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

Optimal result

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

[Out]

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)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3958, 79, 67} \[ \int (a+a \sec (c+d x))^n \sin ^3(c+d x) \, dx=\frac {\cos ^3(c+d x) (a \sec (c+d x)+a)^{n+2}}{3 a^2 d}-\frac {(4-n) (a \sec (c+d x)+a)^{n+2} \operatorname {Hypergeometric2F1}(3,n+2,n+3,\sec (c+d x)+1)}{3 a^2 d (n+2)} \]

[In]

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

[Out]

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

Rule 67

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] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(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] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 3958

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {(-a-a x) (a-a x)^{1+n}}{x^4} \, dx,x,-\sec (c+d x)\right )}{a^2 d} \\ & = \frac {\cos ^3(c+d x) (a+a \sec (c+d x))^{2+n}}{3 a^2 d}+\frac {(4-n) \text {Subst}\left (\int \frac {(a-a x)^{1+n}}{x^3} \, dx,x,-\sec (c+d x)\right )}{3 a d} \\ & = \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)} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

(((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))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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