\(\int \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) [260]

Optimal result

Integrand size = 23, antiderivative size = 82 \[ \int \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=-\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{f}+\frac {\sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{f} \] Output:

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

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.57 \[ \int \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {\sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {2 a+2 b} \sin (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )+\frac {\sqrt {a} \sqrt {-b} \arcsin \left (\frac {\sqrt {-b} \sin (e+f x)}{\sqrt {a}}\right ) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}}}{\sqrt {2 a+b-b \cos (2 (e+f x))}}}{f} \] Input:

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

Output:

(Sqrt[a + b]*ArcTanh[(Sqrt[2*a + 2*b]*Sin[e + f*x])/Sqrt[2*a + b - b*Cos[2 
*(e + f*x)]]] + (Sqrt[a]*Sqrt[-b]*ArcSin[(Sqrt[-b]*Sin[e + f*x])/Sqrt[a]]* 
Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]] 
)/f
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3669, 301, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3669

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

\(\Big \downarrow \) 301

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(-(Sqrt[b]*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]]) + S 
qrt[a + b]*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]]) 
/f
 

Defintions of rubi rules used

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 301

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[b/ 
d   Int[(a + b*x^2)^(p - 1), x], x] - Simp[(b*c - a*d)/d   Int[(a + b*x^2)^ 
(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] 
&& GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4] || (EqQ[p, 2/3] && E 
qQ[b*c + 3*a*d, 0]))
 

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. \(149\) vs. \(2(70)=140\).

Time = 0.38 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.83

method	result	size
default	\(\frac {-\sqrt {b}\, \ln \left (\frac {\sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \sqrt {b}+b \sin \left (f x +e \right )}{\sqrt {b}}\right )+\frac {\sqrt {a +b}\, \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )}{2}-\frac {\sqrt {a +b}\, \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )}{2}}{f}\)	\(150\)

Input:

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

Output:

(-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))+1/ 
2*(a+b)^(1/2)*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))-1/2*(a+b)^(1/2)*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*c 
os(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a)))/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (70) = 140\).

Time = 0.24 (sec) , antiderivative size = 1246, normalized size of antiderivative = 15.20 \[ \int \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/8*(sqrt(b)*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + 2*b^4)*cos(f*x + e 
)^6 + 32*(5*a^2*b^2 + 24*a*b^3 + 24*b^4)*cos(f*x + e)^4 + a^4 + 32*a^3*b + 
 160*a^2*b^2 + 256*a*b^3 + 128*b^4 - 32*(a^3*b + 10*a^2*b^2 + 24*a*b^3 + 1 
6*b^4)*cos(f*x + e)^2 + 8*(16*b^3*cos(f*x + e)^6 - 24*(a*b^2 + 2*b^3)*cos( 
f*x + e)^4 - a^3 - 10*a^2*b - 24*a*b^2 - 16*b^3 + 2*(5*a^2*b + 24*a*b^2 + 
24*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(b)*sin(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)*sq 
rt(-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))/f, -1/8*(4*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))) - sqrt(b)*log( 
128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + 2*b^4)*cos(f*x + e)^6 + 32*(5*a^2*b^ 
2 + 24*a*b^3 + 24*b^4)*cos(f*x + e)^4 + a^4 + 32*a^3*b + 160*a^2*b^2 + 256 
*a*b^3 + 128*b^4 - 32*(a^3*b + 10*a^2*b^2 + 24*a*b^3 + 16*b^4)*cos(f*x + e 
)^2 + 8*(16*b^3*cos(f*x + e)^6 - 24*(a*b^2 + 2*b^3)*cos(f*x + e)^4 - a^3 - 
 10*a^2*b - 24*a*b^2 - 16*b^3 + 2*(5*a^2*b + 24*a*b^2 + 24*b^3)*cos(f*x + 
e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(b)*sin(f*x + e)))/f, 1/4*(sqrt( 
-b)*arctan(1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + 2*b^2)*cos(f*x + e)^2 + a^ 
2 + 8*a*b + 8*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-b)/((2*b^3*cos...
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.54 \[ \int \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=-\frac {2 \, \sqrt {b} \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right ) - \sqrt {a + b} \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right ) - \sqrt {a + b} \operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right )}{2 \, f} \] Input:

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

Output:

-1/2*(2*sqrt(b)*arcsinh(b*sin(f*x + e)/sqrt(a*b)) - sqrt(a + b)*arcsinh(b* 
sin(f*x + e)/(sqrt(a*b)*(sin(f*x + e) + 1)) - a/(sqrt(a*b)*(sin(f*x + e) + 
 1))) - sqrt(a + b)*arcsinh(-b*sin(f*x + e)/(sqrt(a*b)*(sin(f*x + e) - 1)) 
 - a/(sqrt(a*b)*(sin(f*x + e) - 1))))/f
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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