3.2.42 \(\int \frac {(c-c \sec (e+f x))^n}{(a+a \sec (e+f x))^{3/2}} \, dx\) [142]

3.2.42.1 Optimal result

Integrand size = 28, antiderivative size = 205 \[ \int \frac {(c-c \sec (e+f x))^n}{(a+a \sec (e+f x))^{3/2}} \, dx=-\frac {(5-2 n) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+n,\frac {3}{2}+n,\frac {1}{2} (1-\sec (e+f x))\right ) (c-c \sec (e+f x))^n \tan (e+f x)}{4 a f (1+2 n) \sqrt {a+a \sec (e+f x)}}+\frac {2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+n,\frac {3}{2}+n,1-\sec (e+f x)\right ) (c-c \sec (e+f x))^n \tan (e+f x)}{a f (1+2 n) \sqrt {a+a \sec (e+f x)}}-\frac {(c-c \sec (e+f x))^n \tan (e+f x)}{2 a f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)}} \]

output

-1/4*(5-2*n)*hypergeom([1, 1/2+n],[3/2+n],1/2-1/2*sec(f*x+e))*(c-c*sec(f*x 
+e))^n*tan(f*x+e)/a/f/(1+2*n)/(a+a*sec(f*x+e))^(1/2)+2*hypergeom([1, 1/2+n 
],[3/2+n],1-sec(f*x+e))*(c-c*sec(f*x+e))^n*tan(f*x+e)/a/f/(1+2*n)/(a+a*sec 
(f*x+e))^(1/2)-1/2*(c-c*sec(f*x+e))^n*tan(f*x+e)/a/f/(1+sec(f*x+e))/(a+a*s 
ec(f*x+e))^(1/2)
 

3.2.42.2 Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.60 \[ \int \frac {(c-c \sec (e+f x))^n}{(a+a \sec (e+f x))^{3/2}} \, dx=\frac {(c-c \sec (e+f x))^n \left (-2-4 n+(-5+2 n) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+n,\frac {3}{2}+n,\frac {1}{2} (1-\sec (e+f x))\right ) (1+\sec (e+f x))+8 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+n,\frac {3}{2}+n,1-\sec (e+f x)\right ) (1+\sec (e+f x))\right ) \tan (e+f x)}{4 (f+2 f n) (a (1+\sec (e+f x)))^{3/2}} \]

input

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

output

((c - c*Sec[e + f*x])^n*(-2 - 4*n + (-5 + 2*n)*Hypergeometric2F1[1, 1/2 + 
n, 3/2 + n, (1 - Sec[e + f*x])/2]*(1 + Sec[e + f*x]) + 8*Hypergeometric2F1 
[1, 1/2 + n, 3/2 + n, 1 - Sec[e + f*x]]*(1 + Sec[e + f*x]))*Tan[e + f*x])/ 
(4*(f + 2*f*n)*(a*(1 + Sec[e + f*x]))^(3/2))
 

3.2.42.3 Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.95, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4400, 27, 114, 27, 174, 75, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4400

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

\(\displaystyle -\frac {c \tan (e+f x) \left (\frac {1}{4} \left (4 \int \cos (e+f x) (c-c \sec (e+f x))^{n-\frac {1}{2}}d\sec (e+f x)-(5-2 n) \int \frac {(c-c \sec (e+f x))^{n-\frac {1}{2}}}{\sec (e+f x)+1}d\sec (e+f x)\right )+\frac {(c-c \sec (e+f x))^{n+\frac {1}{2}}}{2 c (\sec (e+f x)+1)}\right )}{a f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\)

\(\Big \downarrow \) 75

\(\displaystyle -\frac {c \tan (e+f x) \left (\frac {1}{4} \left (-(5-2 n) \int \frac {(c-c \sec (e+f x))^{n-\frac {1}{2}}}{\sec (e+f x)+1}d\sec (e+f x)-\frac {8 (c-c \sec (e+f x))^{n+\frac {1}{2}} \operatorname {Hypergeometric2F1}\left (1,n+\frac {1}{2},n+\frac {3}{2},1-\sec (e+f x)\right )}{c (2 n+1)}\right )+\frac {(c-c \sec (e+f x))^{n+\frac {1}{2}}}{2 c (\sec (e+f x)+1)}\right )}{a f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\)

\(\Big \downarrow \) 78

\(\displaystyle -\frac {c \tan (e+f x) \left (\frac {1}{4} \left (\frac {(5-2 n) (c-c \sec (e+f x))^{n+\frac {1}{2}} \operatorname {Hypergeometric2F1}\left (1,n+\frac {1}{2},n+\frac {3}{2},\frac {1}{2} (1-\sec (e+f x))\right )}{c (2 n+1)}-\frac {8 (c-c \sec (e+f x))^{n+\frac {1}{2}} \operatorname {Hypergeometric2F1}\left (1,n+\frac {1}{2},n+\frac {3}{2},1-\sec (e+f x)\right )}{c (2 n+1)}\right )+\frac {(c-c \sec (e+f x))^{n+\frac {1}{2}}}{2 c (\sec (e+f x)+1)}\right )}{a f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\)

input

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

output

-((c*((c - c*Sec[e + f*x])^(1/2 + n)/(2*c*(1 + Sec[e + f*x])) + (((5 - 2*n 
)*Hypergeometric2F1[1, 1/2 + n, 3/2 + n, (1 - Sec[e + f*x])/2]*(c - c*Sec[ 
e + f*x])^(1/2 + n))/(c*(1 + 2*n)) - (8*Hypergeometric2F1[1, 1/2 + n, 3/2 
+ n, 1 - Sec[e + f*x]]*(c - c*Sec[e + f*x])^(1/2 + n))/(c*(1 + 2*n)))/4)*T 
an[e + f*x])/(a*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]))
 

3.2.42.3.1 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 75

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

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 114

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e 
 - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

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]
 

3.2.42.4 Maple [F]

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

input

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

output

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

3.2.42.5 Fricas [F]

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

input

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

output

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

3.2.42.6 Sympy [F]

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

input

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

output

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

3.2.42.7 Maxima [F]

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

input

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

output

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

3.2.42.8 Giac [F]

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

input

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

output

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

3.2.42.9 Mupad [F(-1)]

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

input

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

output

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