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

Optimal. Leaf size=124 \[ -\frac {152 a^2 \cos ^5(c+d x)}{3465 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 a d}+\frac {20 \cos ^5(c+d x)}{99 d \sqrt {a \sin (c+d x)+a}}-\frac {38 a \cos ^5(c+d x)}{693 d (a \sin (c+d x)+a)^{3/2}} \]

[Out]

-152/3465*a^2*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-38/693*a*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(3/2)+20/99*cos(d
*x+c)^5/d/(a+a*sin(d*x+c))^(1/2)-2/11*cos(d*x+c)^5*(a+a*sin(d*x+c))^(1/2)/a/d

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Rubi [A]  time = 0.41, antiderivative size = 124, 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 = {2877, 2856, 2674, 2673} \[ -\frac {152 a^2 \cos ^5(c+d x)}{3465 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 a d}+\frac {20 \cos ^5(c+d x)}{99 d \sqrt {a \sin (c+d x)+a}}-\frac {38 a \cos ^5(c+d x)}{693 d (a \sin (c+d x)+a)^{3/2}} \]

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]

(-152*a^2*Cos[c + d*x]^5)/(3465*d*(a + a*Sin[c + d*x])^(5/2)) - (38*a*Cos[c + d*x]^5)/(693*d*(a + a*Sin[c + d*
x])^(3/2)) + (20*Cos[c + d*x]^5)/(99*d*Sqrt[a + a*Sin[c + d*x]]) - (2*Cos[c + d*x]^5*Sqrt[a + a*Sin[c + d*x]])
/(11*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 2877

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

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx &=\frac {\cos ^5(c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {\int \cos ^4(c+d x) \left (-\frac {a}{2}-4 a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{4 a^2}\\ &=\frac {\cos ^5(c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{11 a d}+\frac {19 \int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{88 a}\\ &=\frac {20 \cos ^5(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{11 a d}+\frac {19}{99} \int \frac {\cos ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {38 a \cos ^5(c+d x)}{693 d (a+a \sin (c+d x))^{3/2}}+\frac {20 \cos ^5(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{11 a d}+\frac {1}{693} (76 a) \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {152 a^2 \cos ^5(c+d x)}{3465 d (a+a \sin (c+d x))^{5/2}}-\frac {38 a \cos ^5(c+d x)}{693 d (a+a \sin (c+d x))^{3/2}}+\frac {20 \cos ^5(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{11 a d}\\ \end {align*}

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Mathematica [A]  time = 1.70, size = 143, normalized size = 1.15 \[ -\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \left (5773 \sin \left (\frac {1}{2} (c+d x)\right )+3495 \sin \left (\frac {3}{2} (c+d x)\right )-1505 \sin \left (\frac {5}{2} (c+d x)\right )-315 \sin \left (\frac {7}{2} (c+d x)\right )+5773 \cos \left (\frac {1}{2} (c+d x)\right )-3495 \cos \left (\frac {3}{2} (c+d x)\right )-1505 \cos \left (\frac {5}{2} (c+d x)\right )+315 \cos \left (\frac {7}{2} (c+d x)\right )\right )}{13860 d \sqrt {a (\sin (c+d x)+1)}} \]

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]

-1/13860*((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*(5773*Cos[(c + d*x)/2] - 3495*Cos[(3*(c + d*x))/2] - 1505*Co
s[(5*(c + d*x))/2] + 315*Cos[(7*(c + d*x))/2] + 5773*Sin[(c + d*x)/2] + 3495*Sin[(3*(c + d*x))/2] - 1505*Sin[(
5*(c + d*x))/2] - 315*Sin[(7*(c + d*x))/2]))/(d*Sqrt[a*(1 + Sin[c + d*x])])

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fricas [A]  time = 0.43, size = 155, normalized size = 1.25 \[ -\frac {2 \, {\left (315 \, \cos \left (d x + c\right )^{6} - 35 \, \cos \left (d x + c\right )^{5} - 445 \, \cos \left (d x + c\right )^{4} + 19 \, \cos \left (d x + c\right )^{3} - 38 \, \cos \left (d x + c\right )^{2} + {\left (315 \, \cos \left (d x + c\right )^{5} + 350 \, \cos \left (d x + c\right )^{4} - 95 \, \cos \left (d x + c\right )^{3} - 114 \, \cos \left (d x + c\right )^{2} - 152 \, \cos \left (d x + c\right ) - 304\right )} \sin \left (d x + c\right ) + 152 \, \cos \left (d x + c\right ) + 304\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3465 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \]

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]

-2/3465*(315*cos(d*x + c)^6 - 35*cos(d*x + c)^5 - 445*cos(d*x + c)^4 + 19*cos(d*x + c)^3 - 38*cos(d*x + c)^2 +
(315*cos(d*x + c)^5 + 350*cos(d*x + c)^4 - 95*cos(d*x + c)^3 - 114*cos(d*x + c)^2 - 152*cos(d*x + c) - 304)*s
in(d*x + c) + 152*cos(d*x + c) + 304)*sqrt(a*sin(d*x + c) + a)/(a*d*cos(d*x + c) + a*d*sin(d*x + c) + a*d)

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giac [B]  time = 0.81, size = 342, normalized size = 2.76 \[ \frac {8 \, {\left (\frac {76 \, \sqrt {2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a}} - \frac {\frac {34 \, a^{5}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {187 \, a^{5}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {1155 \, a^{5}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {1287 \, a^{5}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {231 \, a^{5}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {231 \, a^{5}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {1287 \, a^{5}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {1155 \, a^{5}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - 17 \, {\left (\frac {2 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {11 \, a^{5}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {11}{2}}}\right )}}{3465 \, d} \]

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]

8/3465*(76*sqrt(2)*sgn(tan(1/2*d*x + 1/2*c) + 1)/sqrt(a) - (34*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1) + (187*a^5/sg
n(tan(1/2*d*x + 1/2*c) + 1) - (1155*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1) - (1287*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1
) + (231*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1) - (231*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1) + (1287*a^5/sgn(tan(1/2*d*
x + 1/2*c) + 1) - (1155*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1) - 17*(2*a^5*tan(1/2*d*x + 1/2*c)^2/sgn(tan(1/2*d*x +
1/2*c) + 1) + 11*a^5/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x +
1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x +
1/2*c)^2)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(11/2))/d

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maple [A]  time = 1.35, size = 74, normalized size = 0.60 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{3} \left (315 \left (\sin ^{3}\left (d x +c \right )\right )+595 \left (\sin ^{2}\left (d x +c \right )\right )+340 \sin \left (d x +c \right )+136\right )}{3465 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

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]

2/3465*(1+sin(d*x+c))*(sin(d*x+c)-1)^3*(315*sin(d*x+c)^3+595*sin(d*x+c)^2+340*sin(d*x+c)+136)/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 \[ \int \frac {\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2}}{\sqrt {a \sin \left (d x + c\right ) + a}}\,{d x} \]

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

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

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

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

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