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

Optimal. Leaf size=62 \[ -\frac{\cos ^3(c+d x)}{3 a d}+\frac{\cos (c+d x)}{a d}-\frac{\sin (c+d x) \cos (c+d x)}{2 a d}+\frac{x}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.117336, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2839, 2635, 8, 2633} \[ -\frac{\cos ^3(c+d x)}{3 a d}+\frac{\cos (c+d x)}{a d}-\frac{\sin (c+d x) \cos (c+d x)}{2 a d}+\frac{x}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2839

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

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 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

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

Mathematica [A]  time = 0.082831, size = 46, normalized size = 0.74 \[ \frac{-3 \sin (2 (c+d x))+9 \cos (c+d x)-\cos (3 (c+d x))+6 c+6 d x}{12 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*c + 6*d*x + 9*Cos[c + d*x] - Cos[3*(c + d*x)] - 3*Sin[2*(c + d*x)])/(12*a*d)

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Maple [B]  time = 0.072, size = 141, normalized size = 2.3 \begin{align*}{\frac{1}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-{\frac{1}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+{\frac{4}{3\,da} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+{\frac{1}{da}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \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)),x)

[Out]

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

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Maxima [B]  time = 1.65713, size = 211, normalized size = 3.4 \begin{align*} -\frac{\frac{\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{12 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 4}{a + \frac{3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac{3 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{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)),x, algorithm="maxima")

[Out]

-1/3*((3*sin(d*x + c)/(cos(d*x + c) + 1) - 12*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 3*sin(d*x + c)^5/(cos(d*x
+ c) + 1)^5 - 4)/(a + 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a*si
n(d*x + c)^6/(cos(d*x + c) + 1)^6) - 3*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d

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Fricas [A]  time = 1.62002, size = 116, normalized size = 1.87 \begin{align*} -\frac{2 \, \cos \left (d x + c\right )^{3} - 3 \, d x + 3 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, \cos \left (d x + c\right )}{6 \, a 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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 19.6895, size = 563, normalized size = 9.08 \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)),x)

[Out]

Piecewise((3*d*x*tan(c/2 + d*x/2)**6/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2
+ d*x/2)**2 + 6*a*d) + 9*d*x*tan(c/2 + d*x/2)**4/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*
a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 9*d*x*tan(c/2 + d*x/2)**2/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x
/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 3*d*x/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 +
18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 6*tan(c/2 + d*x/2)**5/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x
/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 24*tan(c/2 + d*x/2)**2/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(
c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 6*tan(c/2 + d*x/2)/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d
*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 8/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d
*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d), Ne(d, 0)), (x*sin(c)**2*cos(c)**2/(a*sin(c) + a), True))

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Giac [A]  time = 1.27396, size = 101, normalized size = 1.63 \begin{align*} \frac{\frac{3 \,{\left (d x + c\right )}}{a} + \frac{2 \,{\left (3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, \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 )}^{3} a}}{6 \, 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)),x, algorithm="giac")

[Out]

1/6*(3*(d*x + c)/a + 2*(3*tan(1/2*d*x + 1/2*c)^5 + 12*tan(1/2*d*x + 1/2*c)^2 - 3*tan(1/2*d*x + 1/2*c) + 4)/((t
an(1/2*d*x + 1/2*c)^2 + 1)^3*a))/d