3.183 \(\int \frac{\cos ^9(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=121 \[ \frac{2 (a \sin (c+d x)+a)^{13/2}}{13 a^9 d}-\frac{16 (a \sin (c+d x)+a)^{11/2}}{11 a^8 d}+\frac{16 (a \sin (c+d x)+a)^{9/2}}{3 a^7 d}-\frac{64 (a \sin (c+d x)+a)^{7/2}}{7 a^6 d}+\frac{32 (a \sin (c+d x)+a)^{5/2}}{5 a^5 d} \]

[Out]

(32*(a + a*Sin[c + d*x])^(5/2))/(5*a^5*d) - (64*(a + a*Sin[c + d*x])^(7/2))/(7*a^6*d) + (16*(a + a*Sin[c + d*x
])^(9/2))/(3*a^7*d) - (16*(a + a*Sin[c + d*x])^(11/2))/(11*a^8*d) + (2*(a + a*Sin[c + d*x])^(13/2))/(13*a^9*d)

________________________________________________________________________________________

Rubi [A]  time = 0.0900153, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2667, 43} \[ \frac{2 (a \sin (c+d x)+a)^{13/2}}{13 a^9 d}-\frac{16 (a \sin (c+d x)+a)^{11/2}}{11 a^8 d}+\frac{16 (a \sin (c+d x)+a)^{9/2}}{3 a^7 d}-\frac{64 (a \sin (c+d x)+a)^{7/2}}{7 a^6 d}+\frac{32 (a \sin (c+d x)+a)^{5/2}}{5 a^5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^9/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(32*(a + a*Sin[c + d*x])^(5/2))/(5*a^5*d) - (64*(a + a*Sin[c + d*x])^(7/2))/(7*a^6*d) + (16*(a + a*Sin[c + d*x
])^(9/2))/(3*a^7*d) - (16*(a + a*Sin[c + d*x])^(11/2))/(11*a^8*d) + (2*(a + a*Sin[c + d*x])^(13/2))/(13*a^9*d)

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\cos ^9(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int (a-x)^4 (a+x)^{3/2} \, dx,x,a \sin (c+d x)\right )}{a^9 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (16 a^4 (a+x)^{3/2}-32 a^3 (a+x)^{5/2}+24 a^2 (a+x)^{7/2}-8 a (a+x)^{9/2}+(a+x)^{11/2}\right ) \, dx,x,a \sin (c+d x)\right )}{a^9 d}\\ &=\frac{32 (a+a \sin (c+d x))^{5/2}}{5 a^5 d}-\frac{64 (a+a \sin (c+d x))^{7/2}}{7 a^6 d}+\frac{16 (a+a \sin (c+d x))^{9/2}}{3 a^7 d}-\frac{16 (a+a \sin (c+d x))^{11/2}}{11 a^8 d}+\frac{2 (a+a \sin (c+d x))^{13/2}}{13 a^9 d}\\ \end{align*}

Mathematica [A]  time = 0.290182, size = 64, normalized size = 0.53 \[ \frac{2 \left (1155 \sin ^4(c+d x)-6300 \sin ^3(c+d x)+14210 \sin ^2(c+d x)-16700 \sin (c+d x)+9683\right ) (a (\sin (c+d x)+1))^{5/2}}{15015 a^5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^9/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(2*(a*(1 + Sin[c + d*x]))^(5/2)*(9683 - 16700*Sin[c + d*x] + 14210*Sin[c + d*x]^2 - 6300*Sin[c + d*x]^3 + 1155
*Sin[c + d*x]^4))/(15015*a^5*d)

________________________________________________________________________________________

Maple [A]  time = 0.102, size = 67, normalized size = 0.6 \begin{align*}{\frac{2310\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+12600\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -33040\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-46000\,\sin \left ( dx+c \right ) +50096}{15015\,{a}^{5}d} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^9/(a+a*sin(d*x+c))^(5/2),x)

[Out]

2/15015/a^5*(a+a*sin(d*x+c))^(5/2)*(1155*cos(d*x+c)^4+6300*cos(d*x+c)^2*sin(d*x+c)-16520*cos(d*x+c)^2-23000*si
n(d*x+c)+25048)/d

________________________________________________________________________________________

Maxima [A]  time = 0.942051, size = 120, normalized size = 0.99 \begin{align*} \frac{2 \,{\left (1155 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{13}{2}} - 10920 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{11}{2}} a + 40040 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{9}{2}} a^{2} - 68640 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}} a^{3} + 48048 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a^{4}\right )}}{15015 \, a^{9} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^9/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

2/15015*(1155*(a*sin(d*x + c) + a)^(13/2) - 10920*(a*sin(d*x + c) + a)^(11/2)*a + 40040*(a*sin(d*x + c) + a)^(
9/2)*a^2 - 68640*(a*sin(d*x + c) + a)^(7/2)*a^3 + 48048*(a*sin(d*x + c) + a)^(5/2)*a^4)/(a^9*d)

________________________________________________________________________________________

Fricas [A]  time = 2.30215, size = 247, normalized size = 2.04 \begin{align*} -\frac{2 \,{\left (1155 \, \cos \left (d x + c\right )^{6} - 6230 \, \cos \left (d x + c\right )^{4} - 512 \, \cos \left (d x + c\right )^{2} + 2 \,{\left (1995 \, \cos \left (d x + c\right )^{4} - 1280 \, \cos \left (d x + c\right )^{2} - 2048\right )} \sin \left (d x + c\right ) - 4096\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{15015 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^9/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-2/15015*(1155*cos(d*x + c)^6 - 6230*cos(d*x + c)^4 - 512*cos(d*x + c)^2 + 2*(1995*cos(d*x + c)^4 - 1280*cos(d
*x + c)^2 - 2048)*sin(d*x + c) - 4096)*sqrt(a*sin(d*x + c) + a)/(a^3*d)

________________________________________________________________________________________

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(cos(d*x+c)**9/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.31304, size = 120, normalized size = 0.99 \begin{align*} \frac{2 \,{\left (1155 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{13}{2}} - 10920 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{11}{2}} a + 40040 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{9}{2}} a^{2} - 68640 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}} a^{3} + 48048 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a^{4}\right )}}{15015 \, a^{9} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^9/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

2/15015*(1155*(a*sin(d*x + c) + a)^(13/2) - 10920*(a*sin(d*x + c) + a)^(11/2)*a + 40040*(a*sin(d*x + c) + a)^(
9/2)*a^2 - 68640*(a*sin(d*x + c) + a)^(7/2)*a^3 + 48048*(a*sin(d*x + c) + a)^(5/2)*a^4)/(a^9*d)