\(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{(c-c \sec (e+f x))^{3/2}} \, dx\) [160]

Optimal result

Integrand size = 34, antiderivative size = 74 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{(c-c \sec (e+f x))^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (2,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sec (e+f x))\right ) (a+a \sec (e+f x))^m \tan (e+f x)}{2 c f (1+2 m) \sqrt {c-c \sec (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{(c-c \sec (e+f x))^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (2,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sec (e+f x))\right ) (a (1+\sec (e+f x)))^m \tan (e+f x)}{4 c f \left (\frac {1}{2}+m\right ) \sqrt {c-c \sec (e+f x)}} \] Input:

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

Output:

-1/4*(Hypergeometric2F1[2, 1/2 + m, 3/2 + m, (1 + Sec[e + f*x])/2]*(a*(1 + 
 Sec[e + f*x]))^m*Tan[e + f*x])/(c*f*(1/2 + m)*Sqrt[c - c*Sec[e + f*x]])
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 74, 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, 27, 78}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4449

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 78

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

Input:

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

Output:

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

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 78

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^m/(c-c*sec(f*x+e))^(3/2),x, algorith 
m="fricas")
 

Output:

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

Sympy [F]

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

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**m/(c-c*sec(f*x+e))**(3/2),x)
 

Output:

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

Maxima [F]

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

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^m/(c-c*sec(f*x+e))^(3/2),x, algorith 
m="maxima")
 

Output:

integrate((a*sec(f*x + e) + a)^m*sec(f*x + e)/(-c*sec(f*x + e) + c)^(3/2), 
 x)
 

Giac [F]

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

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^m/(c-c*sec(f*x+e))^(3/2),x, algorith 
m="giac")
 

Output:

integrate((a*sec(f*x + e) + a)^m*sec(f*x + e)/(-c*sec(f*x + e) + c)^(3/2), 
 x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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