3.1.84 \(\int \csc ^3(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [84]

3.1.84.1 Optimal result

Integrand size = 25, antiderivative size = 161 \[ \int \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {b} (3 a+4 b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f}-\frac {\sqrt {a+b} (a+4 b) \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{f}-\frac {\cot (e+f x) \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{2 f} \]

output

-1/2*cot(f*x+e)*csc(f*x+e)*(a+b*sec(f*x+e)^2)^(3/2)/f+1/2*(3*a+4*b)*arctan 
h(sec(f*x+e)*b^(1/2)/(a+b*sec(f*x+e)^2)^(1/2))*b^(1/2)/f-1/2*(a+4*b)*arcta 
nh(sec(f*x+e)*(a+b)^(1/2)/(a+b*sec(f*x+e)^2)^(1/2))*(a+b)^(1/2)/f+b*sec(f* 
x+e)*(a+b*sec(f*x+e)^2)^(1/2)/f
 

3.1.84.2 Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.25 \[ \int \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {\csc ^2(e+f x) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)} \left (\sqrt {2} \sqrt {a+2 b+a \cos (2 (e+f x))} (a+(a+2 b) \cos (2 (e+f x)))-\sqrt {b} (3 a+4 b) \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {b}}\right ) \sin ^2(2 (e+f x))+\sqrt {a+b} (a+4 b) \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {a+b}}\right ) \sin ^2(2 (e+f x))\right )}{4 \sqrt {2} f \sqrt {a+2 b+a \cos (2 (e+f x))}} \]

input

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

output

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

3.1.84.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4622, 369, 403, 27, 398, 224, 219, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4622

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

\(\Big \downarrow \) 369

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

\(\Big \downarrow \) 403

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

input

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

output

((Sec[e + f*x]*(a + b*Sec[e + f*x]^2)^(3/2))/(2*(1 - Sec[e + f*x]^2)) + (S 
qrt[b]*(3*a + 4*b)*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^ 
2]] - Sqrt[a + b]*(a + 4*b)*ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b* 
Sec[e + f*x]^2]] + 2*b*Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/2)/f
 

3.1.84.3.1 Defintions of rubi rules used

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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 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 369

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/(2* 
b*(p + 1))), x] - Simp[e^2/(2*b*(p + 1))   Int[(e*x)^(m - 2)*(a + b*x^2)^(p 
 + 1)*(c + d*x^2)^(q - 1)*Simp[c*(m - 1) + d*(m + 2*q - 1)*x^2, x], x], x] 
/; FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 0 
] && GtQ[m, 1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
 

rule 398

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

rule 403

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( 
x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + 
 q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1))   Int[(a + b*x^2)^p*(c 
+ d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + 
f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
 

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])
 

3.1.84.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3469\) vs. \(2(139)=278\).

Time = 6.92 (sec) , antiderivative size = 3470, normalized size of antiderivative = 21.55

method	result	size
default	\(\text {Expression too large to display}\)	\(3470\)

input

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

output

-1/4/f/b/(a+b)^(7/2)*(14*ln(-4*b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^ 
2)^(1/2)-4*b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*sec(f*x+e)- 
4*sec(f*x+e)*b)*cos(f*x+e)^3*(a+b)^(5/2)*b^(5/2)*a+6*ln(-4*b^(1/2)*((b+a*c 
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos( 
f*x+e))^2)^(1/2)*sec(f*x+e)-4*sec(f*x+e)*b)*cos(f*x+e)^3*(a+b)^(5/2)*b^(3/ 
2)*a^2-14*ln(-4*b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*b^(1 
/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*sec(f*x+e)-4*sec(f*x+e)*b) 
*cos(f*x+e)^2*(a+b)^(5/2)*b^(5/2)*a-6*ln(-4*b^(1/2)*((b+a*cos(f*x+e)^2)/(1 
+cos(f*x+e))^2)^(1/2)-4*b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2 
)*sec(f*x+e)-4*sec(f*x+e)*b)*cos(f*x+e)^2*(a+b)^(5/2)*b^(3/2)*a^2+2*((b+a* 
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)^2*(a+b)^(5/2)*a^2*b+6*((b 
+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)^2*(a+b)^(5/2)*a*b^2-2* 
(a+b)^(5/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^3-4*ln(2/(a+b)^( 
1/2)*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)*cos(f*x+e)+( 
(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-cos(f*x+e)*a+b)/(1+ 
cos(f*x+e)))*cos(f*x+e)^3*b^6-4*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^ 
2)^(1/2)*(a+b)^(1/2)*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2 
)*(a+b)^(1/2)+cos(f*x+e)*a+b)/(-1+cos(f*x+e)))*cos(f*x+e)^3*b^6+4*ln(2/(a+ 
b)^(1/2)*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)*cos(f*x+ 
e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-cos(f*x+e)*a...
 

3.1.84.5 Fricas [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 984, normalized size of antiderivative = 6.11 \[ \int \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]

input

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

output

[1/4*(((a + 4*b)*cos(f*x + e)^3 - (a + 4*b)*cos(f*x + e))*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)) + ((3*a + 4*b)*cos(f*x 
 + e)^3 - (3*a + 4*b)*cos(f*x + e))*sqrt(b)*log((a*cos(f*x + e)^2 + 2*sqrt 
(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f* 
x + e)^2) + 2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt((a*cos(f*x + e)^2 + b)/c 
os(f*x + e)^2))/(f*cos(f*x + e)^3 - f*cos(f*x + e)), 1/4*(2*((a + 4*b)*cos 
(f*x + e)^3 - (a + 4*b)*cos(f*x + e))*sqrt(-a - b)*arctan(sqrt(-a - b)*sqr 
t((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a + b)) + ((3*a + 4 
*b)*cos(f*x + e)^3 - (3*a + 4*b)*cos(f*x + e))*sqrt(b)*log((a*cos(f*x + e) 
^2 + 2*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 
2*b)/cos(f*x + e)^2) + 2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt((a*cos(f*x + 
e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)^3 - f*cos(f*x + e)), -1/4*(2*(( 
3*a + 4*b)*cos(f*x + e)^3 - (3*a + 4*b)*cos(f*x + e))*sqrt(-b)*arctan(sqrt 
(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) - ((a + 4 
*b)*cos(f*x + e)^3 - (a + 4*b)*cos(f*x + e))*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*b)*cos(f*x + e)^2 - b)* 
sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)^3 - f*cos(f*x 
 + e)), 1/2*(((a + 4*b)*cos(f*x + e)^3 - (a + 4*b)*cos(f*x + e))*sqrt(-...
 

3.1.84.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.1.84.7 Maxima [F]

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

input

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

output

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

3.1.84.8 Giac [F(-2)]

Exception generated. \[ \int \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Exception raised: TypeError} \]

input

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

output

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%{[%%%{16384,[5,4]%%%},0]:[1,0,%%%{-1,[1,0]%%%}+%%%{-1,[0 
,1]%%%}]%
 

3.1.84.9 Mupad [F(-1)]

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

input

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

output

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