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

Optimal. Leaf size=92 \[ -\frac {8 a^2 \cos ^3(c+d x)}{105 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}-\frac {2 a \cos ^3(c+d x)}{35 d \sqrt {a \sin (c+d x)+a}} \]

[Out]

-8/105*a^2*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)-2/35*a*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)-2/7*cos(d*x+c)^3
*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.19, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2856, 2674, 2673} \[ -\frac {8 a^2 \cos ^3(c+d x)}{105 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}-\frac {2 a \cos ^3(c+d x)}{35 d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^2*Cos[c + d*x]^3)/(105*d*(a + a*Sin[c + d*x])^(3/2)) - (2*a*Cos[c + d*x]^3)/(35*d*Sqrt[a + a*Sin[c + d*x
]]) - (2*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(7*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]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \sin (c+d x) \sqrt {a+a \sin (c+d x)} \, dx &=-\frac {2 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{7 d}+\frac {1}{7} \int \cos ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {2 a \cos ^3(c+d x)}{35 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{7 d}+\frac {1}{35} (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)}{105 d (a+a \sin (c+d x))^{3/2}}-\frac {2 a \cos ^3(c+d x)}{35 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 89, normalized size = 0.97 \[ -\frac {\sqrt {a (\sin (c+d x)+1)} \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (66 \sin (c+d x)-15 \cos (2 (c+d x))+59)}{105 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/105*((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*Sqrt[a*(1 + Sin[c + d*x])]*(59 - 15*Cos[2*(c + d*x)] + 66*Sin[
c + d*x]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [A]  time = 0.47, size = 111, normalized size = 1.21 \[ -\frac {2 \, {\left (15 \, \cos \left (d x + c\right )^{4} + 18 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2} + {\left (15 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2} - 4 \, \cos \left (d x + c\right ) - 8\right )} \sin \left (d x + c\right ) + 4 \, \cos \left (d x + c\right ) + 8\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{105 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(15*cos(d*x + c)^4 + 18*cos(d*x + c)^3 - cos(d*x + c)^2 + (15*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 = 0.22, size = 129, normalized size = 1.40 \[ -\frac {1}{420} \, \sqrt {2} \sqrt {a} {\left (\frac {21 \, \cos \left (\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {105 \, \cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {15 \, \cos \left (-\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {35 \, \cos \left (-\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/420*sqrt(2)*sqrt(a)*(21*cos(1/4*pi + 5/2*d*x + 5/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 105*cos(1/4*p
i + 1/2*d*x + 1/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 15*cos(-1/4*pi + 7/2*d*x + 7/2*c)*sgn(cos(-1/4*pi
+ 1/2*d*x + 1/2*c))/d + 35*cos(-1/4*pi + 3/2*d*x + 3/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d)

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maple [A]  time = 1.06, size = 65, normalized size = 0.71 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right )^{2} \left (15 \left (\sin ^{2}\left (d x +c \right )\right )+33 \sin \left (d x +c \right )+22\right )}{105 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(1+sin(d*x+c))*a*(sin(d*x+c)-1)^2*(15*sin(d*x+c)^2+33*sin(d*x+c)+22)/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 \[ \int \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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