3.172 \(\int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=30 \[ -\frac{2 a \cos ^5(c+d x)}{5 d (a \sin (c+d x)+a)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0581363, 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} \[ -\frac{2 a \cos ^5(c+d x)}{5 d (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-2*a*Cos[c + d*x]^5)/(5*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]

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=-\frac{2 a \cos ^5(c+d x)}{5 d (a+a \sin (c+d x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0539437, size = 42, normalized size = 1.4 \[ -\frac{2 \cos ^5(c+d x) \sqrt{a (\sin (c+d x)+1)}}{5 a^2 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-2*Cos[c + d*x]^5*Sqrt[a*(1 + Sin[c + d*x])])/(5*a^2*d*(1 + Sin[c + d*x])^3)

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Maple [A]  time = 0.105, 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 ) ^{3}}{5\,ad\cos \left ( dx+c \right ) }{\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)^4/(a+a*sin(d*x+c))^(3/2),x)

[Out]

2/5/a*(1+sin(d*x+c))*(sin(d*x+c)-1)^3/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 )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^4/(a*sin(d*x + c) + a)^(3/2), x)

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Fricas [B]  time = 2.20367, size = 257, normalized size = 8.57 \begin{align*} \frac{2 \,{\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 4\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{5 \,{\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 2.0121, size = 269, normalized size = 8.97 \begin{align*} \frac{\frac{{\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^{8}} - \frac{5 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{8}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{10 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{8}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{10 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{8}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{5 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{8}}\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^{8}}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{5}{2}}} + \frac{4 \, \sqrt{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{\frac{21}{2}}}}{20 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

1/20*((((((sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c)/a^8 - 5*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^8)*tan(1
/2*d*x + 1/2*c) + 10*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^8)*tan(1/2*d*x + 1/2*c) - 10*sgn(tan(1/2*d*x + 1/2*c) + 1
)/a^8)*tan(1/2*d*x + 1/2*c) + 5*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^8)*tan(1/2*d*x + 1/2*c) - sgn(tan(1/2*d*x + 1/
2*c) + 1)/a^8)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(5/2) + 4*sqrt(2)*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^(21/2))/d