\(\int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx\) [1512]

Optimal result

Integrand size = 27, antiderivative size = 89 \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \operatorname {Hypergeometric2F1}\left (3,\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{d (1+n)}+\frac {b \operatorname {Hypergeometric2F1}\left (3,\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{d (2+n)} \]

[Out]

a*hypergeom([3, 1/2+1/2*n],[3/2+1/2*n],sin(d*x+c)^2)*sin(d*x+c)^(1+n)/d/(1+n)+b*hypergeom([3, 1+1/2*n],[1/2*n+
2],sin(d*x+c)^2)*sin(d*x+c)^(2+n)/d/(2+n)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2916, 822, 371} \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (3,\frac {n+1}{2},\frac {n+3}{2},\sin ^2(c+d x)\right )}{d (n+1)}+\frac {b \sin ^{n+2}(c+d x) \operatorname {Hypergeometric2F1}\left (3,\frac {n+2}{2},\frac {n+4}{2},\sin ^2(c+d x)\right )}{d (n+2)} \]

[In]

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

[Out]

(a*Hypergeometric2F1[3, (1 + n)/2, (3 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(1 + n))/(d*(1 + n)) + (b*Hypergeom
etric2F1[3, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(2 + n))/(d*(2 + n))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 822

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 2916

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n (a+x)}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\left (a b^5\right ) \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}+\frac {b^6 \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^{1+n}}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {a \operatorname {Hypergeometric2F1}\left (3,\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{d (1+n)}+\frac {b \operatorname {Hypergeometric2F1}\left (3,\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{d (2+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx=\frac {\sin ^{1+n}(c+d x) \left (a (2+n) \operatorname {Hypergeometric2F1}\left (3,\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right )+b (1+n) \operatorname {Hypergeometric2F1}\left (3,\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin (c+d x)\right )}{d (1+n) (2+n)} \]

[In]

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

[Out]

(Sin[c + d*x]^(1 + n)*(a*(2 + n)*Hypergeometric2F1[3, (1 + n)/2, (3 + n)/2, Sin[c + d*x]^2] + b*(1 + n)*Hyperg
eometric2F1[3, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]))/(d*(1 + n)*(2 + n))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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