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

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

[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))

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Rubi [A]  time = 0.194634, antiderivative size = 95, 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 = {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 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]

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]

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*}

Mathematica [A]  time = 0.663395, 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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Maple [A]  time = 0.16, size = 67, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) ^{6} \left ( 143\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+374\,\sin \left ( dx+c \right ) +263 \right ) }{2145\,{a}^{2}\cos \left ( dx+c \right ) d}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{10}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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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Fricas [B]  time = 2.27415, size = 568, normalized size = 5.98 \begin{align*} -\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 )}} \end{align*}

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

[Out]

Timed out

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Giac [B]  time = 2.47535, size = 648, normalized size = 6.82 \begin{align*} \text{result too large to display} \end{align*}

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]

1/2145*((((((((((((((((263*sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c)/a^19 - 2145*sgn(tan(1/2*d*x + 1/
2*c) + 1)/a^19)*tan(1/2*d*x + 1/2*c) + 7335*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^19)*tan(1/2*d*x + 1/2*c) - 13585*s
gn(tan(1/2*d*x + 1/2*c) + 1)/a^19)*tan(1/2*d*x + 1/2*c) + 15795*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^19)*tan(1/2*d*
x + 1/2*c) - 17589*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^19)*tan(1/2*d*x + 1/2*c) + 29315*sgn(tan(1/2*d*x + 1/2*c) +
1)/a^19)*tan(1/2*d*x + 1/2*c) - 45045*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^19)*tan(1/2*d*x + 1/2*c) + 45045*sgn(ta
n(1/2*d*x + 1/2*c) + 1)/a^19)*tan(1/2*d*x + 1/2*c) - 29315*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^19)*tan(1/2*d*x + 1
/2*c) + 17589*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^19)*tan(1/2*d*x + 1/2*c) - 15795*sgn(tan(1/2*d*x + 1/2*c) + 1)/a
^19)*tan(1/2*d*x + 1/2*c) + 13585*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^19)*tan(1/2*d*x + 1/2*c) - 7335*sgn(tan(1/2*
d*x + 1/2*c) + 1)/a^19)*tan(1/2*d*x + 1/2*c) + 2145*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^19)*tan(1/2*d*x + 1/2*c) -
263*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^19)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(15/2) + 1024*sqrt(2)*sgn(tan(1/2*d*x
+ 1/2*c) + 1)/a^(53/2))/d