\(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx\) [129]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [F]
Sympy [F(-1)]
Maxima [A] (verification not implemented)
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 36, antiderivative size = 145 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=-\frac {a (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac {a^2 \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{c f (c-c \sec (e+f x))^{3/2}}+\frac {a^3 \log (1-\sec (e+f x)) \tan (e+f x)}{c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \] Output:

-1/2*a*(a+a*sec(f*x+e))^(3/2)*tan(f*x+e)/f/(c-c*sec(f*x+e))^(5/2)+a^2*(a+a 
*sec(f*x+e))^(1/2)*tan(f*x+e)/c/f/(c-c*sec(f*x+e))^(3/2)+a^3*ln(1-sec(f*x+ 
e))*tan(f*x+e)/c^2/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.56 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=-\frac {a^3 \left (-\log (1-\sec (e+f x))+\frac {-2+4 \sec (e+f x)}{(-1+\sec (e+f x))^2}\right ) \tan (e+f x)}{c^2 f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \] Input:

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^(5/2))/(c - c*Sec[e + f*x])^( 
5/2),x]
 

Output:

-((a^3*(-Log[1 - Sec[e + f*x]] + (-2 + 4*Sec[e + f*x])/(-1 + Sec[e + f*x]) 
^2)*Tan[e + f*x])/(c^2*f*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x 
]]))
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 4442, 3042, 4442, 3042, 4440}

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.

\(\displaystyle \int \frac {\sec (e+f x) (a \sec (e+f x)+a)^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 4442

\(\displaystyle -\frac {a \int \frac {\sec (e+f x) (\sec (e+f x) a+a)^{3/2}}{(c-c \sec (e+f x))^{3/2}}dx}{c}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{2 f (c-c \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{c}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{2 f (c-c \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 4442

\(\displaystyle -\frac {a \left (-\frac {a \int \frac {\sec (e+f x) \sqrt {\sec (e+f x) a+a}}{\sqrt {c-c \sec (e+f x)}}dx}{c}-\frac {a \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{f (c-c \sec (e+f x))^{3/2}}\right )}{c}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{2 f (c-c \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a \left (-\frac {a \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \sqrt {\csc \left (e+f x+\frac {\pi }{2}\right ) a+a}}{\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx}{c}-\frac {a \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{f (c-c \sec (e+f x))^{3/2}}\right )}{c}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{2 f (c-c \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 4440

\(\displaystyle -\frac {a \left (-\frac {a^2 \tan (e+f x) \log (1-\sec (e+f x))}{c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{f (c-c \sec (e+f x))^{3/2}}\right )}{c}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{2 f (c-c \sec (e+f x))^{5/2}}\)

Input:

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^(5/2))/(c - c*Sec[e + f*x])^(5/2),x 
]
 

Output:

-1/2*(a*(a + a*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(f*(c - c*Sec[e + f*x])^( 
5/2)) - (a*(-((a*Sqrt[a + a*Sec[e + f*x]]*Tan[e + f*x])/(f*(c - c*Sec[e + 
f*x])^(3/2))) - (a^2*Log[1 - Sec[e + f*x]]*Tan[e + f*x])/(c*f*Sqrt[a + a*S 
ec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])))/c
 

Defintions of rubi rules used

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

rule 4440
Int[(csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)])/Sq 
rt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*c*Log[1 + (b/ 
a)*Csc[e + f*x]]*(Cot[e + f*x]/(b*f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc 
[e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && Eq 
Q[a^2 - b^2, 0]
 

rule 4442
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 - 1/2, 0] && LtQ[ 
m, -2^(-1)]
 
Maple [A] (verified)

Time = 2.67 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.78

method result size
default \(-\frac {\left (16 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} \ln \left (-\frac {2 \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )-32 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} \ln \left (-\cot \left (\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {f x}{2}+\frac {e}{2}\right )\right )+16 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} \ln \left (-\frac {2 \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )-3 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+6 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5\right ) \sqrt {\frac {a \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1}}\, a^{2} \sec \left (\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{16 f \sqrt {-\frac {c \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1}}\, c^{2}}\) \(258\)
risch \(\frac {8 i a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{3} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}-\frac {2 i a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}+\frac {i a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\) \(325\)

Input:

int(sec(f*x+e)*(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x,method=_RET 
URNVERBOSE)
 

Output:

-1/16/f*(16*sin(1/2*f*x+1/2*e)^4*ln(-2*(cos(1/2*f*x+1/2*e)-sin(1/2*f*x+1/2 
*e))/(cos(1/2*f*x+1/2*e)+1))-32*sin(1/2*f*x+1/2*e)^4*ln(-cot(1/2*f*x+1/2*e 
)+csc(1/2*f*x+1/2*e))+16*sin(1/2*f*x+1/2*e)^4*ln(-2*(cos(1/2*f*x+1/2*e)+si 
n(1/2*f*x+1/2*e))/(cos(1/2*f*x+1/2*e)+1))-3*cos(1/2*f*x+1/2*e)^4+6*cos(1/2 
*f*x+1/2*e)^2+5)*(a/(2*cos(1/2*f*x+1/2*e)^2-1)*cos(1/2*f*x+1/2*e)^2)^(1/2) 
*a^2/(-c/(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)/c^2*sec(1/ 
2*f*x+1/2*e)*csc(1/2*f*x+1/2*e)^3
 

Fricas [F]

\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sec \left (f x + e\right )}{{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algo 
rithm="fricas")
 

Output:

integral(-(a^2*sec(f*x + e)^3 + 2*a^2*sec(f*x + e)^2 + a^2*sec(f*x + e))*s 
qrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(c^3*sec(f*x + e)^3 - 3* 
c^3*sec(f*x + e)^2 + 3*c^3*sec(f*x + e) - c^3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**(5/2)/(c-c*sec(f*x+e))**(5/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.17 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=-\frac {\frac {2 \, \sqrt {-a} a^{2} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c^{\frac {5}{2}}} + \frac {2 \, \sqrt {-a} a^{2} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac {5}{2}}} - \frac {4 \, \sqrt {-a} a^{2} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{\frac {5}{2}}} + \frac {{\left (\sqrt {-a} a^{2} \sqrt {c} + \frac {2 \, \sqrt {-a} a^{2} \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{4}}{c^{3} \sin \left (f x + e\right )^{4}}}{2 \, f} \] Input:

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algo 
rithm="maxima")
 

Output:

-1/2*(2*sqrt(-a)*a^2*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/c^(5/2) + 2* 
sqrt(-a)*a^2*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^(5/2) - 4*sqrt(-a) 
*a^2*log(sin(f*x + e)/(cos(f*x + e) + 1))/c^(5/2) + (sqrt(-a)*a^2*sqrt(c) 
+ 2*sqrt(-a)*a^2*sqrt(c)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e 
) + 1)^4/(c^3*sin(f*x + e)^4))/f
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algo 
rithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable 
to make series expansion Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{\cos \left (e+f\,x\right )\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((a + a/cos(e + f*x))^(5/2)/(cos(e + f*x)*(c - c/cos(e + f*x))^(5/2)),x 
)
 

Output:

int((a + a/cos(e + f*x))^(5/2)/(cos(e + f*x)*(c - c/cos(e + f*x))^(5/2)), 
x)
 

Reduce [F]

\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, a^{2} \left (-\left (\int \frac {\sqrt {\sec \left (f x +e \right )+1}\, \sqrt {-\sec \left (f x +e \right )+1}\, \sec \left (f x +e \right )^{3}}{\sec \left (f x +e \right )^{3}-3 \sec \left (f x +e \right )^{2}+3 \sec \left (f x +e \right )-1}d x \right )-2 \left (\int \frac {\sqrt {\sec \left (f x +e \right )+1}\, \sqrt {-\sec \left (f x +e \right )+1}\, \sec \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{3}-3 \sec \left (f x +e \right )^{2}+3 \sec \left (f x +e \right )-1}d x \right )-\left (\int \frac {\sqrt {\sec \left (f x +e \right )+1}\, \sqrt {-\sec \left (f x +e \right )+1}\, \sec \left (f x +e \right )}{\sec \left (f x +e \right )^{3}-3 \sec \left (f x +e \right )^{2}+3 \sec \left (f x +e \right )-1}d x \right )\right )}{c^{3}} \] Input:

int(sec(f*x+e)*(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x)
                                                                                    
                                                                                    
 

Output:

(sqrt(c)*sqrt(a)*a**2*( - int((sqrt(sec(e + f*x) + 1)*sqrt( - sec(e + f*x) 
 + 1)*sec(e + f*x)**3)/(sec(e + f*x)**3 - 3*sec(e + f*x)**2 + 3*sec(e + f* 
x) - 1),x) - 2*int((sqrt(sec(e + f*x) + 1)*sqrt( - sec(e + f*x) + 1)*sec(e 
 + f*x)**2)/(sec(e + f*x)**3 - 3*sec(e + f*x)**2 + 3*sec(e + f*x) - 1),x) 
- int((sqrt(sec(e + f*x) + 1)*sqrt( - sec(e + f*x) + 1)*sec(e + f*x))/(sec 
(e + f*x)**3 - 3*sec(e + f*x)**2 + 3*sec(e + f*x) - 1),x)))/c**3