\(\int (a+b \sec (c+d x))^n \tan ^3(c+d x) \, dx\) [351]

Optimal result

Integrand size = 21, antiderivative size = 103 \[ \int (a+b \sec (c+d x))^n \tan ^3(c+d x) \, dx=-\frac {a (a+b \sec (c+d x))^{1+n}}{b^2 d (1+n)}+\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a d (1+n)}+\frac {(a+b \sec (c+d x))^{2+n}}{b^2 d (2+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.78 \[ \int (a+b \sec (c+d x))^n \tan ^3(c+d x) \, dx=\frac {(a+b \sec (c+d x))^{1+n} \left (b^2 (2+n) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b \sec (c+d x)}{a}\right )+a (-a+b (1+n) \sec (c+d x))\right )}{a b^2 d (1+n) (2+n)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 25, 4373, 522, 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.

\(\displaystyle \int \tan ^3(c+d x) (a+b \sec (c+d x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^ndx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^ndx\)

\(\Big \downarrow \) 4373

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

\(\Big \downarrow \) 522

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

rule 522

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

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 4373

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n 
_), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1))   Subst[Int[(b^2 - x^ 
2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, 
 d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
 

Maple [F]

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

Input:

int((a+b*sec(d*x+c))^n*tan(d*x+c)^3,x)
 

Output:

int((a+b*sec(d*x+c))^n*tan(d*x+c)^3,x)
 

Fricas [F]

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

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

Output:

integral((b*sec(d*x + c) + a)^n*tan(d*x + c)^3, x)
 

Sympy [F]

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

integrate((a+b*sec(d*x+c))**n*tan(d*x+c)**3,x)
 

Output:

Integral((a + b*sec(c + d*x))**n*tan(c + d*x)**3, x)
 

Maxima [F]

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

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

Output:

integrate((b*sec(d*x + c) + a)^n*tan(d*x + c)^3, x)
 

Giac [F]

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

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

Output:

integrate((b*sec(d*x + c) + a)^n*tan(d*x + c)^3, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((a+b*sec(d*x+c))^n*tan(d*x+c)^3,x)
 

Output:

((sec(c + d*x)*b + a)**n*tan(c + d*x)**2*n - 2*(sec(c + d*x)*b + a)**n + i 
nt(((sec(c + d*x)*b + a)**n*tan(c + d*x)**3)/(sec(c + d*x)*b*n + 2*sec(c + 
 d*x)*b + a*n + 2*a),x)*a*d*n**3 + 2*int(((sec(c + d*x)*b + a)**n*tan(c + 
d*x)**3)/(sec(c + d*x)*b*n + 2*sec(c + d*x)*b + a*n + 2*a),x)*a*d*n**2 - 2 
*int(((sec(c + d*x)*b + a)**n*tan(c + d*x))/(sec(c + d*x)*b*n + 2*sec(c + 
d*x)*b + a*n + 2*a),x)*a*d*n**2 - 4*int(((sec(c + d*x)*b + a)**n*tan(c + d 
*x))/(sec(c + d*x)*b*n + 2*sec(c + d*x)*b + a*n + 2*a),x)*a*d*n)/(d*n*(n + 
 2))