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

Optimal. Leaf size=76 \[ \frac{3 \cos (c+d x)}{a^3 d}+\frac{3 x}{a^3}+\frac{2 \cos ^3(c+d x)}{a d (a \sin (c+d x)+a)^2}-\frac{\cos ^3(c+d x)}{3 d (a \sin (c+d x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.181297, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2871, 2680, 2682, 8} \[ \frac{3 \cos (c+d x)}{a^3 d}+\frac{3 x}{a^3}+\frac{2 \cos ^3(c+d x)}{a d (a \sin (c+d x)+a)^2}-\frac{\cos ^3(c+d x)}{3 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2871

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*m), x] - Dist[1/g^2, Int[(g*Cos
[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && E
qQ[m + p + 1, 0]

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2682

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e
+ f*x])^(p - 1))/(b*f*(p - 1)), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.665049, size = 96, normalized size = 1.26 \[ \frac{3 \cos (c+d x)-\frac{2 \sin \left (\frac{1}{2} (c+d x)\right ) (13 \sin (c+d x)+11)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}-\frac{2}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+9 c+9 d x}{3 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*c + 9*d*x + 3*Cos[c + d*x] - 2/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (2*Sin[(c + d*x)/2]*(11 + 13*Sin[c
+ d*x]))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)/(3*a^3*d)

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Maple [A]  time = 0.104, size = 106, normalized size = 1.4 \begin{align*} 2\,{\frac{1}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}-{\frac{8}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+4\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}+6\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)+6/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+6/d/a^3/(tan(1/2*d*x+1/2*c)+1)

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Maxima [B]  time = 1.79176, size = 308, normalized size = 4.05 \begin{align*} \frac{2 \,{\left (\frac{\frac{33 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{29 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{27 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{9 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 14}{a^{3} + \frac{3 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{4 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac{9 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*((33*sin(d*x + c)/(cos(d*x + c) + 1) + 29*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 27*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 9*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 14)/(a^3 + 3*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 4*a^
3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*a^3*sin(d*x + c)^4/(cos(
d*x + c) + 1)^4 + a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) + 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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Fricas [A]  time = 1.71524, size = 365, normalized size = 4.8 \begin{align*} \frac{{\left (9 \, d x - 16\right )} \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right )^{3} - 18 \, d x -{\left (9 \, d x + 17\right )} \cos \left (d x + c\right ) -{\left (18 \, d x +{\left (9 \, d x + 19\right )} \cos \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((9*d*x - 16)*cos(d*x + c)^2 + 3*cos(d*x + c)^3 - 18*d*x - (9*d*x + 17)*cos(d*x + c) - (18*d*x + (9*d*x +
19)*cos(d*x + c) + 3*cos(d*x + c)^2 + 2)*sin(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))

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Sympy [A]  time = 60.8184, size = 1246, normalized size = 16.39 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((9*d*x*tan(c/2 + d*x/2)**5/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*
tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 27*d*x*tan(c/2 +
d*x/2)**4/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a
**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 36*d*x*tan(c/2 + d*x/2)**3/(3*a**3*d*tan(c
/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2
+ 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 36*d*x*tan(c/2 + d*x/2)**2/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*
tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2
) + 3*a**3*d) + 27*d*x*tan(c/2 + d*x/2)/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3
*d*tan(c/2 + d*x/2)**3 + 12*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/(3*a**3
*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*
x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 18*tan(c/2 + d*x/2)**4/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**
3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d
*x/2) + 3*a**3*d) + 54*tan(c/2 + d*x/2)**3/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a
**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 58*tan(c/2
+ d*x/2)**2/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12
*a**3*d*tan(c/2 + d*x/2)**2 + 9*a**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 66*tan(c/2 + d*x/2)/(3*a**3*d*tan(c/2 +
d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3*d*tan(c/2 + d*x/2)**3 + 12*a**3*d*tan(c/2 + d*x/2)**2 + 9*a
**3*d*tan(c/2 + d*x/2) + 3*a**3*d) + 28/(3*a**3*d*tan(c/2 + d*x/2)**5 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 12*a**3
*d*tan(c/2 + d*x/2)**3 + 12*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)**2*cos(c)**2/(a*sin(c) + a)**3, True))

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Giac [A]  time = 1.3067, size = 108, normalized size = 1.42 \begin{align*} \frac{\frac{9 \,{\left (d x + c\right )}}{a^{3}} + \frac{6}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} a^{3}} + \frac{2 \,{\left (9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 11\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(9*(d*x + c)/a^3 + 6/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3) + 2*(9*tan(1/2*d*x + 1/2*c)^2 + 24*tan(1/2*d*x + 1
/2*c) + 11)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^3))/d