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]

32/5*(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

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Rubi [A]  time = 0.09, antiderivative size = 121, normalized size of antiderivative = 1.00, 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 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])

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])

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*}

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Mathematica [A]  time = 0.29, 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)

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fricas [A]  time = 0.53, size = 82, normalized size = 0.68 \[ -\frac {2 \, {\left (1155 \, \cos \left (d x + c\right )^{6} - 6230 \, \cos \left (d x + c\right )^{4} - 512 \, \cos \left (d x + c\right )^{2} + 2 \, {\left (1995 \, \cos \left (d x + c\right )^{4} - 1280 \, \cos \left (d x + c\right )^{2} - 2048\right )} \sin \left (d x + c\right ) - 4096\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{15015 \, a^{3} d} \]

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)

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giac [B]  time = 2.95, size = 430, normalized size = 3.55 \[ \frac {2 \, {\left (\frac {9683 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {15015 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {25402 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {90090 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {107393 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {93093 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {183612 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {183612 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {93093 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {107393 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {90090 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {25402 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {9683 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {15015 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{15015 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {13}{2}} d} \]

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*(9683*a^4/sgn(tan(1/2*d*x + 1/2*c) + 1) + (15015*a^4/sgn(tan(1/2*d*x + 1/2*c) + 1) + (25402*a^4/sgn(ta
n(1/2*d*x + 1/2*c) + 1) + (90090*a^4/sgn(tan(1/2*d*x + 1/2*c) + 1) + (107393*a^4/sgn(tan(1/2*d*x + 1/2*c) + 1)
+ (93093*a^4/sgn(tan(1/2*d*x + 1/2*c) + 1) + (183612*a^4/sgn(tan(1/2*d*x + 1/2*c) + 1) + (183612*a^4/sgn(tan(
1/2*d*x + 1/2*c) + 1) + (93093*a^4/sgn(tan(1/2*d*x + 1/2*c) + 1) + (107393*a^4/sgn(tan(1/2*d*x + 1/2*c) + 1) +
(90090*a^4/sgn(tan(1/2*d*x + 1/2*c) + 1) + (25402*a^4/sgn(tan(1/2*d*x + 1/2*c) + 1) + (9683*a^4*tan(1/2*d*x +
1/2*c)/sgn(tan(1/2*d*x + 1/2*c) + 1) + 15015*a^4/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c))*tan(1/2
*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2
*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2
*d*x + 1/2*c))/((a*tan(1/2*d*x + 1/2*c)^2 + a)^(13/2)*d)

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maple [A]  time = 0.17, size = 67, normalized size = 0.55 \[ \frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \left (1155 \left (\cos ^{4}\left (d x +c \right )\right )+6300 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-16520 \left (\cos ^{2}\left (d x +c \right )\right )-23000 \sin \left (d x +c \right )+25048\right )}{15015 a^{5} d} \]

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

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maxima [A]  time = 0.67, size = 89, normalized size = 0.74 \[ \frac {2 \, {\left (1155 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {13}{2}} - 10920 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a + 40040 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{2} - 68640 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} + 48048 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4}\right )}}{15015 \, a^{9} d} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^9}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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