3.159 \(\int \frac{\cos ^4(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=63 \[ -\frac{8 a^2 \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}} \]

[Out]

(-8*a^2*Cos[c + d*x]^5)/(35*d*(a + a*Sin[c + d*x])^(5/2)) - (2*a*Cos[c + d*x]^5)/(7*d*(a + a*Sin[c + d*x])^(3/
2))

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Rubi [A]  time = 0.11114, antiderivative size = 63, normalized size of antiderivative = 1., 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} \[ -\frac{8 a^2 \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-8*a^2*Cos[c + d*x]^5)/(35*d*(a + a*Sin[c + d*x])^(5/2)) - (2*a*Cos[c + d*x]^5)/(7*d*(a + a*Sin[c + d*x])^(3/
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 ^4(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx &=-\frac{2 a \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}+\frac{1}{7} (4 a) \int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac{8 a^2 \cos ^5(c+d x)}{35 d (a+a \sin (c+d x))^{5/2}}-\frac{2 a \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0604323, size = 49, normalized size = 0.78 \[ -\frac{2 (5 \sin (c+d x)+9) \cos ^5(c+d x)}{35 d (\sin (c+d x)+1)^2 \sqrt{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-2*Cos[c + d*x]^5*(9 + 5*Sin[c + d*x]))/(35*d*(1 + Sin[c + d*x])^2*Sqrt[a*(1 + Sin[c + d*x])])

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Maple [A]  time = 0.111, size = 54, normalized size = 0.9 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3} \left ( 5\,\sin \left ( dx+c \right ) +9 \right ) }{35\,d\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))^(1/2),x)

[Out]

2/35*(1+sin(d*x+c))*(sin(d*x+c)-1)^3*(5*sin(d*x+c)+9)/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}}{\sqrt{a \sin \left (d x + c\right ) + a}}\,{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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.92041, size = 306, normalized size = 4.86 \begin{align*} \frac{2 \,{\left (5 \, \cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} +{\left (5 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 8 \, \cos \left (d x + c\right ) + 16\right )} \sin \left (d x + c\right ) - 8 \, \cos \left (d x + c\right ) - 16\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{35 \,{\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a 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))^(1/2),x, algorithm="fricas")

[Out]

2/35*(5*cos(d*x + c)^4 - cos(d*x + c)^3 + 2*cos(d*x + c)^2 + (5*cos(d*x + c)^3 + 6*cos(d*x + c)^2 + 8*cos(d*x
+ c) + 16)*sin(d*x + c) - 8*cos(d*x + c) - 16)*sqrt(a*sin(d*x + c) + a)/(a*d*cos(d*x + c) + a*d*sin(d*x + c) +
a*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{4}{\left (c + d x \right )}}{\sqrt{a \left (\sin{\left (c + d x \right )} + 1\right )}}\, dx \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))**(1/2),x)

[Out]

Integral(cos(c + d*x)**4/sqrt(a*(sin(c + d*x) + 1)), x)

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Giac [B]  time = 1.98357, size = 346, normalized size = 5.49 \begin{align*} \frac{\frac{{\left ({\left ({\left ({\left ({\left ({\left (\frac{9 \, \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^{9}} - \frac{35 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{9}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{49 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{9}}\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^{9}}\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^{9}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{49 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{9}}\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^{9}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{9 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{9}}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{7}{2}}} + \frac{16 \, \sqrt{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{\frac{25}{2}}}}{105 \, 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))^(1/2),x, algorithm="giac")

[Out]

1/105*((((((((9*sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c)/a^9 - 35*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^9)
*tan(1/2*d*x + 1/2*c) + 49*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^9)*tan(1/2*d*x + 1/2*c) - 35*sgn(tan(1/2*d*x + 1/2*
c) + 1)/a^9)*tan(1/2*d*x + 1/2*c) + 35*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^9)*tan(1/2*d*x + 1/2*c) - 49*sgn(tan(1/
2*d*x + 1/2*c) + 1)/a^9)*tan(1/2*d*x + 1/2*c) + 35*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^9)*tan(1/2*d*x + 1/2*c) - 9
*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^9)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(7/2) + 16*sqrt(2)*sgn(tan(1/2*d*x + 1/2*c)
+ 1)/a^(25/2))/d