\(\int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-m} \, dx\) [165]

Optimal result

Integrand size = 34, antiderivative size = 101 \[ \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-m} \, dx=-\frac {2^{\frac {1}{2}-m} c \operatorname {Hypergeometric2F1}\left (\frac {1}{2}+m,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sec (e+f x))\right ) (1-\sec (e+f x))^{\frac {1}{2}+m} (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-1-m} \tan (e+f x)}{f (1+2 m)} \] Output:

-2^(1/2-m)*c*hypergeom([1/2+m, 1/2+m],[3/2+m],1/2+1/2*sec(f*x+e))*(1-sec(f 
*x+e))^(1/2+m)*(a+a*sec(f*x+e))^m*(c-c*sec(f*x+e))^(-1-m)*tan(f*x+e)/f/(1+ 
2*m)
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.03 \[ \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-m} \, dx=-\frac {2^{\frac {1}{2}+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-m,\frac {1}{2}-m,\frac {3}{2}-m,\frac {1}{2} (1-\sec (e+f x))\right ) (1+\sec (e+f x))^{-\frac {1}{2}-m} (a (1+\sec (e+f x)))^m (c-c \sec (e+f x))^{-m} \tan (e+f x)}{f (-1+2 m)} \] Input:

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

Output:

-((2^(1/2 + m)*Hypergeometric2F1[1/2 - m, 1/2 - m, 3/2 - m, (1 - Sec[e + f 
*x])/2]*(1 + Sec[e + f*x])^(-1/2 - m)*(a*(1 + Sec[e + f*x]))^m*Tan[e + f*x 
])/(f*(-1 + 2*m)*(c - c*Sec[e + f*x])^m))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3042, 4449, 80, 79}

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.

\(\displaystyle \int \sec (e+f x) (a \sec (e+f x)+a)^m (c-c \sec (e+f x))^{-m} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4449

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

\(\displaystyle -\frac {c 2^{\frac {1}{2}-m} \tan (e+f x) (1-\sec (e+f x))^{m+\frac {1}{2}} (a \sec (e+f x)+a)^m (c-c \sec (e+f x))^{-m-1} \operatorname {Hypergeometric2F1}\left (m+\frac {1}{2},m+\frac {1}{2},m+\frac {3}{2},\frac {1}{2} (\sec (e+f x)+1)\right )}{f (2 m+1)}\)

Input:

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

Output:

-((2^(1/2 - m)*c*Hypergeometric2F1[1/2 + m, 1/2 + m, 3/2 + m, (1 + Sec[e + 
 f*x])/2]*(1 - Sec[e + f*x])^(1/2 + m)*(a + a*Sec[e + f*x])^m*(c - c*Sec[e 
 + f*x])^(-1 - m)*Tan[e + f*x])/(f*(1 + 2*m)))
 

Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

rule 3042

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

rule 4449

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(c 
sc[(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, 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]
 

Maple [F]

\[\int \sec \left (f x +e \right ) \left (a +a \sec \left (f x +e \right )\right )^{m} \left (-c \left (-1+\sec \left (f x +e \right )\right )\right )^{-m}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^m/((c-c*sec(f*x+e))^m),x, algorithm= 
"fricas")
 

Output:

integral((a*sec(f*x + e) + a)^m*sec(f*x + e)/(-c*sec(f*x + e) + c)^m, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^m/((c-c*sec(f*x+e))^m),x, algorithm= 
"maxima")
 

Output:

integrate((a*sec(f*x + e) + a)^m*sec(f*x + e)/(-c*sec(f*x + e) + c)^m, x)
 

Giac [F]

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

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^m/((c-c*sec(f*x+e))^m),x, algorithm= 
"giac")
 

Output:

integrate((a*sec(f*x + e) + a)^m*sec(f*x + e)/(-c*sec(f*x + e) + c)^m, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(((sec(e + f*x)*a + a)**m*sec(e + f*x))/( - sec(e + f*x)*c + c)**m,x)