3.1.48 \(\int \csc ^5(e+f x) (a+b \tan ^2(e+f x))^2 \, dx\) [48]

3.1.48.1 Optimal result
3.1.48.2 Mathematica [B] (verified)
3.1.48.3 Rubi [A] (verified)
3.1.48.4 Maple [A] (verified)
3.1.48.5 Fricas [B] (verification not implemented)
3.1.48.6 Sympy [F]
3.1.48.7 Maxima [A] (verification not implemented)
3.1.48.8 Giac [B] (verification not implemented)
3.1.48.9 Mupad [B] (verification not implemented)

3.1.48.1 Optimal result

Integrand size = 23, antiderivative size = 123 \[ \int \csc ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=-\frac {\left (3 a^2+24 a b+8 b^2\right ) \text {arctanh}(\cos (e+f x))}{8 f}-\frac {a (a+8 b) \cot (e+f x) \csc (e+f x)}{8 f}+\frac {\left (a^2+8 a b+4 b^2\right ) \sec (e+f x)}{4 f}-\frac {a^2 \csc ^4(e+f x) \sec (e+f x)}{4 f}+\frac {b^2 \sec ^3(e+f x)}{3 f} \]

output
-1/8*(3*a^2+24*a*b+8*b^2)*arctanh(cos(f*x+e))/f-1/8*a*(a+8*b)*cot(f*x+e)*c 
sc(f*x+e)/f+1/4*(a^2+8*a*b+4*b^2)*sec(f*x+e)/f-1/4*a^2*csc(f*x+e)^4*sec(f* 
x+e)/f+1/3*b^2*sec(f*x+e)^3/f
 
3.1.48.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(447\) vs. \(2(123)=246\).

Time = 7.51 (sec) , antiderivative size = 447, normalized size of antiderivative = 3.63 \[ \int \csc ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=\frac {\left (-3 a^2-8 a b\right ) \csc ^2\left (\frac {1}{2} (e+f x)\right )}{32 f}-\frac {a^2 \csc ^4\left (\frac {1}{2} (e+f x)\right )}{64 f}+\frac {\left (-3 a^2-24 a b-8 b^2\right ) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{8 f}+\frac {\left (3 a^2+24 a b+8 b^2\right ) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{8 f}+\frac {\left (3 a^2+8 a b\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{32 f}+\frac {a^2 \sec ^4\left (\frac {1}{2} (e+f x)\right )}{64 f}+\frac {b^2}{12 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {b^2 \sin \left (\frac {1}{2} (e+f x)\right )}{6 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}-\frac {b^2 \sin \left (\frac {1}{2} (e+f x)\right )}{6 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}+\frac {b^2}{12 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {-12 a b \sin \left (\frac {1}{2} (e+f x)\right )-7 b^2 \sin \left (\frac {1}{2} (e+f x)\right )}{6 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}+\frac {12 a b \sin \left (\frac {1}{2} (e+f x)\right )+7 b^2 \sin \left (\frac {1}{2} (e+f x)\right )}{6 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

input
Integrate[Csc[e + f*x]^5*(a + b*Tan[e + f*x]^2)^2,x]
 
output
((-3*a^2 - 8*a*b)*Csc[(e + f*x)/2]^2)/(32*f) - (a^2*Csc[(e + f*x)/2]^4)/(6 
4*f) + ((-3*a^2 - 24*a*b - 8*b^2)*Log[Cos[(e + f*x)/2]])/(8*f) + ((3*a^2 + 
 24*a*b + 8*b^2)*Log[Sin[(e + f*x)/2]])/(8*f) + ((3*a^2 + 8*a*b)*Sec[(e + 
f*x)/2]^2)/(32*f) + (a^2*Sec[(e + f*x)/2]^4)/(64*f) + b^2/(12*f*(Cos[(e + 
f*x)/2] - Sin[(e + f*x)/2])^2) + (b^2*Sin[(e + f*x)/2])/(6*f*(Cos[(e + f*x 
)/2] - Sin[(e + f*x)/2])^3) - (b^2*Sin[(e + f*x)/2])/(6*f*(Cos[(e + f*x)/2 
] + Sin[(e + f*x)/2])^3) + b^2/(12*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) 
^2) + (-12*a*b*Sin[(e + f*x)/2] - 7*b^2*Sin[(e + f*x)/2])/(6*f*(Cos[(e + f 
*x)/2] + Sin[(e + f*x)/2])) + (12*a*b*Sin[(e + f*x)/2] + 7*b^2*Sin[(e + f* 
x)/2])/(6*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))
 
3.1.48.3 Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4147, 25, 366, 360, 25, 1467, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4147

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 366

\(\displaystyle \frac {\frac {1}{4} \int \frac {\sec ^4(e+f x) \left (a^2+8 b a-4 b^2+4 b^2 \sec ^2(e+f x)\right )}{\left (1-\sec ^2(e+f x)\right )^2}d\sec (e+f x)-\frac {a^2 \sec ^5(e+f x)}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\)

\(\Big \downarrow \) 360

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int -\frac {8 b^2 \sec ^4(e+f x)+2 a (a+8 b) \sec ^2(e+f x)+a (a+8 b)}{1-\sec ^2(e+f x)}d\sec (e+f x)+\frac {a (a+8 b) \sec (e+f x)}{2 \left (1-\sec ^2(e+f x)\right )}\right )-\frac {a^2 \sec ^5(e+f x)}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{4} \left (\frac {a (a+8 b) \sec (e+f x)}{2 \left (1-\sec ^2(e+f x)\right )}-\frac {1}{2} \int \frac {8 b^2 \sec ^4(e+f x)+2 a (a+8 b) \sec ^2(e+f x)+a (a+8 b)}{1-\sec ^2(e+f x)}d\sec (e+f x)\right )-\frac {a^2 \sec ^5(e+f x)}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\)

\(\Big \downarrow \) 1467

\(\displaystyle \frac {\frac {1}{4} \left (\frac {a (a+8 b) \sec (e+f x)}{2 \left (1-\sec ^2(e+f x)\right )}-\frac {1}{2} \int \left (-8 b^2 \sec ^2(e+f x)-2 \left (a^2+8 b a+4 b^2\right )+\frac {3 a^2+24 b a+8 b^2}{1-\sec ^2(e+f x)}\right )d\sec (e+f x)\right )-\frac {a^2 \sec ^5(e+f x)}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (-\left (3 a^2+24 a b+8 b^2\right ) \text {arctanh}(\sec (e+f x))+2 \left (a^2+8 a b+4 b^2\right ) \sec (e+f x)+\frac {8}{3} b^2 \sec ^3(e+f x)\right )+\frac {a (a+8 b) \sec (e+f x)}{2 \left (1-\sec ^2(e+f x)\right )}\right )-\frac {a^2 \sec ^5(e+f x)}{4 \left (1-\sec ^2(e+f x)\right )^2}}{f}\)

input
Int[Csc[e + f*x]^5*(a + b*Tan[e + f*x]^2)^2,x]
 
output
(-1/4*(a^2*Sec[e + f*x]^5)/(1 - Sec[e + f*x]^2)^2 + ((a*(a + 8*b)*Sec[e + 
f*x])/(2*(1 - Sec[e + f*x]^2)) + (-((3*a^2 + 24*a*b + 8*b^2)*ArcTanh[Sec[e 
 + f*x]]) + 2*(a^2 + 8*a*b + 4*b^2)*Sec[e + f*x] + (8*b^2*Sec[e + f*x]^3)/ 
3)/2)/4)/f
 

3.1.48.3.1 Defintions of rubi rules used

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

rule 360
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : 
> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p 
+ 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1))   Int[(a + b*x^2)^(p + 1)*Expan 
dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 
- 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; 
FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & 
& (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
 

rule 366
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, 
x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a* 
b^2*e*(p + 1))), x] + Simp[1/(2*a*b^2*(p + 1))   Int[(e*x)^m*(a + b*x^2)^(p 
 + 1)*Simp[(b*c - a*d)^2*(m + 1) + 2*b^2*c^2*(p + 1) + 2*a*b*d^2*(p + 1)*x^ 
2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p 
, -1]
 

rule 1467
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
 x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], 
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e 
 + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4147
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ 
m)   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 
)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( 
m - 1)/2]
 
3.1.48.4 Maple [A] (verified)

Time = 2.21 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.18

method result size
derivativedivides \(\frac {a^{2} \left (\left (-\frac {\csc \left (f x +e \right )^{3}}{4}-\frac {3 \csc \left (f x +e \right )}{8}\right ) \cot \left (f x +e \right )+\frac {3 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{8}\right )+2 a b \left (-\frac {1}{2 \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )}+\frac {3}{2 \cos \left (f x +e \right )}+\frac {3 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{2}\right )+b^{2} \left (\frac {1}{3 \cos \left (f x +e \right )^{3}}+\frac {1}{\cos \left (f x +e \right )}+\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )\right )}{f}\) \(145\)
default \(\frac {a^{2} \left (\left (-\frac {\csc \left (f x +e \right )^{3}}{4}-\frac {3 \csc \left (f x +e \right )}{8}\right ) \cot \left (f x +e \right )+\frac {3 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{8}\right )+2 a b \left (-\frac {1}{2 \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )}+\frac {3}{2 \cos \left (f x +e \right )}+\frac {3 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{2}\right )+b^{2} \left (\frac {1}{3 \cos \left (f x +e \right )^{3}}+\frac {1}{\cos \left (f x +e \right )}+\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )\right )}{f}\) \(145\)
risch \(\frac {{\mathrm e}^{i \left (f x +e \right )} \left (9 a^{2} {\mathrm e}^{12 i \left (f x +e \right )}+72 a b \,{\mathrm e}^{12 i \left (f x +e \right )}+24 b^{2} {\mathrm e}^{12 i \left (f x +e \right )}-6 a^{2} {\mathrm e}^{10 i \left (f x +e \right )}-48 a b \,{\mathrm e}^{10 i \left (f x +e \right )}-16 b^{2} {\mathrm e}^{10 i \left (f x +e \right )}-105 a^{2} {\mathrm e}^{8 i \left (f x +e \right )}-72 a b \,{\mathrm e}^{8 i \left (f x +e \right )}-152 b^{2} {\mathrm e}^{8 i \left (f x +e \right )}-180 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+96 a b \,{\mathrm e}^{6 i \left (f x +e \right )}+288 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}-105 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-72 a b \,{\mathrm e}^{4 i \left (f x +e \right )}-152 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-6 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-48 a b \,{\mathrm e}^{2 i \left (f x +e \right )}-16 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+9 a^{2}+72 a b +24 b^{2}\right )}{12 f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{8 f}+\frac {3 a \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b}{f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b^{2}}{f}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{8 f}-\frac {3 a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b}{f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b^{2}}{f}\) \(460\)

input
int(csc(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
 
output
1/f*(a^2*((-1/4*csc(f*x+e)^3-3/8*csc(f*x+e))*cot(f*x+e)+3/8*ln(csc(f*x+e)- 
cot(f*x+e)))+2*a*b*(-1/2/sin(f*x+e)^2/cos(f*x+e)+3/2/cos(f*x+e)+3/2*ln(csc 
(f*x+e)-cot(f*x+e)))+b^2*(1/3/cos(f*x+e)^3+1/cos(f*x+e)+ln(csc(f*x+e)-cot( 
f*x+e))))
 
3.1.48.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (113) = 226\).

Time = 0.30 (sec) , antiderivative size = 284, normalized size of antiderivative = 2.31 \[ \int \csc ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=\frac {6 \, {\left (3 \, a^{2} + 24 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{6} - 10 \, {\left (3 \, a^{2} + 24 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 16 \, {\left (6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + 16 \, b^{2} - 3 \, {\left ({\left (3 \, a^{2} + 24 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{7} - 2 \, {\left (3 \, a^{2} + 24 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{5} + {\left (3 \, a^{2} + 24 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + 3 \, {\left ({\left (3 \, a^{2} + 24 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{7} - 2 \, {\left (3 \, a^{2} + 24 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{5} + {\left (3 \, a^{2} + 24 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{48 \, {\left (f \cos \left (f x + e\right )^{7} - 2 \, f \cos \left (f x + e\right )^{5} + f \cos \left (f x + e\right )^{3}\right )}} \]

input
integrate(csc(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")
 
output
1/48*(6*(3*a^2 + 24*a*b + 8*b^2)*cos(f*x + e)^6 - 10*(3*a^2 + 24*a*b + 8*b 
^2)*cos(f*x + e)^4 + 16*(6*a*b + b^2)*cos(f*x + e)^2 + 16*b^2 - 3*((3*a^2 
+ 24*a*b + 8*b^2)*cos(f*x + e)^7 - 2*(3*a^2 + 24*a*b + 8*b^2)*cos(f*x + e) 
^5 + (3*a^2 + 24*a*b + 8*b^2)*cos(f*x + e)^3)*log(1/2*cos(f*x + e) + 1/2) 
+ 3*((3*a^2 + 24*a*b + 8*b^2)*cos(f*x + e)^7 - 2*(3*a^2 + 24*a*b + 8*b^2)* 
cos(f*x + e)^5 + (3*a^2 + 24*a*b + 8*b^2)*cos(f*x + e)^3)*log(-1/2*cos(f*x 
 + e) + 1/2))/(f*cos(f*x + e)^7 - 2*f*cos(f*x + e)^5 + f*cos(f*x + e)^3)
 
3.1.48.6 Sympy [F]

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

input
integrate(csc(f*x+e)**5*(a+b*tan(f*x+e)**2)**2,x)
 
output
Integral((a + b*tan(e + f*x)**2)**2*csc(e + f*x)**5, x)
 
3.1.48.7 Maxima [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.33 \[ \int \csc ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=-\frac {3 \, {\left (3 \, a^{2} + 24 \, a b + 8 \, b^{2}\right )} \log \left (\cos \left (f x + e\right ) + 1\right ) - 3 \, {\left (3 \, a^{2} + 24 \, a b + 8 \, b^{2}\right )} \log \left (\cos \left (f x + e\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (3 \, a^{2} + 24 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{6} - 5 \, {\left (3 \, a^{2} + 24 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, b^{2}\right )}}{\cos \left (f x + e\right )^{7} - 2 \, \cos \left (f x + e\right )^{5} + \cos \left (f x + e\right )^{3}}}{48 \, f} \]

input
integrate(csc(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")
 
output
-1/48*(3*(3*a^2 + 24*a*b + 8*b^2)*log(cos(f*x + e) + 1) - 3*(3*a^2 + 24*a* 
b + 8*b^2)*log(cos(f*x + e) - 1) - 2*(3*(3*a^2 + 24*a*b + 8*b^2)*cos(f*x + 
 e)^6 - 5*(3*a^2 + 24*a*b + 8*b^2)*cos(f*x + e)^4 + 8*(6*a*b + b^2)*cos(f* 
x + e)^2 + 8*b^2)/(cos(f*x + e)^7 - 2*cos(f*x + e)^5 + cos(f*x + e)^3))/f
 
3.1.48.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (113) = 226\).

Time = 0.69 (sec) , antiderivative size = 391, normalized size of antiderivative = 3.18 \[ \int \csc ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=-\frac {\frac {24 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {48 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {3 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 12 \, {\left (3 \, a^{2} + 24 \, a b + 8 \, b^{2}\right )} \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right ) + \frac {3 \, {\left (a^{2} - \frac {8 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {16 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {18 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {144 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {48 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) - 1\right )}^{2}} - \frac {256 \, {\left (3 \, a b + 2 \, b^{2} + \frac {6 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {3 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {3 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}{{\left (\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1\right )}^{3}}}{192 \, f} \]

input
integrate(csc(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")
 
output
-1/192*(24*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 48*a*b*(cos(f*x + e 
) - 1)/(cos(f*x + e) + 1) - 3*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^ 
2 - 12*(3*a^2 + 24*a*b + 8*b^2)*log(abs(-cos(f*x + e) + 1)/abs(cos(f*x + e 
) + 1)) + 3*(a^2 - 8*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 16*a*b*(c 
os(f*x + e) - 1)/(cos(f*x + e) + 1) + 18*a^2*(cos(f*x + e) - 1)^2/(cos(f*x 
 + e) + 1)^2 + 144*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 48*b^2* 
(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1)^2/(cos(f*x + 
 e) - 1)^2 - 256*(3*a*b + 2*b^2 + 6*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 
 1) + 3*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 3*a*b*(cos(f*x + e) - 
1)^2/(cos(f*x + e) + 1)^2 + 3*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^ 
2)/((cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1)^3)/f
 
3.1.48.9 Mupad [B] (verification not implemented)

Time = 10.29 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.98 \[ \int \csc ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=\frac {a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{64\,f}+\frac {\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (\frac {3\,a^2}{8}+3\,a\,b+b^2\right )}{f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {5\,a^2}{4}+4\,b\,a\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (2\,a^2+68\,a\,b+64\,b^2\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {21\,a^2}{4}+76\,a\,b+\frac {128\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (\frac {23\,a^2}{4}+140\,a\,b+64\,b^2\right )+\frac {a^2}{4}}{f\,\left (-16\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+48\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-48\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {a^2}{8}+\frac {b\,a}{4}\right )}{f} \]

input
int((a + b*tan(e + f*x)^2)^2/sin(e + f*x)^5,x)
 
output
(a^2*tan(e/2 + (f*x)/2)^4)/(64*f) + (log(tan(e/2 + (f*x)/2))*(3*a*b + (3*a 
^2)/8 + b^2))/f - (tan(e/2 + (f*x)/2)^2*(4*a*b + (5*a^2)/4) - tan(e/2 + (f 
*x)/2)^8*(68*a*b + 2*a^2 + 64*b^2) - tan(e/2 + (f*x)/2)^4*(76*a*b + (21*a^ 
2)/4 + (128*b^2)/3) + tan(e/2 + (f*x)/2)^6*(140*a*b + (23*a^2)/4 + 64*b^2) 
 + a^2/4)/(f*(16*tan(e/2 + (f*x)/2)^4 - 48*tan(e/2 + (f*x)/2)^6 + 48*tan(e 
/2 + (f*x)/2)^8 - 16*tan(e/2 + (f*x)/2)^10)) + (tan(e/2 + (f*x)/2)^2*((a*b 
)/4 + a^2/8))/f