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

Optimal. Leaf size=92 \[ \frac{8 a^2 \cos ^5(c+d x)}{315 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 \cos ^5(c+d x)}{9 d \sqrt{a \sin (c+d x)+a}}+\frac{2 a \cos ^5(c+d x)}{63 d (a \sin (c+d x)+a)^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.19254, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2856, 2674, 2673} \[ \frac{8 a^2 \cos ^5(c+d x)}{315 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 \cos ^5(c+d x)}{9 d \sqrt{a \sin (c+d x)+a}}+\frac{2 a \cos ^5(c+d x)}{63 d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 (c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx &=-\frac{2 \cos ^5(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}-\frac{1}{9} \int \frac{\cos ^4(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=\frac{2 a \cos ^5(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac{2 \cos ^5(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}-\frac{1}{63} (4 a) \int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=\frac{8 a^2 \cos ^5(c+d x)}{315 d (a+a \sin (c+d x))^{5/2}}+\frac{2 a \cos ^5(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac{2 \cos ^5(c+d x)}{9 d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.56883, size = 87, normalized size = 0.95 \[ -\frac{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (130 \sin (c+d x)-35 \cos (2 (c+d x))+87)}{315 d \sqrt{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(87 - 35*Cos[2*(c + d*x)] + 13
0*Sin[c + d*x]))/(315*d*Sqrt[a*(1 + Sin[c + d*x])])

________________________________________________________________________________________

Maple [A]  time = 0.756, size = 64, 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}+65\,\sin \left ( dx+c \right ) +26 \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)^4*sin(d*x+c)/(a+a*sin(d*x+c))^(1/2),x)

[Out]

2/315*(1+sin(d*x+c))*(sin(d*x+c)-1)^3*(35*sin(d*x+c)^2+65*sin(d*x+c)+26)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

________________________________________________________________________________________

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 )}{\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)^4*sin(d*x+c)/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.05287, size = 365, normalized size = 3.97 \begin{align*} -\frac{2 \,{\left (35 \, \cos \left (d x + c\right )^{5} + 40 \, \cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} -{\left (35 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{3} - 6 \, \cos \left (d x + c\right )^{2} - 8 \, \cos \left (d x + c\right ) - 16\right )} \sin \left (d x + c\right ) - 8 \, \cos \left (d x + c\right ) - 16\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)^4*sin(d*x+c)/(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 - cos(d*x + c)^3 + 2*cos(d*x + c)^2 - (35*cos(d*x + c)^4 - 5*cos
(d*x + c)^3 - 6*cos(d*x + c)^2 - 8*cos(d*x + c) - 16)*sin(d*x + c) - 8*cos(d*x + c) - 16)*sqrt(a*sin(d*x + c)
+ a)/(a*d*cos(d*x + c) + a*d*sin(d*x + c) + a*d)

________________________________________________________________________________________

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))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 2.37807, size = 351, normalized size = 3.82 \begin{align*} -\frac{\frac{{\left ({\left ({\left ({\left ({\left ({\left (\frac{13 \, \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{99 \, \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{105 \, \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 ) - \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{105 \, \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{99 \, \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{13 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{11}}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{9}{2}}} - \frac{8 \, \sqrt{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{\frac{31}{2}}}}{630 \, 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))^(1/2),x, algorithm="giac")

[Out]

-1/630*((((((((13*sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c)^2/a^11 - 99*sgn(tan(1/2*d*x + 1/2*c) + 1)
/a^11)*tan(1/2*d*x + 1/2*c) + 105*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) - 63*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)*tan(1/2*d*x + 1/2*c) - 105
*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)*tan(1/2*d*x + 1/2*c) + 99*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)*tan(1/2*d*x
+ 1/2*c)^2 - 13*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) - 8*sqrt(2)*sgn(tan(
1/2*d*x + 1/2*c) + 1)/a^(31/2))/d