3.2.4 \(\int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [104]

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

[Out]

-64/315*a^3*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-16/63*a^2*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(3/2)-2/9*a*cos(d*
x+c)^5/d/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 95, normalized size of antiderivative = 1.00, 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 = {2753, 2752} \begin {gather*} -\frac {64 a^3 \cos ^5(c+d x)}{315 d (a \sin (c+d x)+a)^{5/2}}-\frac {16 a^2 \cos ^5(c+d x)}{63 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}} \end {gather*}

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 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2753

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 \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*}

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Mathematica [A]
time = 0.48, size = 89, normalized size = 0.94 \begin {gather*} -\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \sqrt {a (1+\sin (c+d x))} (249-35 \cos (2 (c+d x))+220 \sin (c+d x))}{315 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/315*((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*Si
n[c + d*x]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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Maple [A]
time = 0.36, size = 65, normalized size = 0.68

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right )^{3} \left (35 \left (\sin ^{2}\left (d x +c \right )\right )+110 \sin \left (d x +c \right )+107\right )}{315 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(65\)

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,method=_RETURNVERBOSE)

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.36, size = 132, normalized size = 1.39 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \cos ^{4}{\left (c + d x \right )}\, dx \end {gather*}

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 [A]
time = 6.14, size = 147, normalized size = 1.55 \begin {gather*} \frac {\sqrt {2} {\left (1890 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 420 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 252 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 45 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 35 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )} \sqrt {a}}{2520 \, d} \end {gather*}

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]

1/2520*sqrt(2)*(1890*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) - 420*sgn(cos(-1/4*pi
+ 1/2*d*x + 1/2*c))*sin(-3/4*pi + 3/2*d*x + 3/2*c) - 252*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-5/4*pi + 5/2
*d*x + 5/2*c) + 45*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-7/4*pi + 7/2*d*x + 7/2*c) + 35*sgn(cos(-1/4*pi + 1
/2*d*x + 1/2*c))*sin(-9/4*pi + 9/2*d*x + 9/2*c))*sqrt(a)/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^4\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(a + a*sin(c + d*x))^(1/2),x)

[Out]

int(cos(c + d*x)^4*(a + a*sin(c + d*x))^(1/2), x)

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