\(\int \frac {1}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [437]

Optimal result

Integrand size = 16, antiderivative size = 125 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{5/2} f}-\frac {b \tan (e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (5 a+3 b) \tan (e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}} \] Output:

arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(5/2)/f-1/3*b*tan( 
f*x+e)/a/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(3/2)-1/3*b*(5*a+3*b)*tan(f*x+e)/a^2 
/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 6.29 (sec) , antiderivative size = 1927, normalized size of antiderivative = 15.42 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

Integrate[(a + b*Sec[e + f*x]^2)^(-5/2),x]
 

Output:

(3*(a + b)*AppellF1[1/2, -2, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/ 
(a + b)]*Cos[e + f*x]^4*Sin[e + f*x])/(4*Sqrt[2]*f*(a + b*Sec[e + f*x]^2)^ 
(5/2)*(a + b - a*Sin[e + f*x]^2)^(5/2)*(3*(a + b)*AppellF1[1/2, -2, 5/2, 3 
/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (5*a*AppellF1[3/2, -2, 7 
/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 4*(a + b)*AppellF1[ 
3/2, -1, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f* 
x]^2)*((15*a*(a + b)*AppellF1[1/2, -2, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e 
+ f*x]^2)/(a + b)]*Cos[e + f*x]^5*Sin[e + f*x]^2)/(4*Sqrt[2]*(a + b - a*Si 
n[e + f*x]^2)^(7/2)*(3*(a + b)*AppellF1[1/2, -2, 5/2, 3/2, Sin[e + f*x]^2, 
 (a*Sin[e + f*x]^2)/(a + b)] + (5*a*AppellF1[3/2, -2, 7/2, 5/2, Sin[e + f* 
x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 4*(a + b)*AppellF1[3/2, -1, 5/2, 5/2, 
Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f*x]^2)) + (3*(a + b) 
*AppellF1[1/2, -2, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*C 
os[e + f*x]^5)/(4*Sqrt[2]*(a + b - a*Sin[e + f*x]^2)^(5/2)*(3*(a + b)*Appe 
llF1[1/2, -2, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (5*a 
*AppellF1[3/2, -2, 7/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 
 4*(a + b)*AppellF1[3/2, -1, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/ 
(a + b)])*Sin[e + f*x]^2)) - (3*(a + b)*AppellF1[1/2, -2, 5/2, 3/2, Sin[e 
+ f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^3*Sin[e + f*x]^2)/(Sqrt 
[2]*(a + b - a*Sin[e + f*x]^2)^(5/2)*(3*(a + b)*AppellF1[1/2, -2, 5/2, ...
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {3042, 4616, 316, 402, 27, 291, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4616

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {3 (a+b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2}}-\frac {b (5 a+3 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

Input:

Int[(a + b*Sec[e + f*x]^2)^(-5/2),x]
 

Output:

(-1/3*(b*Tan[e + f*x])/(a*(a + b)*(a + b + b*Tan[e + f*x]^2)^(3/2)) + ((3* 
(a + b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/a^( 
3/2) - (b*(5*a + 3*b)*Tan[e + f*x])/(a*(a + b)*Sqrt[a + b + b*Tan[e + f*x] 
^2]))/(3*a*(a + b)))/f
 

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[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 4616

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = 
FreeFactors[Tan[e + f*x], x]}, Simp[ff/f   Subst[Int[(a + b + b*ff^2*x^2)^p 
/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] 
&& NeQ[a + b, 0] && NeQ[p, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(860\) vs. \(2(111)=222\).

Time = 6.07 (sec) , antiderivative size = 861, normalized size of antiderivative = 6.89

method	result	size
default	\(\text {Expression too large to display}\)	\(861\)

Input:

int(1/(a+b*sec(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/f*(1/3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^4*ln(4*(-a)^(1/2)*( 
(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*c 
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*(3+3*sec(f*x+e))+1/3* 
((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b*ln(4*(-a)^(1/2)*((b+a*co 
s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+ 
e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*(6+6*sec(f*x+e)+6*sec(f*x+e) 
^2+6*sec(f*x+e)^3)+1/3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b^2 
*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4* 
(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*(3+ 
3*sec(f*x+e)+12*sec(f*x+e)^2+12*sec(f*x+e)^3+3*sec(f*x+e)^4+3*sec(f*x+e)^5 
)+1/3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b^3*ln(4*(-a)^(1/2)*(( 
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*co 
s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*(6*sec(f*x+e)^2+6*sec( 
f*x+e)^3+6*sec(f*x+e)^4+6*sec(f*x+e)^5)+1/3*((b+a*cos(f*x+e)^2)/(1+cos(f*x 
+e))^2)^(1/2)*b^4*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1 
/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4* 
sin(f*x+e)*a)*(3*sec(f*x+e)^4+3*sec(f*x+e)^5)-2*(-a)^(1/2)*a^3*b*tan(f*x+e 
)+1/3*(-a)^(1/2)*a^2*b^2*(-4*tan(f*x+e)-11*sec(f*x+e)^2*tan(f*x+e))+1/3*(- 
7*cos(f*x+e)^2-5)*(-a)^(1/2)*a*b^3*tan(f*x+e)*sec(f*x+e)^4-(-a)^(1/2)*b^4* 
tan(f*x+e)*sec(f*x+e)^4)/(a+b)^2/a^2/(-a)^(1/2)/(a+b*sec(f*x+e)^2)^(5/2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (111) = 222\).

Time = 0.55 (sec) , antiderivative size = 881, normalized size of antiderivative = 7.05 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(1/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")
 

Output:

[-1/24*(3*((a^4 + 2*a^3*b + a^2*b^2)*cos(f*x + e)^4 + a^2*b^2 + 2*a*b^3 + 
b^4 + 2*(a^3*b + 2*a^2*b^2 + a*b^3)*cos(f*x + e)^2)*sqrt(-a)*log(128*a^4*c 
os(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 
5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 
 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 + 8*(16*a^3*cos(f*x 
 + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2) 
*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*s 
qrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 8*(2*(3*a^3*b + 
 2*a^2*b^2)*cos(f*x + e)^3 + (5*a^2*b^2 + 3*a*b^3)*cos(f*x + e))*sqrt((a*c 
os(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^7 + 2*a^6*b + a^5*b^2 
)*f*cos(f*x + e)^4 + 2*(a^6*b + 2*a^5*b^2 + a^4*b^3)*f*cos(f*x + e)^2 + (a 
^5*b^2 + 2*a^4*b^3 + a^3*b^4)*f), -1/12*(3*((a^4 + 2*a^3*b + a^2*b^2)*cos( 
f*x + e)^4 + a^2*b^2 + 2*a*b^3 + b^4 + 2*(a^3*b + 2*a^2*b^2 + a*b^3)*cos(f 
*x + e)^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f* 
x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^ 
2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a 
^2*b)*cos(f*x + e)^2)*sin(f*x + e))) + 4*(2*(3*a^3*b + 2*a^2*b^2)*cos(f*x 
+ e)^3 + (5*a^2*b^2 + 3*a*b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/c 
os(f*x + e)^2)*sin(f*x + e))/((a^7 + 2*a^6*b + a^5*b^2)*f*cos(f*x + e)^4 + 
 2*(a^6*b + 2*a^5*b^2 + a^4*b^3)*f*cos(f*x + e)^2 + (a^5*b^2 + 2*a^4*b^...
 

Sympy [F]

\[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(a+b*sec(f*x+e)**2)**(5/2),x)
 

Output:

Integral((a + b*sec(e + f*x)**2)**(-5/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(1/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^(-5/2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}}{\sec \left (f x +e \right )^{6} b^{3}+3 \sec \left (f x +e \right )^{4} a \,b^{2}+3 \sec \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:

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

Output:

int(sqrt(sec(e + f*x)**2*b + a)/(sec(e + f*x)**6*b**3 + 3*sec(e + f*x)**4* 
a*b**2 + 3*sec(e + f*x)**2*a**2*b + a**3),x)