3.349 \(\int \sec ^5(c+d x) (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=51 \[ -\frac{a^2 (a \sin (c+d x)+a)^{m-2} \, _2F_1\left (3,m-2;m-1;\frac{1}{2} (\sin (c+d x)+1)\right )}{8 d (2-m)} \]

[Out]

-(a^2*Hypergeometric2F1[3, -2 + m, -1 + m, (1 + Sin[c + d*x])/2]*(a + a*Sin[c + d*x])^(-2 + m))/(8*d*(2 - m))

________________________________________________________________________________________

Rubi [A]  time = 0.0561663, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2667, 68} \[ -\frac{a^2 (a \sin (c+d x)+a)^{m-2} \, _2F_1\left (3,m-2;m-1;\frac{1}{2} (\sin (c+d x)+1)\right )}{8 d (2-m)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*Hypergeometric2F1[3, -2 + m, -1 + m, (1 + Sin[c + d*x])/2]*(a + a*Sin[c + d*x])^(-2 + m))/(8*d*(2 - m))

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \sec ^5(c+d x) (a+a \sin (c+d x))^m \, dx &=\frac{a^5 \operatorname{Subst}\left (\int \frac{(a+x)^{-3+m}}{(a-x)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{a^2 \, _2F_1\left (3,-2+m;-1+m;\frac{1}{2} (1+\sin (c+d x))\right ) (a+a \sin (c+d x))^{-2+m}}{8 d (2-m)}\\ \end{align*}

Mathematica [B]  time = 0.710449, size = 163, normalized size = 3.2 \[ \frac{(a (\sin (c+d x)+1))^m \left (\frac{6 (\sin (c+d x)+1) \, _2F_1\left (1,m+1;m+2;\frac{1}{2} (\sin (c+d x)+1)\right )}{m+1}+\frac{3 (\sin (c+d x)+1) \, _2F_1\left (2,m+1;m+2;\frac{1}{2} (\sin (c+d x)+1)\right )}{m+1}+\frac{(\sin (c+d x)+1) \, _2F_1\left (3,m+1;m+2;\frac{1}{2} (\sin (c+d x)+1)\right )}{m+1}+\frac{12}{(m-1) (\sin (c+d x)+1)}+\frac{8}{(m-2) (\sin (c+d x)+1)^2}+\frac{12}{m}\right )}{64 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[c + d*x]))^m*(12/m + 8/((-2 + m)*(1 + Sin[c + d*x])^2) + 12/((-1 + m)*(1 + Sin[c + d*x])) + (6*Hy
pergeometric2F1[1, 1 + m, 2 + m, (1 + Sin[c + d*x])/2]*(1 + Sin[c + d*x]))/(1 + m) + (3*Hypergeometric2F1[2, 1
+ m, 2 + m, (1 + Sin[c + d*x])/2]*(1 + Sin[c + d*x]))/(1 + m) + (Hypergeometric2F1[3, 1 + m, 2 + m, (1 + Sin[
c + d*x])/2]*(1 + Sin[c + d*x]))/(1 + m)))/(64*d)

________________________________________________________________________________________

Maple [F]  time = 0.134, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{5} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{m} \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*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*(a+a*sin(d*x+c))**m,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{m} \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*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

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