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

Optimal. Leaf size=67 \[ \frac{\sin ^3(c+d x)}{3 a d}-\frac{\sin ^2(c+d x)}{2 a d}+\frac{\sin (c+d x)}{a d}-\frac{\log (\sin (c+d x)+1)}{a d} \]

[Out]

-(Log[1 + Sin[c + d*x]]/(a*d)) + Sin[c + d*x]/(a*d) - Sin[c + d*x]^2/(2*a*d) + Sin[c + d*x]^3/(3*a*d)

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Rubi [A]  time = 0.0781027, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2833, 12, 43} \[ \frac{\sin ^3(c+d x)}{3 a d}-\frac{\sin ^2(c+d x)}{2 a d}+\frac{\sin (c+d x)}{a d}-\frac{\log (\sin (c+d x)+1)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[1 + Sin[c + d*x]]/(a*d)) + Sin[c + d*x]/(a*d) - Sin[c + d*x]^2/(2*a*d) + Sin[c + d*x]^3/(3*a*d)

Rule 2833

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.103932, size = 50, normalized size = 0.75 \[ \frac{2 \sin ^3(c+d x)-3 \sin ^2(c+d x)+6 \sin (c+d x)-6 \log (\sin (c+d x)+1)}{6 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*Log[1 + Sin[c + d*x]] + 6*Sin[c + d*x] - 3*Sin[c + d*x]^2 + 2*Sin[c + d*x]^3)/(6*a*d)

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Maple [A]  time = 0.023, size = 64, normalized size = 1. \begin{align*} -{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{da}}+{\frac{\sin \left ( dx+c \right ) }{da}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,da}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.1263, size = 72, normalized size = 1.07 \begin{align*} \frac{\frac{2 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right )}{a} - \frac{6 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.48379, size = 127, normalized size = 1.9 \begin{align*} \frac{3 \, \cos \left (d x + c\right )^{2} - 2 \,{\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 6 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{6 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*cos(d*x + c)^2 - 2*(cos(d*x + c)^2 - 4)*sin(d*x + c) - 6*log(sin(d*x + c) + 1))/(a*d)

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Sympy [A]  time = 3.40311, size = 66, normalized size = 0.99 \begin{align*} \begin{cases} - \frac{\log{\left (\sin{\left (c + d x \right )} + 1 \right )}}{a d} + \frac{\sin ^{3}{\left (c + d x \right )}}{3 a d} + \frac{\sin{\left (c + d x \right )}}{a d} + \frac{\cos ^{2}{\left (c + d x \right )}}{2 a d} & \text{for}\: d \neq 0 \\\frac{x \sin ^{3}{\left (c \right )} \cos{\left (c \right )}}{a \sin{\left (c \right )} + a} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-log(sin(c + d*x) + 1)/(a*d) + sin(c + d*x)**3/(3*a*d) + sin(c + d*x)/(a*d) + cos(c + d*x)**2/(2*a*
d), Ne(d, 0)), (x*sin(c)**3*cos(c)/(a*sin(c) + a), True))

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Giac [A]  time = 1.23687, size = 86, normalized size = 1.28 \begin{align*} -\frac{\frac{6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac{2 \, a^{2} \sin \left (d x + c\right )^{3} - 3 \, a^{2} \sin \left (d x + c\right )^{2} + 6 \, a^{2} \sin \left (d x + c\right )}{a^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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