3.2.47 \(\int (a+a \sec (c+d x))^n \sin ^3(c+d x) \, dx\) [147]

3.2.47.1 Optimal result

Integrand size = 21, antiderivative size = 83 \[ \int (a+a \sec (c+d x))^n \sin ^3(c+d x) \, dx=\frac {\cos ^3(c+d x) (a+a \sec (c+d x))^{2+n}}{3 a^2 d}-\frac {(4-n) \operatorname {Hypergeometric2F1}(3,2+n,3+n,1+\sec (c+d x)) (a+a \sec (c+d x))^{2+n}}{3 a^2 d (2+n)} \]

output

1/3*cos(d*x+c)^3*(a+a*sec(d*x+c))^(2+n)/a^2/d-1/3*(4-n)*hypergeom([3, 2+n] 
,[3+n],1+sec(d*x+c))*(a+a*sec(d*x+c))^(2+n)/a^2/d/(2+n)
 

3.2.47.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.81 \[ \int (a+a \sec (c+d x))^n \sin ^3(c+d x) \, dx=\frac {\left ((2+n) \cos ^3(c+d x)+(-4+n) \operatorname {Hypergeometric2F1}(3,2+n,3+n,1+\sec (c+d x))\right ) (1+\sec (c+d x))^2 (a (1+\sec (c+d x)))^n}{3 d (2+n)} \]

input

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

output

(((2 + n)*Cos[c + d*x]^3 + (-4 + n)*Hypergeometric2F1[3, 2 + n, 3 + n, 1 + 
 Sec[c + d*x]])*(1 + Sec[c + d*x])^2*(a*(1 + Sec[c + d*x]))^n)/(3*d*(2 + n 
))
 

3.2.47.3 Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4361, 25, 27, 87, 75}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4361

\(\displaystyle -\frac {\int -a \cos ^4(c+d x) (1-\sec (c+d x)) (\sec (c+d x) a+a)^{n+1}d(-\sec (c+d x))}{a^2 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int a \cos ^4(c+d x) (1-\sec (c+d x)) (\sec (c+d x) a+a)^{n+1}d(-\sec (c+d x))}{a^2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \cos ^4(c+d x) (1-\sec (c+d x)) (\sec (c+d x) a+a)^{n+1}d(-\sec (c+d x))}{a d}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {\frac {1}{3} (4-n) \int -\cos ^3(c+d x) (\sec (c+d x) a+a)^{n+1}d(-\sec (c+d x))+\frac {\cos ^3(c+d x) (a \sec (c+d x)+a)^{n+2}}{3 a}}{a d}\)

\(\Big \downarrow \) 75

\(\displaystyle \frac {\frac {\cos ^3(c+d x) (a \sec (c+d x)+a)^{n+2}}{3 a}-\frac {(4-n) (a \sec (c+d x)+a)^{n+2} \operatorname {Hypergeometric2F1}(3,n+2,n+3,\sec (c+d x)+1)}{3 a (n+2)}}{a d}\)

input

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

output

((Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(2 + n))/(3*a) - ((4 - n)*Hypergeome 
tric2F1[3, 2 + n, 3 + n, 1 + Sec[c + d*x]]*(a + a*Sec[c + d*x])^(2 + n))/( 
3*a*(2 + n)))/(a*d)
 

3.2.47.3.1 Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 75

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x 
)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + 
 d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 3042

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

rule 4361

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m 
_), x_Symbol] :> Simp[-(f*b^(p - 1))^(-1)   Subst[Int[(-a + b*x)^((p - 1)/2 
)*((a + b*x)^(m + (p - 1)/2)/x^(p + 1)), x], x, Csc[e + f*x]], x] /; FreeQ[ 
{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
 

3.2.47.4 Maple [F]

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

input

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

output

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

3.2.47.5 Fricas [F]

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

input

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

output

integral(-(cos(d*x + c)^2 - 1)*(a*sec(d*x + c) + a)^n*sin(d*x + c), x)
 

3.2.47.6 Sympy [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^n \sin ^3(c+d x) \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.2.47.7 Maxima [F]

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

input

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

output

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

3.2.47.8 Giac [F(-2)]

Exception generated. \[ \int (a+a \sec (c+d x))^n \sin ^3(c+d x) \, dx=\text {Exception raised: RuntimeError} \]

input

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

output

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Value
 

3.2.47.9 Mupad [F(-1)]

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

input

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

output

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