\(\int (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)} \, dx\) [230]

Optimal result

Integrand size = 23, antiderivative size = 114 \[ \int (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)} \, dx=\frac {2^{\frac {5}{2}+n} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2}+n,1,\frac {7}{4},-\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{\frac {3}{2}+n} (a+a \sec (c+d x))^n \tan ^{\frac {3}{2}}(c+d x)}{3 d} \] Output:

1/3*2^(5/2+n)*AppellF1(3/4,1,1/2+n,7/4,(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)))^(3/2+n)*(a+a*sec(d*x 
+c))^n*tan(d*x+c)^(3/2)/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(238\) vs. \(2(114)=228\).

Time = 4.59 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.09 \[ \int (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)} \, dx=\frac {56 \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2}+n,1,\frac {7}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) (a (1+\sec (c+d x)))^n \sin \left (\frac {1}{2} (c+d x)\right ) \sqrt {\tan (c+d x)}}{d \left (6 \left (2 \operatorname {AppellF1}\left (\frac {7}{4},\frac {1}{2}+n,2,\frac {11}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-(1+2 n) \operatorname {AppellF1}\left (\frac {7}{4},\frac {3}{2}+n,1,\frac {11}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+21 \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2}+n,1,\frac {7}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right )} \] Input:

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

Output:

(56*AppellF1[3/4, 1/2 + n, 1, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2 
]*Cos[(c + d*x)/2]^3*(a*(1 + Sec[c + d*x]))^n*Sin[(c + d*x)/2]*Sqrt[Tan[c 
+ d*x]])/(d*(6*(2*AppellF1[7/4, 1/2 + n, 2, 11/4, Tan[(c + d*x)/2]^2, -Tan 
[(c + d*x)/2]^2] - (1 + 2*n)*AppellF1[7/4, 3/2 + n, 1, 11/4, Tan[(c + d*x) 
/2]^2, -Tan[(c + d*x)/2]^2])*(-1 + Cos[c + d*x]) + 21*AppellF1[3/4, 1/2 + 
n, 1, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x])))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 114, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4377

\(\displaystyle \frac {2^{n+\frac {5}{2}} \tan ^{\frac {3}{2}}(c+d x) \left (\frac {1}{\sec (c+d x)+1}\right )^{n+\frac {3}{2}} (a \sec (c+d x)+a)^n \operatorname {AppellF1}\left (\frac {3}{4},n+\frac {1}{2},1,\frac {7}{4},-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{3 d}\)

Input:

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

Output:

(2^(5/2 + n)*AppellF1[3/4, 1/2 + n, 1, 7/4, -((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))^(3/2 + n)*(a + a*Sec[c + d*x])^n*Tan[c + d*x]^(3/2))/(3*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]

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

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]

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

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

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

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

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

Output:

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

Reduce [F]

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