3 + 15*a^4*b^2)) + (3*(-a*b)^(1/2)*(a - b)*(((9*a^11*b)/2 - (3*a 
^2*b^10)/2 - (15*a^3*b^9)/2 - 6*a^4*b^8 + 42*a^5*b^7 + 147*a^6*b^6 + 231*a 
^7*b^5 + 210*a^8*b^4 + 114*a^9*b^3 + (69*a^10*b^2)/2)/(9*a*b^8 + 9*a^8*b + 
 a^9 + b^9 + 36*a^2*b^7 + 84*a^3*b^6 + 126*a^4*b^5 + 126*a^5*b^4 + 84*a^6* 
b^3 + 36*a^7*b^2) - (3*cos(e + f*x)*(-a*b)^(1/2)*(a - b)*(1792*a^10*b + 25 
6*a^11 - 256*a^2*b^9 - 1792*a^3*b^8 - 5120*a^4*b^7 - 7168*a^5*b^6 - 3584*a 
^6*b^5 + 3584*a^7*b^4 + 7168*a^8*b^3 + 5120*a^9*b^2))/(128*(4*a*b^3 + 4*a^ 
3*b + a^4 + b^4 + 6*a^2*b^2)*(6*a*b^5 + 6*a^5*b + a^6 + b^6 + 15*a^2*b^4 + 
 20*a^3*b^3 + 15*a^4*b^2))))/(4*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2 
)))*(a - b)*3i)/(4*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2)) + ((-a*b)^ 
(1/2)*((cos(e + f*x)*(9*a^7 - 108*a^6*b + 153*a^3*b^4 - 396*a^4*b^3 + 486* 
a^5*b^2))/(32*(6*a*b^5 + 6*a^5*b + a^6 + b^6 + 15*a^2*b^4 + 20*a^3*b^3 + 1 
5*a^4*b^2)) - (3*(-a*b)^(1/2)*(a - b)*(((9*a^11*b)/2 - (3*a^2*b^10)/2 - (1 
5*a^3*b^9)/2 - 6*a^4*b^8 + 42*a^5*b^7 + 147*a^6*b^6 + 231*a^7*b^5 + 210*a^ 
8*b^4 + 114*a^9*b^3 + (69*a^10*b^2)/2)/(9*a*b^8 + 9*a^8*b + a^9 + b^9 + 36 
*a^2*b^7 + 84*a^3*b^6 + 126*a^4*b^5 + 126*a^5*b^4 + 84*a^6*b^3 + 36*a^7*b^ 
2) + (3*cos(e + f*x)*(-a*b)^(1/2)*(a - b)*(1792*a^10*b + 256*a^11 - 256*a^ 
2*b^9 - 1792*a^3*b^8 - 5120*a^4*b^7 - 7168*a^5*b^6 - 3584*a^6*b^5 + 358...
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 895, normalized size of antiderivative = 4.54 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 96*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt( 
b))*sin(e + f*x)**6*a**2 + 96*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f 
*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**6*a*b + 96*sqrt(b)*sqrt(a)*atan(( 
sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**2 - 96 
*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*si 
n(e + f*x)**4*b**2 + 96*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) 
 + sqrt(a))/sqrt(b))*sin(e + f*x)**6*a**2 - 96*sqrt(b)*sqrt(a)*atan((sqrt( 
a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**6*a*b - 96*sqrt( 
b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + 
f*x)**4*a**2 + 96*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqr 
t(a))/sqrt(b))*sin(e + f*x)**4*b**2 - 24*cos(e + f*x)*sin(e + f*x)**4*a**3 
 + 48*cos(e + f*x)*sin(e + f*x)**4*a**2*b + 72*cos(e + f*x)*sin(e + f*x)** 
4*a*b**2 + 8*cos(e + f*x)*sin(e + f*x)**2*a**3 - 24*cos(e + f*x)*sin(e + f 
*x)**2*a**2*b - 72*cos(e + f*x)*sin(e + f*x)**2*a*b**2 - 40*cos(e + f*x)*s 
in(e + f*x)**2*b**3 + 16*cos(e + f*x)*a**3 + 48*cos(e + f*x)*a**2*b + 48*c 
os(e + f*x)*a*b**2 + 16*cos(e + f*x)*b**3 + 24*log(tan((e + f*x)/2))*sin(e 
 + f*x)**6*a**3 - 144*log(tan((e + f*x)/2))*sin(e + f*x)**6*a**2*b + 24*lo 
g(tan((e + f*x)/2))*sin(e + f*x)**6*a*b**2 - 24*log(tan((e + f*x)/2))*sin( 
e + f*x)**4*a**3 + 120*log(tan((e + f*x)/2))*sin(e + f*x)**4*a**2*b + 120* 
log(tan((e + f*x)/2))*sin(e + f*x)**4*a*b**2 - 24*log(tan((e + f*x)/2))...