3.186 \(\int \frac{\cos ^6(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=30 \[ -\frac{2 a \cos ^7(c+d x)}{7 d (a \sin (c+d x)+a)^{7/2}} \]

[Out]

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

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Rubi [A]  time = 0.0571351, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.043, Rules used = {2673} \[ -\frac{2 a \cos ^7(c+d x)}{7 d (a \sin (c+d x)+a)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-2*a*Cos[c + d*x]^7)/(7*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]

Rubi steps

\begin{align*} \int \frac{\cos ^6(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=-\frac{2 a \cos ^7(c+d x)}{7 d (a+a \sin (c+d x))^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.111325, size = 42, normalized size = 1.4 \[ -\frac{2 \cos ^7(c+d x) \sqrt{a (\sin (c+d x)+1)}}{7 a^3 d (\sin (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-2*Cos[c + d*x]^7*Sqrt[a*(1 + Sin[c + d*x])])/(7*a^3*d*(1 + Sin[c + d*x])^4)

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Maple [A]  time = 0.104, size = 47, normalized size = 1.6 \begin{align*} -{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) ^{4}}{7\,{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)^6/(a+a*sin(d*x+c))^(5/2),x)

[Out]

-2/7/a^2*(1+sin(d*x+c))*(sin(d*x+c)-1)^4/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 )^{6}}{{\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)^6/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^6/(a*sin(d*x + c) + a)^(5/2), x)

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Fricas [B]  time = 2.15058, size = 309, normalized size = 10.3 \begin{align*} -\frac{2 \,{\left (\cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{3} - 8 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} - 4 \, \cos \left (d x + c\right ) - 8\right )} \sin \left (d x + c\right ) + 4 \, \cos \left (d x + c\right ) + 8\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{7 \,{\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)^6/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-2/7*(cos(d*x + c)^4 - 3*cos(d*x + c)^3 - 8*cos(d*x + c)^2 + (cos(d*x + c)^3 + 4*cos(d*x + c)^2 - 4*cos(d*x +
c) - 8)*sin(d*x + c) + 4*cos(d*x + c) + 8)*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)**6/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 2.16901, size = 344, normalized size = 11.47 \begin{align*} \frac{\frac{{\left ({\left ({\left ({\left ({\left ({\left (\frac{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{11}} - \frac{7 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{11}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{21 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{11}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{35 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{11}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{35 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{11}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{21 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{11}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{7 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{11}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{11}}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{7}{2}}} + \frac{8 \, \sqrt{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{\frac{29}{2}}}}{21 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/21*((((((((sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c)/a^11 - 7*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)*t
an(1/2*d*x + 1/2*c) + 21*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)*tan(1/2*d*x + 1/2*c) - 35*sgn(tan(1/2*d*x + 1/2*c
) + 1)/a^11)*tan(1/2*d*x + 1/2*c) + 35*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)*tan(1/2*d*x + 1/2*c) - 21*sgn(tan(1
/2*d*x + 1/2*c) + 1)/a^11)*tan(1/2*d*x + 1/2*c) + 7*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)*tan(1/2*d*x + 1/2*c) -
sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(7/2) + 8*sqrt(2)*sgn(tan(1/2*d*x + 1/2*c)
+ 1)/a^(29/2))/d