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

Optimal. Leaf size=61 \[ -\frac {7 \cos (c+d x)}{3 a^3 d (\sin (c+d x)+1)}-\frac {x}{a^3}+\frac {2 \cos (c+d x)}{3 a d (a \sin (c+d x)+a)^2} \]

[Out]

-x/a^3-7/3*cos(d*x+c)/a^3/d/(1+sin(d*x+c))+2/3*cos(d*x+c)/a/d/(a+a*sin(d*x+c))^2

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2857, 2735, 2648} \[ -\frac {7 \cos (c+d x)}{3 a^3 d (\sin (c+d x)+1)}-\frac {x}{a^3}+\frac {2 \cos (c+d x)}{3 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a^3) - (7*Cos[c + d*x])/(3*a^3*d*(1 + Sin[c + d*x])) + (2*Cos[c + d*x])/(3*a*d*(a + a*Sin[c + d*x])^2)

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2735

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

Rule 2857

Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_
)]), x_Symbol] :> Simp[(2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(2*m + 3)), x] + Dist[
1/(b^3*(2*m + 3)), Int[(a + b*Sin[e + f*x])^(m + 2)*(b*c + 2*a*d*(m + 1) - b*d*(2*m + 3)*Sin[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -3/2]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {2 \cos (c+d x)}{3 a d (a+a \sin (c+d x))^2}-\frac {\int \frac {-4 a+3 a \sin (c+d x)}{a+a \sin (c+d x)} \, dx}{3 a^3}\\ &=-\frac {x}{a^3}+\frac {2 \cos (c+d x)}{3 a d (a+a \sin (c+d x))^2}+\frac {7 \int \frac {1}{a+a \sin (c+d x)} \, dx}{3 a^2}\\ &=-\frac {x}{a^3}+\frac {2 \cos (c+d x)}{3 a d (a+a \sin (c+d x))^2}-\frac {7 \cos (c+d x)}{3 d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.42, size = 145, normalized size = 2.38 \[ -\frac {180 d x \sin \left (c+\frac {d x}{2}\right )+60 d x \sin \left (c+\frac {3 d x}{2}\right )+3 \sin \left (2 c+\frac {3 d x}{2}\right )-351 \cos \left (c+\frac {d x}{2}\right )+277 \cos \left (c+\frac {3 d x}{2}\right )-60 d x \cos \left (2 c+\frac {3 d x}{2}\right )-471 \sin \left (\frac {d x}{2}\right )+180 d x \cos \left (\frac {d x}{2}\right )}{120 a^3 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/120*(180*d*x*Cos[(d*x)/2] - 351*Cos[c + (d*x)/2] + 277*Cos[c + (3*d*x)/2] - 60*d*x*Cos[2*c + (3*d*x)/2] - 4
71*Sin[(d*x)/2] + 180*d*x*Sin[c + (d*x)/2] + 60*d*x*Sin[c + (3*d*x)/2] + 3*Sin[2*c + (3*d*x)/2])/(a^3*d*(Cos[c
/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)

________________________________________________________________________________________

fricas [B]  time = 0.51, size = 124, normalized size = 2.03 \[ -\frac {{\left (3 \, d x - 7\right )} \cos \left (d x + c\right )^{2} - 6 \, d x - {\left (3 \, d x + 5\right )} \cos \left (d x + c\right ) - {\left (6 \, d x + {\left (3 \, d x + 7\right )} \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) + 2}{3 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d \cos \left (d x + c\right ) - 2 \, a^{3} d - {\left (a^{3} d \cos \left (d x + c\right ) + 2 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((3*d*x - 7)*cos(d*x + c)^2 - 6*d*x - (3*d*x + 5)*cos(d*x + c) - (6*d*x + (3*d*x + 7)*cos(d*x + c) + 2)*s
in(d*x + c) + 2)/(a^3*d*cos(d*x + c)^2 - a^3*d*cos(d*x + c) - 2*a^3*d - (a^3*d*cos(d*x + c) + 2*a^3*d)*sin(d*x
+ c))

________________________________________________________________________________________

giac [A]  time = 0.22, size = 60, normalized size = 0.98 \[ -\frac {\frac {3 \, {\left (d x + c\right )}}{a^{3}} + \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*(d*x + c)/a^3 + 2*(3*tan(1/2*d*x + 1/2*c)^2 + 12*tan(1/2*d*x + 1/2*c) + 5)/(a^3*(tan(1/2*d*x + 1/2*c)
+ 1)^3))/d

________________________________________________________________________________________

maple [A]  time = 0.42, size = 83, normalized size = 1.36 \[ -\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}+\frac {8}{3 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {4}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/a^3*arctan(tan(1/2*d*x+1/2*c))+8/3/d/a^3/(tan(1/2*d*x+1/2*c)+1)^3-4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^2-2/d/a^
3/(tan(1/2*d*x+1/2*c)+1)

________________________________________________________________________________________

maxima [B]  time = 0.44, size = 142, normalized size = 2.33 \[ -\frac {2 \, {\left (\frac {\frac {12 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 5}{a^{3} + \frac {3 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}} + \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*((12*sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 5)/(a^3 + 3*a^3*sin(d*x +
c)/(cos(d*x + c) + 1) + 3*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3) +
3*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

________________________________________________________________________________________

mupad [B]  time = 8.77, size = 54, normalized size = 0.89 \[ -\frac {x}{a^3}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {10}{3}}{a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- x/a^3 - (8*tan(c/2 + (d*x)/2) + 2*tan(c/2 + (d*x)/2)^2 + 10/3)/(a^3*d*(tan(c/2 + (d*x)/2) + 1)^3)

________________________________________________________________________________________

sympy [A]  time = 22.21, size = 529, normalized size = 8.67 \[ \begin {cases} - \frac {3 d x \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{3} d} - \frac {9 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{3} d} - \frac {9 d x \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{3} d} - \frac {3 d x}{3 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{3} d} - \frac {6 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{3} d} - \frac {24 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{3} d} - \frac {10}{3 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \sin {\relax (c )} \cos ^{2}{\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-3*d*x*tan(c/2 + d*x/2)**3/(3*a**3*d*tan(c/2 + d*x/2)**3 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*
tan(c/2 + d*x/2) + 3*a**3*d) - 9*d*x*tan(c/2 + d*x/2)**2/(3*a**3*d*tan(c/2 + d*x/2)**3 + 9*a**3*d*tan(c/2 + d*
x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) - 9*d*x*tan(c/2 + d*x/2)/(3*a**3*d*tan(c/2 + d*x/2)**3 + 9*a**
3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) - 3*d*x/(3*a**3*d*tan(c/2 + d*x/2)**3 + 9*a**3
*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) - 6*tan(c/2 + d*x/2)**2/(3*a**3*d*tan(c/2 + d*x
/2)**3 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) - 24*tan(c/2 + d*x/2)/(3*a**3*d*
tan(c/2 + d*x/2)**3 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) - 10/(3*a**3*d*tan(
c/2 + d*x/2)**3 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d), Ne(d, 0)), (x*sin(c)*c
os(c)**2/(a*sin(c) + a)**3, True))

________________________________________________________________________________________