\(\int \frac {\csc ^5(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\) [133]

Optimal result

Integrand size = 25, antiderivative size = 187 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {3 (a-5 b) (a-b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{8 a^{7/2} f}-\frac {5 (a-b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {(13 a-15 b) b \sec (e+f x)}{8 a^3 f \sqrt {a-b+b \sec ^2(e+f x)}} \] Output:

-3/8*(a-5*b)*(a-b)*arctanh(a^(1/2)*sec(f*x+e)/(a-b+b*sec(f*x+e)^2)^(1/2))/ 
a^(7/2)/f-5/8*(a-b)*cot(f*x+e)*csc(f*x+e)/a^2/f/(a-b+b*sec(f*x+e)^2)^(1/2) 
-1/4*cot(f*x+e)^3*csc(f*x+e)/a/f/(a-b+b*sec(f*x+e)^2)^(1/2)-1/8*(13*a-15*b 
)*b*sec(f*x+e)/a^3/f/(a-b+b*sec(f*x+e)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 3.47 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.84 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {\left (\left (-8 a^2+52 a b-60 b^2\right ) \cos (2 (e+f x))+(a-b) (-11 a-45 b+3 (a-5 b) \cos (4 (e+f x)))\right ) \csc ^4(e+f x) \sec (e+f x)}{4 \sqrt {2} a^3 \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}+\frac {3 (a-5 b) (a-b) \cos (e+f x) \left (2 \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right )+\log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{2 a^{7/2} \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}}}{8 f} \] Input:

Integrate[Csc[e + f*x]^5/(a + b*Tan[e + f*x]^2)^(3/2),x]
 

Output:

((((-8*a^2 + 52*a*b - 60*b^2)*Cos[2*(e + f*x)] + (a - b)*(-11*a - 45*b + 3 
*(a - 5*b)*Cos[4*(e + f*x)]))*Csc[e + f*x]^4*Sec[e + f*x])/(4*Sqrt[2]*a^3* 
Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]) + (3*(a - 5*b)*(a 
 - b)*Cos[e + f*x]*(2*ArcTanh[Tan[(e + f*x)/2]^2 - Sqrt[4*b*Tan[(e + f*x)/ 
2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]/Sqrt[a]] + Log[a - 2*b - a*Tan[(e + 
f*x)/2]^2 + Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2] 
^2)^2]])*Sec[(e + f*x)/2]^2*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e 
+ f*x]^2])/(2*a^(7/2)*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x 
)/2]^4]))/(8*f)
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4147, 25, 372, 402, 25, 27, 402, 27, 291, 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.

\(\displaystyle \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x)^5 \left (a+b \tan (e+f x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 4147

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {5 (a-b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {(a-b) \left (\frac {b (13 a-15 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\int -\frac {3 (a-5 b) (a-b)}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{a (a-b)}\right )}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \sqrt {a+b \sec ^2(e+f x)-b}}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {5 (a-b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {(a-b) \left (\frac {3 (a-5 b) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{a}+\frac {b (13 a-15 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \sqrt {a+b \sec ^2(e+f x)-b}}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\frac {5 (a-b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {(a-b) \left (\frac {3 (a-5 b) \int \frac {1}{1-\frac {a \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}}{a}+\frac {b (13 a-15 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \sqrt {a+b \sec ^2(e+f x)-b}}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {5 (a-b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {(a-b) \left (\frac {3 (a-5 b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{a^{3/2}}+\frac {b (13 a-15 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \sqrt {a+b \sec ^2(e+f x)-b}}}{f}\)

Input:

Int[Csc[e + f*x]^5/(a + b*Tan[e + f*x]^2)^(3/2),x]
 

Output:

(-1/4*Sec[e + f*x]/(a*(1 - Sec[e + f*x]^2)^2*Sqrt[a - b + b*Sec[e + f*x]^2 
]) + ((5*(a - b)*Sec[e + f*x])/(2*a*(1 - Sec[e + f*x]^2)*Sqrt[a - b + b*Se 
c[e + f*x]^2]) - ((a - b)*((3*(a - 5*b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqr 
t[a - b + b*Sec[e + f*x]^2]])/a^(3/2) + ((13*a - 15*b)*b*Sec[e + f*x])/(a* 
(a - b)*Sqrt[a - b + b*Sec[e + f*x]^2])))/(2*a))/(4*a))/f
 

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 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 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 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 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 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]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1413\) vs. \(2(167)=334\).

Time = 6.31 (sec) , antiderivative size = 1414, normalized size of antiderivative = 7.56

method	result	size
default	\(\text {Expression too large to display}\)	\(1414\)

Input:

int(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/64/f/a^(9/2)*(((1-cos(f*x+e))^10*csc(f*x+e)^10+7*(cos(f*x+e)-1)^8*csc(f* 
x+e)^8-8*(1-cos(f*x+e))^6*csc(f*x+e)^6-8*(1-cos(f*x+e))^4*csc(f*x+e)^4+7*( 
1-cos(f*x+e))^2*csc(f*x+e)^2+1)*a^(7/2)+(-10*(cos(f*x+e)-1)^8*csc(f*x+e)^8 
+114*(1-cos(f*x+e))^6*csc(f*x+e)^6+114*(1-cos(f*x+e))^4*csc(f*x+e)^4-10*(1 
-cos(f*x+e))^2*csc(f*x+e)^2)*b*a^(5/2)+(-120*(1-cos(f*x+e))^6*csc(f*x+e)^6 
-120*(1-cos(f*x+e))^4*csc(f*x+e)^4)*b^2*a^(3/2)+24*ln(2/a^(1/2)*(a^(1/2)*( 
(a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)+((a*cos 
(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+cos 
(f*x+e)*b+b)/(cos(f*x+e)+1))*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+ 
1)^2)^(1/2)*a^3*(1-cos(f*x+e))^4*csc(f*x+e)^4-144*ln(2/a^(1/2)*(a^(1/2)*(( 
a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)+((a*cos( 
f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+cos( 
f*x+e)*b+b)/(cos(f*x+e)+1))*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1 
)^2)^(1/2)*a^2*b*(1-cos(f*x+e))^4*csc(f*x+e)^4+120*ln(2/a^(1/2)*(a^(1/2)*( 
(a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)+((a*cos 
(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+cos 
(f*x+e)*b+b)/(cos(f*x+e)+1))*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+ 
1)^2)^(1/2)*a*b^2*(1-cos(f*x+e))^4*csc(f*x+e)^4+24*ln(2/(1-cos(f*x+e))^2*( 
-a*(1-cos(f*x+e))^2+2*(1-cos(f*x+e))^2*b+2*((a*cos(f*x+e)^2+b*sin(f*x+e)^2 
)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)*sin(f*x+e)^2+a*sin(f*x+e)^2))*((a*cos...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 356 vs. \(2 (167) = 334\).

Time = 0.30 (sec) , antiderivative size = 722, normalized size of antiderivative = 3.86 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[1/16*(3*((a^3 - 7*a^2*b + 11*a*b^2 - 5*b^3)*cos(f*x + e)^6 - (2*a^3 - 15* 
a^2*b + 28*a*b^2 - 15*b^3)*cos(f*x + e)^4 + a^2*b - 6*a*b^2 + 5*b^3 + (a^3 
 - 9*a^2*b + 23*a*b^2 - 15*b^3)*cos(f*x + e)^2)*sqrt(a)*log(-2*((a - b)*co 
s(f*x + e)^2 - 2*sqrt(a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2) 
*cos(f*x + e) + a + b)/(cos(f*x + e)^2 - 1)) + 2*(3*(a^3 - 6*a^2*b + 5*a*b 
^2)*cos(f*x + e)^5 - (5*a^3 - 31*a^2*b + 30*a*b^2)*cos(f*x + e)^3 - (13*a^ 
2*b - 15*a*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + 
e)^2))/((a^5 - a^4*b)*f*cos(f*x + e)^6 + a^4*b*f - (2*a^5 - 3*a^4*b)*f*cos 
(f*x + e)^4 + (a^5 - 3*a^4*b)*f*cos(f*x + e)^2), -1/8*(3*((a^3 - 7*a^2*b + 
 11*a*b^2 - 5*b^3)*cos(f*x + e)^6 - (2*a^3 - 15*a^2*b + 28*a*b^2 - 15*b^3) 
*cos(f*x + e)^4 + a^2*b - 6*a*b^2 + 5*b^3 + (a^3 - 9*a^2*b + 23*a*b^2 - 15 
*b^3)*cos(f*x + e)^2)*sqrt(-a)*arctan(-sqrt(-a)*sqrt(((a - b)*cos(f*x + e) 
^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/((a - b)*cos(f*x + e)^2 + b)) - (3*(a 
^3 - 6*a^2*b + 5*a*b^2)*cos(f*x + e)^5 - (5*a^3 - 31*a^2*b + 30*a*b^2)*cos 
(f*x + e)^3 - (13*a^2*b - 15*a*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + 
e)^2 + b)/cos(f*x + e)^2))/((a^5 - a^4*b)*f*cos(f*x + e)^6 + a^4*b*f - (2* 
a^5 - 3*a^4*b)*f*cos(f*x + e)^4 + (a^5 - 3*a^4*b)*f*cos(f*x + e)^2)]
 

Sympy [F]

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

integrate(csc(f*x+e)**5/(a+b*tan(f*x+e)**2)**(3/2),x)
 

Output:

Integral(csc(e + f*x)**5/(a + b*tan(e + f*x)**2)**(3/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1250 vs. \(2 (167) = 334\).

Time = 1.87 (sec) , antiderivative size = 1250, normalized size of antiderivative = 6.68 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

1/64*(((((a^8*b - a^7*b^2)*tan(1/2*f*x + 1/2*e)^2/(a^9*b*sgn(tan(1/2*f*x + 
 1/2*e)^2 - 1) - a^8*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)) + (7*a^8*b - 17* 
a^7*b^2 + 10*a^6*b^3)/(a^9*b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - a^8*b^2*sgn 
(tan(1/2*f*x + 1/2*e)^2 - 1)))*tan(1/2*f*x + 1/2*e)^2 - (17*a^8*b - 145*a^ 
7*b^2 + 248*a^6*b^3 - 120*a^5*b^4)/(a^9*b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) 
- a^8*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)))*tan(1/2*f*x + 1/2*e)^2 + (9*a^ 
8*b + 41*a^7*b^2 - 114*a^6*b^3 + 64*a^5*b^4)/(a^9*b*sgn(tan(1/2*f*x + 1/2* 
e)^2 - 1) - a^8*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)))/sqrt(a*tan(1/2*f*x + 
 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) - 
 24*(a^2 - 6*a*b + 5*b^2)*arctan(-(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a 
*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1 
/2*e)^2 + a))/sqrt(-a))/(sqrt(-a)*a^3*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)) - 1 
2*(a^2 - 6*a*b + 5*b^2)*log(abs((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*t 
an(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2 
*e)^2 + a))*sqrt(a) - a + 2*b))/(a^(7/2)*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)) 
+ 4*(4*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2 
*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*a^2 - 16*(s 
qrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/ 
2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*a*b + 14*(sqrt(a)*ta 
n(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x ...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Hanged} \] Input:

int(1/(sin(e + f*x)^5*(a + b*tan(e + f*x)^2)^(3/2)),x)
 

Output:

\text{Hanged}
 

Reduce [F]

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

int(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x)
 

Output:

int((sqrt(tan(e + f*x)**2*b + a)*csc(e + f*x)**5)/(tan(e + f*x)**4*b**2 + 
2*tan(e + f*x)**2*a*b + a**2),x)