\(\int \frac {(e \tan (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx\) [217]

Optimal result

Integrand size = 25, antiderivative size = 131 \[ \int \frac {(e \tan (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {2^{\frac {1}{2}+m} \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2}+m,1,\frac {3+m}{2},-\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{\frac {1}{2}+m} (e \tan (c+d x))^{1+m}}{d e (1+m) \sqrt {a+a \sec (c+d x)}} \] Output:

2^(1/2+m)*AppellF1(1/2+1/2*m,1,-1/2+m,3/2+1/2*m,(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+m)*(e* 
tan(d*x+c))^(1+m)/d/e/(1+m)/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4377

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

Input:

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

Output:

(2^(1/2 + m)*AppellF1[(1 + m)/2, -1/2 + m, 1, (3 + m)/2, -((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 + m)*(e*Tan[c + d*x])^(1 + m))/(d*e*(1 + m)*S 
qrt[a + a*Sec[c + d*x]])
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(e**m*sqrt(a)*int((tan(c + d*x)**m*sqrt(sec(c + d*x) + 1))/(sec(c + d*x) + 
 1),x))/a