3.1.58 \(\int \frac {\csc ^3(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [58]

3.1.58.1 Optimal result
3.1.58.2 Mathematica [C] (warning: unable to verify)
3.1.58.3 Rubi [A] (verified)
3.1.58.4 Maple [A] (verified)
3.1.58.5 Fricas [B] (verification not implemented)
3.1.58.6 Sympy [F(-1)]
3.1.58.7 Maxima [B] (verification not implemented)
3.1.58.8 Giac [B] (verification not implemented)
3.1.58.9 Mupad [B] (verification not implemented)

3.1.58.1 Optimal result

Integrand size = 23, antiderivative size = 213 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\sqrt {b} \left (15 a^2-10 a b-b^2\right ) \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{8 a^{3/2} (a+b)^4 f}-\frac {(a-5 b) \text {arctanh}(\cos (e+f x))}{2 (a+b)^4 f}-\frac {(2 a-b) b \cos (e+f x)}{4 a (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )^2}+\frac {\left (4 a^2-9 a b-b^2\right ) \cos (e+f x)}{8 a (a+b)^3 f \left (b+a \cos ^2(e+f x)\right )}-\frac {\cos (e+f x) \cot ^2(e+f x)}{2 (a+b) f \left (b+a \cos ^2(e+f x)\right )^2} \]

output
-1/2*(a-5*b)*arctanh(cos(f*x+e))/(a+b)^4/f-1/4*(2*a-b)*b*cos(f*x+e)/a/(a+b 
)^2/f/(b+a*cos(f*x+e)^2)^2+1/8*(4*a^2-9*a*b-b^2)*cos(f*x+e)/a/(a+b)^3/f/(b 
+a*cos(f*x+e)^2)-1/2*cos(f*x+e)*cot(f*x+e)^2/(a+b)/f/(b+a*cos(f*x+e)^2)^2+ 
1/8*(15*a^2-10*a*b-b^2)*arctan(cos(f*x+e)*a^(1/2)/b^(1/2))*b^(1/2)/a^(3/2) 
/(a+b)^4/f
 
3.1.58.2 Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 5.36 (sec) , antiderivative size = 532, normalized size of antiderivative = 2.50 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^5(e+f x) \left (\frac {8 b^2 (a+b)^2}{a}-\frac {2 b (a+b) (9 a+b) (a+2 b+a \cos (2 (e+f x)))}{a}-\frac {\sqrt {b} \left (-15 a^2+10 a b+b^2\right ) \arctan \left (\frac {\left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right ) (a+2 b+a \cos (2 (e+f x)))^2 \sec (e+f x)}{a^{3/2}}-\frac {\sqrt {b} \left (-15 a^2+10 a b+b^2\right ) \arctan \left (\frac {\left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right ) (a+2 b+a \cos (2 (e+f x)))^2 \sec (e+f x)}{a^{3/2}}-(a+b) (a+2 b+a \cos (2 (e+f x)))^2 \csc ^2\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x)-4 (a-5 b) (a+2 b+a \cos (2 (e+f x)))^2 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right ) \sec (e+f x)+4 (a-5 b) (a+2 b+a \cos (2 (e+f x)))^2 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sec (e+f x)+(a+b) (a+2 b+a \cos (2 (e+f x)))^2 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x)\right )}{64 (a+b)^4 f \left (a+b \sec ^2(e+f x)\right )^3} \]

input
Integrate[Csc[e + f*x]^3/(a + b*Sec[e + f*x]^2)^3,x]
 
output
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^5*((8*b^2*(a + b)^2)/a - (2*b 
*(a + b)*(9*a + b)*(a + 2*b + a*Cos[2*(e + f*x)]))/a - (Sqrt[b]*(-15*a^2 + 
 10*a*b + b^2)*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*S 
in[e])^2]*Tan[(f*x)/2]))/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2*Sec[e + 
 f*x])/a^(3/2) - (Sqrt[b]*(-15*a^2 + 10*a*b + b^2)*ArcTan[((-Sqrt[a] + I*S 
qrt[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*Sec[e + f*x])/a^(3/2) - (a + b)*(a + 2*b + a* 
Cos[2*(e + f*x)])^2*Csc[(e + f*x)/2]^2*Sec[e + f*x] - 4*(a - 5*b)*(a + 2*b 
 + a*Cos[2*(e + f*x)])^2*Log[Cos[(e + f*x)/2]]*Sec[e + f*x] + 4*(a - 5*b)* 
(a + 2*b + a*Cos[2*(e + f*x)])^2*Log[Sin[(e + f*x)/2]]*Sec[e + f*x] + (a + 
 b)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Sec[(e + f*x)/2]^2*Sec[e + f*x]))/(64 
*(a + b)^4*f*(a + b*Sec[e + f*x]^2)^3)
 
3.1.58.3 Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 4621, 372, 440, 27, 402, 25, 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.

\(\displaystyle \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x)^3 \left (a+b \sec (e+f x)^2\right )^3}dx\)

\(\Big \downarrow \) 4621

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 440

\(\displaystyle -\frac {\frac {\cos ^3(e+f x)}{2 (a+b) \left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )^2}-\frac {\frac {\int \frac {2 \left ((2 a-b) b-\left (2 a^2-8 b a-b^2\right ) \cos ^2(e+f x)\right )}{\left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )^2}d\cos (e+f x)}{4 a (a+b)}-\frac {b (2 a-b) \cos (e+f x)}{2 a (a+b) \left (a \cos ^2(e+f x)+b\right )^2}}{2 (a+b)}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\cos ^3(e+f x)}{2 (a+b) \left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )^2}-\frac {\frac {\int \frac {(2 a-b) b-\left (2 a^2-8 b a-b^2\right ) \cos ^2(e+f x)}{\left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )^2}d\cos (e+f x)}{2 a (a+b)}-\frac {b (2 a-b) \cos (e+f x)}{2 a (a+b) \left (a \cos ^2(e+f x)+b\right )^2}}{2 (a+b)}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle -\frac {\frac {\cos ^3(e+f x)}{2 (a+b) \left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {\left (4 a^2-9 a b-b^2\right ) \cos (e+f x)}{2 (a+b) \left (a \cos ^2(e+f x)+b\right )}-\frac {\int -\frac {b \left ((11 a-b) b-\left (4 a^2-9 b a-b^2\right ) \cos ^2(e+f x)\right )}{\left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )}d\cos (e+f x)}{2 b (a+b)}}{2 a (a+b)}-\frac {b (2 a-b) \cos (e+f x)}{2 a (a+b) \left (a \cos ^2(e+f x)+b\right )^2}}{2 (a+b)}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {\cos ^3(e+f x)}{2 (a+b) \left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {\int \frac {b \left ((11 a-b) b-\left (4 a^2-9 b a-b^2\right ) \cos ^2(e+f x)\right )}{\left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )}d\cos (e+f x)}{2 b (a+b)}+\frac {\left (4 a^2-9 a b-b^2\right ) \cos (e+f x)}{2 (a+b) \left (a \cos ^2(e+f x)+b\right )}}{2 a (a+b)}-\frac {b (2 a-b) \cos (e+f x)}{2 a (a+b) \left (a \cos ^2(e+f x)+b\right )^2}}{2 (a+b)}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\cos ^3(e+f x)}{2 (a+b) \left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {\int \frac {(11 a-b) b-\left (4 a^2-9 b a-b^2\right ) \cos ^2(e+f x)}{\left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )}d\cos (e+f x)}{2 (a+b)}+\frac {\left (4 a^2-9 a b-b^2\right ) \cos (e+f x)}{2 (a+b) \left (a \cos ^2(e+f x)+b\right )}}{2 a (a+b)}-\frac {b (2 a-b) \cos (e+f x)}{2 a (a+b) \left (a \cos ^2(e+f x)+b\right )^2}}{2 (a+b)}}{f}\)

\(\Big \downarrow \) 397

\(\displaystyle -\frac {\frac {\cos ^3(e+f x)}{2 (a+b) \left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {\frac {b \left (15 a^2-10 a b-b^2\right ) \int \frac {1}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{a+b}-\frac {4 a (a-5 b) \int \frac {1}{1-\cos ^2(e+f x)}d\cos (e+f x)}{a+b}}{2 (a+b)}+\frac {\left (4 a^2-9 a b-b^2\right ) \cos (e+f x)}{2 (a+b) \left (a \cos ^2(e+f x)+b\right )}}{2 a (a+b)}-\frac {b (2 a-b) \cos (e+f x)}{2 a (a+b) \left (a \cos ^2(e+f x)+b\right )^2}}{2 (a+b)}}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {\frac {\cos ^3(e+f x)}{2 (a+b) \left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {\frac {\sqrt {b} \left (15 a^2-10 a b-b^2\right ) \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{\sqrt {a} (a+b)}-\frac {4 a (a-5 b) \int \frac {1}{1-\cos ^2(e+f x)}d\cos (e+f x)}{a+b}}{2 (a+b)}+\frac {\left (4 a^2-9 a b-b^2\right ) \cos (e+f x)}{2 (a+b) \left (a \cos ^2(e+f x)+b\right )}}{2 a (a+b)}-\frac {b (2 a-b) \cos (e+f x)}{2 a (a+b) \left (a \cos ^2(e+f x)+b\right )^2}}{2 (a+b)}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\frac {\cos ^3(e+f x)}{2 (a+b) \left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {\frac {\sqrt {b} \left (15 a^2-10 a b-b^2\right ) \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{\sqrt {a} (a+b)}-\frac {4 a (a-5 b) \text {arctanh}(\cos (e+f x))}{a+b}}{2 (a+b)}+\frac {\left (4 a^2-9 a b-b^2\right ) \cos (e+f x)}{2 (a+b) \left (a \cos ^2(e+f x)+b\right )}}{2 a (a+b)}-\frac {b (2 a-b) \cos (e+f x)}{2 a (a+b) \left (a \cos ^2(e+f x)+b\right )^2}}{2 (a+b)}}{f}\)

input
Int[Csc[e + f*x]^3/(a + b*Sec[e + f*x]^2)^3,x]
 
output
-((Cos[e + f*x]^3/(2*(a + b)*(1 - Cos[e + f*x]^2)*(b + a*Cos[e + f*x]^2)^2 
) - (-1/2*((2*a - b)*b*Cos[e + f*x])/(a*(a + b)*(b + a*Cos[e + f*x]^2)^2) 
+ (((Sqrt[b]*(15*a^2 - 10*a*b - b^2)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b] 
])/(Sqrt[a]*(a + b)) - (4*a*(a - 5*b)*ArcTanh[Cos[e + f*x]])/(a + b))/(2*( 
a + b)) + ((4*a^2 - 9*a*b - b^2)*Cos[e + f*x])/(2*(a + b)*(b + a*Cos[e + f 
*x]^2)))/(2*a*(a + b)))/(2*(a + b)))/f)
 

3.1.58.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 440
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[g*(b*e - a*f)*(g*x)^(m - 1)*(a + 
 b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] - Simp[ 
g^2/(2*b*(b*c - a*d)*(p + 1))   Int[(g*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + 
d*x^2)^q*Simp[c*(b*e - a*f)*(m - 1) + (d*(b*e - a*f)*(m + 2*q + 1) - b*2*(c 
*f - d*e)*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, q}, x] && 
 LtQ[p, -1] && GtQ[m, 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]
 
3.1.58.4 Maple [A] (verified)

Time = 1.76 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.93

method result size
derivativedivides \(\frac {\frac {1}{4 \left (a +b \right )^{3} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (a -5 b \right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{4 \left (a +b \right )^{4}}+\frac {b \left (\frac {\left (-\frac {9}{8} a^{2}-\frac {5}{4} a b -\frac {1}{8} b^{2}\right ) \cos \left (f x +e \right )^{3}-\frac {b \left (7 a^{2}+6 a b -b^{2}\right ) \cos \left (f x +e \right )}{8 a}}{\left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}-10 a b -b^{2}\right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{8 a \sqrt {a b}}\right )}{\left (a +b \right )^{4}}+\frac {1}{4 \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-a +5 b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{4 \left (a +b \right )^{4}}}{f}\) \(198\)
default \(\frac {\frac {1}{4 \left (a +b \right )^{3} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (a -5 b \right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{4 \left (a +b \right )^{4}}+\frac {b \left (\frac {\left (-\frac {9}{8} a^{2}-\frac {5}{4} a b -\frac {1}{8} b^{2}\right ) \cos \left (f x +e \right )^{3}-\frac {b \left (7 a^{2}+6 a b -b^{2}\right ) \cos \left (f x +e \right )}{8 a}}{\left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}-10 a b -b^{2}\right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{8 a \sqrt {a b}}\right )}{\left (a +b \right )^{4}}+\frac {1}{4 \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-a +5 b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{4 \left (a +b \right )^{4}}}{f}\) \(198\)
risch \(-\frac {-4 a^{3} {\mathrm e}^{11 i \left (f x +e \right )}+9 a^{2} b \,{\mathrm e}^{11 i \left (f x +e \right )}+a \,b^{2} {\mathrm e}^{11 i \left (f x +e \right )}-20 a^{3} {\mathrm e}^{9 i \left (f x +e \right )}-23 a^{2} b \,{\mathrm e}^{9 i \left (f x +e \right )}+29 a \,b^{2} {\mathrm e}^{9 i \left (f x +e \right )}-4 b^{3} {\mathrm e}^{9 i \left (f x +e \right )}-40 a^{3} {\mathrm e}^{7 i \left (f x +e \right )}-114 a^{2} b \,{\mathrm e}^{7 i \left (f x +e \right )}-94 a \,b^{2} {\mathrm e}^{7 i \left (f x +e \right )}+4 b^{3} {\mathrm e}^{7 i \left (f x +e \right )}-40 a^{3} {\mathrm e}^{5 i \left (f x +e \right )}-114 a^{2} b \,{\mathrm e}^{5 i \left (f x +e \right )}-94 a \,b^{2} {\mathrm e}^{5 i \left (f x +e \right )}+4 b^{3} {\mathrm e}^{5 i \left (f x +e \right )}-20 a^{3} {\mathrm e}^{3 i \left (f x +e \right )}-23 a^{2} b \,{\mathrm e}^{3 i \left (f x +e \right )}+29 a \,b^{2} {\mathrm e}^{3 i \left (f x +e \right )}-4 b^{3} {\mathrm e}^{3 i \left (f x +e \right )}-4 a^{3} {\mathrm e}^{i \left (f x +e \right )}+9 a^{2} b \,{\mathrm e}^{i \left (f x +e \right )}+a \,b^{2} {\mathrm e}^{i \left (f x +e \right )}}{4 a f \left (a +b \right )^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2} \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 )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) a}{2 f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {5 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b}{2 f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) a}{2 f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}+\frac {5 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b}{2 f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {15 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 )}{16 \left (a +b \right )^{4} f}+\frac {5 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}{8 a \left (a +b \right )^{4} f}+\frac {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^{2}}{16 a^{2} \left (a +b \right )^{4} f}+\frac {15 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 )}{16 \left (a +b \right )^{4} f}-\frac {5 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}{8 a \left (a +b \right )^{4} f}-\frac {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^{2}}{16 a^{2} \left (a +b \right )^{4} f}\) \(882\)

input
int(csc(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)
 
output
1/f*(1/4/(a+b)^3/(-1+cos(f*x+e))+1/4*(a-5*b)/(a+b)^4*ln(-1+cos(f*x+e))+b/( 
a+b)^4*(((-9/8*a^2-5/4*a*b-1/8*b^2)*cos(f*x+e)^3-1/8*b*(7*a^2+6*a*b-b^2)/a 
*cos(f*x+e))/(b+a*cos(f*x+e)^2)^2+1/8*(15*a^2-10*a*b-b^2)/a/(a*b)^(1/2)*ar 
ctan(a*cos(f*x+e)/(a*b)^(1/2)))+1/4/(a+b)^3/(1+cos(f*x+e))+1/4/(a+b)^4*(-a 
+5*b)*ln(1+cos(f*x+e)))
 
3.1.58.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 649 vs. \(2 (195) = 390\).

Time = 0.43 (sec) , antiderivative size = 1332, normalized size of antiderivative = 6.25 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]

input
integrate(csc(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
 
output
[1/16*(2*(4*a^4 - 5*a^3*b - 10*a^2*b^2 - a*b^3)*cos(f*x + e)^5 + 2*(17*a^3 
*b + 11*a^2*b^2 - 5*a*b^3 + b^4)*cos(f*x + e)^3 - ((15*a^4 - 10*a^3*b - a^ 
2*b^2)*cos(f*x + e)^6 - (15*a^4 - 40*a^3*b + 19*a^2*b^2 + 2*a*b^3)*cos(f*x 
 + e)^4 - 15*a^2*b^2 + 10*a*b^3 + b^4 - (30*a^3*b - 35*a^2*b^2 + 8*a*b^3 + 
 b^4)*cos(f*x + e)^2)*sqrt(-b/a)*log((a*cos(f*x + e)^2 - 2*a*sqrt(-b/a)*co 
s(f*x + e) - b)/(a*cos(f*x + e)^2 + b)) + 2*(11*a^2*b^2 + 10*a*b^3 - b^4)* 
cos(f*x + e) - 4*((a^4 - 5*a^3*b)*cos(f*x + e)^6 - (a^4 - 7*a^3*b + 10*a^2 
*b^2)*cos(f*x + e)^4 - a^2*b^2 + 5*a*b^3 - (2*a^3*b - 11*a^2*b^2 + 5*a*b^3 
)*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2) + 4*((a^4 - 5*a^3*b)*cos(f*x 
 + e)^6 - (a^4 - 7*a^3*b + 10*a^2*b^2)*cos(f*x + e)^4 - a^2*b^2 + 5*a*b^3 
- (2*a^3*b - 11*a^2*b^2 + 5*a*b^3)*cos(f*x + e)^2)*log(-1/2*cos(f*x + e) + 
 1/2))/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^6 
 - (a^7 + 2*a^6*b - 2*a^5*b^2 - 8*a^4*b^3 - 7*a^3*b^4 - 2*a^2*b^5)*f*cos(f 
*x + e)^4 - (2*a^6*b + 7*a^5*b^2 + 8*a^4*b^3 + 2*a^3*b^4 - 2*a^2*b^5 - a*b 
^6)*f*cos(f*x + e)^2 - (a^5*b^2 + 4*a^4*b^3 + 6*a^3*b^4 + 4*a^2*b^5 + a*b^ 
6)*f), 1/8*((4*a^4 - 5*a^3*b - 10*a^2*b^2 - a*b^3)*cos(f*x + e)^5 + (17*a^ 
3*b + 11*a^2*b^2 - 5*a*b^3 + b^4)*cos(f*x + e)^3 + ((15*a^4 - 10*a^3*b - a 
^2*b^2)*cos(f*x + e)^6 - (15*a^4 - 40*a^3*b + 19*a^2*b^2 + 2*a*b^3)*cos(f* 
x + e)^4 - 15*a^2*b^2 + 10*a*b^3 + b^4 - (30*a^3*b - 35*a^2*b^2 + 8*a*b^3 
+ b^4)*cos(f*x + e)^2)*sqrt(b/a)*arctan(a*sqrt(b/a)*cos(f*x + e)/b) + (...
 
3.1.58.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]

input
integrate(csc(f*x+e)**3/(a+b*sec(f*x+e)**2)**3,x)
 
output
Timed out
 
3.1.58.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 399 vs. \(2 (195) = 390\).

Time = 0.27 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.87 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {2 \, {\left (a - 5 \, b\right )} \log \left (\cos \left (f x + e\right ) + 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac {2 \, {\left (a - 5 \, b\right )} \log \left (\cos \left (f x + e\right ) - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac {{\left (15 \, a^{2} b - 10 \, a b^{2} - b^{3}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} \sqrt {a b}} - \frac {{\left (4 \, a^{3} - 9 \, a^{2} b - a b^{2}\right )} \cos \left (f x + e\right )^{5} + {\left (17 \, a^{2} b - 6 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (11 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )}{{\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \cos \left (f x + e\right )^{6} - a^{4} b^{2} - 3 \, a^{3} b^{3} - 3 \, a^{2} b^{4} - a b^{5} - {\left (a^{6} + a^{5} b - 3 \, a^{4} b^{2} - 5 \, a^{3} b^{3} - 2 \, a^{2} b^{4}\right )} \cos \left (f x + e\right )^{4} - {\left (2 \, a^{5} b + 5 \, a^{4} b^{2} + 3 \, a^{3} b^{3} - a^{2} b^{4} - a b^{5}\right )} \cos \left (f x + e\right )^{2}}}{8 \, f} \]

input
integrate(csc(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
 
output
-1/8*(2*(a - 5*b)*log(cos(f*x + e) + 1)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b 
^3 + b^4) - 2*(a - 5*b)*log(cos(f*x + e) - 1)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 
 4*a*b^3 + b^4) - (15*a^2*b - 10*a*b^2 - b^3)*arctan(a*cos(f*x + e)/sqrt(a 
*b))/((a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*sqrt(a*b)) - ((4*a^3 
 - 9*a^2*b - a*b^2)*cos(f*x + e)^5 + (17*a^2*b - 6*a*b^2 + b^3)*cos(f*x + 
e)^3 + (11*a*b^2 - b^3)*cos(f*x + e))/((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^ 
3)*cos(f*x + e)^6 - a^4*b^2 - 3*a^3*b^3 - 3*a^2*b^4 - a*b^5 - (a^6 + a^5*b 
 - 3*a^4*b^2 - 5*a^3*b^3 - 2*a^2*b^4)*cos(f*x + e)^4 - (2*a^5*b + 5*a^4*b^ 
2 + 3*a^3*b^3 - a^2*b^4 - a*b^5)*cos(f*x + e)^2))/f
 
3.1.58.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 750 vs. \(2 (195) = 390\).

Time = 0.52 (sec) , antiderivative size = 750, normalized size of antiderivative = 3.52 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {2 \, {\left (a - 5 \, b\right )} \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac {{\left (15 \, a^{2} b - 10 \, a b^{2} - b^{3}\right )} \arctan \left (-\frac {a \cos \left (f x + e\right ) - b}{\sqrt {a b} \cos \left (f x + e\right ) + \sqrt {a b}}\right )}{{\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} \sqrt {a b}} + \frac {{\left (a + b - \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {10 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} {\left (\cos \left (f x + e\right ) - 1\right )}} - \frac {\cos \left (f x + e\right ) - 1}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}} - \frac {2 \, {\left (9 \, a^{3} b + 17 \, a^{2} b^{2} + 7 \, a b^{3} - b^{4} + \frac {27 \, a^{3} b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a^{2} b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {23 \, a b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {3 \, b^{4} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {27 \, a^{3} b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {21 \, a^{2} b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {29 \, a b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {3 \, b^{4} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {9 \, a^{3} b {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {5 \, a^{2} b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {13 \, a b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {b^{4} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{{\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} {\left (a + b + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {2 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}^{2}}}{8 \, f} \]

input
integrate(csc(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
 
output
1/8*(2*(a - 5*b)*log(abs(-cos(f*x + e) + 1)/abs(cos(f*x + e) + 1))/(a^4 + 
4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - (15*a^2*b - 10*a*b^2 - b^3)*arctan( 
-(a*cos(f*x + e) - b)/(sqrt(a*b)*cos(f*x + e) + sqrt(a*b)))/((a^5 + 4*a^4* 
b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*sqrt(a*b)) + (a + b - 2*a*(cos(f*x + e) 
 - 1)/(cos(f*x + e) + 1) + 10*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1))*(co 
s(f*x + e) + 1)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*(cos(f*x + e) 
 - 1)) - (cos(f*x + e) - 1)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(cos(f*x + e) 
 + 1)) - 2*(9*a^3*b + 17*a^2*b^2 + 7*a*b^3 - b^4 + 27*a^3*b*(cos(f*x + e) 
- 1)/(cos(f*x + e) + 1) + a^2*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 
23*a*b^3*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 3*b^4*(cos(f*x + e) - 1)/ 
(cos(f*x + e) + 1) + 27*a^3*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 
21*a^2*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 29*a*b^3*(cos(f*x + 
 e) - 1)^2/(cos(f*x + e) + 1)^2 - 3*b^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) 
 + 1)^2 + 9*a^3*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 5*a^2*b^2*(c 
os(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 13*a*b^3*(cos(f*x + e) - 1)^3/(c 
os(f*x + e) + 1)^3 + b^4*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3)/((a^5 
+ 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*(a + b + 2*a*(cos(f*x + e) - 1) 
/(cos(f*x + e) + 1) - 2*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(cos(f 
*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b*(cos(f*x + e) - 1)^2/(cos(f*x + e) 
 + 1)^2)^2))/f
 
3.1.58.9 Mupad [B] (verification not implemented)

Time = 19.87 (sec) , antiderivative size = 2728, normalized size of antiderivative = 12.81 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]

input
int(1/(sin(e + f*x)^3*(a + b/cos(e + f*x)^2)^3),x)
 
output
- ((cos(e + f*x)^3*(17*a^2*b - 6*a*b^2 + b^3))/(8*(a*b^3 + 3*a^3*b + a^4 + 
 3*a^2*b^2)) - (cos(e + f*x)^5*(9*a*b - 4*a^2 + b^2))/(8*(3*a*b^2 + 3*a^2* 
b + a^3 + b^3)) + (b^2*cos(e + f*x)*(11*a - b))/(8*(a*b^3 + 3*a^3*b + a^4 
+ 3*a^2*b^2)))/(f*(b^2 - cos(e + f*x)^4*(2*a*b - a^2) + cos(e + f*x)^2*(2* 
a*b - b^2) - a^2*cos(e + f*x)^6)) - (log(cos(e + f*x) - 1)*((3*b)/(2*(a + 
b)^4) - 1/(4*(a + b)^3)))/f - (log(cos(e + f*x) + 1)*(a - 5*b))/(4*f*(a + 
b)^4) - (atan(((((cos(e + f*x)*(20*a*b^5 - 160*a^5*b + 16*a^6 + b^6 + 70*a 
^2*b^4 - 300*a^3*b^3 + 625*a^4*b^2))/(32*(a*b^6 + 6*a^6*b + a^7 + 6*a^2*b^ 
5 + 15*a^3*b^4 + 20*a^4*b^3 + 15*a^5*b^2)) + ((-a^3*b)^(1/2)*(((11*a^11*b) 
/2 - (a^2*b^10)/2 + (3*a^3*b^9)/2 + 30*a^4*b^8 + 126*a^5*b^7 + 273*a^6*b^6 
 + 357*a^7*b^5 + 294*a^8*b^4 + 150*a^9*b^3 + (87*a^10*b^2)/2)/(a*b^9 + 9*a 
^9*b + a^10 + 9*a^2*b^8 + 36*a^3*b^7 + 84*a^4*b^6 + 126*a^5*b^5 + 126*a^6* 
b^4 + 84*a^7*b^3 + 36*a^8*b^2) - (cos(e + f*x)*(-a^3*b)^(1/2)*(10*a*b - 15 
*a^2 + b^2)*(1792*a^11*b + 256*a^12 - 256*a^3*b^9 - 1792*a^4*b^8 - 5120*a^ 
5*b^7 - 7168*a^6*b^6 - 3584*a^7*b^5 + 3584*a^8*b^4 + 7168*a^9*b^3 + 5120*a 
^10*b^2))/(512*(4*a^6*b + a^7 + a^3*b^4 + 4*a^4*b^3 + 6*a^5*b^2)*(a*b^6 + 
6*a^6*b + a^7 + 6*a^2*b^5 + 15*a^3*b^4 + 20*a^4*b^3 + 15*a^5*b^2)))*(10*a* 
b - 15*a^2 + b^2))/(16*(4*a^6*b + a^7 + a^3*b^4 + 4*a^4*b^3 + 6*a^5*b^2))) 
*(-a^3*b)^(1/2)*(10*a*b - 15*a^2 + b^2)*1i)/(16*(4*a^6*b + a^7 + a^3*b^4 + 
 4*a^4*b^3 + 6*a^5*b^2)) + (((cos(e + f*x)*(20*a*b^5 - 160*a^5*b + 16*a...