3.182 \(\int \frac {\cos ^{10}(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=95 \[ -\frac {64 a^3 \cos ^{11}(c+d x)}{2145 d (a \sin (c+d x)+a)^{11/2}}-\frac {16 a^2 \cos ^{11}(c+d x)}{195 d (a \sin (c+d x)+a)^{9/2}}-\frac {2 a \cos ^{11}(c+d x)}{15 d (a \sin (c+d x)+a)^{7/2}} \]

[Out]

-64/2145*a^3*cos(d*x+c)^11/d/(a+a*sin(d*x+c))^(11/2)-16/195*a^2*cos(d*x+c)^11/d/(a+a*sin(d*x+c))^(9/2)-2/15*a*
cos(d*x+c)^11/d/(a+a*sin(d*x+c))^(7/2)

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Rubi [A]  time = 0.19, antiderivative size = 95, 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 = {2674, 2673} \[ -\frac {16 a^2 \cos ^{11}(c+d x)}{195 d (a \sin (c+d x)+a)^{9/2}}-\frac {64 a^3 \cos ^{11}(c+d x)}{2145 d (a \sin (c+d x)+a)^{11/2}}-\frac {2 a \cos ^{11}(c+d x)}{15 d (a \sin (c+d x)+a)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^10/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-64*a^3*Cos[c + d*x]^11)/(2145*d*(a + a*Sin[c + d*x])^(11/2)) - (16*a^2*Cos[c + d*x]^11)/(195*d*(a + a*Sin[c
+ d*x])^(9/2)) - (2*a*Cos[c + d*x]^11)/(15*d*(a + a*Sin[c + d*x])^(7/2))

Rule 2673

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2674

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^{10}(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=-\frac {2 a \cos ^{11}(c+d x)}{15 d (a+a \sin (c+d x))^{7/2}}+\frac {1}{15} (8 a) \int \frac {\cos ^{10}(c+d x)}{(a+a \sin (c+d x))^{7/2}} \, dx\\ &=-\frac {16 a^2 \cos ^{11}(c+d x)}{195 d (a+a \sin (c+d x))^{9/2}}-\frac {2 a \cos ^{11}(c+d x)}{15 d (a+a \sin (c+d x))^{7/2}}+\frac {1}{195} \left (32 a^2\right ) \int \frac {\cos ^{10}(c+d x)}{(a+a \sin (c+d x))^{9/2}} \, dx\\ &=-\frac {64 a^3 \cos ^{11}(c+d x)}{2145 d (a+a \sin (c+d x))^{11/2}}-\frac {16 a^2 \cos ^{11}(c+d x)}{195 d (a+a \sin (c+d x))^{9/2}}-\frac {2 a \cos ^{11}(c+d x)}{15 d (a+a \sin (c+d x))^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.77, size = 59, normalized size = 0.62 \[ -\frac {2 \left (143 \sin ^2(c+d x)+374 \sin (c+d x)+263\right ) \cos ^{11}(c+d x)}{2145 d (\sin (c+d x)+1)^3 (a (\sin (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^10/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-2*Cos[c + d*x]^11*(263 + 374*Sin[c + d*x] + 143*Sin[c + d*x]^2))/(2145*d*(1 + Sin[c + d*x])^3*(a*(1 + Sin[c
+ d*x]))^(5/2))

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fricas [B]  time = 0.77, size = 201, normalized size = 2.12 \[ -\frac {2 \, {\left (143 \, \cos \left (d x + c\right )^{8} - 341 \, \cos \left (d x + c\right )^{7} - 736 \, \cos \left (d x + c\right )^{6} + 28 \, \cos \left (d x + c\right )^{5} - 40 \, \cos \left (d x + c\right )^{4} + 64 \, \cos \left (d x + c\right )^{3} - 128 \, \cos \left (d x + c\right )^{2} + {\left (143 \, \cos \left (d x + c\right )^{7} + 484 \, \cos \left (d x + c\right )^{6} - 252 \, \cos \left (d x + c\right )^{5} - 280 \, \cos \left (d x + c\right )^{4} - 320 \, \cos \left (d x + c\right )^{3} - 384 \, \cos \left (d x + c\right )^{2} - 512 \, \cos \left (d x + c\right ) - 1024\right )} \sin \left (d x + c\right ) + 512 \, \cos \left (d x + c\right ) + 1024\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{2145 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^10/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-2/2145*(143*cos(d*x + c)^8 - 341*cos(d*x + c)^7 - 736*cos(d*x + c)^6 + 28*cos(d*x + c)^5 - 40*cos(d*x + c)^4
+ 64*cos(d*x + c)^3 - 128*cos(d*x + c)^2 + (143*cos(d*x + c)^7 + 484*cos(d*x + c)^6 - 252*cos(d*x + c)^5 - 280
*cos(d*x + c)^4 - 320*cos(d*x + c)^3 - 384*cos(d*x + c)^2 - 512*cos(d*x + c) - 1024)*sin(d*x + c) + 512*cos(d*
x + c) + 1024)*sqrt(a*sin(d*x + c) + a)/(a^3*d*cos(d*x + c) + a^3*d*sin(d*x + c) + a^3*d)

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giac [B]  time = 2.36, size = 526, normalized size = 5.54 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^10/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

2/2145*(1024*sqrt(2)*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^(5/2) - (263*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1) - (2145*a^
5/sgn(tan(1/2*d*x + 1/2*c) + 1) - (7335*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1) - (13585*a^5/sgn(tan(1/2*d*x + 1/2*c
) + 1) - (15795*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1) - (17589*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1) - (29315*a^5/sgn(
tan(1/2*d*x + 1/2*c) + 1) - (45045*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1) - (45045*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1
) - (29315*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1) - (17589*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1) - (15795*a^5/sgn(tan(1
/2*d*x + 1/2*c) + 1) - (13585*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1) - (7335*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1) + (2
63*a^5*tan(1/2*d*x + 1/2*c)/sgn(tan(1/2*d*x + 1/2*c) + 1) - 2145*a^5/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))*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)^(
15/2))/d

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maple [A]  time = 0.20, size = 67, normalized size = 0.71 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{6} \left (143 \left (\sin ^{2}\left (d x +c \right )\right )+374 \sin \left (d x +c \right )+263\right )}{2145 a^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^10/(a+a*sin(d*x+c))^(5/2),x)

[Out]

-2/2145/a^2*(1+sin(d*x+c))*(sin(d*x+c)-1)^6*(143*sin(d*x+c)^2+374*sin(d*x+c)+263)/cos(d*x+c)/(a+a*sin(d*x+c))^
(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{10}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^10/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^10/(a*sin(d*x + c) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^{10}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^10/(a + a*sin(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^10/(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)**10/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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