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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 62, normalized size of antiderivative = 1.00, 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 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]

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 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]

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 = 0.09, 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)

________________________________________________________________________________________

fricas [A]  time = 0.49, size = 45, normalized size = 0.73 \[ -\frac {2 \, \cos \left (d x + c\right )^{3} - 3 \, d x + 3 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, \cos \left (d x + c\right )}{6 \, a d} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.14, size = 75, normalized size = 1.21 \[ \frac {\frac {3 \, {\left (d x + c\right )}}{a} + \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a}}{6 \, d} \]

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

________________________________________________________________________________________

maple [B]  time = 0.28, size = 141, normalized size = 2.27 \[ \frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4}{3 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d} \]

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

________________________________________________________________________________________

maxima [B]  time = 0.42, size = 156, normalized size = 2.52 \[ -\frac {\frac {\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {12 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 4}{a + \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{3 \, d} \]

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

________________________________________________________________________________________

mupad [B]  time = 10.46, size = 66, normalized size = 1.06 \[ \frac {x}{2\,a}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {4}{3}}{a\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 11.35, size = 563, normalized size = 9.08 \[ \begin {cases} \frac {3 d x \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} + \frac {9 d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} + \frac {9 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} + \frac {3 d x}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} + \frac {6 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} + \frac {24 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} - \frac {6 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} + \frac {8}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{2}{\relax (c )} \cos ^{2}{\relax (c )}}{a \sin {\relax (c )} + a} & \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)**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))

________________________________________________________________________________________