3.3.30 \(\int (c (d \sec (e+f x))^p)^n (a+a \sec (e+f x))^m \, dx\) [230]

3.3.30.1 Optimal result

Integrand size = 27, antiderivative size = 106 \[ \int \left (c (d \sec (e+f x))^p\right )^n (a+a \sec (e+f x))^m \, dx=-\frac {\operatorname {AppellF1}\left (n p,\frac {1}{2},\frac {1}{2}-m,1+n p,\sec (e+f x),-\sec (e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n (1+\sec (e+f x))^{-\frac {1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)}{f n p \sqrt {1-\sec (e+f x)}} \]

output

-AppellF1(n*p,1/2-m,1/2,n*p+1,-sec(f*x+e),sec(f*x+e))*(c*(d*sec(f*x+e))^p) 
^n*(1+sec(f*x+e))^(-1/2-m)*(a+a*sec(f*x+e))^m*tan(f*x+e)/f/n/p/(1-sec(f*x+ 
e))^(1/2)
 

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

Leaf count is larger than twice the leaf count of optimal. \(2425\) vs. \(2(106)=212\).

Time = 13.88 (sec) , antiderivative size = 2425, normalized size of antiderivative = 22.88 \[ \int \left (c (d \sec (e+f x))^p\right )^n (a+a \sec (e+f x))^m \, dx=\text {Result too large to show} \]

input

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

output

(3*2^(1 + m)*AppellF1[1/2, m + n*p, 1 - n*p, 3/2, Tan[(e + f*x)/2]^2, -Tan 
[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^(-1 + n*p)*(Cos[(e + f*x)/2]^2*Sec[e 
 + f*x])^(m + n*p)*(c*(d*Sec[e + f*x])^p)^n*(a*(1 + Sec[e + f*x]))^m*Tan[( 
e + f*x)/2])/(f*(3*AppellF1[1/2, m + n*p, 1 - n*p, 3/2, Tan[(e + f*x)/2]^2 
, -Tan[(e + f*x)/2]^2] + 2*((-1 + n*p)*AppellF1[3/2, m + n*p, 2 - n*p, 5/2 
, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (m + n*p)*AppellF1[3/2, 1 + m 
 + n*p, 1 - n*p, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f 
*x)/2]^2)*((3*2^m*AppellF1[1/2, m + n*p, 1 - n*p, 3/2, Tan[(e + f*x)/2]^2, 
 -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^(n*p)*(Cos[(e + f*x)/2]^2*Sec[e 
 + f*x])^(m + n*p))/(3*AppellF1[1/2, m + n*p, 1 - n*p, 3/2, Tan[(e + f*x)/ 
2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + n*p)*AppellF1[3/2, m + n*p, 2 - n*p, 
 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (m + n*p)*AppellF1[3/2, 1 
 + m + n*p, 1 - n*p, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e 
 + f*x)/2]^2) + (3*2^(1 + m)*(-1 + n*p)*AppellF1[1/2, m + n*p, 1 - n*p, 3/ 
2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^(-1 + n*p 
)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(m + n*p)*Tan[(e + f*x)/2]^2)/(3*Appel 
lF1[1/2, m + n*p, 1 - n*p, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 
 2*((-1 + n*p)*AppellF1[3/2, m + n*p, 2 - n*p, 5/2, Tan[(e + f*x)/2]^2, -T 
an[(e + f*x)/2]^2] + (m + n*p)*AppellF1[3/2, 1 + m + n*p, 1 - n*p, 5/2, Ta 
n[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2) + (3*2^(1 +...
 

3.3.30.3 Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 4436, 3042, 4315, 3042, 4314, 150}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sec (e+f x)+a)^m \left (c (d \sec (e+f x))^p\right )^ndx\)

\(\Big \downarrow \) 4436

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4315

\(\displaystyle (\sec (e+f x)+1)^{-m} (a \sec (e+f x)+a)^m (d \sec (e+f x))^{-n p} \left (c (d \sec (e+f x))^p\right )^n \int (d \sec (e+f x))^{n p} (\sec (e+f x)+1)^mdx\)

\(\Big \downarrow \) 3042

\(\displaystyle (\sec (e+f x)+1)^{-m} (a \sec (e+f x)+a)^m (d \sec (e+f x))^{-n p} \left (c (d \sec (e+f x))^p\right )^n \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{n p} \left (\csc \left (e+f x+\frac {\pi }{2}\right )+1\right )^mdx\)

\(\Big \downarrow \) 4314

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

\(\Big \downarrow \) 150

\(\displaystyle -\frac {\tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \operatorname {AppellF1}\left (n p,\frac {1}{2},\frac {1}{2}-m,n p+1,\sec (e+f x),-\sec (e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{f n p \sqrt {1-\sec (e+f x)}}\)

input

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

output

-((AppellF1[n*p, 1/2, 1/2 - m, 1 + n*p, Sec[e + f*x], -Sec[e + f*x]]*(c*(d 
*Sec[e + f*x])^p)^n*(1 + Sec[e + f*x])^(-1/2 - m)*(a + a*Sec[e + f*x])^m*T 
an[e + f*x])/(f*n*p*Sqrt[1 - Sec[e + f*x]]))
 

3.3.30.3.1 Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 3042

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

rule 4314

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Simp[a^2*d*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x 
]]*Sqrt[a - b*Csc[e + f*x]]))   Subst[Int[(d*x)^(n - 1)*((a + b*x)^(m - 1/2 
)/Sqrt[a - b*x]), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, 
x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] && GtQ[a, 0]
 

rule 4315

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Simp[a^IntPart[m]*((a + b*Csc[e + f*x])^FracPart[m 
]/(1 + (b/a)*Csc[e + f*x])^FracPart[m])   Int[(1 + (b/a)*Csc[e + f*x])^m*(d 
*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^ 
2, 0] &&  !IntegerQ[m] &&  !GtQ[a, 0]
 

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]
 

3.3.30.4 Maple [F]

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

input

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

output

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

3.3.30.5 Fricas [F]

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

input

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

output

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

3.3.30.6 Sympy [F]

\[ \int \left (c (d \sec (e+f x))^p\right )^n (a+a \sec (e+f x))^m \, dx=\int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{m} \left (c \left (d \sec {\left (e + f x \right )}\right )^{p}\right )^{n}\, dx \]

input

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

output

Integral((a*(sec(e + f*x) + 1))**m*(c*(d*sec(e + f*x))**p)**n, x)
 

3.3.30.7 Maxima [F]

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

input

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

output

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

3.3.30.8 Giac [F]

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

input

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

output

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

3.3.30.9 Mupad [F(-1)]

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

input

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

output

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