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

Optimal. Leaf size=95 \[ -\frac{16 a^2 \cos ^5(c+d x)}{63 d (a \sin (c+d x)+a)^{3/2}}-\frac{64 a^3 \cos ^5(c+d x)}{315 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 a \cos ^5(c+d x)}{9 d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(-64*a^3*Cos[c + d*x]^5)/(315*d*(a + a*Sin[c + d*x])^(5/2)) - (16*a^2*Cos[c + d*x]^5)/(63*d*(a + a*Sin[c + d*x
])^(3/2)) - (2*a*Cos[c + d*x]^5)/(9*d*Sqrt[a + a*Sin[c + d*x]])

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Rubi [A]  time = 0.178518, 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} \[ -\frac{16 a^2 \cos ^5(c+d x)}{63 d (a \sin (c+d x)+a)^{3/2}}-\frac{64 a^3 \cos ^5(c+d x)}{315 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 a \cos ^5(c+d x)}{9 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*a^3*Cos[c + d*x]^5)/(315*d*(a + a*Sin[c + d*x])^(5/2)) - (16*a^2*Cos[c + d*x]^5)/(63*d*(a + a*Sin[c + d*x
])^(3/2)) - (2*a*Cos[c + d*x]^5)/(9*d*Sqrt[a + a*Sin[c + d*x]])

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 \cos ^4(c+d x) \sqrt{a+a \sin (c+d x)} \, dx &=-\frac{2 a \cos ^5(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}+\frac{1}{9} (8 a) \int \frac{\cos ^4(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=-\frac{16 a^2 \cos ^5(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac{2 a \cos ^5(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}+\frac{1}{63} \left (32 a^2\right ) \int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac{64 a^3 \cos ^5(c+d x)}{315 d (a+a \sin (c+d x))^{5/2}}-\frac{16 a^2 \cos ^5(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac{2 a \cos ^5(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.759997, size = 89, normalized size = 0.94 \[ -\frac{\sqrt{a (\sin (c+d x)+1)} \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5 (220 \sin (c+d x)-35 \cos (2 (c+d x))+249)}{315 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*Sqrt[a*(1 + Sin[c + d*x])]*(249 - 35*Cos[2*(c + d*x)] + 220*Sin[c +
d*x]))/(315*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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Maple [A]  time = 0.109, size = 65, normalized size = 0.7 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ) a \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3} \left ( 35\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+110\,\sin \left ( dx+c \right ) +107 \right ) }{315\,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]

2/315*(1+sin(d*x+c))*a*(sin(d*x+c)-1)^3*(35*sin(d*x+c)^2+110*sin(d*x+c)+107)/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 \sqrt{a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4}\,{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(sqrt(a*sin(d*x + c) + a)*cos(d*x + c)^4, x)

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Fricas [A]  time = 1.68447, size = 367, normalized size = 3.86 \begin{align*} -\frac{2 \,{\left (35 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{4} + 8 \, \cos \left (d x + c\right )^{3} - 16 \, \cos \left (d x + c\right )^{2} -{\left (35 \, \cos \left (d x + c\right )^{4} + 40 \, \cos \left (d x + c\right )^{3} + 48 \, \cos \left (d x + c\right )^{2} + 64 \, \cos \left (d x + c\right ) + 128\right )} \sin \left (d x + c\right ) + 64 \, \cos \left (d x + c\right ) + 128\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{315 \,{\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + 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]

-2/315*(35*cos(d*x + c)^5 - 5*cos(d*x + c)^4 + 8*cos(d*x + c)^3 - 16*cos(d*x + c)^2 - (35*cos(d*x + c)^4 + 40*
cos(d*x + c)^3 + 48*cos(d*x + c)^2 + 64*cos(d*x + c) + 128)*sin(d*x + c) + 64*cos(d*x + c) + 128)*sqrt(a*sin(d
*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4}\,{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="giac")

[Out]

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