3.1279 \(\int \frac{\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=76 \[ \frac{a^2 \sin (c+d x)}{b^3 d}-\frac{a^3 \log (a+b \sin (c+d x))}{b^4 d}-\frac{a \sin ^2(c+d x)}{2 b^2 d}+\frac{\sin ^3(c+d x)}{3 b d} \]

[Out]

-((a^3*Log[a + b*Sin[c + d*x]])/(b^4*d)) + (a^2*Sin[c + d*x])/(b^3*d) - (a*Sin[c + d*x]^2)/(2*b^2*d) + Sin[c +
d*x]^3/(3*b*d)

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Rubi [A]  time = 0.0922599, antiderivative size = 76, 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} \[ \frac{a^2 \sin (c+d x)}{b^3 d}-\frac{a^3 \log (a+b \sin (c+d x))}{b^4 d}-\frac{a \sin ^2(c+d x)}{2 b^2 d}+\frac{\sin ^3(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.201359, size = 66, normalized size = 0.87 \[ \frac{6 a^2 b \sin (c+d x)-6 a^3 \log (a+b \sin (c+d x))-3 a b^2 \sin ^2(c+d x)+2 b^3 \sin ^3(c+d x)}{6 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a^3*Log[a + b*Sin[c + d*x]] + 6*a^2*b*Sin[c + d*x] - 3*a*b^2*Sin[c + d*x]^2 + 2*b^3*Sin[c + d*x]^3)/(6*b^4
*d)

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Maple [A]  time = 0.027, size = 73, normalized size = 1. \begin{align*} -{\frac{{a}^{3}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{{b}^{4}d}}+{\frac{{a}^{2}\sin \left ( dx+c \right ) }{{b}^{3}d}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}a}{2\,{b}^{2}d}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,bd}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^3*ln(a+b*sin(d*x+c))/b^4/d+a^2*sin(d*x+c)/b^3/d-1/2*a*sin(d*x+c)^2/b^2/d+1/3*sin(d*x+c)^3/b/d

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Maxima [A]  time = 0.975544, size = 90, normalized size = 1.18 \begin{align*} -\frac{\frac{6 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{4}} - \frac{2 \, b^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{2} + 6 \, a^{2} \sin \left (d x + c\right )}{b^{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+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.46896, size = 167, normalized size = 2.2 \begin{align*} \frac{3 \, a b^{2} \cos \left (d x + c\right )^{2} - 6 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \,{\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{6 \, b^{4} 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+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.56747, size = 128, normalized size = 1.68 \begin{align*} \begin{cases} \frac{x \sin ^{3}{\left (c \right )} \cos{\left (c \right )}}{a} & \text{for}\: b = 0 \wedge d = 0 \\\frac{x \sin ^{3}{\left (c \right )} \cos{\left (c \right )}}{a + b \sin{\left (c \right )}} & \text{for}\: d = 0 \\\frac{- \frac{\sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} - \frac{\cos ^{4}{\left (c + d x \right )}}{4 d}}{a} & \text{for}\: b = 0 \\- \frac{a^{3} \log{\left (\frac{a}{b} + \sin{\left (c + d x \right )} \right )}}{b^{4} d} + \frac{a^{2} \sin{\left (c + d x \right )}}{b^{3} d} + \frac{a \cos ^{2}{\left (c + d x \right )}}{2 b^{2} d} + \frac{\sin ^{3}{\left (c + d x \right )}}{3 b d} & \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+b*sin(d*x+c)),x)

[Out]

Piecewise((x*sin(c)**3*cos(c)/a, Eq(b, 0) & Eq(d, 0)), (x*sin(c)**3*cos(c)/(a + b*sin(c)), Eq(d, 0)), ((-sin(c
+ d*x)**2*cos(c + d*x)**2/(2*d) - cos(c + d*x)**4/(4*d))/a, Eq(b, 0)), (-a**3*log(a/b + sin(c + d*x))/(b**4*d
) + a**2*sin(c + d*x)/(b**3*d) + a*cos(c + d*x)**2/(2*b**2*d) + sin(c + d*x)**3/(3*b*d), True))

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Giac [A]  time = 1.1695, size = 92, normalized size = 1.21 \begin{align*} -\frac{\frac{6 \, a^{3} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{4}} - \frac{2 \, b^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{2} + 6 \, a^{2} \sin \left (d x + c\right )}{b^{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+b*sin(d*x+c)),x, algorithm="giac")

[Out]

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