3.443 \(\int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=156 \[ -\frac{368 a^2 \cos ^5(c+d x)}{9009 d (a \sin (c+d x)+a)^{3/2}}-\frac{1472 a^3 \cos ^5(c+d x)}{45045 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 a d}+\frac{20 \cos ^5(c+d x) \sqrt{a \sin (c+d x)+a}}{143 d}-\frac{46 a \cos ^5(c+d x)}{1287 d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(-1472*a^3*Cos[c + d*x]^5)/(45045*d*(a + a*Sin[c + d*x])^(5/2)) - (368*a^2*Cos[c + d*x]^5)/(9009*d*(a + a*Sin[
c + d*x])^(3/2)) - (46*a*Cos[c + d*x]^5)/(1287*d*Sqrt[a + a*Sin[c + d*x]]) + (20*Cos[c + d*x]^5*Sqrt[a + a*Sin
[c + d*x]])/(143*d) - (2*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(13*a*d)

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Rubi [A]  time = 0.424233, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {2878, 2856, 2674, 2673} \[ -\frac{368 a^2 \cos ^5(c+d x)}{9009 d (a \sin (c+d x)+a)^{3/2}}-\frac{1472 a^3 \cos ^5(c+d x)}{45045 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 a d}+\frac{20 \cos ^5(c+d x) \sqrt{a \sin (c+d x)+a}}{143 d}-\frac{46 a \cos ^5(c+d x)}{1287 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-1472*a^3*Cos[c + d*x]^5)/(45045*d*(a + a*Sin[c + d*x])^(5/2)) - (368*a^2*Cos[c + d*x]^5)/(9009*d*(a + a*Sin[
c + d*x])^(3/2)) - (46*a*Cos[c + d*x]^5)/(1287*d*Sqrt[a + a*Sin[c + d*x]]) + (20*Cos[c + d*x]^5*Sqrt[a + a*Sin
[c + d*x]])/(143*d) - (2*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(13*a*d)

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]

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 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) \sin ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx &=-\frac{2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 a d}+\frac{2 \int \cos ^4(c+d x) \left (\frac{3 a}{2}-5 a \sin (c+d x)\right ) \sqrt{a+a \sin (c+d x)} \, dx}{13 a}\\ &=\frac{20 \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{143 d}-\frac{2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 a d}+\frac{23}{143} \int \cos ^4(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{46 a \cos ^5(c+d x)}{1287 d \sqrt{a+a \sin (c+d x)}}+\frac{20 \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{143 d}-\frac{2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 a d}+\frac{(184 a) \int \frac{\cos ^4(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx}{1287}\\ &=-\frac{368 a^2 \cos ^5(c+d x)}{9009 d (a+a \sin (c+d x))^{3/2}}-\frac{46 a \cos ^5(c+d x)}{1287 d \sqrt{a+a \sin (c+d x)}}+\frac{20 \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{143 d}-\frac{2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 a d}+\frac{\left (736 a^2\right ) \int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{9009}\\ &=-\frac{1472 a^3 \cos ^5(c+d x)}{45045 d (a+a \sin (c+d x))^{5/2}}-\frac{368 a^2 \cos ^5(c+d x)}{9009 d (a+a \sin (c+d x))^{3/2}}-\frac{46 a \cos ^5(c+d x)}{1287 d \sqrt{a+a \sin (c+d x)}}+\frac{20 \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{143 d}-\frac{2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 a d}\\ \end{align*}

Mathematica [A]  time = 3.73492, size = 109, normalized size = 0.7 \[ -\frac{\sqrt{a (\sin (c+d x)+1)} \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5 (119780 \sin (c+d x)-21420 \sin (3 (c+d x))-62440 \cos (2 (c+d x))+3465 \cos (4 (c+d x))+81183)}{180180 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^2*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])]*(81183 - 62440*Cos[2*(c + d*x)] + 3465*Co
s[4*(c + d*x)] + 119780*Sin[c + d*x] - 21420*Sin[3*(c + d*x)]))/(180180*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]
))

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Maple [A]  time = 0.632, size = 85, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ) a \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3} \left ( 3465\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}+10710\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+12145\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+6940\,\sin \left ( dx+c \right ) +2776 \right ) }{45045\,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*sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x)

[Out]

2/45045*(1+sin(d*x+c))*a*(sin(d*x+c)-1)^3*(3465*sin(d*x+c)^4+10710*sin(d*x+c)^3+12145*sin(d*x+c)^2+6940*sin(d*
x+c)+2776)/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} \sin \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*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)^4*sin(d*x + c)^2, x)

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Fricas [A]  time = 1.1321, size = 509, normalized size = 3.26 \begin{align*} \frac{2 \,{\left (3465 \, \cos \left (d x + c\right )^{7} - 315 \, \cos \left (d x + c\right )^{6} - 4585 \, \cos \left (d x + c\right )^{5} + 115 \, \cos \left (d x + c\right )^{4} - 184 \, \cos \left (d x + c\right )^{3} + 368 \, \cos \left (d x + c\right )^{2} -{\left (3465 \, \cos \left (d x + c\right )^{6} + 3780 \, \cos \left (d x + c\right )^{5} - 805 \, \cos \left (d x + c\right )^{4} - 920 \, \cos \left (d x + c\right )^{3} - 1104 \, \cos \left (d x + c\right )^{2} - 1472 \, \cos \left (d x + c\right ) - 2944\right )} \sin \left (d x + c\right ) - 1472 \, \cos \left (d x + c\right ) - 2944\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{45045 \,{\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*sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/45045*(3465*cos(d*x + c)^7 - 315*cos(d*x + c)^6 - 4585*cos(d*x + c)^5 + 115*cos(d*x + c)^4 - 184*cos(d*x + c
)^3 + 368*cos(d*x + c)^2 - (3465*cos(d*x + c)^6 + 3780*cos(d*x + c)^5 - 805*cos(d*x + c)^4 - 920*cos(d*x + c)^
3 - 1104*cos(d*x + c)^2 - 1472*cos(d*x + c) - 2944)*sin(d*x + c) - 1472*cos(d*x + c) - 2944)*sqrt(a*sin(d*x +
c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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} \sin \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(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*sin(d*x + c)^2, x)