\(\int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [286]

Optimal result

Integrand size = 25, antiderivative size = 142 \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) f \sqrt {\frac {a+b \sin ^2(e+f x)}{a}}}+\frac {\operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right ) \sqrt {\frac {a+b \sin ^2(e+f x)}{a}}}{f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{(a+b) f} \] Output:

-EllipticE(sin(f*x+e),(-b/a)^(1/2))*(a+b*sin(f*x+e)^2)^(1/2)/(a+b)/f/((a+b 
*sin(f*x+e)^2)/a)^(1/2)+InverseJacobiAM(f*x+e,(-b/a)^(1/2))*((a+b*sin(f*x+ 
e)^2)/a)^(1/2)/f/(a+b*sin(f*x+e)^2)^(1/2)+(a+b*sin(f*x+e)^2)^(1/2)*tan(f*x 
+e)/(a+b)/f
 

Mathematica [A] (verified)

Time = 1.11 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99 \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {-2 a \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+2 (a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )+\sqrt {2} (2 a+b-b \cos (2 (e+f x))) \tan (e+f x)}{2 (a+b) f \sqrt {2 a+b-b \cos (2 (e+f x))}} \] Input:

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

Output:

(-2*a*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] + 
2*(a + b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a) 
] + Sqrt[2]*(2*a + b - b*Cos[2*(e + f*x)])*Tan[e + f*x])/(2*(a + b)*f*Sqrt 
[2*a + b - b*Cos[2*(e + f*x)]])
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.30, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3671, 316, 27, 326, 323, 321, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3671

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 326

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

\(\Big \downarrow \) 323

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

\(\Big \downarrow \) 321

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {b \left (\frac {(a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}\right )}{a+b}+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) \sqrt {1-\sin ^2(e+f x)}}\right )}{f}\)

\(\Big \downarrow \) 330

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {b \left (\frac {(a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\sqrt {a+b \sin ^2(e+f x)} \int \frac {\sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}\right )}{a+b}+\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) \sqrt {1-\sin ^2(e+f x)}}\right )}{f}\)

\(\Big \downarrow \) 327

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

Input:

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

Output:

(Sqrt[Cos[e + f*x]^2]*Sec[e + f*x]*((Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^ 
2])/((a + b)*Sqrt[1 - Sin[e + f*x]^2]) + (b*(-((EllipticE[ArcSin[Sin[e + f 
*x]], -(b/a)]*Sqrt[a + b*Sin[e + f*x]^2])/(b*Sqrt[1 + (b*Sin[e + f*x]^2)/a 
])) + ((a + b)*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sqrt[1 + (b*Sin[e + 
 f*x]^2)/a])/(b*Sqrt[a + b*Sin[e + f*x]^2])))/(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 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 321

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[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] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
 

rule 323

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

rule 326

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

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 3042

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

rule 3671

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*(Sqrt[ 
Cos[e + f*x]^2]/(f*Cos[e + f*x]))   Subst[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/2] &&  !IntegerQ[p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(277\) vs. \(2(137)=274\).

Time = 2.30 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.96

method	result	size
default	\(\frac {\sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}\, \left (a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )+b \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \cos \left (f x +e \right )^{2}}{a}+\frac {a +b}{a}}\, \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-\cos \left (f x +e \right )^{2} b \sin \left (f x +e \right )+a \sin \left (f x +e \right )+b \sin \left (f x +e \right )\right )}{\left (a +b \right ) \sqrt {-\left (a +b \sin \left (f x +e \right )^{2}\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, \cos \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}\, f}\)	\(278\)

Input:

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

Output:

(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(a*(cos(f*x+e)^2)^(1/2)*(-b/a*c 
os(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-b/a)^(1/2))+b*(cos(f*x+e 
)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-b/a)^( 
1/2))-a*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE(s 
in(f*x+e),(-b/a)^(1/2))-cos(f*x+e)^2*b*sin(f*x+e)+a*sin(f*x+e)+b*sin(f*x+e 
))/(a+b)/(-(a+b*sin(f*x+e)^2)*(sin(f*x+e)-1)*(1+sin(f*x+e)))^(1/2)/cos(f*x 
+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 632, normalized size of antiderivative = 4.45 \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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