\(\int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx\) [146]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 21, antiderivative size = 123 \[ \int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx=\frac {(12-n) \cos ^4(c+d x) (a+a \sec (c+d x))^{3+n}}{20 a^3 d}-\frac {\cos ^5(c+d x) (a+a \sec (c+d x))^{3+n}}{5 a^3 d}+\frac {\left (32-13 n+n^2\right ) \operatorname {Hypergeometric2F1}(4,3+n,4+n,1+\sec (c+d x)) (a+a \sec (c+d x))^{3+n}}{20 a^3 d (3+n)} \]

[Out]

1/20*(12-n)*cos(d*x+c)^4*(a+a*sec(d*x+c))^(3+n)/a^3/d-1/5*cos(d*x+c)^5*(a+a*sec(d*x+c))^(3+n)/a^3/d+1/20*(n^2-
13*n+32)*hypergeom([4, 3+n],[4+n],1+sec(d*x+c))*(a+a*sec(d*x+c))^(3+n)/a^3/d/(3+n)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3958, 91, 79, 67} \[ \int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx=\frac {\left (n^2-13 n+32\right ) (a \sec (c+d x)+a)^{n+3} \operatorname {Hypergeometric2F1}(4,n+3,n+4,\sec (c+d x)+1)}{20 a^3 d (n+3)}-\frac {\cos ^5(c+d x) (a \sec (c+d x)+a)^{n+3}}{5 a^3 d}+\frac {(12-n) \cos ^4(c+d x) (a \sec (c+d x)+a)^{n+3}}{20 a^3 d} \]

[In]

Int[(a + a*Sec[c + d*x])^n*Sin[c + d*x]^5,x]

[Out]

((12 - n)*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(3 + n))/(20*a^3*d) - (Cos[c + d*x]^5*(a + a*Sec[c + d*x])^(3 +
n))/(5*a^3*d) + ((32 - 13*n + n^2)*Hypergeometric2F1[4, 3 + n, 4 + n, 1 + Sec[c + d*x]]*(a + a*Sec[c + d*x])^(
3 + n))/(20*a^3*d*(3 + 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 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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)^2 (a-a x)^{2+n}}{x^6} \, dx,x,-\sec (c+d x)\right )}{a^4 d} \\ & = -\frac {\cos ^5(c+d x) (a+a \sec (c+d x))^{3+n}}{5 a^3 d}-\frac {\text {Subst}\left (\int \frac {(a-a x)^{2+n} \left (a^3 (12-n)+5 a^3 x\right )}{x^5} \, dx,x,-\sec (c+d x)\right )}{5 a^5 d} \\ & = \frac {(12-n) \cos ^4(c+d x) (a+a \sec (c+d x))^{3+n}}{20 a^3 d}-\frac {\cos ^5(c+d x) (a+a \sec (c+d x))^{3+n}}{5 a^3 d}-\frac {\left (32-13 n+n^2\right ) \text {Subst}\left (\int \frac {(a-a x)^{2+n}}{x^4} \, dx,x,-\sec (c+d x)\right )}{20 a^2 d} \\ & = \frac {(12-n) \cos ^4(c+d x) (a+a \sec (c+d x))^{3+n}}{20 a^3 d}-\frac {\cos ^5(c+d x) (a+a \sec (c+d x))^{3+n}}{5 a^3 d}+\frac {\left (32-13 n+n^2\right ) \operatorname {Hypergeometric2F1}(4,3+n,4+n,1+\sec (c+d x)) (a+a \sec (c+d x))^{3+n}}{20 a^3 d (3+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.68 \[ \int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx=-\frac {\left ((3+n) \cos ^4(c+d x) (-12+n+4 \cos (c+d x))-\left (32-13 n+n^2\right ) \operatorname {Hypergeometric2F1}(4,3+n,4+n,1+\sec (c+d x))\right ) (1+\sec (c+d x))^3 (a (1+\sec (c+d x)))^n}{20 d (3+n)} \]

[In]

Integrate[(a + a*Sec[c + d*x])^n*Sin[c + d*x]^5,x]

[Out]

-1/20*(((3 + n)*Cos[c + d*x]^4*(-12 + n + 4*Cos[c + d*x]) - (32 - 13*n + n^2)*Hypergeometric2F1[4, 3 + n, 4 +
n, 1 + Sec[c + d*x]])*(1 + Sec[c + d*x])^3*(a*(1 + Sec[c + d*x]))^n)/(d*(3 + n))

Maple [F]

\[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \sin \left (d x +c \right )^{5}d x\]

[In]

int((a+a*sec(d*x+c))^n*sin(d*x+c)^5,x)

[Out]

int((a+a*sec(d*x+c))^n*sin(d*x+c)^5,x)

Fricas [F]

\[ \int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{5} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^n*sin(d*x+c)^5,x, algorithm="fricas")

[Out]

integral((cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*(a*sec(d*x + c) + a)^n*sin(d*x + c), x)

Sympy [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))**n*sin(d*x+c)**5,x)

[Out]

Timed out

Maxima [F]

\[ \int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{5} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^n*sin(d*x+c)^5,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*sin(d*x + c)^5, x)

Giac [F]

\[ \int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{5} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^n*sin(d*x+c)^5,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*sin(d*x + c)^5, x)

Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^n \sin ^5(c+d x) \, dx=\int {\sin \left (c+d\,x\right )}^5\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]

[In]

int(sin(c + d*x)^5*(a + a/cos(c + d*x))^n,x)

[Out]

int(sin(c + d*x)^5*(a + a/cos(c + d*x))^n, x)