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

[Out]

-1472/45045*a^3*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-368/9009*a^2*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(3/2)-2/13*
cos(d*x+c)^5*(a+a*sin(d*x+c))^(3/2)/a/d-46/1287*a*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(1/2)+20/143*cos(d*x+c)^5*(a
+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.42, antiderivative size = 156, normalized size of antiderivative = 1.00, 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 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 ^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*}

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Mathematica [A]  time = 3.91, size = 109, normalized size = 0.70 \[ -\frac {\sqrt {a (\sin (c+d x)+1)} \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 (119780 \sin (c+d x)-21420 \sin (3 (c+d x))-62440 \cos (2 (c+d x))+3465 \cos (4 (c+d x))+81183)}{180180 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

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]

-1/180180*((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*Cos[4*(c + d*x)] + 119780*Sin[c + d*x] - 21420*Sin[3*(c + d*x)]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/
2]))

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fricas [A]  time = 0.44, size = 172, normalized size = 1.10 \[ \frac {2 \, {\left (3465 \, \cos \left (d x + c\right )^{7} - 315 \, \cos \left (d x + c\right )^{6} - 4585 \, \cos \left (d x + c\right )^{5} + 115 \, \cos \left (d x + c\right )^{4} - 184 \, \cos \left (d x + c\right )^{3} + 368 \, \cos \left (d x + c\right )^{2} - {\left (3465 \, \cos \left (d x + c\right )^{6} + 3780 \, \cos \left (d x + c\right )^{5} - 805 \, \cos \left (d x + c\right )^{4} - 920 \, \cos \left (d x + c\right )^{3} - 1104 \, \cos \left (d x + c\right )^{2} - 1472 \, \cos \left (d x + c\right ) - 2944\right )} \sin \left (d x + c\right ) - 1472 \, \cos \left (d x + c\right ) - 2944\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{45045 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

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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giac [A]  time = 0.29, size = 219, normalized size = 1.40 \[ -\frac {1}{1441440} \, \sqrt {2} \sqrt {a} {\left (\frac {4095 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {11}{2} \, d x + \frac {11}{2} \, c\right )}{d} + \frac {12870 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} - \frac {15015 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {3465 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {13}{2} \, d x + \frac {13}{2} \, c\right )}{d} + \frac {10010 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} - \frac {9009 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} - \frac {180180 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-1/1441440*sqrt(2)*sqrt(a)*(4095*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 11/2*d*x + 11/2*c)/d + 12870
*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 7/2*d*x + 7/2*c)/d - 15015*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c
))*sin(1/4*pi + 3/2*d*x + 3/2*c)/d + 3465*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 13/2*d*x + 13/2*c)
/d + 10010*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 9/2*d*x + 9/2*c)/d - 9009*sgn(cos(-1/4*pi + 1/2*d
*x + 1/2*c))*sin(-1/4*pi + 5/2*d*x + 5/2*c)/d - 180180*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 = 0.98, size = 85, normalized size = 0.54 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right )^{3} \left (3465 \left (\sin ^{4}\left (d x +c \right )\right )+10710 \left (\sin ^{3}\left (d x +c \right )\right )+12145 \left (\sin ^{2}\left (d x +c \right )\right )+6940 \sin \left (d x +c \right )+2776\right )}{45045 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^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 = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(c + d*x)^4*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 \[ \int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}\, dx \]

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]

Integral(sqrt(a*(sin(c + d*x) + 1))*sin(c + d*x)**2*cos(c + d*x)**4, x)

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