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]

-ln(1+sin(d*x+c))/a/d+sin(d*x+c)/a/d-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 = 67, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.11, 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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fricas [A]  time = 0.53, size = 48, normalized size = 0.72 \[ \frac {3 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 6 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{6 \, a d} \]

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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giac [A]  time = 0.16, size = 64, normalized size = 0.96 \[ -\frac {\frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{3} - 3 \, a^{2} \sin \left (d x + c\right )^{2} + 6 \, a^{2} \sin \left (d x + c\right )}{a^{3}}}{6 \, d} \]

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

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maple [A]  time = 0.12, size = 64, normalized size = 0.96 \[ -\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{a d}+\frac {\sin \left (d x +c \right )}{a d}-\frac {\sin ^{2}\left (d x +c \right )}{2 a d}+\frac {\sin ^{3}\left (d x +c \right )}{3 d a} \]

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

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maxima [A]  time = 0.31, size = 53, normalized size = 0.79 \[ \frac {\frac {2 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right )}{a} - \frac {6 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a}}{6 \, d} \]

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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mupad [B]  time = 0.06, size = 56, normalized size = 0.84 \[ -\frac {\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{a}-\frac {\sin \left (c+d\,x\right )}{a}+\frac {{\sin \left (c+d\,x\right )}^2}{2\,a}-\frac {{\sin \left (c+d\,x\right )}^3}{3\,a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.96, size = 66, normalized size = 0.99 \[ \begin {cases} - \frac {\log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a d} + \frac {\sin ^{3}{\left (c + d x \right )}}{3 a d} - \frac {\sin ^{2}{\left (c + d x \right )}}{2 a d} + \frac {\sin {\left (c + d x \right )}}{a d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{3}{\relax (c )} \cos {\relax (c )}}{a \sin {\relax (c )} + a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*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)**2/(2*a*d) + sin(c + d*x)/(a*
d), Ne(d, 0)), (x*sin(c)**3*cos(c)/(a*sin(c) + a), True))

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