3.3.41 \(\int \frac {(c (d \sec (e+f x))^p)^n}{(a+b \sec (e+f x))^2} \, dx\) [241]

3.3.41.1 Optimal result

Integrand size = 27, antiderivative size = 322 \[ \int \frac {\left (c (d \sec (e+f x))^p\right )^n}{(a+b \sec (e+f x))^2} \, dx=-\frac {2 a b \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-2+n p),2,\frac {3}{2},\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos ^2(e+f x)^{\frac {n p}{2}} \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{\left (a^2-b^2\right )^2 f}+\frac {a^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-3+n p),2,\frac {3}{2},\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (-1+n p)} \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{\left (a^2-b^2\right )^2 f}+\frac {b^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-1+n p),2,\frac {3}{2},\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (-1+n p)} \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{\left (a^2-b^2\right )^2 f} \]

output

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

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

Leaf count is larger than twice the leaf count of optimal. \(14108\) vs. \(2(322)=644\).

Time = 46.81 (sec) , antiderivative size = 14108, normalized size of antiderivative = 43.81 \[ \int \frac {\left (c (d \sec (e+f x))^p\right )^n}{(a+b \sec (e+f x))^2} \, dx=\text {Result too large to show} \]

input

Integrate[(c*(d*Sec[e + f*x])^p)^n/(a + b*Sec[e + f*x])^2,x]
 

output

Result too large to show
 

3.3.41.3 Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 4436, 3042, 4356, 3042, 3303, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4436

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4356

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3303

\(\displaystyle \cos ^{n p}(e+f x) \left (c (d \sec (e+f x))^p\right )^n \int \left (\frac {b^2 \cos ^{2-n p}(e+f x)}{\left (b^2-a^2 \cos ^2(e+f x)\right )^2}-\frac {2 a b \cos ^{3-n p}(e+f x)}{\left (b^2-a^2 \cos ^2(e+f x)\right )^2}+\frac {a^2 \cos ^{4-n p}(e+f x)}{\left (a^2 \cos ^2(e+f x)-b^2\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

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

input

Int[(c*(d*Sec[e + f*x])^p)^n/(a + b*Sec[e + f*x])^2,x]
 

output

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

3.3.41.3.1 Defintions of rubi rules used

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 3303

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

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]
 

3.3.41.4 Maple [F]

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

input

int((c*(d*sec(f*x+e))^p)^n/(a+b*sec(f*x+e))^2,x)
 

output

int((c*(d*sec(f*x+e))^p)^n/(a+b*sec(f*x+e))^2,x)
 

3.3.41.5 Fricas [F]

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

input

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

output

integral(((d*sec(f*x + e))^p*c)^n/(b^2*sec(f*x + e)^2 + 2*a*b*sec(f*x + e) 
 + a^2), x)
 

3.3.41.6 Sympy [F]

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

input

integrate((c*(d*sec(f*x+e))**p)**n/(a+b*sec(f*x+e))**2,x)
 

output

Integral((c*(d*sec(e + f*x))**p)**n/(a + b*sec(e + f*x))**2, x)
 

3.3.41.7 Maxima [F]

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

input

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

output

integrate(((d*sec(f*x + e))^p*c)^n/(b*sec(f*x + e) + a)^2, x)
 

3.3.41.8 Giac [F]

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

input

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

output

integrate(((d*sec(f*x + e))^p*c)^n/(b*sec(f*x + e) + a)^2, x)
 

3.3.41.9 Mupad [F(-1)]

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

input

int((c*(d/cos(e + f*x))^p)^n/(a + b/cos(e + f*x))^2,x)
 

output

int((c*(d/cos(e + f*x))^p)^n/(a + b/cos(e + f*x))^2, x)