3.1512 \(\int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=89 \[ \frac{a \sin ^{n+1}(c+d x) \, _2F_1\left (3,\frac{n+1}{2};\frac{n+3}{2};\sin ^2(c+d x)\right )}{d (n+1)}+\frac{b \sin ^{n+2}(c+d x) \, _2F_1\left (3,\frac{n+2}{2};\frac{n+4}{2};\sin ^2(c+d x)\right )}{d (n+2)} \]

[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))

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Rubi [A]  time = 0.113608, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2837, 808, 364} \[ \frac{a \sin ^{n+1}(c+d x) \, _2F_1\left (3,\frac{n+1}{2};\frac{n+3}{2};\sin ^2(c+d x)\right )}{d (n+1)}+\frac{b \sin ^{n+2}(c+d x) \, _2F_1\left (3,\frac{n+2}{2};\frac{n+4}{2};\sin ^2(c+d x)\right )}{d (n+2)} \]

Antiderivative was successfully verified.

[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 2837

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*x)/b)^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]

Rule 808

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx &=\frac{b^5 \operatorname{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 ) \operatorname{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 \operatorname{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 \, _2F_1\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 \, _2F_1\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]  time = 0.1109, size = 89, normalized size = 1. \[ \frac{\sin ^{n+1}(c+d x) \left (a (n+2) \, _2F_1\left (3,\frac{n+1}{2};\frac{n+3}{2};\sin ^2(c+d x)\right )+b (n+1) \sin (c+d x) \, _2F_1\left (3,\frac{n+2}{2};\frac{n+4}{2};\sin ^2(c+d x)\right )\right )}{d (n+1) (n+2)} \]

Antiderivative was successfully verified.

[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))

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Maple [F]  time = 0.885, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{n} \left ( a+b\sin \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sec \left (d x + c\right )^{5} \sin \left (d x + c\right ) + a \sec \left (d x + c\right )^{5}\right )} \sin \left (d x + c\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)