\(\int (-\sec (e+f x))^n (a+a \sec (e+f x))^m \, dx\) [335]

Optimal result

Integrand size = 23, antiderivative size = 113 \[ \int (-\sec (e+f x))^n (a+a \sec (e+f x))^m \, dx=-\frac {2^{\frac {1}{2}+m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2}-m,1-n,\frac {3}{2},\frac {1}{2} (1-\sec (e+f x)),1-\sec (e+f x)\right ) (-\sec (e+f x))^{-1+n} \sec ^{2-n}(e+f x) (1+\sec (e+f x))^{-\frac {1}{2}-m} (a+a \sec (e+f x))^m \sin (e+f x)}{f} \] Output:

-2^(1/2+m)*AppellF1(1/2,1-n,1/2-m,3/2,1-sec(f*x+e),1/2-1/2*sec(f*x+e))*(-s 
ec(f*x+e))^(-1+n)*sec(f*x+e)^(2-n)*(1+sec(f*x+e))^(-1/2-m)*(a+a*sec(f*x+e) 
)^m*sin(f*x+e)/f
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(825\) vs. \(2(113)=226\).

Time = 5.25 (sec) , antiderivative size = 825, normalized size of antiderivative = 7.30 \[ \int (-\sec (e+f x))^n (a+a \sec (e+f x))^m \, dx =\text {Too large to display} \] Input:

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

Output:

(30*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2] 
^2]*Cos[(e + f*x)/2]*(-Sec[e + f*x])^n*(a*(1 + Sec[e + f*x]))^m*Sin[(e + f 
*x)/2])/(f*(15*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[( 
e + f*x)/2]^2] + 30*(-1 + n)*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x 
)/2]^2, -Tan[(e + f*x)/2]^2]*Sin[(e + f*x)/2]^2 + 10*((-1 + n)*AppellF1[3/ 
2, m + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (m + n)*A 
ppellF1[3/2, 1 + m + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^ 
2])*Tan[(e + f*x)/2]^2 - (18*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x 
)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2*(5*(-1 + n)*AppellF1[3/2, 
m + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 5*(m + n)*Ap 
pellF1[3/2, 1 + m + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2 
] + 2*((2 - 3*n + n^2)*AppellF1[5/2, m + n, 3 - n, 7/2, Tan[(e + f*x)/2]^2 
, -Tan[(e + f*x)/2]^2] + (m + n)*(2*(-1 + n)*AppellF1[5/2, 1 + m + n, 2 - 
n, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (1 + m + n)*AppellF1[5/ 
2, 2 + m + n, 1 - n, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]))*Tan[( 
e + f*x)/2]^2))/(3*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -T 
an[(e + f*x)/2]^2] + 2*((-1 + n)*AppellF1[3/2, m + n, 2 - n, 5/2, Tan[(e + 
 f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (m + n)*AppellF1[3/2, 1 + m + n, 1 - n, 
 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2) + 15*( 
m + n)*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f...
 

Rubi [A] (warning: unable to verify)

Time = 0.39 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.77, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4315, 3042, 4313, 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.

\(\displaystyle \int (-\sec (e+f x))^n (a \sec (e+f x)+a)^m \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4315

\(\displaystyle (\sec (e+f x)+1)^{-m} (a \sec (e+f x)+a)^m \int (-\sec (e+f x))^n (\sec (e+f x)+1)^mdx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4313

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

\(\Big \downarrow \) 150

\(\displaystyle \frac {\sqrt {2} \tan (e+f x) (a \sec (e+f x)+a)^m \operatorname {AppellF1}\left (m+\frac {1}{2},1-n,\frac {1}{2},m+\frac {3}{2},\sec (e+f x)+1,\frac {1}{2} (\sec (e+f x)+1)\right )}{f (2 m+1) \sqrt {1-\sec (e+f x)}}\)

Input:

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

Output:

(Sqrt[2]*AppellF1[1/2 + m, 1 - n, 1/2, 3/2 + m, 1 + Sec[e + f*x], (1 + Sec 
[e + f*x])/2]*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/(f*(1 + 2*m)*Sqrt[1 - S 
ec[e + f*x]])
 

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 4313

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

Maple [F]

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

Input:

int((-sec(f*x+e))^n*(a+a*sec(f*x+e))^m,x)
 

Output:

int((-sec(f*x+e))^n*(a+a*sec(f*x+e))^m,x)
 

Fricas [F]

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

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

Output:

integral((a*sec(f*x + e) + a)^m*(-sec(f*x + e))^n, x)
 

Sympy [F]

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

integrate((-sec(f*x+e))**n*(a+a*sec(f*x+e))**m,x)
 

Output:

Integral((-sec(e + f*x))**n*(a*(sec(e + f*x) + 1))**m, x)
 

Maxima [F]

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

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

Output:

integrate((a*sec(f*x + e) + a)^m*(-sec(f*x + e))^n, x)
 

Giac [F]

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

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

Output:

integrate((a*sec(f*x + e) + a)^m*(-sec(f*x + e))^n, x)
 

Mupad [F(-1)]

Timed out. \[ \int (-\sec (e+f x))^n (a+a \sec (e+f x))^m \, dx=\int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (-\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \] Input:

int((a + a/cos(e + f*x))^m*(-1/cos(e + f*x))^n,x)
 

Output:

int((a + a/cos(e + f*x))^m*(-1/cos(e + f*x))^n, x)
 

Reduce [F]

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

int((-sec(f*x+e))^n*(a+a*sec(f*x+e))^m,x)
 

Output:

( - 1)**n*int(sec(e + f*x)**n*(sec(e + f*x)*a + a)**m,x)