3.170 \(\int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=63 \[ -\frac {8 a^2 \cos ^7(c+d x)}{63 d (a \sin (c+d x)+a)^{7/2}}-\frac {2 a \cos ^7(c+d x)}{9 d (a \sin (c+d x)+a)^{5/2}} \]

[Out]

-8/63*a^2*cos(d*x+c)^7/d/(a+a*sin(d*x+c))^(7/2)-2/9*a*cos(d*x+c)^7/d/(a+a*sin(d*x+c))^(5/2)

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Rubi [A]  time = 0.12, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2674, 2673} \[ -\frac {8 a^2 \cos ^7(c+d x)}{63 d (a \sin (c+d x)+a)^{7/2}}-\frac {2 a \cos ^7(c+d x)}{9 d (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-8*a^2*Cos[c + d*x]^7)/(63*d*(a + a*Sin[c + d*x])^(7/2)) - (2*a*Cos[c + d*x]^7)/(9*d*(a + a*Sin[c + d*x])^(5/
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 ^6(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=-\frac {2 a \cos ^7(c+d x)}{9 d (a+a \sin (c+d x))^{5/2}}+\frac {1}{9} (4 a) \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\\ &=-\frac {8 a^2 \cos ^7(c+d x)}{63 d (a+a \sin (c+d x))^{7/2}}-\frac {2 a \cos ^7(c+d x)}{9 d (a+a \sin (c+d x))^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 49, normalized size = 0.78 \[ -\frac {2 (7 \sin (c+d x)+11) \cos ^7(c+d x)}{63 d (\sin (c+d x)+1)^2 (a (\sin (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-2*Cos[c + d*x]^7*(11 + 7*Sin[c + d*x]))/(63*d*(1 + Sin[c + d*x])^2*(a*(1 + Sin[c + d*x]))^(3/2))

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fricas [B]  time = 0.76, size = 142, normalized size = 2.25 \[ \frac {2 \, {\left (7 \, \cos \left (d x + c\right )^{5} + 17 \, \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} - {\left (7 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{3} - 12 \, \cos \left (d x + c\right )^{2} - 16 \, \cos \left (d x + c\right ) - 32\right )} \sin \left (d x + c\right ) - 16 \, \cos \left (d x + c\right ) - 32\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{63 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2/63*(7*cos(d*x + c)^5 + 17*cos(d*x + c)^4 - 2*cos(d*x + c)^3 + 4*cos(d*x + c)^2 - (7*cos(d*x + c)^4 - 10*cos(
d*x + c)^3 - 12*cos(d*x + c)^2 - 16*cos(d*x + c) - 32)*sin(d*x + c) - 16*cos(d*x + c) - 32)*sqrt(a*sin(d*x + c
) + a)/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) + a^2*d)

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giac [B]  time = 1.72, size = 340, normalized size = 5.40 \[ \frac {2 \, {\left (\frac {32 \, \sqrt {2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}}} - \frac {\frac {11 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {63 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {144 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {168 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {126 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {126 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {168 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {144 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {11 \, a^{3} \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 {63 \, a^{3}}{\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 )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {9}{2}}}\right )}}{63 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

2/63*(32*sqrt(2)*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^(3/2) - (11*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (63*a^3/sgn(t
an(1/2*d*x + 1/2*c) + 1) - (144*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (168*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (
126*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (126*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (168*a^3/sgn(tan(1/2*d*x + 1/
2*c) + 1) - (144*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) + (11*a^3*tan(1/2*d*x + 1/2*c)/sgn(tan(1/2*d*x + 1/2*c) + 1
) - 63*a^3/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))*ta
n(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)^(9/2))/d

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maple [A]  time = 0.20, size = 57, normalized size = 0.90 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{4} \left (7 \sin \left (d x +c \right )+11\right )}{63 a \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6/(a+a*sin(d*x+c))^(3/2),x)

[Out]

-2/63/a*(1+sin(d*x+c))*(sin(d*x+c)-1)^4*(7*sin(d*x+c)+11)/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 \[ \int \frac {\cos \left (d x + c\right )^{6}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^6/(a*sin(d*x + c) + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\cos \left (c+d\,x\right )}^6}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6/(a + a*sin(c + d*x))^(3/2),x)

[Out]

int(cos(c + d*x)^6/(a + a*sin(c + d*x))^(3/2), 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)**6/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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