3.473 \(\int \frac{\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=92 \[ -\frac{2 \cos ^5(c+d x)}{9 a d \sqrt{a \sin (c+d x)+a}}+\frac{20 \cos ^5(c+d x)}{63 d (a \sin (c+d x)+a)^{3/2}}-\frac{46 a \cos ^5(c+d x)}{315 d (a \sin (c+d x)+a)^{5/2}} \]

[Out]

(-46*a*Cos[c + d*x]^5)/(315*d*(a + a*Sin[c + d*x])^(5/2)) + (20*Cos[c + d*x]^5)/(63*d*(a + a*Sin[c + d*x])^(3/
2)) - (2*Cos[c + d*x]^5)/(9*a*d*Sqrt[a + a*Sin[c + d*x]])

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Rubi [A]  time = 0.369968, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {2877, 2856, 2674, 2673} \[ -\frac{2 \cos ^5(c+d x)}{9 a d \sqrt{a \sin (c+d x)+a}}+\frac{20 \cos ^5(c+d x)}{63 d (a \sin (c+d x)+a)^{3/2}}-\frac{46 a \cos ^5(c+d x)}{315 d (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-46*a*Cos[c + d*x]^5)/(315*d*(a + a*Sin[c + d*x])^(5/2)) + (20*Cos[c + d*x]^5)/(63*d*(a + a*Sin[c + d*x])^(3/
2)) - (2*Cos[c + d*x]^5)/(9*a*d*Sqrt[a + a*Sin[c + d*x]])

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]

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 \frac{\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac{\cos ^5(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac{\int \frac{\cos ^4(c+d x) \left (-\frac{3 a}{2}-2 a \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx}{2 a^2}\\ &=\frac{\cos ^5(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac{2 \cos ^5(c+d x)}{9 a d \sqrt{a+a \sin (c+d x)}}+\frac{23 \int \frac{\cos ^4(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx}{36 a}\\ &=\frac{20 \cos ^5(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac{2 \cos ^5(c+d x)}{9 a d \sqrt{a+a \sin (c+d x)}}+\frac{23}{63} \int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac{46 a \cos ^5(c+d x)}{315 d (a+a \sin (c+d x))^{5/2}}+\frac{20 \cos ^5(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac{2 \cos ^5(c+d x)}{9 a d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 3.61044, size = 92, normalized size = 1. \[ -\frac{\sqrt{a (\sin (c+d x)+1)} \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5 (40 \sin (c+d x)-35 \cos (2 (c+d x))+51)}{315 a^2 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

-((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*Sqrt[a*(1 + Sin[c + d*x])]*(51 - 35*Cos[2*(c + d*x)] + 40*Sin[c + d*
x]))/(315*a^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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Maple [A]  time = 0.736, size = 67, normalized size = 0.7 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3} \left ( 35\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+20\,\sin \left ( dx+c \right ) +8 \right ) }{315\,ad\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))^(3/2),x)

[Out]

2/315/a*(1+sin(d*x+c))*(sin(d*x+c)-1)^3*(35*sin(d*x+c)^2+20*sin(d*x+c)+8)/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 )^{4} \sin \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{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))^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^4*sin(d*x + c)^2/(a*sin(d*x + c) + a)^(3/2), x)

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Fricas [A]  time = 1.08125, size = 386, normalized size = 4.2 \begin{align*} -\frac{2 \,{\left (35 \, \cos \left (d x + c\right )^{5} + 85 \, \cos \left (d x + c\right )^{4} - 73 \, \cos \left (d x + c\right )^{3} - 169 \, \cos \left (d x + c\right )^{2} -{\left (35 \, \cos \left (d x + c\right )^{4} - 50 \, \cos \left (d x + c\right )^{3} - 123 \, \cos \left (d x + c\right )^{2} + 46 \, \cos \left (d x + c\right ) + 92\right )} \sin \left (d x + c\right ) + 46 \, \cos \left (d x + c\right ) + 92\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{315 \,{\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} 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))^(3/2),x, algorithm="fricas")

[Out]

-2/315*(35*cos(d*x + c)^5 + 85*cos(d*x + c)^4 - 73*cos(d*x + c)^3 - 169*cos(d*x + c)^2 - (35*cos(d*x + c)^4 -
50*cos(d*x + c)^3 - 123*cos(d*x + c)^2 + 46*cos(d*x + c) + 92)*sin(d*x + c) + 46*cos(d*x + c) + 92)*sqrt(a*sin
(d*x + c) + a)/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) + a^2*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))**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 2.26862, size = 351, normalized size = 3.82 \begin{align*} -\frac{\frac{{\left ({\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^{12}} + \frac{9 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{12}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{105 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{12}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{252 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{12}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{252 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{12}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{105 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{12}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{9 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{12}}\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^{12}}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{9}{2}}} + \frac{23 \, \sqrt{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{\frac{33}{2}}}}{80640 \, d} \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))^(3/2),x, algorithm="giac")

[Out]

-1/80640*((((((((2*sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c)^2/a^12 + 9*sgn(tan(1/2*d*x + 1/2*c) + 1)
/a^12)*tan(1/2*d*x + 1/2*c) - 105*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^12)*tan(1/2*d*x + 1/2*c) + 252*sgn(tan(1/2*d
*x + 1/2*c) + 1)/a^12)*tan(1/2*d*x + 1/2*c) - 252*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^12)*tan(1/2*d*x + 1/2*c) + 1
05*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^12)*tan(1/2*d*x + 1/2*c) - 9*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^12)*tan(1/2*d*
x + 1/2*c)^2 - 2*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^12)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(9/2) + 23*sqrt(2)*sgn(tan
(1/2*d*x + 1/2*c) + 1)/a^(33/2))/d