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

Optimal. Leaf size=69 \[ -\frac{2 \cos (c+d x)}{a^2 d}+\frac{\sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac{2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}-\frac{5 x}{2 a^2} \]

[Out]

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

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Rubi [A]  time = 0.271632, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2874, 2950, 2709, 2638, 2635, 8, 2648} \[ -\frac{2 \cos (c+d x)}{a^2 d}+\frac{\sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac{2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}-\frac{5 x}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2874

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[n, 0])

Rule 2950

Int[sin[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[a^n*c^n, Int[Tan[e + f*x]^p*(a + b*Sin[e + f*x])^(m - n), x], x] /; FreeQ[{a,
b, c, d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[p + 2*n, 0] && IntegerQ[n]

Rule 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

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

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]

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\int \frac{\sin ^2(c+d x) (a-a \sin (c+d x))}{a+a \sin (c+d x)} \, dx}{a^2}\\ &=\frac{\int (a-a \sin (c+d x))^2 \tan ^2(c+d x) \, dx}{a^4}\\ &=\frac{\int \left (-2+2 \sin (c+d x)-\sin ^2(c+d x)+\frac{2}{1+\sin (c+d x)}\right ) \, dx}{a^2}\\ &=-\frac{2 x}{a^2}-\frac{\int \sin ^2(c+d x) \, dx}{a^2}+\frac{2 \int \sin (c+d x) \, dx}{a^2}+\frac{2 \int \frac{1}{1+\sin (c+d x)} \, dx}{a^2}\\ &=-\frac{2 x}{a^2}-\frac{2 \cos (c+d x)}{a^2 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac{2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))}-\frac{\int 1 \, dx}{2 a^2}\\ &=-\frac{5 x}{2 a^2}-\frac{2 \cos (c+d x)}{a^2 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac{2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.146948, size = 69, normalized size = 1. \[ \frac{-10 (c+d x)+\sin (2 (c+d x))-8 \cos (c+d x)+\frac{16 \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}}{4 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*(c + d*x) - 8*Cos[c + d*x] + (16*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + Sin[2*(c + d*x
)])/(4*a^2*d)

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Maple [B]  time = 0.095, size = 163, normalized size = 2.4 \begin{align*} -{\frac{1}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{1}{d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-4\,{\frac{1}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-5\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}-4\,{\frac{1}{d{a}^{2} \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))^2,x)

[Out]

-1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)
^2+1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)-4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2-5/d/a^2*arctan(tan
(1/2*d*x+1/2*c))-4/d/a^2/(tan(1/2*d*x+1/2*c)+1)

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Maxima [B]  time = 1.61373, size = 305, normalized size = 4.42 \begin{align*} -\frac{\frac{\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{11 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 8}{a^{2} + \frac{a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{2 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac{5 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{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))^2,x, algorithm="maxima")

[Out]

-((3*sin(d*x + c)/(cos(d*x + c) + 1) + 11*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 5*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + 5*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 8)/(a^2 + a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 2*a^2*sin(d*
x + c)^2/(cos(d*x + c) + 1)^2 + 2*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + a^2*sin(d*x + c)^4/(cos(d*x + c) +
1)^4 + a^2*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) + 5*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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Fricas [A]  time = 1.69242, size = 259, normalized size = 3.75 \begin{align*} -\frac{\cos \left (d x + c\right )^{3} + 5 \, d x +{\left (5 \, d x + 7\right )} \cos \left (d x + c\right ) + 4 \, \cos \left (d x + c\right )^{2} +{\left (5 \, d x - \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) + 4}{2 \,{\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)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/2*(cos(d*x + c)^3 + 5*d*x + (5*d*x + 7)*cos(d*x + c) + 4*cos(d*x + c)^2 + (5*d*x - cos(d*x + c)^2 + 3*cos(d
*x + c) - 4)*sin(d*x + c) + 4)/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) + a^2*d)

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Sympy [A]  time = 28.2271, size = 1358, normalized size = 19.68 \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))**2,x)

[Out]

Piecewise((-35*d*x*tan(c/2 + d*x/2)**5/(14*a**2*d*tan(c/2 + d*x/2)**5 + 14*a**2*d*tan(c/2 + d*x/2)**4 + 28*a**
2*d*tan(c/2 + d*x/2)**3 + 28*a**2*d*tan(c/2 + d*x/2)**2 + 14*a**2*d*tan(c/2 + d*x/2) + 14*a**2*d) - 35*d*x*tan
(c/2 + d*x/2)**4/(14*a**2*d*tan(c/2 + d*x/2)**5 + 14*a**2*d*tan(c/2 + d*x/2)**4 + 28*a**2*d*tan(c/2 + d*x/2)**
3 + 28*a**2*d*tan(c/2 + d*x/2)**2 + 14*a**2*d*tan(c/2 + d*x/2) + 14*a**2*d) - 70*d*x*tan(c/2 + d*x/2)**3/(14*a
**2*d*tan(c/2 + d*x/2)**5 + 14*a**2*d*tan(c/2 + d*x/2)**4 + 28*a**2*d*tan(c/2 + d*x/2)**3 + 28*a**2*d*tan(c/2
+ d*x/2)**2 + 14*a**2*d*tan(c/2 + d*x/2) + 14*a**2*d) - 70*d*x*tan(c/2 + d*x/2)**2/(14*a**2*d*tan(c/2 + d*x/2)
**5 + 14*a**2*d*tan(c/2 + d*x/2)**4 + 28*a**2*d*tan(c/2 + d*x/2)**3 + 28*a**2*d*tan(c/2 + d*x/2)**2 + 14*a**2*
d*tan(c/2 + d*x/2) + 14*a**2*d) - 35*d*x*tan(c/2 + d*x/2)/(14*a**2*d*tan(c/2 + d*x/2)**5 + 14*a**2*d*tan(c/2 +
d*x/2)**4 + 28*a**2*d*tan(c/2 + d*x/2)**3 + 28*a**2*d*tan(c/2 + d*x/2)**2 + 14*a**2*d*tan(c/2 + d*x/2) + 14*a
**2*d) - 35*d*x/(14*a**2*d*tan(c/2 + d*x/2)**5 + 14*a**2*d*tan(c/2 + d*x/2)**4 + 28*a**2*d*tan(c/2 + d*x/2)**3
+ 28*a**2*d*tan(c/2 + d*x/2)**2 + 14*a**2*d*tan(c/2 + d*x/2) + 14*a**2*d) + 30*tan(c/2 + d*x/2)**5/(14*a**2*d
*tan(c/2 + d*x/2)**5 + 14*a**2*d*tan(c/2 + d*x/2)**4 + 28*a**2*d*tan(c/2 + d*x/2)**3 + 28*a**2*d*tan(c/2 + d*x
/2)**2 + 14*a**2*d*tan(c/2 + d*x/2) + 14*a**2*d) - 40*tan(c/2 + d*x/2)**4/(14*a**2*d*tan(c/2 + d*x/2)**5 + 14*
a**2*d*tan(c/2 + d*x/2)**4 + 28*a**2*d*tan(c/2 + d*x/2)**3 + 28*a**2*d*tan(c/2 + d*x/2)**2 + 14*a**2*d*tan(c/2
+ d*x/2) + 14*a**2*d) - 10*tan(c/2 + d*x/2)**3/(14*a**2*d*tan(c/2 + d*x/2)**5 + 14*a**2*d*tan(c/2 + d*x/2)**4
+ 28*a**2*d*tan(c/2 + d*x/2)**3 + 28*a**2*d*tan(c/2 + d*x/2)**2 + 14*a**2*d*tan(c/2 + d*x/2) + 14*a**2*d) - 9
4*tan(c/2 + d*x/2)**2/(14*a**2*d*tan(c/2 + d*x/2)**5 + 14*a**2*d*tan(c/2 + d*x/2)**4 + 28*a**2*d*tan(c/2 + d*x
/2)**3 + 28*a**2*d*tan(c/2 + d*x/2)**2 + 14*a**2*d*tan(c/2 + d*x/2) + 14*a**2*d) - 12*tan(c/2 + d*x/2)/(14*a**
2*d*tan(c/2 + d*x/2)**5 + 14*a**2*d*tan(c/2 + d*x/2)**4 + 28*a**2*d*tan(c/2 + d*x/2)**3 + 28*a**2*d*tan(c/2 +
d*x/2)**2 + 14*a**2*d*tan(c/2 + d*x/2) + 14*a**2*d) - 82/(14*a**2*d*tan(c/2 + d*x/2)**5 + 14*a**2*d*tan(c/2 +
d*x/2)**4 + 28*a**2*d*tan(c/2 + d*x/2)**3 + 28*a**2*d*tan(c/2 + d*x/2)**2 + 14*a**2*d*tan(c/2 + d*x/2) + 14*a*
*2*d), Ne(d, 0)), (x*sin(c)**2*cos(c)**2/(a*sin(c) + a)**2, True))

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Giac [A]  time = 1.42527, size = 123, normalized size = 1.78 \begin{align*} -\frac{\frac{5 \,{\left (d x + c\right )}}{a^{2}} + \frac{2 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac{8}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}}}{2 \, 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))^2,x, algorithm="giac")

[Out]

-1/2*(5*(d*x + c)/a^2 + 2*(tan(1/2*d*x + 1/2*c)^3 + 4*tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c) + 4)/((tan
(1/2*d*x + 1/2*c)^2 + 1)^2*a^2) + 8/(a^2*(tan(1/2*d*x + 1/2*c) + 1)))/d