\(\int \frac {\cos ^9(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\) [183]

Optimal result

Integrand size = 23, antiderivative size = 121 \[ \int \frac {\cos ^9(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {32 (a+a \sin (c+d x))^{5/2}}{5 a^5 d}-\frac {64 (a+a \sin (c+d x))^{7/2}}{7 a^6 d}+\frac {16 (a+a \sin (c+d x))^{9/2}}{3 a^7 d}-\frac {16 (a+a \sin (c+d x))^{11/2}}{11 a^8 d}+\frac {2 (a+a \sin (c+d x))^{13/2}}{13 a^9 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2746, 45} \[ \int \frac {\cos ^9(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {2 (a \sin (c+d x)+a)^{13/2}}{13 a^9 d}-\frac {16 (a \sin (c+d x)+a)^{11/2}}{11 a^8 d}+\frac {16 (a \sin (c+d x)+a)^{9/2}}{3 a^7 d}-\frac {64 (a \sin (c+d x)+a)^{7/2}}{7 a^6 d}+\frac {32 (a \sin (c+d x)+a)^{5/2}}{5 a^5 d} \]

[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 45

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])

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 {\text {Subst}\left (\int (a-x)^4 (a+x)^{3/2} \, dx,x,a \sin (c+d x)\right )}{a^9 d} \\ & = \frac {\text {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] (verified)

Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.53 \[ \int \frac {\cos ^9(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {2 (a (1+\sin (c+d x)))^{5/2} \left (9683-16700 \sin (c+d x)+14210 \sin ^2(c+d x)-6300 \sin ^3(c+d x)+1155 \sin ^4(c+d x)\right )}{15015 a^5 d} \]

[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] (verified)

Time = 0.60 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {13}{2}}}{13}-\frac {16 a \left (a +a \sin \left (d x +c \right )\right )^{\frac {11}{2}}}{11}+\frac {16 a^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{3}-\frac {64 a^{3} \left (a +a \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {32 a^{4} \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}}{d \,a^{9}}\)	\(90\)
default	\(\frac {\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {13}{2}}}{13}-\frac {16 a \left (a +a \sin \left (d x +c \right )\right )^{\frac {11}{2}}}{11}+\frac {16 a^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{3}-\frac {64 a^{3} \left (a +a \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {32 a^{4} \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}}{d \,a^{9}}\)	\(90\)

[In]

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

[Out]

2/d/a^9*(1/13*(a+a*sin(d*x+c))^(13/2)-8/11*a*(a+a*sin(d*x+c))^(11/2)+8/3*a^2*(a+a*sin(d*x+c))^(9/2)-32/7*a^3*(
a+a*sin(d*x+c))^(7/2)+16/5*a^4*(a+a*sin(d*x+c))^(5/2))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.68 \[ \int \frac {\cos ^9(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (1155 \, \cos \left (d x + c\right )^{6} - 6230 \, \cos \left (d x + c\right )^{4} - 512 \, \cos \left (d x + c\right )^{2} + 2 \, {\left (1995 \, \cos \left (d x + c\right )^{4} - 1280 \, \cos \left (d x + c\right )^{2} - 2048\right )} \sin \left (d x + c\right ) - 4096\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{15015 \, a^{3} d} \]

[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)]

Timed out. \[ \int \frac {\cos ^9(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.74 \[ \int \frac {\cos ^9(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {2 \, {\left (1155 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {13}{2}} - 10920 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a + 40040 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{2} - 68640 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} + 48048 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4}\right )}}{15015 \, a^{9} d} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^9(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {128 \, {\left (1155 \, \sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 5460 \, \sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 10010 \, \sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 8580 \, \sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3003 \, \sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )}}{15015 \, a^{3} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \]

[In]

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

[Out]

128/15015*(1155*sqrt(2)*sqrt(a)*cos(-1/4*pi + 1/2*d*x + 1/2*c)^13 - 5460*sqrt(2)*sqrt(a)*cos(-1/4*pi + 1/2*d*x
+ 1/2*c)^11 + 10010*sqrt(2)*sqrt(a)*cos(-1/4*pi + 1/2*d*x + 1/2*c)^9 - 8580*sqrt(2)*sqrt(a)*cos(-1/4*pi + 1/2
*d*x + 1/2*c)^7 + 3003*sqrt(2)*sqrt(a)*cos(-1/4*pi + 1/2*d*x + 1/2*c)^5)/(a^3*d*sgn(cos(-1/4*pi + 1/2*d*x + 1/
2*c)))

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^9(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^9}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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