\(\int (a+a \sec (c+d x))^n \tan ^4(c+d x) \, dx\) [225]

Optimal result

Integrand size = 21, antiderivative size = 106 \[ \int (a+a \sec (c+d x))^n \tan ^4(c+d x) \, dx=\frac {2^{5+n} \operatorname {AppellF1}\left (\frac {5}{2},4+n,1,\frac {7}{2},-\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{5+n} (a+a \sec (c+d x))^n \tan ^5(c+d x)}{5 d} \] Output:

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

Mathematica [F]

\[ \int (a+a \sec (c+d x))^n \tan ^4(c+d x) \, dx=\int (a+a \sec (c+d x))^n \tan ^4(c+d x) \, dx \] Input:

Integrate[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^4,x]
 

Output:

Integrate[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^4, x]
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.

\(\displaystyle \int \tan ^4(c+d x) (a \sec (c+d x)+a)^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cot \left (c+d x+\frac {\pi }{2}\right )^4 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^ndx\)

\(\Big \downarrow \) 4377

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

Input:

Int[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^4,x]
 

Output:

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

\[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \tan \left (d x +c \right )^{4}d x\]

Input:

int((a+a*sec(d*x+c))^n*tan(d*x+c)^4,x)
 

Output:

int((a+a*sec(d*x+c))^n*tan(d*x+c)^4,x)
 

Fricas [F]

\[ \int (a+a \sec (c+d x))^n \tan ^4(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{4} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^4,x, algorithm="fricas")
 

Output:

integral((a*sec(d*x + c) + a)^n*tan(d*x + c)^4, x)
 

Sympy [F]

\[ \int (a+a \sec (c+d x))^n \tan ^4(c+d x) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \tan ^{4}{\left (c + d x \right )}\, dx \] Input:

integrate((a+a*sec(d*x+c))**n*tan(d*x+c)**4,x)
 

Output:

Integral((a*(sec(c + d*x) + 1))**n*tan(c + d*x)**4, x)
 

Maxima [F]

\[ \int (a+a \sec (c+d x))^n \tan ^4(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{4} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^4,x, algorithm="maxima")
 

Output:

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^4, x)
 

Giac [F]

\[ \int (a+a \sec (c+d x))^n \tan ^4(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{4} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^4,x, algorithm="giac")
 

Output:

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^n \tan ^4(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \] Input:

int(tan(c + d*x)^4*(a + a/cos(c + d*x))^n,x)
 

Output:

int(tan(c + d*x)^4*(a + a/cos(c + d*x))^n, x)
 

Reduce [F]

\[ \int (a+a \sec (c+d x))^n \tan ^4(c+d x) \, dx=\int \left (\sec \left (d x +c \right ) a +a \right )^{n} \tan \left (d x +c \right )^{4}d x \] Input:

int((a+a*sec(d*x+c))^n*tan(d*x+c)^4,x)
 

Output:

int((sec(c + d*x)*a + a)**n*tan(c + d*x)**4,x)