\(\int \frac {\csc ^3(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\) [144]

Optimal result

Integrand size = 25, antiderivative size = 177 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a-5 b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 a^{7/2} f}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {5 b \sec (e+f x)}{6 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(13 a-15 b) b \sec (e+f x)}{6 a^3 (a-b) f \sqrt {a-b+b \sec ^2(e+f x)}} \] Output:

-1/2*(a-5*b)*arctanh(a^(1/2)*sec(f*x+e)/(a-b+b*sec(f*x+e)^2)^(1/2))/a^(7/2 
)/f-1/2*cot(f*x+e)*csc(f*x+e)/a/f/(a-b+b*sec(f*x+e)^2)^(3/2)-5/6*b*sec(f*x 
+e)/a^2/f/(a-b+b*sec(f*x+e)^2)^(3/2)-1/6*(13*a-15*b)*b*sec(f*x+e)/a^3/(a-b 
)/f/(a-b+b*sec(f*x+e)^2)^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(380\) vs. \(2(177)=354\).

Time = 3.53 (sec) , antiderivative size = 380, normalized size of antiderivative = 2.15 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {\sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (8 a b^2 \cos (e+f x)-24 (a-b) b \cos (e+f x) (a+b+(a-b) \cos (2 (e+f x)))-3 (a-b) (a+b+(a-b) \cos (2 (e+f x)))^2 \cot (e+f x) \csc (e+f x)\right )}{3 a^3 (a-b) (a+b+(a-b) \cos (2 (e+f x)))^2}+\frac {(a-5 b) \cos (e+f x) \left (2 \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right )+\log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{2 a^{7/2} \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}}}{2 f} \] Input:

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

Output:

((Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(8*a*b^2 
*Cos[e + f*x] - 24*(a - b)*b*Cos[e + f*x]*(a + b + (a - b)*Cos[2*(e + f*x) 
]) - 3*(a - b)*(a + b + (a - b)*Cos[2*(e + f*x)])^2*Cot[e + f*x]*Csc[e + f 
*x]))/(3*a^3*(a - b)*(a + b + (a - b)*Cos[2*(e + f*x)])^2) + ((a - 5*b)*Co 
s[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]) 
)/(2*f)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4147, 373, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4147

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

\(\Big \downarrow \) 373

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(Sec[e + f*x]/(2*a*(1 - Sec[e + f*x]^2)*(a - b + b*Sec[e + f*x]^2)^(3/2)) 
- ((5*b*Sec[e + f*x])/(3*a*(a - b + b*Sec[e + f*x]^2)^(3/2)) + ((3*(a - 5* 
b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[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 
]))/(3*a))/(2*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 373

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 
1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1))   Int[(e 
*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 
 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[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. \(3761\) vs. \(2(157)=314\).

Time = 174.00 (sec) , antiderivative size = 3762, normalized size of antiderivative = 21.25

method	result	size
default	\(\text {Expression too large to display}\)	\(3762\)

Input:

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

Output:

-1/12/f/a^(17/2)/(a^2-2*a*b+b^2)/(2*(-b*(a-b))^(1/2)+a-2*b)^3/(2*(-b*(a-b) 
)^(1/2)-a+2*b)^3*(6*a^(37/2)*cos(f*x+e)^10+cos(f*x+e)^8*(-66*cos(f*x+e)^2+ 
54)*a^(35/2)*b+cos(f*x+e)^6*(270*cos(f*x+e)^4-416*cos(f*x+e)^2+152)*a^(33/ 
2)*b^2+cos(f*x+e)^4*(570*cos(f*x+e)^4-704*cos(f*x+e)^2+192)*sin(f*x+e)^2*a 
^(31/2)*b^3+cos(f*x+e)^2*(690*cos(f*x+e)^4-636*cos(f*x+e)^2+114)*sin(f*x+e 
)^4*a^(29/2)*b^4+(486*cos(f*x+e)^4-296*cos(f*x+e)^2+26)*sin(f*x+e)^6*a^(27 
/2)*b^5+(186*cos(f*x+e)^2-56)*sin(f*x+e)^8*a^(25/2)*b^6+30*sin(f*x+e)^10*a 
^(23/2)*b^7+cos(f*x+e)^8*(3*cos(f*x+e)+3)*sin(f*x+e)^2*((a*cos(f*x+e)^2+b* 
sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*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*s 
in(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^18+cos(f*x+e)^6*(-33*cos(f*x+e)^3-33*cos(f*x+e)^2+12*cos 
(f*x+e)+12)*sin(f*x+e)^2*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2 
)^(1/2)*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^17*b+cos(f*x 
+e)^4*(135*cos(f*x+e)^5+135*cos(f*x+e)^4-120*cos(f*x+e)^3-120*cos(f*x+e)^2 
+18*cos(f*x+e)+18)*sin(f*x+e)^2*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+ 
e)+1)^2)^(1/2)*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*...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 448 vs. \(2 (157) = 314\).

Time = 0.30 (sec) , antiderivative size = 906, normalized size of antiderivative = 5.12 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/12*(3*((a^4 - 8*a^3*b + 18*a^2*b^2 - 16*a*b^3 + 5*b^4)*cos(f*x + e)^6 
- (a^4 - 10*a^3*b + 32*a^2*b^2 - 38*a*b^3 + 15*b^4)*cos(f*x + e)^4 - a^2*b 
^2 + 6*a*b^3 - 5*b^4 - (2*a^3*b - 15*a^2*b^2 + 28*a*b^3 - 15*b^4)*cos(f*x 
+ e)^2)*sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 + 2*sqrt(a)*sqrt(((a - b)*c 
os(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^4 - 7*a^3*b + 11*a^2*b^2 - 5*a*b^3)*cos(f*x + e)^5 + 2*(9*a 
^3*b - 23*a^2*b^2 + 15*a*b^3)*cos(f*x + e)^3 + (13*a^2*b^2 - 15*a*b^3)*cos 
(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 - 3*a^ 
6*b + 3*a^5*b^2 - a^4*b^3)*f*cos(f*x + e)^6 - (a^7 - 5*a^6*b + 7*a^5*b^2 - 
 3*a^4*b^3)*f*cos(f*x + e)^4 - (2*a^6*b - 5*a^5*b^2 + 3*a^4*b^3)*f*cos(f*x 
 + e)^2 - (a^5*b^2 - a^4*b^3)*f), -1/6*(3*((a^4 - 8*a^3*b + 18*a^2*b^2 - 1 
6*a*b^3 + 5*b^4)*cos(f*x + e)^6 - (a^4 - 10*a^3*b + 32*a^2*b^2 - 38*a*b^3 
+ 15*b^4)*cos(f*x + e)^4 - a^2*b^2 + 6*a*b^3 - 5*b^4 - (2*a^3*b - 15*a^2*b 
^2 + 28*a*b^3 - 15*b^4)*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^4 - 7*a^3*b + 11*a^2*b^2 - 5*a*b^3)*cos(f*x + e)^5 + 2* 
(9*a^3*b - 23*a^2*b^2 + 15*a*b^3)*cos(f*x + e)^3 + (13*a^2*b^2 - 15*a*b^3) 
*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 - 
3*a^6*b + 3*a^5*b^2 - a^4*b^3)*f*cos(f*x + e)^6 - (a^7 - 5*a^6*b + 7*a^5*b 
^2 - 3*a^4*b^3)*f*cos(f*x + e)^4 - (2*a^6*b - 5*a^5*b^2 + 3*a^4*b^3)*f*...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1111 vs. \(2 (157) = 314\).

Time = 2.16 (sec) , antiderivative size = 1111, normalized size of antiderivative = 6.28 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/24*(((((3*(a^12*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 2*a^11*b^3*sgn(tan 
(1/2*f*x + 1/2*e)^2 - 1) + a^10*b^4*sgn(tan(1/2*f*x + 1/2*e)^2 - 1))*tan(1 
/2*f*x + 1/2*e)^2/(a^13*b^2 - 2*a^12*b^3 + a^11*b^4) - 4*(3*a^12*b^2*sgn(t 
an(1/2*f*x + 1/2*e)^2 - 1) - 24*a^11*b^3*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 
 41*a^10*b^4*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 20*a^9*b^5*sgn(tan(1/2*f*x 
+ 1/2*e)^2 - 1))/(a^13*b^2 - 2*a^12*b^3 + a^11*b^4))*tan(1/2*f*x + 1/2*e)^ 
2 + 6*(3*a^12*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 22*a^11*b^3*sgn(tan(1/ 
2*f*x + 1/2*e)^2 - 1) + 71*a^10*b^4*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 92*a 
^9*b^5*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 40*a^8*b^6*sgn(tan(1/2*f*x + 1/2* 
e)^2 - 1))/(a^13*b^2 - 2*a^12*b^3 + a^11*b^4))*tan(1/2*f*x + 1/2*e)^2 - 12 
*(a^12*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 17*a^10*b^4*sgn(tan(1/2*f*x + 
 1/2*e)^2 - 1) + 32*a^9*b^5*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 16*a^8*b^6*s 
gn(tan(1/2*f*x + 1/2*e)^2 - 1))/(a^13*b^2 - 2*a^12*b^3 + a^11*b^4))*tan(1/ 
2*f*x + 1/2*e)^2 + (3*a^12*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 42*a^11*b 
^3*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 101*a^10*b^4*sgn(tan(1/2*f*x + 1/2*e) 
^2 - 1) + 56*a^9*b^5*sgn(tan(1/2*f*x + 1/2*e)^2 - 1))/(a^13*b^2 - 2*a^12*b 
^3 + a^11*b^4))/(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)^(3/2) - 12*(a - 5*b)*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...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^3\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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