3.1.76 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^{3/2}} \, dx\) [76]

3.1.76.1 Optimal result
3.1.76.2 Mathematica [C] (verified)
3.1.76.3 Rubi [A] (verified)
3.1.76.4 Maple [A] (verified)
3.1.76.5 Fricas [A] (verification not implemented)
3.1.76.6 Sympy [F]
3.1.76.7 Maxima [F]
3.1.76.8 Giac [A] (verification not implemented)
3.1.76.9 Mupad [F(-1)]

3.1.76.1 Optimal result

Integrand size = 34, antiderivative size = 113 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^{3/2}} \, dx=\frac {3 \sqrt {2} a^2 \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{c^{3/2} f}-\frac {2 a^2 \tan (e+f x)}{f (c-c \sec (e+f x))^{3/2}}-\frac {2 a^2 \tan (e+f x)}{c f \sqrt {c-c \sec (e+f x)}} \]

output
3*a^2*arctan(1/2*c^(1/2)*tan(f*x+e)*2^(1/2)/(c-c*sec(f*x+e))^(1/2))*2^(1/2 
)/c^(3/2)/f-2*a^2*tan(f*x+e)/f/(c-c*sec(f*x+e))^(3/2)-2*a^2*tan(f*x+e)/c/f 
/(c-c*sec(f*x+e))^(1/2)
 
3.1.76.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.42 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.57 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^{3/2}} \, dx=-\frac {a^2 \operatorname {Hypergeometric2F1}\left (2,\frac {5}{2},\frac {7}{2},\frac {1}{2} (1+\sec (e+f x))\right ) (1+\sec (e+f x))^2 \tan (e+f x)}{10 c f \sqrt {c-c \sec (e+f x)}} \]

input
Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^2)/(c - c*Sec[e + f*x])^(3/2) 
,x]
 
output
-1/10*(a^2*Hypergeometric2F1[2, 5/2, 7/2, (1 + Sec[e + f*x])/2]*(1 + Sec[e 
 + f*x])^2*Tan[e + f*x])/(c*f*Sqrt[c - c*Sec[e + f*x]])
 
3.1.76.3 Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3042, 4445, 3042, 4444, 3042, 4282, 216}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^2}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 4445

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 a \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )}{\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx}{2 c}-\frac {\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4444

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 a \left (2 a \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx+\frac {2 a \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}\right )}{2 c}-\frac {\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle -\frac {3 a \left (\frac {2 a \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}-\frac {4 a \int \frac {1}{\frac {c^2 \tan ^2(e+f x)}{c-c \sec (e+f x)}+2 c}d\frac {c \tan (e+f x)}{\sqrt {c-c \sec (e+f x)}}}{f}\right )}{2 c}-\frac {\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))^{3/2}}-\frac {3 a \left (\frac {2 a \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}-\frac {2 \sqrt {2} a \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {c} f}\right )}{2 c}\)

input
Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^2)/(c - c*Sec[e + f*x])^(3/2),x]
 
output
-(((a^2 + a^2*Sec[e + f*x])*Tan[e + f*x])/(f*(c - c*Sec[e + f*x])^(3/2))) 
- (3*a*((-2*Sqrt[2]*a*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Se 
c[e + f*x]])])/(Sqrt[c]*f) + (2*a*Tan[e + f*x])/(f*Sqrt[c - c*Sec[e + f*x] 
])))/(2*c)
 

3.1.76.3.1 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 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 4444
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.))/ 
Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*d*Cot[e + 
f*x]*((c + d*Csc[e + f*x])^(n - 1)/(f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), 
 x] + Simp[2*c*((2*n - 1)/(2*n - 1))   Int[Csc[e + f*x]*((c + d*Csc[e + f*x 
])^(n - 1)/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x 
] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]
 

rule 4445
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp[2*a*c*Cot[e + 
f*x]*(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^(n - 1)/(b*f*(2*m + 1))), 
 x] - Simp[d*((2*n - 1)/(b*(2*m + 1)))   Int[Csc[e + f*x]*(a + b*Csc[e + f* 
x])^(m + 1)*(c + d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f 
}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0] && LtQ[m, -2^ 
(-1)] && IntegerQ[2*m]
 
3.1.76.4 Maple [A] (verified)

Time = 4.84 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.06

method result size
default \(\frac {a^{2} \sqrt {2}\, \left (3 \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \tan \left (f x +e \right )+2 \sqrt {2}\, \cot \left (f x +e \right )+\sqrt {2}\, \csc \left (f x +e \right )-\sqrt {2}\, \sec \left (f x +e \right ) \csc \left (f x +e \right )\right )}{c f \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}}\) \(120\)
parts \(-\frac {a^{2} \sqrt {2}\, \left (1-\cos \left (f x +e \right )\right ) \left (\left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{\frac {3}{2}}-\sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\, \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-\arctan \left (\frac {1}{\sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\right ) \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}\right ) \csc \left (f x +e \right )}{4 f \left (\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right )^{\frac {3}{2}} \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{\frac {3}{2}}}-\frac {a^{2} \sqrt {2}\, \left (1-\cos \left (f x +e \right )\right ) \left (7 \arctan \left (\frac {1}{\sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\right ) \left (1-\cos \left (f x +e \right )\right )^{2} \sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\, \csc \left (f x +e \right )^{2}-9 \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1\right ) \csc \left (f x +e \right )}{4 f \left (\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right )^{\frac {3}{2}} \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{2}}+\frac {a^{2} \sqrt {2}\, \left (1-\cos \left (f x +e \right )\right ) \left (-3 \arctan \left (\frac {1}{\sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\right ) \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+\sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right ) \csc \left (f x +e \right )}{2 f \left (\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right )^{\frac {3}{2}} \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{\frac {3}{2}}}\) \(597\)

input
int(sec(f*x+e)*(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^(3/2),x,method=_RETURNV 
ERBOSE)
 
output
a^2/c/f*2^(1/2)/(-c*(sec(f*x+e)-1))^(1/2)*(3*arctan(1/2*2^(1/2)/(-cos(f*x+ 
e)/(cos(f*x+e)+1))^(1/2))*(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*tan(f*x+e)+2* 
2^(1/2)*cot(f*x+e)+2^(1/2)*csc(f*x+e)-2^(1/2)*sec(f*x+e)*csc(f*x+e))
 
3.1.76.5 Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 372, normalized size of antiderivative = 3.29 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^{3/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (a^{2} c \cos \left (f x + e\right ) - a^{2} c\right )} \sqrt {-\frac {1}{c}} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sqrt {-\frac {1}{c}} + {\left (3 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 4 \, {\left (2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2} \cos \left (f x + e\right ) - a^{2}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{2 \, {\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}, -\frac {\frac {3 \, \sqrt {2} {\left (a^{2} c \cos \left (f x + e\right ) - a^{2} c\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right )}{\sqrt {c}} - 2 \, {\left (2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2} \cos \left (f x + e\right ) - a^{2}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{{\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}\right ] \]

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

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

input
integrate(sec(f*x+e)*(a+a*sec(f*x+e))**2/(c-c*sec(f*x+e))**(3/2),x)
 
output
a**2*(Integral(sec(e + f*x)/(-c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x) + c 
*sqrt(-c*sec(e + f*x) + c)), x) + Integral(2*sec(e + f*x)**2/(-c*sqrt(-c*s 
ec(e + f*x) + c)*sec(e + f*x) + c*sqrt(-c*sec(e + f*x) + c)), x) + Integra 
l(sec(e + f*x)**3/(-c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x) + c*sqrt(-c*s 
ec(e + f*x) + c)), x))
 
3.1.76.7 Maxima [F]

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

input
integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^(3/2),x, algorith 
m="maxima")
 
output
integrate((a*sec(f*x + e) + a)^2*sec(f*x + e)/(-c*sec(f*x + e) + c)^(3/2), 
 x)
 
3.1.76.8 Giac [A] (verification not implemented)

Time = 1.23 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.96 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^{3/2}} \, dx=-\frac {a^{2} {\left (\frac {3 \, \sqrt {2} \arctan \left (\frac {\sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}}{\sqrt {c}}\right )}{c^{\frac {3}{2}}} + \frac {\sqrt {2} {\left (3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}}{{\left ({\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {3}{2}} + \sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c} c\right )} c}\right )}}{f} \]

input
integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^(3/2),x, algorith 
m="giac")
 
output
-a^2*(3*sqrt(2)*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/c^(3/2) 
 + sqrt(2)*(3*c*tan(1/2*f*x + 1/2*e)^2 - c)/(((c*tan(1/2*f*x + 1/2*e)^2 - 
c)^(3/2) + sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c)*c))/f
 
3.1.76.9 Mupad [F(-1)]

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

input
int((a + a/cos(e + f*x))^2/(cos(e + f*x)*(c - c/cos(e + f*x))^(3/2)),x)
 
output
int((a + a/cos(e + f*x))^2/(cos(e + f*x)*(c - c/cos(e + f*x))^(3/2)), x)