\(\int \sec ^5(c+d x) (a+a \sin (c+d x))^m \, dx\) [349]

Optimal result

Integrand size = 21, antiderivative size = 51 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^m \, dx=-\frac {a^2 \operatorname {Hypergeometric2F1}\left (3,-2+m,-1+m,\frac {1}{2} (1+\sin (c+d x))\right ) (a+a \sin (c+d x))^{-2+m}}{8 d (2-m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2746, 70} \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^m \, dx=-\frac {a^2 (a \sin (c+d x)+a)^{m-2} \operatorname {Hypergeometric2F1}\left (3,m-2,m-1,\frac {1}{2} (\sin (c+d x)+1)\right )}{8 d (2-m)} \]

[In]

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

[Out]

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

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 2746

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])

Rubi steps \begin{align*} \text {integral}& = \frac {a^5 \text {Subst}\left (\int \frac {(a+x)^{-3+m}}{(a-x)^3} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {a^2 \operatorname {Hypergeometric2F1}\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] (verified)

Leaf count is larger than twice the leaf count of optimal. \(163\) vs. \(2(51)=102\).

Time = 0.44 (sec) , antiderivative size = 163, normalized size of antiderivative = 3.20 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {(a (1+\sin (c+d x)))^m \left (\frac {12}{m}+\frac {8}{(-2+m) (1+\sin (c+d x))^2}+\frac {12}{(-1+m) (1+\sin (c+d x))}+\frac {6 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {1}{2} (1+\sin (c+d x))\right ) (1+\sin (c+d x))}{1+m}+\frac {3 \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {1}{2} (1+\sin (c+d x))\right ) (1+\sin (c+d x))}{1+m}+\frac {\operatorname {Hypergeometric2F1}\left (3,1+m,2+m,\frac {1}{2} (1+\sin (c+d x))\right ) (1+\sin (c+d x))}{1+m}\right )}{64 d} \]

[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]

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

[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)

Fricas [F]

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

[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)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[In]

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

[Out]

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