\(\int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx\) [231]

Optimal result

Integrand size = 23, antiderivative size = 111 \[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\frac {2^{\frac {3}{2}+n} \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2}+n,1,\frac {5}{4},-\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{\frac {1}{2}+n} (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)}}{d} \] Output:

2^(3/2+n)*AppellF1(1/4,1,-1/2+n,5/4,(a-a*sec(d*x+c))/(a+a*sec(d*x+c)),-(a- 
a*sec(d*x+c))/(a+a*sec(d*x+c)))*(1/(1+sec(d*x+c)))^(1/2+n)*(a+a*sec(d*x+c) 
)^n*tan(d*x+c)^(1/2)/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(229\) vs. \(2(111)=222\).

Time = 8.35 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.06 \[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\frac {10 \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2}+n,1,\frac {5}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos (c+d x) (1+\cos (c+d x)) (a (1+\sec (c+d x)))^n \sqrt {\tan (c+d x)}}{d \left (2 \left (2 \operatorname {AppellF1}\left (\frac {5}{4},-\frac {1}{2}+n,2,\frac {9}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+(1-2 n) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2}+n,1,\frac {9}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+5 \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2}+n,1,\frac {5}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right )} \] Input:

Integrate[(a + a*Sec[c + d*x])^n/Sqrt[Tan[c + d*x]],x]
 

Output:

(10*AppellF1[1/4, -1/2 + n, 1, 5/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^ 
2]*Cos[c + d*x]*(1 + Cos[c + d*x])*(a*(1 + Sec[c + d*x]))^n*Sqrt[Tan[c + d 
*x]])/(d*(2*(2*AppellF1[5/4, -1/2 + n, 2, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c 
 + d*x)/2]^2] + (1 - 2*n)*AppellF1[5/4, 1/2 + n, 1, 9/4, Tan[(c + d*x)/2]^ 
2, -Tan[(c + d*x)/2]^2])*(-1 + Cos[c + d*x]) + 5*AppellF1[1/4, -1/2 + n, 1 
, 5/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x])))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3042, 4377}

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.

\(\displaystyle \int \frac {(a \sec (c+d x)+a)^n}{\sqrt {\tan (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^n}{\sqrt {-\cot \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 4377

\(\displaystyle \frac {2^{n+\frac {3}{2}} \sqrt {\tan (c+d x)} \left (\frac {1}{\sec (c+d x)+1}\right )^{n+\frac {1}{2}} (a \sec (c+d x)+a)^n \operatorname {AppellF1}\left (\frac {1}{4},n-\frac {1}{2},1,\frac {5}{4},-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{d}\)

Input:

Int[(a + a*Sec[c + d*x])^n/Sqrt[Tan[c + d*x]],x]
 

Output:

(2^(3/2 + n)*AppellF1[1/4, -1/2 + n, 1, 5/4, -((a - a*Sec[c + d*x])/(a + a 
*Sec[c + d*x])), (a - a*Sec[c + d*x])/(a + a*Sec[c + d*x])]*((1 + Sec[c + 
d*x])^(-1))^(1/2 + n)*(a + a*Sec[c + d*x])^n*Sqrt[Tan[c + d*x]])/d
 

Defintions of rubi rules used

rule 3042

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

rule 4377

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( 
a_))^(n_), x_Symbol] :> Simp[(-2^(m + n + 1))*(e*Cot[c + d*x])^(m + 1)*((a 
+ b*Csc[c + d*x])^n/(d*e*(m + 1)))*(a/(a + b*Csc[c + d*x]))^(m + n + 1)*App 
ellF1[(m + 1)/2, m + n, 1, (m + 3)/2, -(a - b*Csc[c + d*x])/(a + b*Csc[c + 
d*x]), (a - b*Csc[c + d*x])/(a + b*Csc[c + d*x])], x] /; FreeQ[{a, b, c, d, 
 e, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]
 

Maple [F]

\[\int \frac {\left (a +a \sec \left (d x +c \right )\right )^{n}}{\sqrt {\tan \left (d x +c \right )}}d x\]

Input:

int((a+a*sec(d*x+c))^n/tan(d*x+c)^(1/2),x)
 

Output:

int((a+a*sec(d*x+c))^n/tan(d*x+c)^(1/2),x)
 

Fricas [F]

\[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\tan \left (d x + c\right )}} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^n/tan(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

integral((a*sec(d*x + c) + a)^n/sqrt(tan(d*x + c)), x)
 

Sympy [F]

\[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n}}{\sqrt {\tan {\left (c + d x \right )}}}\, dx \] Input:

integrate((a+a*sec(d*x+c))**n/tan(d*x+c)**(1/2),x)
 

Output:

Integral((a*(sec(c + d*x) + 1))**n/sqrt(tan(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\tan \left (d x + c\right )}} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^n/tan(d*x+c)^(1/2),x, algorithm="maxima")
 

Output:

integrate((a*sec(d*x + c) + a)^n/sqrt(tan(d*x + c)), x)
 

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\tan \left (d x + c\right )}} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^n/tan(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

integrate((a*sec(d*x + c) + a)^n/sqrt(tan(d*x + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}} \,d x \] Input:

int((a + a/cos(c + d*x))^n/tan(c + d*x)^(1/2),x)
 

Output:

int((a + a/cos(c + d*x))^n/tan(c + d*x)^(1/2), x)
 

Reduce [F]

\[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {\sqrt {\tan \left (d x +c \right )}\, \left (\sec \left (d x +c \right ) a +a \right )^{n}}{\tan \left (d x +c \right )}d x \] Input:

int((a+a*sec(d*x+c))^n/tan(d*x+c)^(1/2),x)
 

Output:

int((sqrt(tan(c + d*x))*(sec(c + d*x)*a + a)**n)/tan(c + d*x),x)