3.3.40 \(\int \frac {(c (d \sec (e+f x))^p)^n}{a+b \sec (e+f x)} \, dx\) [240]

3.3.40.1 Optimal result

Integrand size = 27, antiderivative size = 206 \[ \int \frac {\left (c (d \sec (e+f x))^p\right )^n}{a+b \sec (e+f x)} \, dx=-\frac {b \operatorname {AppellF1}\left (\frac {1}{2},\frac {n p}{2},1,\frac {3}{2},\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos ^2(e+f x)^{\frac {n p}{2}} \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{\left (a^2-b^2\right ) f}+\frac {a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-1+n p),1,\frac {3}{2},\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (-1+n p)} \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{\left (a^2-b^2\right ) f} \]

output

-b*AppellF1(1/2,1/2*n*p,1,3/2,sin(f*x+e)^2,a^2*sin(f*x+e)^2/(a^2-b^2))*(co 
s(f*x+e)^2)^(1/2*n*p)*(c*(d*sec(f*x+e))^p)^n*sin(f*x+e)/(a^2-b^2)/f+a*Appe 
llF1(1/2,1/2*n*p-1/2,1,3/2,sin(f*x+e)^2,a^2*sin(f*x+e)^2/(a^2-b^2))*cos(f* 
x+e)*(cos(f*x+e)^2)^(1/2*n*p-1/2)*(c*(d*sec(f*x+e))^p)^n*sin(f*x+e)/(a^2-b 
^2)/f
 

3.3.40.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(5411\) vs. \(2(206)=412\).

Time = 32.62 (sec) , antiderivative size = 5411, normalized size of antiderivative = 26.27 \[ \int \frac {\left (c (d \sec (e+f x))^p\right )^n}{a+b \sec (e+f x)} \, dx=\text {Result too large to show} \]

input

Integrate[(c*(d*Sec[e + f*x])^p)^n/(a + b*Sec[e + f*x]),x]
 

output

Result too large to show
 

3.3.40.3 Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 4436, 3042, 4356, 3042, 3302, 3042, 3668, 25, 333}

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.

\(\displaystyle \int \frac {\left (c (d \sec (e+f x))^p\right )^n}{a+b \sec (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (c (d \sec (e+f x))^p\right )^n}{a+b \sec (e+f x)}dx\)

\(\Big \downarrow \) 4436

\(\displaystyle (d \sec (e+f x))^{-n p} \left (c (d \sec (e+f x))^p\right )^n \int \frac {(d \sec (e+f x))^{n p}}{a+b \sec (e+f x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (d \sec (e+f x))^{-n p} \left (c (d \sec (e+f x))^p\right )^n \int \frac {\left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{n p}}{a+b \csc \left (e+f x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 4356

\(\displaystyle \cos ^{n p}(e+f x) \left (c (d \sec (e+f x))^p\right )^n \int \frac {\cos ^{1-n p}(e+f x)}{b+a \cos (e+f x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \cos ^{n p}(e+f x) \left (c (d \sec (e+f x))^p\right )^n \int \frac {\sin \left (e+f x+\frac {\pi }{2}\right )^{1-n p}}{b+a \sin \left (e+f x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 3302

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

\(\Big \downarrow \) 3042

\(\displaystyle \cos ^{n p}(e+f x) \left (c (d \sec (e+f x))^p\right )^n \left (b \int \frac {\sin \left (e+f x+\frac {\pi }{2}\right )^{1-n p}}{b^2-a^2 \sin \left (e+f x+\frac {\pi }{2}\right )^2}dx-a \int \frac {\sin \left (e+f x+\frac {\pi }{2}\right )^{2-n p}}{b^2-a^2 \sin \left (e+f x+\frac {\pi }{2}\right )^2}dx\right )\)

\(\Big \downarrow \) 3668

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 333

\(\displaystyle \cos ^{n p}(e+f x) \left (c (d \sec (e+f x))^p\right )^n \left (\frac {a \sin (e+f x) \cos ^{1-n p}(e+f x) \cos ^2(e+f x)^{\frac {1}{2} (n p-1)} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (n p-1),1,\frac {3}{2},\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )}-\frac {b \sin (e+f x) \cos ^{-n p}(e+f x) \cos ^2(e+f x)^{\frac {n p}{2}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {n p}{2},1,\frac {3}{2},\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )}\right )\)

input

Int[(c*(d*Sec[e + f*x])^p)^n/(a + b*Sec[e + f*x]),x]
 

output

Cos[e + f*x]^(n*p)*(c*(d*Sec[e + f*x])^p)^n*(-((b*AppellF1[1/2, (n*p)/2, 1 
, 3/2, Sin[e + f*x]^2, (a^2*Sin[e + f*x]^2)/(a^2 - b^2)]*(Cos[e + f*x]^2)^ 
((n*p)/2)*Sin[e + f*x])/((a^2 - b^2)*f*Cos[e + f*x]^(n*p))) + (a*AppellF1[ 
1/2, (-1 + n*p)/2, 1, 3/2, Sin[e + f*x]^2, (a^2*Sin[e + f*x]^2)/(a^2 - b^2 
)]*Cos[e + f*x]^(1 - n*p)*(Cos[e + f*x]^2)^((-1 + n*p)/2)*Sin[e + f*x])/(( 
a^2 - b^2)*f))
 

3.3.40.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 3042

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

rule 3302

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( 
x_)]), x_Symbol] :> Simp[a   Int[(d*Sin[e + f*x])^n/(a^2 - b^2*Sin[e + f*x] 
^2), x], x] - Simp[b/d   Int[(d*Sin[e + f*x])^(n + 1)/(a^2 - b^2*Sin[e + f* 
x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0]
 

rule 3668

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[( 
-ff)*d^(2*IntPart[(m - 1)/2] + 1)*((d*Sin[e + f*x])^(2*FracPart[(m - 1)/2]) 
/(f*(Sin[e + f*x]^2)^FracPart[(m - 1)/2]))   Subst[Int[(1 - ff^2*x^2)^((m - 
 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, 
 d, e, f, m, p}, x] &&  !IntegerQ[m]
 

rule 4356

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_.), x_Symbol] :> Simp[Sin[e + f*x]^n*(d*Csc[e + f*x])^n   Int[(b + 
 a*Sin[e + f*x])^m/Sin[e + f*x]^(m + n), x], x] /; FreeQ[{a, b, d, e, f, n} 
, x] && NeQ[a^2 - b^2, 0] && IntegerQ[m]
 

rule 4436

Int[((c_.)*((d_.)*sec[(e_.) + (f_.)*(x_)])^(p_))^(n_)*((a_.) + (b_.)*sec[(e 
_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Simp[c^IntPart[n]*((c*(d*Sec[e + f*x 
])^p)^FracPart[n]/(d*Sec[e + f*x])^(p*FracPart[n]))   Int[(a + b*Sec[e + f* 
x])^m*(d*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, 
x] &&  !IntegerQ[n]
 

3.3.40.4 Maple [F]

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

input

int((c*(d*sec(f*x+e))^p)^n/(a+b*sec(f*x+e)),x)
 

output

int((c*(d*sec(f*x+e))^p)^n/(a+b*sec(f*x+e)),x)
 

3.3.40.5 Fricas [F]

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

input

integrate((c*(d*sec(f*x+e))^p)^n/(a+b*sec(f*x+e)),x, algorithm="fricas")
 

output

integral(((d*sec(f*x + e))^p*c)^n/(b*sec(f*x + e) + a), x)
 

3.3.40.6 Sympy [F]

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

input

integrate((c*(d*sec(f*x+e))**p)**n/(a+b*sec(f*x+e)),x)
 

output

Integral((c*(d*sec(e + f*x))**p)**n/(a + b*sec(e + f*x)), x)
 

3.3.40.7 Maxima [F]

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

input

integrate((c*(d*sec(f*x+e))^p)^n/(a+b*sec(f*x+e)),x, algorithm="maxima")
 

output

integrate(((d*sec(f*x + e))^p*c)^n/(b*sec(f*x + e) + a), x)
 

3.3.40.8 Giac [F]

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

input

integrate((c*(d*sec(f*x+e))^p)^n/(a+b*sec(f*x+e)),x, algorithm="giac")
 

output

integrate(((d*sec(f*x + e))^p*c)^n/(b*sec(f*x + e) + a), x)
 

3.3.40.9 Mupad [F(-1)]

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

input

int((c*(d/cos(e + f*x))^p)^n/(a + b/cos(e + f*x)),x)
 

output

int((c*(d/cos(e + f*x))^p)^n/(a + b/cos(e + f*x)), x)