\(\int \frac {\csc ^3(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [110]

Optimal result

Integrand size = 25, antiderivative size = 126 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(a-2 b) \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 (a+b)^{5/2} f}-\frac {\cot (e+f x) \csc (e+f x)}{2 (a+b) f \sqrt {a+b \sec ^2(e+f x)}}-\frac {3 b \sec (e+f x)}{2 (a+b)^2 f \sqrt {a+b \sec ^2(e+f x)}} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.77 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left ((a+b) \csc ^2(e+f x)-(a-2 b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1-\frac {a \sin ^2(e+f x)}{a+b}\right )\right ) \sec ^3(e+f x)}{4 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \] Input:

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

Output:

-1/4*((a + 2*b + a*Cos[2*(e + f*x)])*((a + b)*Csc[e + f*x]^2 - (a - 2*b)*H 
ypergeometric2F1[-1/2, 1, 1/2, 1 - (a*Sin[e + f*x]^2)/(a + b)])*Sec[e + f* 
x]^3)/((a + b)^2*f*(a + b*Sec[e + f*x]^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4622, 373, 402, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4622

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

\(\Big \downarrow \) 373

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(Sec[e + f*x]/(2*(a + b)*(1 - Sec[e + f*x]^2)*Sqrt[a + b*Sec[e + f*x]^2]) 
- (((a - 2*b)*ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2 
]])/(a + b)^(3/2) + (3*b*Sec[e + f*x])/((a + b)*Sqrt[a + b*Sec[e + f*x]^2] 
))/(2*(a + b)))/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 4622

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si 
mp[1/(f*ff^m)   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p 
/x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x 
] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1189\) vs. \(2(110)=220\).

Time = 1.51 (sec) , antiderivative size = 1190, normalized size of antiderivative = 9.44

method	result	size
default	\(\text {Expression too large to display}\)	\(1190\)

Input:

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

Output:

1/8/f/(a+b)^(9/2)*((a+b)^(5/2)*b+(a+b)^(5/2)*a+((1-cos(f*x+e))^6*csc(f*x+e 
)^6-(1-cos(f*x+e))^4*csc(f*x+e)^4-(1-cos(f*x+e))^2*csc(f*x+e)^2)*a*(a+b)^( 
5/2)+((1-cos(f*x+e))^6*csc(f*x+e)^6+11*(1-cos(f*x+e))^4*csc(f*x+e)^4+11*(1 
-cos(f*x+e))^2*csc(f*x+e)^2)*b*(a+b)^(5/2)+4*((b+a*cos(f*x+e)^2)/(1+cos(f* 
x+e))^2)^(1/2)*ln(-4*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1 
/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+cos 
(f*x+e)*a+b)/(-1+cos(f*x+e)))*a^3*(1-cos(f*x+e))^2*csc(f*x+e)^2-12*b^2*ln( 
-4*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+(a+ 
b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+cos(f*x+e)*a+b)/(-1+c 
os(f*x+e)))*a*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(1-cos(f*x+e))^2 
*csc(f*x+e)^2-8*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*((a+b)^( 
1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*(( 
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+cos(f*x+e)*a+b)/(-1+cos(f*x+e))) 
*b^3*(1-cos(f*x+e))^2*csc(f*x+e)^2+4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2) 
^(1/2)*ln(2/(a+b)^(1/2)*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2) 
^(1/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)- 
cos(f*x+e)*a+b)/(1+cos(f*x+e)))*a^3*(1-cos(f*x+e))^2*csc(f*x+e)^2-12*b^2*l 
n(2/(a+b)^(1/2)*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*c 
os(f*x+e)+(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-cos(f*x+ 
e)*a+b)/(1+cos(f*x+e)))*a*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (110) = 220\).

Time = 0.22 (sec) , antiderivative size = 556, normalized size of antiderivative = 4.41 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {{\left ({\left (a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 2 \, b^{2}\right )} \sqrt {a + b} \log \left (\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {a + b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left ({\left (a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left ({\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{4} + 2 \, a^{3} b - 2 \, a b^{3} - b^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} f\right )}}, \frac {{\left ({\left (a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 2 \, b^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-a - b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a \cos \left (f x + e\right )^{2} + b}\right ) + {\left ({\left (a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left ({\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{4} + 2 \, a^{3} b - 2 \, a b^{3} - b^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} f\right )}}\right ] \] Input:

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

Output:

[-1/4*(((a^2 - 2*a*b)*cos(f*x + e)^4 - (a^2 - 3*a*b + 2*b^2)*cos(f*x + e)^ 
2 - a*b + 2*b^2)*sqrt(a + b)*log(2*(a*cos(f*x + e)^2 + 2*sqrt(a + b)*sqrt( 
(a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + 2*b)/(cos(f*x + 
e)^2 - 1)) - 2*((a^2 - a*b - 2*b^2)*cos(f*x + e)^3 + 3*(a*b + b^2)*cos(f*x 
 + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^4 + 3*a^3*b + 3*a^ 
2*b^2 + a*b^3)*f*cos(f*x + e)^4 - (a^4 + 2*a^3*b - 2*a*b^3 - b^4)*f*cos(f* 
x + e)^2 - (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*f), 1/2*(((a^2 - 2*a*b)*cos 
(f*x + e)^4 - (a^2 - 3*a*b + 2*b^2)*cos(f*x + e)^2 - a*b + 2*b^2)*sqrt(-a 
- b)*arctan(sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f 
*x + e)/(a*cos(f*x + e)^2 + b)) + ((a^2 - a*b - 2*b^2)*cos(f*x + e)^3 + 3* 
(a*b + b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a 
^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^4 - (a^4 + 2*a^3*b - 2*a* 
b^3 - b^4)*f*cos(f*x + e)^2 - (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

integrate(csc(f*x+e)^3/(a+b*sec(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. 903 vs. \(2 (110) = 220\).

Time = 0.88 (sec) , antiderivative size = 903, normalized size of antiderivative = 7.17 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/8*((((a^5*b*sgn(cos(f*x + e)) + 4*a^4*b^2*sgn(cos(f*x + e)) + 6*a^3*b^3 
*sgn(cos(f*x + e)) + 4*a^2*b^4*sgn(cos(f*x + e)) + a*b^5*sgn(cos(f*x + e)) 
)*tan(1/2*f*x + 1/2*e)^2/(a^6*b + 5*a^5*b^2 + 10*a^4*b^3 + 10*a^3*b^4 + 5* 
a^2*b^5 + a*b^6) - 2*(a^5*b*sgn(cos(f*x + e)) - 2*a^4*b^2*sgn(cos(f*x + e) 
) - 12*a^3*b^3*sgn(cos(f*x + e)) - 14*a^2*b^4*sgn(cos(f*x + e)) - 5*a*b^5* 
sgn(cos(f*x + e)))/(a^6*b + 5*a^5*b^2 + 10*a^4*b^3 + 10*a^3*b^4 + 5*a^2*b^ 
5 + a*b^6))*tan(1/2*f*x + 1/2*e)^2 + (a^5*b*sgn(cos(f*x + e)) + 12*a^4*b^2 
*sgn(cos(f*x + e)) + 30*a^3*b^3*sgn(cos(f*x + e)) + 28*a^2*b^4*sgn(cos(f*x 
 + e)) + 9*a*b^5*sgn(cos(f*x + e)))/(a^6*b + 5*a^5*b^2 + 10*a^4*b^3 + 10*a 
^3*b^4 + 5*a^2*b^5 + a*b^6))/sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x 
 + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a 
+ b) - 4*(a - 2*b)*arctan(-(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*ta 
n(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e) 
^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))/sqrt(-a - b))/((a^2 + 2*a*b + b^ 
2)*sqrt(-a - b)*sgn(cos(f*x + e))) - 2*(a - 2*b)*log(abs((sqrt(a + b)*tan( 
1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e 
)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))*sq 
rt(a + b) - a + b))/((a^2 + 2*a*b + b^2)*sqrt(a + b)*sgn(cos(f*x + e))) + 
2*((sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b 
*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x ...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sec(e + f*x)**2*b + a)*csc(e + f*x)**3)/(sec(e + f*x)**4*b**2 + 
2*sec(e + f*x)**2*a*b + a**2),x)