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

Optimal. Leaf size=37 \[ \frac {\sin ^2(c+d x)}{2 a d}-\frac {\sin ^3(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2835, 2564, 30} \[ \frac {\sin ^2(c+d x)}{2 a d}-\frac {\sin ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2835

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 28, normalized size = 0.76 \[ \frac {(3-2 \sin (c+d x)) \sin ^2(c+d x)}{6 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 37, normalized size = 1.00 \[ -\frac {3 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right )}{6 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 29, normalized size = 0.78 \[ -\frac {2 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2}}{6 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 30, normalized size = 0.81 \[ -\frac {\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 29, normalized size = 0.78 \[ -\frac {2 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2}}{6 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.54, size = 26, normalized size = 0.70 \[ -\frac {{\sin \left (c+d\,x\right )}^2\,\left (2\,\sin \left (c+d\,x\right )-3\right )}{6\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 10.74, size = 224, normalized size = 6.05 \[ \begin {cases} \frac {6 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} - \frac {8 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {6 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} & \text {for}\: d \neq 0 \\\frac {x \sin {\relax (c )} \cos ^{3}{\relax (c )}}{a \sin {\relax (c )} + a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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