3.2.24 \(\int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx\) [124]

3.2.24.1 Optimal result
3.2.24.2 Mathematica [A] (verified)
3.2.24.3 Rubi [A] (verified)
3.2.24.4 Maple [A] (warning: unable to verify)
3.2.24.5 Fricas [F]
3.2.24.6 Sympy [F(-1)]
3.2.24.7 Maxima [F(-2)]
3.2.24.8 Giac [A] (verification not implemented)
3.2.24.9 Mupad [F(-1)]

3.2.24.1 Optimal result

Integrand size = 30, antiderivative size = 220 \[ \int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\frac {c^4 \log (\cos (e+f x)) \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {2 c^4 \log (1+\sec (e+f x)) \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {4 c^4 \tan (e+f x)}{a^2 f (1+\sec (e+f x))^2 \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {4 c^4 \tan (e+f x)}{a^2 f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

output
c^4*ln(cos(f*x+e))*tan(f*x+e)/a^2/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e) 
)^(1/2)+2*c^4*ln(1+sec(f*x+e))*tan(f*x+e)/a^2/f/(a+a*sec(f*x+e))^(1/2)/(c- 
c*sec(f*x+e))^(1/2)-4*c^4*tan(f*x+e)/a^2/f/(1+sec(f*x+e))^2/(a+a*sec(f*x+e 
))^(1/2)/(c-c*sec(f*x+e))^(1/2)+4*c^4*tan(f*x+e)/a^2/f/(1+sec(f*x+e))/(a+a 
*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)
 
3.2.24.2 Mathematica [A] (verified)

Time = 1.78 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.39 \[ \int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=-\frac {c^4 \left (-\log (\cos (e+f x))-2 \log (1+\sec (e+f x))-\frac {4 \sec (e+f x)}{(1+\sec (e+f x))^2}\right ) \tan (e+f x)}{a^2 f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]

input
Integrate[(c - c*Sec[e + f*x])^(7/2)/(a + a*Sec[e + f*x])^(5/2),x]
 
output
-((c^4*(-Log[Cos[e + f*x]] - 2*Log[1 + Sec[e + f*x]] - (4*Sec[e + f*x])/(1 
 + Sec[e + f*x])^2)*Tan[e + f*x])/(a^2*f*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c 
 - c*Sec[e + f*x]]))
 
3.2.24.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.40, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 4400, 27, 99, 2009}

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.

\(\displaystyle \int \frac {(c-c \sec (e+f x))^{7/2}}{(a \sec (e+f x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{7/2}}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\)

\(\Big \downarrow \) 4400

\(\displaystyle -\frac {a c \tan (e+f x) \int \frac {c^3 \cos (e+f x) (1-\sec (e+f x))^3}{a^3 (\sec (e+f x)+1)^3}d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 99

\(\displaystyle -\frac {c^4 \tan (e+f x) \int \left (\cos (e+f x)-\frac {2}{\sec (e+f x)+1}+\frac {4}{(\sec (e+f x)+1)^2}-\frac {8}{(\sec (e+f x)+1)^3}\right )d\sec (e+f x)}{a^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\)

\(\Big \downarrow \) 2009

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

input
Int[(c - c*Sec[e + f*x])^(7/2)/(a + a*Sec[e + f*x])^(5/2),x]
 
output
-((c^4*(Log[Sec[e + f*x]] - 2*Log[1 + Sec[e + f*x]] + 4/(1 + Sec[e + f*x]) 
^2 - 4/(1 + Sec[e + f*x]))*Tan[e + f*x])/(a^2*f*Sqrt[a + a*Sec[e + f*x]]*S 
qrt[c - c*Sec[e + f*x]]))
 

3.2.24.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 99
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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

rule 4400
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( 
d_.) + (c_))^(n_.), x_Symbol] :> Simp[a*c*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e 
 + f*x]]*Sqrt[c + d*Csc[e + f*x]]))   Subst[Int[(a + b*x)^(m - 1/2)*((c + d 
*x)^(n - 1/2)/x), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n 
}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
 
3.2.24.4 Maple [A] (warning: unable to verify)

Time = 2.36 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.98

method result size
default \(-\frac {\sqrt {2}\, \sqrt {-\frac {2 a}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\, \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{4} \left (\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right )^{\frac {7}{2}} \sin \left (f x +e \right )^{7} \left (\left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+\ln \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1\right )+\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )\right )}{2 f \,a^{3} \left (1-\cos \left (f x +e \right )\right )^{7}}\) \(216\)
risch \(-\frac {c^{3} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (-4 i \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) {\mathrm e}^{3 i \left (f x +e \right )}+8 i {\mathrm e}^{i \left (f x +e \right )}+{\mathrm e}^{4 i \left (f x +e \right )} f x +2 \,{\mathrm e}^{4 i \left (f x +e \right )} e -4 i {\mathrm e}^{i \left (f x +e \right )} \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )+16 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) {\mathrm e}^{3 i \left (f x +e \right )}+4 \,{\mathrm e}^{3 i \left (f x +e \right )} f x +8 \,{\mathrm e}^{3 i \left (f x +e \right )} e +16 i {\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+4 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) {\mathrm e}^{4 i \left (f x +e \right )}+6 \,{\mathrm e}^{2 i \left (f x +e \right )} f x +12 \,{\mathrm e}^{2 i \left (f x +e \right )} e +24 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+8 i {\mathrm e}^{3 i \left (f x +e \right )}+4 \,{\mathrm e}^{i \left (f x +e \right )} f x -i {\mathrm e}^{4 i \left (f x +e \right )} \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )+8 \,{\mathrm e}^{i \left (f x +e \right )} e -i \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )-6 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )+4 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+f x +2 e \right )}{a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\) \(450\)

input
int((c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
 
output
-1/2/f*2^(1/2)/a^3*(-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)^4*(c*(1-cos(f*x+e))^2/((1-cos(f*x+e))^2*csc(f*x+ 
e)^2-1)*csc(f*x+e)^2)^(7/2)/(1-cos(f*x+e))^7*sin(f*x+e)^7*((1-cos(f*x+e))^ 
4*csc(f*x+e)^4+ln(-cot(f*x+e)+csc(f*x+e)+1)+ln((1-cos(f*x+e))^2*csc(f*x+e) 
^2+1)+ln(-cot(f*x+e)+csc(f*x+e)-1))
 
3.2.24.5 Fricas [F]

\[ \int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\int { \frac {{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

input
integrate((c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="fric 
as")
 
output
integral(-(c^3*sec(f*x + e)^3 - 3*c^3*sec(f*x + e)^2 + 3*c^3*sec(f*x + e) 
- c^3)*sqrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(a^3*sec(f*x + e 
)^3 + 3*a^3*sec(f*x + e)^2 + 3*a^3*sec(f*x + e) + a^3), x)
 
3.2.24.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\text {Timed out} \]

input
integrate((c-c*sec(f*x+e))**(7/2)/(a+a*sec(f*x+e))**(5/2),x)
 
output
Timed out
 
3.2.24.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

input
integrate((c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="maxi 
ma")
 
output
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 
3.2.24.8 Giac [A] (verification not implemented)

Time = 1.46 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.42 \[ \int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=-\frac {{\left ({\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} \sqrt {-a c} a^{2} {\left | c \right |} + 2 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} \sqrt {-a c} a^{2} c {\left | c \right |}\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{a^{5} f} \]

input
integrate((c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="giac 
")
 
output
-((c*tan(1/2*f*x + 1/2*e)^2 - c)^2*sqrt(-a*c)*a^2*abs(c) + 2*(c*tan(1/2*f* 
x + 1/2*e)^2 - c)*sqrt(-a*c)*a^2*c*abs(c))*sgn(tan(1/2*f*x + 1/2*e)^3 + ta 
n(1/2*f*x + 1/2*e))/(a^5*f)
 
3.2.24.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\int \frac {{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{7/2}}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

input
int((c - c/cos(e + f*x))^(7/2)/(a + a/cos(e + f*x))^(5/2),x)
 
output
int((c - c/cos(e + f*x))^(7/2)/(a + a/cos(e + f*x))^(5/2), x)