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

Optimal. Leaf size=74 \[ \frac{\sin (c+d x)}{a^3 d}-\frac{3}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{3 \log (\sin (c+d x)+1)}{a^3 d}+\frac{1}{2 a d (a \sin (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.0896497, antiderivative size = 74, 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} \[ \frac{\sin (c+d x)}{a^3 d}-\frac{3}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{3 \log (\sin (c+d x)+1)}{a^3 d}+\frac{1}{2 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.402829, size = 70, normalized size = 0.95 \[ \frac{\frac{\sin ^2(c+d x)}{(\sin (c+d x)+1)^2}+4 \sin (c+d x)+\frac{-10 \sin (c+d x)-9}{(\sin (c+d x)+1)^2}-12 \log (\sin (c+d x)+1)}{4 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*Log[1 + Sin[c + d*x]] + 4*Sin[c + d*x] + (-9 - 10*Sin[c + d*x])/(1 + Sin[c + d*x])^2 + Sin[c + d*x]^2/(1
+ Sin[c + d*x])^2)/(4*a^3*d)

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Maple [A]  time = 0.036, size = 68, normalized size = 0.9 \begin{align*}{\frac{\sin \left ( dx+c \right ) }{{a}^{3}d}}+{\frac{1}{2\,{a}^{3}d \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-3\,{\frac{1}{{a}^{3}d \left ( 1+\sin \left ( dx+c \right ) \right ) }}-3\,{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{{a}^{3}d}} \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))^3,x)

[Out]

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

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Maxima [A]  time = 1.12152, size = 96, normalized size = 1.3 \begin{align*} -\frac{\frac{6 \, \sin \left (d x + c\right ) + 5}{a^{3} \sin \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) + a^{3}} + \frac{6 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} - \frac{2 \, \sin \left (d x + c\right )}{a^{3}}}{2 \, 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))^3,x, algorithm="maxima")

[Out]

-1/2*((6*sin(d*x + c) + 5)/(a^3*sin(d*x + c)^2 + 2*a^3*sin(d*x + c) + a^3) + 6*log(sin(d*x + c) + 1)/a^3 - 2*s
in(d*x + c)/a^3)/d

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Fricas [A]  time = 1.41297, size = 251, normalized size = 3.39 \begin{align*} \frac{4 \, \cos \left (d x + c\right )^{2} - 6 \,{\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (\cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) + 1}{2 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \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))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.69133, size = 303, normalized size = 4.09 \begin{align*} \begin{cases} - \frac{6 \log{\left (\sin{\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} - \frac{12 \log{\left (\sin{\left (c + d x \right )} + 1 \right )} \sin{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} - \frac{6 \log{\left (\sin{\left (c + d x \right )} + 1 \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} + \frac{2 \sin ^{3}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} - \frac{12 \sin{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} - \frac{9}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} & \text{for}\: d \neq 0 \\\frac{x \sin ^{3}{\left (c \right )} \cos{\left (c \right )}}{\left (a \sin{\left (c \right )} + a\right )^{3}} & \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))**3,x)

[Out]

Piecewise((-6*log(sin(c + d*x) + 1)*sin(c + d*x)**2/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3
*d) - 12*log(sin(c + d*x) + 1)*sin(c + d*x)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) - 6*
log(sin(c + d*x) + 1)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) + 2*sin(c + d*x)**3/(2*a**
3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) - 12*sin(c + d*x)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d
*sin(c + d*x) + 2*a**3*d) - 9/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d), Ne(d, 0)), (x*sin
(c)**3*cos(c)/(a*sin(c) + a)**3, True))

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Giac [A]  time = 1.2596, size = 76, normalized size = 1.03 \begin{align*} -\frac{\frac{6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac{2 \, \sin \left (d x + c\right )}{a^{3}} + \frac{6 \, \sin \left (d x + c\right ) + 5}{a^{3}{\left (\sin \left (d x + c\right ) + 1\right )}^{2}}}{2 \, 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))^3,x, algorithm="giac")

[Out]

-1/2*(6*log(abs(sin(d*x + c) + 1))/a^3 - 2*sin(d*x + c)/a^3 + (6*sin(d*x + c) + 5)/(a^3*(sin(d*x + c) + 1)^2))
/d