3.3.40 \(\int \frac {\sec (e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx\) [240]

3.3.40.1 Optimal result
3.3.40.2 Mathematica [A] (verified)
3.3.40.3 Rubi [A] (verified)
3.3.40.4 Maple [B] (warning: unable to verify)
3.3.40.5 Fricas [A] (verification not implemented)
3.3.40.6 Sympy [F]
3.3.40.7 Maxima [F]
3.3.40.8 Giac [F(-2)]
3.3.40.9 Mupad [F(-1)]

3.3.40.1 Optimal result

Integrand size = 33, antiderivative size = 122 \[ \int \frac {\sec (e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} (c-d) f}-\frac {2 \sqrt {d} \arctan \left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {c+d} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} (c-d) \sqrt {c+d} f} \]

output
arctan(1/2*a^(1/2)*tan(f*x+e)*2^(1/2)/(a+a*sec(f*x+e))^(1/2))*2^(1/2)/(c-d 
)/f/a^(1/2)-2*arctan(a^(1/2)*d^(1/2)*tan(f*x+e)/(c+d)^(1/2)/(a+a*sec(f*x+e 
))^(1/2))*d^(1/2)/(c-d)/f/a^(1/2)/(c+d)^(1/2)
 
3.3.40.2 Mathematica [A] (verified)

Time = 1.43 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.18 \[ \int \frac {\sec (e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\frac {2 \left (\sqrt {-c-d} \arcsin \left (\tan \left (\frac {1}{2} (e+f x)\right )\right )+\sqrt {2} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {-c-d} \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )\right )}{\sqrt {-c-d} (c-d) f \sqrt {\cos (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )} \sqrt {a (1+\sec (e+f x))}} \]

input
Integrate[Sec[e + f*x]/(Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])),x]
 
output
(2*(Sqrt[-c - d]*ArcSin[Tan[(e + f*x)/2]] + Sqrt[2]*Sqrt[d]*ArcTanh[(Sqrt[ 
d]*Tan[(e + f*x)/2])/(Sqrt[-c - d]*Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])])] 
))/(Sqrt[-c - d]*(c - d)*f*Sqrt[Cos[e + f*x]*Sec[(e + f*x)/2]^2]*Sqrt[a*(1 
 + Sec[e + f*x])])
 
3.3.40.3 Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 4460, 3042, 4282, 216, 4455, 218}

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.

\(\displaystyle \int \frac {\sec (e+f x)}{\sqrt {a \sec (e+f x)+a} (c+d \sec (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\sqrt {a \csc \left (e+f x+\frac {\pi }{2}\right )+a} \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 4460

\(\displaystyle \frac {\int \frac {\sec (e+f x)}{\sqrt {\sec (e+f x) a+a}}dx}{c-d}-\frac {d \int \frac {\sec (e+f x) \sqrt {\sec (e+f x) a+a}}{c+d \sec (e+f x)}dx}{a (c-d)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (e+f x+\frac {\pi }{2}\right ) a+a}}dx}{c-d}-\frac {d \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \sqrt {\csc \left (e+f x+\frac {\pi }{2}\right ) a+a}}{c+d \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{a (c-d)}\)

\(\Big \downarrow \) 4282

\(\displaystyle -\frac {2 \int \frac {1}{\frac {a^2 \tan ^2(e+f x)}{\sec (e+f x) a+a}+2 a}d\left (-\frac {a \tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )}{f (c-d)}-\frac {d \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \sqrt {\csc \left (e+f x+\frac {\pi }{2}\right ) a+a}}{c+d \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{a (c-d)}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a}}\right )}{\sqrt {a} f (c-d)}-\frac {d \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \sqrt {\csc \left (e+f x+\frac {\pi }{2}\right ) a+a}}{c+d \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{a (c-d)}\)

\(\Big \downarrow \) 4455

\(\displaystyle \frac {2 d \int \frac {1}{\frac {a^2 d \tan ^2(e+f x)}{\sec (e+f x) a+a}+a (c+d)}d\left (-\frac {a \tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )}{f (c-d)}+\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a}}\right )}{\sqrt {a} f (c-d)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a}}\right )}{\sqrt {a} f (c-d)}-\frac {2 \sqrt {d} \arctan \left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {c+d} \sqrt {a \sec (e+f x)+a}}\right )}{\sqrt {a} f (c-d) \sqrt {c+d}}\)

input
Int[Sec[e + f*x]/(Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])),x]
 
output
(Sqrt[2]*ArcTan[(Sqrt[a]*Tan[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])] 
)/(Sqrt[a]*(c - d)*f) - (2*Sqrt[d]*ArcTan[(Sqrt[a]*Sqrt[d]*Tan[e + f*x])/( 
Sqrt[c + d]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[a]*(c - d)*Sqrt[c + d]*f)
 

3.3.40.3.1 Defintions of rubi rules used

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 218
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4282
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2/f   Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ 
a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4455
Int[(csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)])/(c 
sc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[-2*(b/f)   Subst[In 
t[1/(b*c + a*d + d*x^2), x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], 
 x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 
0]
 

rule 4460
Int[csc[(e_.) + (f_.)*(x_)]/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(cs 
c[(e_.) + (f_.)*(x_)]*(d_.) + (c_))), x_Symbol] :> Simp[b/(b*c - a*d)   Int 
[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] - Simp[d/(b*c - a*d)   Int[C 
sc[e + f*x]*(Sqrt[a + b*Csc[e + f*x]]/(c + d*Csc[e + f*x])), x], x] /; Free 
Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && (EqQ[a^2 - b^2, 0] || EqQ[ 
c^2 - d^2, 0])
 
3.3.40.4 Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(503\) vs. \(2(99)=198\).

Time = 19.78 (sec) , antiderivative size = 504, normalized size of antiderivative = 4.13

method result size
default \(\frac {\left (2 \sqrt {\left (c +d \right ) \left (c -d \right )}\, \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )+\sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right ) \sqrt {\frac {d}{c -d}}+d \sqrt {2}\, \ln \left (-\frac {2 \left (-\sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\, \sqrt {2}\, \sqrt {\frac {d}{c -d}}\, c +\sqrt {2}\, \sqrt {\frac {d}{c -d}}\, \sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\, d +\sqrt {\left (c +d \right ) \left (c -d \right )}\, \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )+c -d \right )}{c \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )-\left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ) d +\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )-d \sqrt {2}\, \ln \left (-\frac {2 \left (\sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\, \sqrt {2}\, \sqrt {\frac {d}{c -d}}\, c -\sqrt {2}\, \sqrt {\frac {d}{c -d}}\, \sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\, d +\sqrt {\left (c +d \right ) \left (c -d \right )}\, \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )-c +d \right )}{-c \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )+\left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ) d +\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )\right ) \sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\, \sqrt {-\frac {2 a}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}}{2 f \sqrt {\frac {d}{c -d}}\, \left (c -d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}\, a}\) \(504\)

input
int(sec(f*x+e)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x,method=_RETURNVER 
BOSE)
 
output
1/2/f/(d/(c-d))^(1/2)/(c-d)/((c+d)*(c-d))^(1/2)/a*(2*((c+d)*(c-d))^(1/2)*l 
n(csc(f*x+e)-cot(f*x+e)+((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2))*(d/(c-d)) 
^(1/2)+d*2^(1/2)*ln(-2*(-((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*2^(1/2)*( 
d/(c-d))^(1/2)*c+2^(1/2)*(d/(c-d))^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1) 
^(1/2)*d+((c+d)*(c-d))^(1/2)*(-cot(f*x+e)+csc(f*x+e))+c-d)/(c*(-cot(f*x+e) 
+csc(f*x+e))-(-cot(f*x+e)+csc(f*x+e))*d+((c+d)*(c-d))^(1/2)))-d*2^(1/2)*ln 
(-2*(((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*2^(1/2)*(d/(c-d))^(1/2)*c-2^( 
1/2)*(d/(c-d))^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*d+((c+d)*(c-d 
))^(1/2)*(-cot(f*x+e)+csc(f*x+e))-c+d)/(-c*(-cot(f*x+e)+csc(f*x+e))+(-cot( 
f*x+e)+csc(f*x+e))*d+((c+d)*(c-d))^(1/2))))*((1-cos(f*x+e))^2*csc(f*x+e)^2 
-1)^(1/2)*(-2*a/((1-cos(f*x+e))^2*csc(f*x+e)^2-1))^(1/2)
 
3.3.40.5 Fricas [A] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 963, normalized size of antiderivative = 7.89 \[ \int \frac {\sec (e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\text {Too large to display} \]

input
integrate(sec(f*x+e)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm= 
"fricas")
 
output
[-1/2*(sqrt(2)*sqrt(-1/a)*log((2*sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x 
 + e))*sqrt(-1/a)*cos(f*x + e)*sin(f*x + e) + 3*cos(f*x + e)^2 + 2*cos(f*x 
 + e) - 1)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) + sqrt(-d/(a*c + a*d))*l 
og(-((c^2 + 8*c*d + 8*d^2)*cos(f*x + e)^3 + (c^2 + 2*c*d)*cos(f*x + e)^2 - 
 4*((c^2 + 3*c*d + 2*d^2)*cos(f*x + e)^2 - (c*d + d^2)*cos(f*x + e))*sqrt( 
-d/(a*c + a*d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) + d^2 
 - (6*c*d + 7*d^2)*cos(f*x + e))/(c^2*cos(f*x + e)^3 + (c^2 + 2*c*d)*cos(f 
*x + e)^2 + d^2 + (2*c*d + d^2)*cos(f*x + e))))/((c - d)*f), -1/2*(sqrt(2) 
*sqrt(-1/a)*log((2*sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(-1 
/a)*cos(f*x + e)*sin(f*x + e) + 3*cos(f*x + e)^2 + 2*cos(f*x + e) - 1)/(co 
s(f*x + e)^2 + 2*cos(f*x + e) + 1)) + 2*sqrt(d/(a*c + a*d))*arctan(2*(c + 
d)*sqrt(d/(a*c + a*d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e 
)*sin(f*x + e)/((c + 2*d)*cos(f*x + e)^2 + (c + d)*cos(f*x + e) - d)))/((c 
 - d)*f), -1/2*(sqrt(-d/(a*c + a*d))*log(-((c^2 + 8*c*d + 8*d^2)*cos(f*x + 
 e)^3 + (c^2 + 2*c*d)*cos(f*x + e)^2 - 4*((c^2 + 3*c*d + 2*d^2)*cos(f*x + 
e)^2 - (c*d + d^2)*cos(f*x + e))*sqrt(-d/(a*c + a*d))*sqrt((a*cos(f*x + e) 
 + a)/cos(f*x + e))*sin(f*x + e) + d^2 - (6*c*d + 7*d^2)*cos(f*x + e))/(c^ 
2*cos(f*x + e)^3 + (c^2 + 2*c*d)*cos(f*x + e)^2 + d^2 + (2*c*d + d^2)*cos( 
f*x + e))) + 2*sqrt(2)*arctan(sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + 
e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e)))/sqrt(a))/((c - d)*f), -(sqrt(d...
 
3.3.40.6 Sympy [F]

\[ \int \frac {\sec (e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\int \frac {\sec {\left (e + f x \right )}}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \left (c + d \sec {\left (e + f x \right )}\right )}\, dx \]

input
integrate(sec(f*x+e)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))**(1/2),x)
 
output
Integral(sec(e + f*x)/(sqrt(a*(sec(e + f*x) + 1))*(c + d*sec(e + f*x))), x 
)
 
3.3.40.7 Maxima [F]

\[ \int \frac {\sec (e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\int { \frac {\sec \left (f x + e\right )}{\sqrt {a \sec \left (f x + e\right ) + a} {\left (d \sec \left (f x + e\right ) + c\right )}} \,d x } \]

input
integrate(sec(f*x+e)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm= 
"maxima")
 
output
integrate(sec(f*x + e)/(sqrt(a*sec(f*x + e) + a)*(d*sec(f*x + e) + c)), x)
 
3.3.40.8 Giac [F(-2)]

Exception generated. \[ \int \frac {\sec (e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\text {Exception raised: TypeError} \]

input
integrate(sec(f*x+e)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm= 
"giac")
 
output
Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 
3.3.40.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\int \frac {1}{\cos \left (e+f\,x\right )\,\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )} \,d x \]

input
int(1/(cos(e + f*x)*(a + a/cos(e + f*x))^(1/2)*(c + d/cos(e + f*x))),x)
 
output
int(1/(cos(e + f*x)*(a + a/cos(e + f*x))^(1/2)*(c + d/cos(e + f*x))), x)