\(\int \frac {\sec ^3(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [271]

Optimal result

Integrand size = 25, antiderivative size = 152 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {a \sin (e+f x)}{b (a+b) f \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}+\frac {E\left (\arcsin (\sin (e+f x))\left |\frac {a}{a+b}\right .\right ) \left (a+b-a \sin ^2(e+f x)\right )}{b (a+b) f \sqrt {\cos ^2(e+f x)} \sqrt {\frac {a+b-a \sin ^2(e+f x)}{a+b}} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}} \] Output:

-a*sin(f*x+e)/b/(a+b)/f/(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)+Elliptic 
E(sin(f*x+e),(a/(a+b))^(1/2))*(a+b-a*sin(f*x+e)^2)/b/(a+b)/f/(cos(f*x+e)^2 
)^(1/2)/((a+b-a*sin(f*x+e)^2)/(a+b))^(1/2)/(sec(f*x+e)^2*(a+b-a*sin(f*x+e) 
^2))^(1/2)
 

Mathematica [A] (verified)

Time = 2.72 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.74 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^3(e+f x) \left (\sqrt {2} (a+b) \sqrt {\frac {a+2 b+a \cos (2 (e+f x))}{a+b}} E\left (e+f x\left |\frac {a}{a+b}\right .\right )-a \sin (2 (e+f x))\right )}{4 b (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.23, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4636, 2057, 2058, 316, 25, 330, 327}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4636

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

\(\Big \downarrow \) 2057

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

\(\Big \downarrow \) 2058

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 330

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

\(\Big \downarrow \) 327

\(\displaystyle \frac {\sqrt {-a \sin ^2(e+f x)+a+b} \left (\frac {\sqrt {-a \sin ^2(e+f x)+a+b} E\left (\arcsin (\sin (e+f x))\left |\frac {a}{a+b}\right .\right )}{b (a+b) \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}-\frac {a \sin (e+f x) \sqrt {1-\sin ^2(e+f x)}}{b (a+b) \sqrt {-a \sin ^2(e+f x)+a+b}}\right )}{f \sqrt {1-\sin ^2(e+f x)} \sqrt {\frac {-a \sin ^2(e+f x)+a+b}{1-\sin ^2(e+f x)}}}\)

Input:

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

Output:

(Sqrt[a + b - a*Sin[e + f*x]^2]*(-((a*Sin[e + f*x]*Sqrt[1 - Sin[e + f*x]^2 
])/(b*(a + b)*Sqrt[a + b - a*Sin[e + f*x]^2])) + (EllipticE[ArcSin[Sin[e + 
 f*x]], a/(a + b)]*Sqrt[a + b - a*Sin[e + f*x]^2])/(b*(a + b)*Sqrt[1 - (a* 
Sin[e + f*x]^2)/(a + b)])))/(f*Sqrt[1 - Sin[e + f*x]^2]*Sqrt[(a + b - a*Si 
n[e + f*x]^2)/(1 - Sin[e + f*x]^2)])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 330

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
Sqrt[a + b*x^2]/Sqrt[1 + (b/a)*x^2]   Int[Sqrt[1 + (b/a)*x^2]/Sqrt[c + d*x^ 
2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ[a, 
0]
 

rule 2057

Int[(u_.)*((a_) + (b_.)/((c_) + (d_.)*(x_)^(n_)))^(p_), x_Symbol] :> Int[u* 
((b + a*c + a*d*x^n)/(c + d*x^n))^p, x] /; FreeQ[{a, b, c, d, n, p}, x]
 

rule 2058

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_.)*((c_) + (d_.)*(x_)^(n_))^ 
(r_.))^(p_), x_Symbol] :> Simp[Simp[(e*(a + b*x^n)^q*(c + d*x^n)^r)^p/((a + 
 b*x^n)^(p*q)*(c + d*x^n)^(p*r))]   Int[u*(a + b*x^n)^(p*q)*(c + d*x^n)^(p* 
r), x], x] /; FreeQ[{a, b, c, d, e, n, p, q, r}, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4636

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

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 5.40 (sec) , antiderivative size = 1783, normalized size of antiderivative = 11.73

method	result	size
default	\(\text {Expression too large to display}\)	\(1783\)

Input:

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

Output:

-1/f/(2*I*a^(1/2)*b^(1/2)-a+b)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/(a+ 
b)/b*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a* 
(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)/((1-c 
os(f*x+e))^2*csc(f*x+e)^2-1)^3/(a+b*sec(f*x+e)^2)^(3/2)*(I*(2*cot(f*x+e)-2 
*csc(f*x+e)+2*(1-cos(f*x+e))^3*csc(f*x+e)^3)*((I*b^(1/2)+a^(1/2))^2/(a+b)) 
^(1/2)*b^(1/2)*a^(3/2)+I*(-2*cot(f*x+e)+2*csc(f*x+e)+2*(1-cos(f*x+e))^3*cs 
c(f*x+e)^3)*((I*b^(1/2)+a^(1/2))^2/(a+b))^(1/2)*b^(3/2)*a^(1/2)+(2*cot(f*x 
+e)-2*csc(f*x+e))*b*a*((I*b^(1/2)+a^(1/2))^2/(a+b))^(1/2)+(-(1-cos(f*x+e)) 
^3*csc(f*x+e)^3-cot(f*x+e)+csc(f*x+e))*a^2*((I*b^(1/2)+a^(1/2))^2/(a+b))^( 
1/2)+((1-cos(f*x+e))^3*csc(f*x+e)^3-cot(f*x+e)+csc(f*x+e))*b^2*((I*b^(1/2) 
+a^(1/2))^2/(a+b))^(1/2)+4*I*a^(1/2)*b^(3/2)*(1/(a+b)*(I*a^(1/2)*b^(1/2)*c 
os(f*x+e)-I*a^(1/2)*b^(1/2)+cos(f*x+e)*a+b)/(1+cos(f*x+e)))^(1/2)*((-I*a^( 
1/2)*b^(1/2)*cos(f*x+e)+I*a^(1/2)*b^(1/2)+cos(f*x+e)*a+b)/(a+b)/(1+cos(f*x 
+e)))^(1/2)*EllipticF(((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*(csc(f*x+e)- 
cot(f*x+e)),(1/(a+b)^2*(-4*I*a^(3/2)*b^(1/2)+4*I*a^(1/2)*b^(3/2)+a^2-6*a*b 
+b^2))^(1/2))+4*I*a^(3/2)*b^(1/2)*(1/(a+b)*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I 
*a^(1/2)*b^(1/2)+cos(f*x+e)*a+b)/(1+cos(f*x+e)))^(1/2)*((-I*a^(1/2)*b^(1/2 
)*cos(f*x+e)+I*a^(1/2)*b^(1/2)+cos(f*x+e)*a+b)/(a+b)/(1+cos(f*x+e)))^(1/2) 
*EllipticF(((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*(csc(f*x+e)-cot(f*x+e)) 
,(1/(a+b)^2*(-4*I*a^(3/2)*b^(1/2)+4*I*a^(1/2)*b^(3/2)+a^2-6*a*b+b^2))^(...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 894, normalized size of antiderivative = 5.88 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(2*a^3*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)^2*sin 
(f*x + e) - (2*(-I*a^3*cos(f*x + e)^2 - I*a^2*b)*sqrt(a)*sqrt((a*b + b^2)/ 
a^2) - (-I*a^2*b - 2*I*a*b^2 + (-I*a^3 - 2*I*a^2*b)*cos(f*x + e)^2)*sqrt(a 
))*sqrt((2*a*sqrt((a*b + b^2)/a^2) - a - 2*b)/a)*elliptic_e(arcsin(sqrt((2 
*a*sqrt((a*b + b^2)/a^2) - a - 2*b)/a)*(cos(f*x + e) + I*sin(f*x + e))), ( 
a^2 + 8*a*b + 8*b^2 + 4*(a^2 + 2*a*b)*sqrt((a*b + b^2)/a^2))/a^2) - (2*(I* 
a^3*cos(f*x + e)^2 + I*a^2*b)*sqrt(a)*sqrt((a*b + b^2)/a^2) - (I*a^2*b + 2 
*I*a*b^2 + (I*a^3 + 2*I*a^2*b)*cos(f*x + e)^2)*sqrt(a))*sqrt((2*a*sqrt((a* 
b + b^2)/a^2) - a - 2*b)/a)*elliptic_e(arcsin(sqrt((2*a*sqrt((a*b + b^2)/a 
^2) - a - 2*b)/a)*(cos(f*x + e) - I*sin(f*x + e))), (a^2 + 8*a*b + 8*b^2 + 
 4*(a^2 + 2*a*b)*sqrt((a*b + b^2)/a^2))/a^2) + 2*(2*(I*a^2*b*cos(f*x + e)^ 
2 + I*a*b^2)*sqrt(a)*sqrt((a*b + b^2)/a^2) + (I*a^2*b + 3*I*a*b^2 + 2*I*b^ 
3 + (I*a^3 + 3*I*a^2*b + 2*I*a*b^2)*cos(f*x + e)^2)*sqrt(a))*sqrt((2*a*sqr 
t((a*b + b^2)/a^2) - a - 2*b)/a)*elliptic_f(arcsin(sqrt((2*a*sqrt((a*b + b 
^2)/a^2) - a - 2*b)/a)*(cos(f*x + e) + I*sin(f*x + e))), (a^2 + 8*a*b + 8* 
b^2 + 4*(a^2 + 2*a*b)*sqrt((a*b + b^2)/a^2))/a^2) + 2*(2*(-I*a^2*b*cos(f*x 
 + e)^2 - I*a*b^2)*sqrt(a)*sqrt((a*b + b^2)/a^2) + (-I*a^2*b - 3*I*a*b^2 - 
 2*I*b^3 + (-I*a^3 - 3*I*a^2*b - 2*I*a*b^2)*cos(f*x + e)^2)*sqrt(a))*sqrt( 
(2*a*sqrt((a*b + b^2)/a^2) - a - 2*b)/a)*elliptic_f(arcsin(sqrt((2*a*sqrt( 
(a*b + b^2)/a^2) - a - 2*b)/a)*(cos(f*x + e) - I*sin(f*x + e))), (a^2 +...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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