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

Optimal. Leaf size=60 \[ \frac{6 a \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}} \]

[Out]

(6*a*Cos[c + d*x]^5)/(35*d*(a + a*Sin[c + d*x])^(5/2)) - (2*Cos[c + d*x]^5)/(7*d*(a + a*Sin[c + d*x])^(3/2))

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Rubi [A]  time = 0.150723, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {2856, 2673} \[ \frac{6 a \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*Cos[c + d*x]^5)/(35*d*(a + a*Sin[c + d*x])^(5/2)) - (2*Cos[c + d*x]^5)/(7*d*(a + a*Sin[c + d*x])^(3/2))

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 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 (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=-\frac{2 \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}-\frac{3}{7} \int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=\frac{6 a \cos ^5(c+d x)}{35 d (a+a \sin (c+d x))^{5/2}}-\frac{2 \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.85613, size = 82, normalized size = 1.37 \[ -\frac{2 (5 \sin (c+d x)+2) \sqrt{a (\sin (c+d x)+1)} \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5}{35 a^2 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.8, size = 57, normalized size = 1. \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3} \left ( 5\,\sin \left ( dx+c \right ) +2 \right ) }{35\,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)/(a+a*sin(d*x+c))^(3/2),x)

[Out]

2/35/a*(1+sin(d*x+c))*(sin(d*x+c)-1)^3*(5*sin(d*x+c)+2)/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 )}{{\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)/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.04614, size = 320, normalized size = 5.33 \begin{align*} \frac{2 \,{\left (5 \, \cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{3} - 19 \, \cos \left (d x + c\right )^{2} +{\left (5 \, \cos \left (d x + c\right )^{3} + 13 \, \cos \left (d x + c\right )^{2} - 6 \, \cos \left (d x + c\right ) - 12\right )} \sin \left (d x + c\right ) + 6 \, \cos \left (d x + c\right ) + 12\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{35 \,{\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)/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2/35*(5*cos(d*x + c)^4 - 8*cos(d*x + c)^3 - 19*cos(d*x + c)^2 + (5*cos(d*x + c)^3 + 13*cos(d*x + c)^2 - 6*cos(
d*x + c) - 12)*sin(d*x + c) + 6*cos(d*x + c) + 12)*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)/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 2.22917, size = 274, normalized size = 4.57 \begin{align*} -\frac{\frac{{\left ({\left ({\left ({\left (\frac{\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^{10}} - \frac{14 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{10}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{35 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{10}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{35 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{10}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{14 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{10}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{10}}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{7}{2}}} - \frac{6 \, \sqrt{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{\frac{27}{2}}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/840*((((((sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c)^2/a^10 - 14*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^10
)*tan(1/2*d*x + 1/2*c) + 35*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^10)*tan(1/2*d*x + 1/2*c) - 35*sgn(tan(1/2*d*x + 1/
2*c) + 1)/a^10)*tan(1/2*d*x + 1/2*c) + 14*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^10)*tan(1/2*d*x + 1/2*c)^2 - sgn(tan
(1/2*d*x + 1/2*c) + 1)/a^10)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(7/2) - 6*sqrt(2)*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^
(27/2))/d