3.3.41 \(\int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx\) [241]

3.3.41.1 Optimal result
3.3.41.2 Mathematica [A] (verified)
3.3.41.3 Rubi [A] (verified)
3.3.41.4 Maple [B] (warning: unable to verify)
3.3.41.5 Fricas [A] (verification not implemented)
3.3.41.6 Sympy [F]
3.3.41.7 Maxima [F]
3.3.41.8 Giac [F(-2)]
3.3.41.9 Mupad [F(-1)]

3.3.41.1 Optimal result

Integrand size = 35, antiderivative size = 124 \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx=-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} (c-d) f}+\frac {2 c \arctan \left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {c+d} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} (c-d) \sqrt {d} \sqrt {c+d} f} \]

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

Time = 0.72 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.14 \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx=-\frac {2 \left (\sqrt {d} \sqrt {c+d} \arctan \left (\frac {\sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\cos (e+f x)}}\right )-\sqrt {2} c \arctan \left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d} \sqrt {\cos (e+f x)}}\right )\right ) \cos \left (\frac {1}{2} (e+f x)\right )}{(c-d) \sqrt {d} \sqrt {c+d} f \sqrt {\cos (e+f x)} \sqrt {a (1+\sec (e+f x))}} \]

input
Integrate[Sec[e + f*x]^2/(Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])),x 
]
 
output
(-2*(Sqrt[d]*Sqrt[c + d]*ArcTan[Sin[(e + f*x)/2]/Sqrt[Cos[e + f*x]]] - Sqr 
t[2]*c*ArcTan[(Sqrt[2]*Sqrt[d]*Sin[(e + f*x)/2])/(Sqrt[c + d]*Sqrt[Cos[e + 
 f*x]])])*Cos[(e + f*x)/2])/((c - d)*Sqrt[d]*Sqrt[c + d]*f*Sqrt[Cos[e + f* 
x]]*Sqrt[a*(1 + Sec[e + f*x])])
 
3.3.41.3 Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4464, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4464

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4282

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 4455

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

\(\Big \downarrow \) 218

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

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

3.3.41.3.1 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 4464
Int[csc[(e_.) + (f_.)*(x_)]^2/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*( 
csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))), x_Symbol] :> Simp[-a/(b*c - a*d) 
Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[c/(b*c - a*d)   In 
t[Csc[e + f*x]*(Sqrt[a + b*Csc[e + f*x]]/(c + d*Csc[e + f*x])), x], x] /; F 
reeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && (EqQ[a^2 - b^2, 0] || E 
qQ[c^2 - d^2, 0])
 
3.3.41.4 Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(503\) vs. \(2(101)=202\).

Time = 18.95 (sec) , antiderivative size = 504, normalized size of antiderivative = 4.06

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

input
int(sec(f*x+e)^2/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x,method=_RETURNV 
ERBOSE)
 
output
-1/2/f/a/((c+d)*(c-d))^(1/2)/(c-d)/(d/(c-d))^(1/2)*(-2*a/((1-cos(f*x+e))^2 
*csc(f*x+e)^2-1))^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*(2*((c+d)* 
(c-d))^(1/2)*ln(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)-c*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*cs 
c(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)) 
)+c*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))))
 
3.3.41.5 Fricas [A] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 1041, normalized size of antiderivative = 8.40 \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\text {Too large to display} \]

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

\[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\int \frac {\sec ^{2}{\left (e + f x \right )}}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \left (c + d \sec {\left (e + f x \right )}\right )}\, dx \]

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

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

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

Exception generated. \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\text {Exception raised: TypeError} \]

input
integrate(sec(f*x+e)^2/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorith 
m="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
 
3.3.41.9 Mupad [F(-1)]

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

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