3.323 \(\int \cos ^2(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=124 \[ -\frac {8 a^2 \cos ^3(c+d x)}{63 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 a d}+\frac {4 \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{21 d}-\frac {2 a \cos ^3(c+d x)}{21 d \sqrt {a \sin (c+d x)+a}} \]

[Out]

-8/63*a^2*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)-2/9*cos(d*x+c)^3*(a+a*sin(d*x+c))^(3/2)/a/d-2/21*a*cos(d*x+c)^
3/d/(a+a*sin(d*x+c))^(1/2)+4/21*cos(d*x+c)^3*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.36, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2878, 2856, 2674, 2673} \[ -\frac {8 a^2 \cos ^3(c+d x)}{63 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 a d}+\frac {4 \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{21 d}-\frac {2 a \cos ^3(c+d x)}{21 d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*Sin[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-8*a^2*Cos[c + d*x]^3)/(63*d*(a + a*Sin[c + d*x])^(3/2)) - (2*a*Cos[c + d*x]^3)/(21*d*Sqrt[a + a*Sin[c + d*x]
]) + (4*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(21*d) - (2*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2))/(9*a*d
)

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]

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 2856

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p + 1
, 0]

Rule 2878

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> -Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*g*(m + p + 2)), x] + Dist[1/
(b*(m + p + 2)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m*(b*(m + 1) - a*(p + 1)*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 2, 0]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx &=-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 a d}+\frac {2 \int \cos ^2(c+d x) \left (\frac {3 a}{2}-3 a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{9 a}\\ &=\frac {4 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{21 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 a d}+\frac {5}{21} \int \cos ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {2 a \cos ^3(c+d x)}{21 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{21 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 a d}+\frac {1}{21} (4 a) \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {8 a^2 \cos ^3(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac {2 a \cos ^3(c+d x)}{21 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{21 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 a d}\\ \end {align*}

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Mathematica [A]  time = 0.60, size = 99, normalized size = 0.80 \[ \frac {\sqrt {a (\sin (c+d x)+1)} \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (-69 \sin (c+d x)+7 \sin (3 (c+d x))+30 \cos (2 (c+d x))-62)}{126 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*Sin[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*Sqrt[a*(1 + Sin[c + d*x])]*(-62 + 30*Cos[2*(c + d*x)] - 69*Sin[c + d*
x] + 7*Sin[3*(c + d*x)]))/(126*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [A]  time = 0.48, size = 130, normalized size = 1.05 \[ \frac {2 \, {\left (7 \, \cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{4} - 11 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - {\left (7 \, \cos \left (d x + c\right )^{4} + 8 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2} - 4 \, \cos \left (d x + c\right ) - 8\right )} \sin \left (d x + c\right ) - 4 \, \cos \left (d x + c\right ) - 8\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{63 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/63*(7*cos(d*x + c)^5 - cos(d*x + c)^4 - 11*cos(d*x + c)^3 + cos(d*x + c)^2 - (7*cos(d*x + c)^4 + 8*cos(d*x +
c)^3 - 3*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)/(d*
cos(d*x + c) + d*sin(d*x + c) + d)

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giac [A]  time = 0.27, size = 99, normalized size = 0.80 \[ -\frac {1}{504} \, \sqrt {2} \sqrt {a} {\left (\frac {9 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {7 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} - \frac {126 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-1/504*sqrt(2)*sqrt(a)*(9*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 7/2*d*x + 7/2*c)/d + 7*sgn(cos(-1/4
*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 9/2*d*x + 9/2*c)/d - 126*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi
+ 1/2*d*x + 1/2*c)/d)

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maple [A]  time = 1.02, size = 75, normalized size = 0.60 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right )^{2} \left (7 \left (\sin ^{3}\left (d x +c \right )\right )+15 \left (\sin ^{2}\left (d x +c \right )\right )+12 \sin \left (d x +c \right )+8\right )}{63 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/63*(1+sin(d*x+c))*a*(sin(d*x+c)-1)^2*(7*sin(d*x+c)^3+15*sin(d*x+c)^2+12*sin(d*x+c)+8)/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 \[ \int \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(d*x + c) + a)*cos(d*x + c)^2*sin(d*x + c)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^2\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*sin(d*x+c)**2*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(c + d*x) + 1))*sin(c + d*x)**2*cos(c + d*x)**2, x)

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