\(\int \frac {\sec ^3(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [291]

Optimal result

Integrand size = 25, antiderivative size = 134 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {(a+4 b) \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 (a+b)^{5/2} f}-\frac {(a-2 b) b \sin (e+f x)}{2 a (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sec (e+f x) \tan (e+f x)}{2 (a+b) f \sqrt {a+b \sin ^2(e+f x)}} \] Output:

1/2*(a+4*b)*arctanh((a+b)^(1/2)*sin(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2))/(a+b) 
^(5/2)/f-1/2*(a-2*b)*b*sin(f*x+e)/a/(a+b)^2/f/(a+b*sin(f*x+e)^2)^(1/2)+1/2 
*sec(f*x+e)*tan(f*x+e)/(a+b)/f/(a+b*sin(f*x+e)^2)^(1/2)
 

Mathematica [C] (warning: unable to verify)

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

Time = 0.51 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.67 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\sec ^5(e+f x) \left (16 (a+b) \, _3F_2\left (2,2,3;1,\frac {9}{2};-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2+16 (a+b) \operatorname {Hypergeometric2F1}\left (2,3,\frac {9}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \left (4 a^2+7 a b \sin ^2(e+f x)+3 b^2 \sin ^4(e+f x)\right )-7 a \cos ^2(e+f x) \operatorname {Hypergeometric2F1}\left (1,2,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \left (15 a^2+20 a b \sin ^2(e+f x)+8 b^2 \sin ^4(e+f x)\right )\right ) \tan (e+f x)}{105 a^4 f \sqrt {a+b \sin ^2(e+f x)}} \] Input:

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

Output:

-1/105*(Sec[e + f*x]^5*(16*(a + b)*HypergeometricPFQ[{2, 2, 3}, {1, 9/2}, 
-(((a + b)*Tan[e + f*x]^2)/a)]*Sin[e + f*x]^2*(a + b*Sin[e + f*x]^2)^2 + 1 
6*(a + b)*Hypergeometric2F1[2, 3, 9/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sin[ 
e + f*x]^2*(4*a^2 + 7*a*b*Sin[e + f*x]^2 + 3*b^2*Sin[e + f*x]^4) - 7*a*Cos 
[e + f*x]^2*Hypergeometric2F1[1, 2, 7/2, -(((a + b)*Tan[e + f*x]^2)/a)]*(1 
5*a^2 + 20*a*b*Sin[e + f*x]^2 + 8*b^2*Sin[e + f*x]^4))*Tan[e + f*x])/(a^4* 
f*Sqrt[a + b*Sin[e + f*x]^2])
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 142, 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, 3669, 316, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3669

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(Sin[e + f*x]/(2*(a + b)*(1 - Sin[e + f*x]^2)*Sqrt[a + b*Sin[e + f*x]^2]) 
+ (((a + 4*b)*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2 
]])/(a + b)^(3/2) - ((a - 2*b)*b*Sin[e + f*x])/(a*(a + b)*Sqrt[a + b*Sin[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 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 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 3669

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f   S 
ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] 
/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3218\) vs. \(2(118)=236\).

Time = 0.63 (sec) , antiderivative size = 3219, normalized size of antiderivative = 24.02

method	result	size
default	\(\text {Expression too large to display}\)	\(3219\)

Input:

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

Output:

1/4/(a+b)^(1/2)/a/b^5/cos(f*x+e)^2/(a^4*b^2*cos(f*x+e)^4+4*a^3*b^3*cos(f*x 
+e)^4+6*a^2*b^4*cos(f*x+e)^4+4*a*b^5*cos(f*x+e)^4+b^6*cos(f*x+e)^4-2*a^5*b 
*cos(f*x+e)^2-10*a^4*b^2*cos(f*x+e)^2-20*a^3*b^3*cos(f*x+e)^2-20*a^2*b^4*c 
os(f*x+e)^2-10*a*b^5*cos(f*x+e)^2-2*b^6*cos(f*x+e)^2+a^6+6*a^5*b+15*a^4*b^ 
2+20*a^3*b^3+15*b^4*a^2+6*b^5*a+b^6)*(-a*(8*b^(19/2)*(a+b)^(1/2)*ln(((a+b- 
b*cos(f*x+e)^2)^(1/2)*b^(1/2)+b*sin(f*x+e))/b^(1/2))-8*b^(19/2)*(a+b)^(1/2 
)*ln(((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+b^2*sin(f*x+e))/b^(3 
/2))+24*b^(17/2)*(a+b)^(1/2)*ln(((a+b-b*cos(f*x+e)^2)^(1/2)*b^(1/2)+b*sin( 
f*x+e))/b^(1/2))*a-24*b^(17/2)*(a+b)^(1/2)*ln(((-b*cos(f*x+e)^2+(a*b^2+b^3 
)/b^2)^(1/2)*b^(3/2)+b^2*sin(f*x+e))/b^(3/2))*a+24*b^(15/2)*(a+b)^(1/2)*ln 
(((a+b-b*cos(f*x+e)^2)^(1/2)*b^(1/2)+b*sin(f*x+e))/b^(1/2))*a^2-24*b^(15/2 
)*(a+b)^(1/2)*ln(((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+b^2*sin( 
f*x+e))/b^(3/2))*a^2+8*b^(13/2)*(a+b)^(1/2)*ln(((a+b-b*cos(f*x+e)^2)^(1/2) 
*b^(1/2)+b*sin(f*x+e))/b^(1/2))*a^3-8*b^(13/2)*(a+b)^(1/2)*ln(((-b*cos(f*x 
+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^(3/2)+b^2*sin(f*x+e))/b^(3/2))*a^3-ln(2/(si 
n(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^5*b 
^5-8*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x 
+e)+a))*a^4*b^6-22*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^( 
1/2)+b*sin(f*x+e)+a))*a^3*b^7-28*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*c 
os(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^2*b^8-17*ln(2/(sin(f*x+e)-1)*((a+...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (118) = 236\).

Time = 0.36 (sec) , antiderivative size = 625, normalized size of antiderivative = 4.66 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {{\left ({\left (a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a^{3} + 5 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a + b} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2} - {\left (a^{2} b - a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{8 \, {\left ({\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2}\right )}}, -\frac {{\left ({\left (a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a^{3} + 5 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) + 2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2} - {\left (a^{2} b - a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{4 \, {\left ({\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2}\right )}}\right ] \] Input:

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

Output:

[1/8*(((a^2*b + 4*a*b^2)*cos(f*x + e)^4 - (a^3 + 5*a^2*b + 4*a*b^2)*cos(f* 
x + e)^2)*sqrt(a + b)*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 8*(a^2 + 
 3*a*b + 2*b^2)*cos(f*x + e)^2 - 4*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)* 
sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b)*sin(f*x + e) + 8*a^2 + 16*a*b 
+ 8*b^2)/cos(f*x + e)^4) - 4*(a^3 + 2*a^2*b + a*b^2 - (a^2*b - a*b^2 - 2*b 
^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a^4*b 
+ 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*f*cos(f*x + e)^4 - (a^5 + 4*a^4*b + 6*a^3 
*b^2 + 4*a^2*b^3 + a*b^4)*f*cos(f*x + e)^2), -1/4*(((a^2*b + 4*a*b^2)*cos( 
f*x + e)^4 - (a^3 + 5*a^2*b + 4*a*b^2)*cos(f*x + e)^2)*sqrt(-a - b)*arctan 
(1/2*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b 
)*sqrt(-a - b)/(((a*b + b^2)*cos(f*x + e)^2 - a^2 - 2*a*b - b^2)*sin(f*x + 
 e))) + 2*(a^3 + 2*a^2*b + a*b^2 - (a^2*b - a*b^2 - 2*b^3)*cos(f*x + e)^2) 
*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a^4*b + 3*a^3*b^2 + 3*a^2 
*b^3 + a*b^4)*f*cos(f*x + e)^4 - (a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + 
a*b^4)*f*cos(f*x + e)^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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