3.336 \(\int \frac{\cos ^2(c+d x) \sin ^3(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=158 \[ -\frac{4 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{105 a^2 d}+\frac{2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt{a \sin (c+d x)+a}}-\frac{2 \sin ^3(c+d x) \cos (c+d x)}{63 d \sqrt{a \sin (c+d x)+a}}+\frac{8 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{315 a d}-\frac{4 \cos (c+d x)}{45 d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(-4*Cos[c + d*x])/(45*d*Sqrt[a + a*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sin[c + d*x]^3)/(63*d*Sqrt[a + a*Sin[c + d
*x]]) + (2*Cos[c + d*x]*Sin[c + d*x]^4)/(9*d*Sqrt[a + a*Sin[c + d*x]]) + (8*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*
x]])/(315*a*d) - (4*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(105*a^2*d)

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Rubi [A]  time = 0.403586, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2879, 2981, 2770, 2759, 2751, 2646} \[ -\frac{4 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{105 a^2 d}+\frac{2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt{a \sin (c+d x)+a}}-\frac{2 \sin ^3(c+d x) \cos (c+d x)}{63 d \sqrt{a \sin (c+d x)+a}}+\frac{8 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{315 a d}-\frac{4 \cos (c+d x)}{45 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*Sin[c + d*x]^3)/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-4*Cos[c + d*x])/(45*d*Sqrt[a + a*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sin[c + d*x]^3)/(63*d*Sqrt[a + a*Sin[c + d
*x]]) + (2*Cos[c + d*x]*Sin[c + d*x]^4)/(9*d*Sqrt[a + a*Sin[c + d*x]]) + (8*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*
x]])/(315*a*d) - (4*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(105*a^2*d)

Rule 2879

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[2*m, 2*n]

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rule 2770

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(2*n*(b*c + a*d)
)/(b*(2*n + 1)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}
, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]

Rule 2759

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

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2646

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(-2*b*Cos[c + d*x])/(d*Sqrt[a + b*Sin[c + d*
x]]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x) \sin ^3(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx &=\frac{\int \sin ^3(c+d x) (a-a \sin (c+d x)) \sqrt{a+a \sin (c+d x)} \, dx}{a^2}\\ &=\frac{2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}+\frac{\int \sin ^3(c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{9 a}\\ &=-\frac{2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt{a+a \sin (c+d x)}}+\frac{2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}+\frac{2 \int \sin ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{21 a}\\ &=-\frac{2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt{a+a \sin (c+d x)}}+\frac{2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}-\frac{4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a^2 d}+\frac{4 \int \left (\frac{3 a}{2}-a \sin (c+d x)\right ) \sqrt{a+a \sin (c+d x)} \, dx}{105 a^2}\\ &=-\frac{2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt{a+a \sin (c+d x)}}+\frac{2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}+\frac{8 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{315 a d}-\frac{4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a^2 d}+\frac{2 \int \sqrt{a+a \sin (c+d x)} \, dx}{45 a}\\ &=-\frac{4 \cos (c+d x)}{45 d \sqrt{a+a \sin (c+d x)}}-\frac{2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt{a+a \sin (c+d x)}}+\frac{2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}+\frac{8 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{315 a d}-\frac{4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a^2 d}\\ \end{align*}

Mathematica [A]  time = 1.23475, size = 97, normalized size = 0.61 \[ \frac{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (-201 \sin (c+d x)+35 \sin (3 (c+d x))+60 \cos (2 (c+d x))-124)}{630 d \sqrt{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*Sin[c + d*x]^3)/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-124 + 60*Cos[2*(c + d*x)] - 2
01*Sin[c + d*x] + 35*Sin[3*(c + d*x)]))/(630*d*Sqrt[a*(1 + Sin[c + d*x])])

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Maple [A]  time = 0.67, size = 74, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2} \left ( 35\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+30\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+24\,\sin \left ( dx+c \right ) +16 \right ) }{315\,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)^2*sin(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),x)

[Out]

-2/315*(1+sin(d*x+c))*(sin(d*x+c)-1)^2*(35*sin(d*x+c)^3+30*sin(d*x+c)^2+24*sin(d*x+c)+16)/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 \frac{\cos \left (d x + c\right )^{2} \sin \left (d x + c\right )^{3}}{\sqrt{a \sin \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^2*sin(d*x + c)^3/sqrt(a*sin(d*x + c) + a), x)

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Fricas [A]  time = 1.62542, size = 373, normalized size = 2.36 \begin{align*} \frac{2 \,{\left (35 \, \cos \left (d x + c\right )^{5} + 40 \, \cos \left (d x + c\right )^{4} - 64 \, \cos \left (d x + c\right )^{3} - 82 \, \cos \left (d x + c\right )^{2} -{\left (35 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{3} - 69 \, \cos \left (d x + c\right )^{2} + 13 \, \cos \left (d x + c\right ) + 26\right )} \sin \left (d x + c\right ) + 13 \, \cos \left (d x + c\right ) + 26\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{315 \,{\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*cos(d*x + c)^5 + 40*cos(d*x + c)^4 - 64*cos(d*x + c)^3 - 82*cos(d*x + c)^2 - (35*cos(d*x + c)^4 - 5*
cos(d*x + c)^3 - 69*cos(d*x + c)^2 + 13*cos(d*x + c) + 26)*sin(d*x + c) + 13*cos(d*x + c) + 26)*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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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)**2*sin(d*x+c)**3/(a+a*sin(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 2.53172, size = 282, normalized size = 1.78 \begin{align*} -\frac{\frac{4 \,{\left ({\left ({\left ({\left ({\left (\frac{2 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{11}} + \frac{9 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{11}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{63 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{11}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{63 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{11}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{9 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{11}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{2 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{11}}\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{9}{2}}} - \frac{13 \, \sqrt{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{\frac{31}{2}}}}{161280 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-1/161280*(4*(((((2*sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c)^2/a^11 + 9*sgn(tan(1/2*d*x + 1/2*c) + 1
)/a^11)*tan(1/2*d*x + 1/2*c)^2 - 63*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)*tan(1/2*d*x + 1/2*c) + 63*sgn(tan(1/2*
d*x + 1/2*c) + 1)/a^11)*tan(1/2*d*x + 1/2*c)^2 - 9*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)*tan(1/2*d*x + 1/2*c)^2
- 2*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(9/2) - 13*sqrt(2)*sgn(tan(1/2*d*x + 1/
2*c) + 1)/a^(31/2))/d