\(\int (d \sec (e+f x))^n (1+\sec (e+f x))^m \, dx\) [337]

Optimal result

Integrand size = 21, antiderivative size = 97 \[ \int (d \sec (e+f x))^n (1+\sec (e+f x))^m \, dx=\frac {2^{\frac {1}{2}+m} d \operatorname {AppellF1}\left (\frac {1}{2},1-n,\frac {1}{2}-m,\frac {3}{2},1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right ) \sec ^{2-n}(e+f x) (d \sec (e+f x))^{-1+n} \sin (e+f x)}{f \sqrt {1+\sec (e+f x)}} \] Output:

2^(1/2+m)*d*AppellF1(1/2,1-n,1/2-m,3/2,1-sec(f*x+e),1/2-1/2*sec(f*x+e))*se 
c(f*x+e)^(2-n)*(d*sec(f*x+e))^(-1+n)*sin(f*x+e)/f/(1+sec(f*x+e))^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(823\) vs. \(2(97)=194\).

Time = 5.05 (sec) , antiderivative size = 823, normalized size of antiderivative = 8.48 \[ \int (d \sec (e+f x))^n (1+\sec (e+f x))^m \, dx =\text {Too large to display} \] Input:

Integrate[(d*Sec[e + f*x])^n*(1 + 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]*(d*Sec[e + f*x])^n*(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)*Appe 
llF1[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)*Appel 
lF1[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, -Tan[ 
(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*x)...
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.81, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {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.

\(\displaystyle \int (\sec (e+f x)+1)^m (d \sec (e+f x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (\csc \left (e+f x+\frac {\pi }{2}\right )+1\right )^m \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^ndx\)

\(\Big \downarrow \) 4314

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

\(\Big \downarrow \) 150

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

Input:

Int[(d*Sec[e + f*x])^n*(1 + Sec[e + f*x])^m,x]
 

Output:

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

Maple [F]

\[\int \left (d \sec \left (f x +e \right )\right )^{n} \left (1+\sec \left (f x +e \right )\right )^{m}d x\]

Input:

int((d*sec(f*x+e))^n*(1+sec(f*x+e))^m,x)
 

Output:

int((d*sec(f*x+e))^n*(1+sec(f*x+e))^m,x)
 

Fricas [F]

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

integrate((d*sec(f*x+e))^n*(1+sec(f*x+e))^m,x, algorithm="fricas")
 

Output:

integral((d*sec(f*x + e))^n*(sec(f*x + e) + 1)^m, x)
 

Sympy [F]

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

integrate((d*sec(f*x+e))**n*(1+sec(f*x+e))**m,x)
 

Output:

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

Maxima [F]

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

integrate((d*sec(f*x+e))^n*(1+sec(f*x+e))^m,x, algorithm="maxima")
 

Output:

integrate((d*sec(f*x + e))^n*(sec(f*x + e) + 1)^m, x)
 

Giac [F]

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

integrate((d*sec(f*x+e))^n*(1+sec(f*x+e))^m,x, algorithm="giac")
 

Output:

integrate((d*sec(f*x + e))^n*(sec(f*x + e) + 1)^m, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((d*sec(f*x+e))^n*(1+sec(f*x+e))^m,x)
 

Output:

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