\(\int (a+b (c \sec (e+f x))^n)^p \tan ^3(e+f x) \, dx\) [464]

Optimal result

Integrand size = 25, antiderivative size = 145 \[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^3(e+f x) \, dx=\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {a+b (c \sec (e+f x))^n}{a}\right ) \left (a+b (c \sec (e+f x))^n\right )^{1+p}}{a f n (1+p)}+\frac {\operatorname {Hypergeometric2F1}\left (\frac {2}{n},-p,\frac {2+n}{n},-\frac {b (c \sec (e+f x))^n}{a}\right ) \sec ^2(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac {a+b (c \sec (e+f x))^n}{a}\right )^{-p}}{2 f} \] Output:

hypergeom([1, p+1],[2+p],(a+b*(c*sec(f*x+e))^n)/a)*(a+b*(c*sec(f*x+e))^n)^ 
(p+1)/a/f/n/(p+1)+1/2*hypergeom([-p, 2/n],[(2+n)/n],-b*(c*sec(f*x+e))^n/a) 
*sec(f*x+e)^2*(a+b*(c*sec(f*x+e))^n)^p/f/(((a+b*(c*sec(f*x+e))^n)/a)^p)
 

Mathematica [A] (warning: unable to verify)

Time = 3.97 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.12 \[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^3(e+f x) \, dx=\frac {\left (a+b (c \sec (e+f x))^n\right )^p \left (\frac {2 \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}\right ) \left (a+b \left (c \sqrt {\sec ^2(e+f x)}\right )^n\right )}{a n (1+p)}+\operatorname {Hypergeometric2F1}\left (\frac {2}{n},-p,\frac {2+n}{n},-\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}\right ) \sec ^2(e+f x) \left (1+\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}\right )^{-p}\right )}{2 f} \] Input:

Integrate[(a + b*(c*Sec[e + f*x])^n)^p*Tan[e + f*x]^3,x]
 

Output:

((a + b*(c*Sec[e + f*x])^n)^p*((2*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + ( 
b*(c*Sqrt[Sec[e + f*x]^2])^n)/a]*(a + b*(c*Sqrt[Sec[e + f*x]^2])^n))/(a*n* 
(1 + p)) + (Hypergeometric2F1[2/n, -p, (2 + n)/n, -((b*(c*Sqrt[Sec[e + f*x 
]^2])^n)/a)]*Sec[e + f*x]^2)/(1 + (b*(c*Sqrt[Sec[e + f*x]^2])^n)/a)^p))/(2 
*f)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4627, 25, 7293, 2009}

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.

\(\displaystyle \int \tan ^3(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x)^3 \left (a+b (c \sec (e+f x))^n\right )^pdx\)

\(\Big \downarrow \) 4627

\(\displaystyle \frac {\int -\cos (e+f x) \left (1-\sec ^2(e+f x)\right ) \left (b (c \sec (e+f x))^n+a\right )^pd\sec (e+f x)}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \cos (e+f x) \left (1-\sec ^2(e+f x)\right ) \left (b (c \sec (e+f x))^n+a\right )^pd\sec (e+f x)}{f}\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {\int \left (\cos (e+f x) \left (b (c \sec (e+f x))^n+a\right )^p-\sec (e+f x) \left (b (c \sec (e+f x))^n+a\right )^p\right )d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{2} \sec ^2(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac {b (c \sec (e+f x))^n}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {2}{n},-p,\frac {n+2}{n},-\frac {b (c \sec (e+f x))^n}{a}\right )+\frac {\left (a+b (c \sec (e+f x))^n\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b (c \sec (e+f x))^n}{a}+1\right )}{a n (p+1)}}{f}\)

Input:

Int[(a + b*(c*Sec[e + f*x])^n)^p*Tan[e + f*x]^3,x]
 

Output:

((Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*(c*Sec[e + f*x])^n)/a]*(a + b* 
(c*Sec[e + f*x])^n)^(1 + p))/(a*n*(1 + p)) + (Hypergeometric2F1[2/n, -p, ( 
2 + n)/n, -((b*(c*Sec[e + f*x])^n)/a)]*Sec[e + f*x]^2*(a + b*(c*Sec[e + f* 
x])^n)^p)/(2*(1 + (b*(c*Sec[e + f*x])^n)/a)^p))/f
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4627

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si 
mp[1/f   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), x] 
, x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[( 
m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4] || IGtQ[p, 0] || Integers 
Q[2*n, p])
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [F]

\[\int \left (a +b \left (c \sec \left (f x +e \right )\right )^{n}\right )^{p} \tan \left (f x +e \right )^{3}d x\]

Input:

int((a+b*(c*sec(f*x+e))^n)^p*tan(f*x+e)^3,x)
 

Output:

int((a+b*(c*sec(f*x+e))^n)^p*tan(f*x+e)^3,x)
 

Fricas [F]

\[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^3(e+f x) \, dx=\int { {\left (\left (c \sec \left (f x + e\right )\right )^{n} b + a\right )}^{p} \tan \left (f x + e\right )^{3} \,d x } \] Input:

integrate((a+b*(c*sec(f*x+e))^n)^p*tan(f*x+e)^3,x, algorithm="fricas")
 

Output:

integral(((c*sec(f*x + e))^n*b + a)^p*tan(f*x + e)^3, x)
 

Sympy [F]

\[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^3(e+f x) \, dx=\int \left (a + b \left (c \sec {\left (e + f x \right )}\right )^{n}\right )^{p} \tan ^{3}{\left (e + f x \right )}\, dx \] Input:

integrate((a+b*(c*sec(f*x+e))**n)**p*tan(f*x+e)**3,x)
 

Output:

Integral((a + b*(c*sec(e + f*x))**n)**p*tan(e + f*x)**3, x)
 

Maxima [F]

\[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^3(e+f x) \, dx=\int { {\left (\left (c \sec \left (f x + e\right )\right )^{n} b + a\right )}^{p} \tan \left (f x + e\right )^{3} \,d x } \] Input:

integrate((a+b*(c*sec(f*x+e))^n)^p*tan(f*x+e)^3,x, algorithm="maxima")
 

Output:

integrate(((c*sec(f*x + e))^n*b + a)^p*tan(f*x + e)^3, x)
 

Giac [F]

\[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^3(e+f x) \, dx=\int { {\left (\left (c \sec \left (f x + e\right )\right )^{n} b + a\right )}^{p} \tan \left (f x + e\right )^{3} \,d x } \] Input:

integrate((a+b*(c*sec(f*x+e))^n)^p*tan(f*x+e)^3,x, algorithm="giac")
 

Output:

integrate(((c*sec(f*x + e))^n*b + a)^p*tan(f*x + e)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^3(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^3\,{\left (a+b\,{\left (\frac {c}{\cos \left (e+f\,x\right )}\right )}^n\right )}^p \,d x \] Input:

int(tan(e + f*x)^3*(a + b*(c/cos(e + f*x))^n)^p,x)
 

Output:

int(tan(e + f*x)^3*(a + b*(c/cos(e + f*x))^n)^p, x)
 

Reduce [F]

\[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^3(e+f x) \, dx=\frac {\left (c^{n} \sec \left (f x +e \right )^{n} b +a \right )^{p} \tan \left (f x +e \right )^{2} n p -2 \left (c^{n} \sec \left (f x +e \right )^{n} b +a \right )^{p}+\left (\int \frac {\left (c^{n} \sec \left (f x +e \right )^{n} b +a \right )^{p} \tan \left (f x +e \right )^{3}}{c^{n} \sec \left (f x +e \right )^{n} b n p +2 c^{n} \sec \left (f x +e \right )^{n} b +a n p +2 a}d x \right ) a f \,n^{3} p^{3}+2 \left (\int \frac {\left (c^{n} \sec \left (f x +e \right )^{n} b +a \right )^{p} \tan \left (f x +e \right )^{3}}{c^{n} \sec \left (f x +e \right )^{n} b n p +2 c^{n} \sec \left (f x +e \right )^{n} b +a n p +2 a}d x \right ) a f \,n^{2} p^{2}-2 \left (\int \frac {\left (c^{n} \sec \left (f x +e \right )^{n} b +a \right )^{p} \tan \left (f x +e \right )}{c^{n} \sec \left (f x +e \right )^{n} b n p +2 c^{n} \sec \left (f x +e \right )^{n} b +a n p +2 a}d x \right ) a f \,n^{2} p^{2}-4 \left (\int \frac {\left (c^{n} \sec \left (f x +e \right )^{n} b +a \right )^{p} \tan \left (f x +e \right )}{c^{n} \sec \left (f x +e \right )^{n} b n p +2 c^{n} \sec \left (f x +e \right )^{n} b +a n p +2 a}d x \right ) a f n p}{f n p \left (n p +2\right )} \] Input:

int((a+b*(c*sec(f*x+e))^n)^p*tan(f*x+e)^3,x)
 

Output:

((c**n*sec(e + f*x)**n*b + a)**p*tan(e + f*x)**2*n*p - 2*(c**n*sec(e + f*x 
)**n*b + a)**p + int(((c**n*sec(e + f*x)**n*b + a)**p*tan(e + f*x)**3)/(c* 
*n*sec(e + f*x)**n*b*n*p + 2*c**n*sec(e + f*x)**n*b + a*n*p + 2*a),x)*a*f* 
n**3*p**3 + 2*int(((c**n*sec(e + f*x)**n*b + a)**p*tan(e + f*x)**3)/(c**n* 
sec(e + f*x)**n*b*n*p + 2*c**n*sec(e + f*x)**n*b + a*n*p + 2*a),x)*a*f*n** 
2*p**2 - 2*int(((c**n*sec(e + f*x)**n*b + a)**p*tan(e + f*x))/(c**n*sec(e 
+ f*x)**n*b*n*p + 2*c**n*sec(e + f*x)**n*b + a*n*p + 2*a),x)*a*f*n**2*p**2 
 - 4*int(((c**n*sec(e + f*x)**n*b + a)**p*tan(e + f*x))/(c**n*sec(e + f*x) 
**n*b*n*p + 2*c**n*sec(e + f*x)**n*b + a*n*p + 2*a),x)*a*f*n*p)/(f*n*p*(n* 
p + 2))