3.243 \(\int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=60 \[ \frac {2}{d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {\log (\sin (c+d x)+1)}{a^3 d}-\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 60, 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} \[ \frac {2}{d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {\log (\sin (c+d x)+1)}{a^3 d}-\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]

[Out]

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

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

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Mathematica [A]  time = 0.65, size = 65, normalized size = 1.08 \[ \frac {\frac {16 \sin (c+d x)}{(\sin (c+d x)+1)^2}+8 \log (\sin (c+d x)+1)+\frac {12}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}}{8 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]

[Out]

(8*Log[1 + Sin[c + d*x]] + 12/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + (16*Sin[c + d*x])/(1 + Sin[c + d*x])^2
)/(8*a^3*d)

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fricas [A]  time = 0.53, size = 75, normalized size = 1.25 \[ \frac {2 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, \sin \left (d x + c\right ) - 3}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 45, normalized size = 0.75 \[ \frac {\frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} + \frac {4 \, \sin \left (d x + c\right ) + 3}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.22, size = 54, normalized size = 0.90 \[ -\frac {1}{2 d \,a^{3} \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{a^{3} d}+\frac {2}{d \,a^{3} \left (1+\sin \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x)

[Out]

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

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maxima [A]  time = 0.41, size = 60, normalized size = 1.00 \[ \frac {\frac {4 \, \sin \left (d x + c\right ) + 3}{a^{3} \sin \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) + a^{3}} + \frac {2 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.06, size = 44, normalized size = 0.73 \[ \frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3\,d}+\frac {2\,\sin \left (c+d\,x\right )+\frac {3}{2}}{a^3\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*sin(c + d*x)^2)/(a + a*sin(c + d*x))^3,x)

[Out]

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

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sympy [A]  time = 1.93, size = 257, normalized size = 4.28 \[ \begin {cases} \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {4 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {4 \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {3}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{2}{\relax (c )} \cos {\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*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) + 4*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) + 2*lo
g(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) + 4*sin(c + d*x)/(2*a**3*d*s
in(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) + 3/(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)**2*cos(c)/(a*sin(c) + a)**3, True))

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