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

Optimal. Leaf size=49 \[ \frac{\sin ^2(c+d x)}{2 a d}-\frac{\sin (c+d x)}{a d}+\frac{\log (\sin (c+d x)+1)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.0699884, antiderivative size = 49, 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} \[ \frac{\sin ^2(c+d x)}{2 a d}-\frac{\sin (c+d x)}{a d}+\frac{\log (\sin (c+d x)+1)}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*Sin[c + d*x]^2)/(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)

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

Mathematica [A]  time = 0.0591142, size = 38, normalized size = 0.78 \[ \frac{\sin ^2(c+d x)-2 \sin (c+d x)+2 \log (\sin (c+d x)+1)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.02, size = 48, 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}} \end{align*}

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)),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

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Maxima [A]  time = 0.976022, size = 55, normalized size = 1.12 \begin{align*} \frac{\frac{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )}{a} + \frac{2 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a}}{2 \, d} \end{align*}

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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.33008, size = 97, normalized size = 1.98 \begin{align*} -\frac{\cos \left (d x + c\right )^{2} - 2 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, \sin \left (d x + c\right )}{2 \, a d} \end{align*}

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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.45616, size = 53, normalized size = 1.08 \begin{align*} \begin{cases} \frac{\log{\left (\sin{\left (c + d x \right )} + 1 \right )}}{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 ^{2}{\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)**2/(a+a*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.25012, size = 61, normalized size = 1.24 \begin{align*} \frac{\frac{2 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} + \frac{a \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{a^{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)^2/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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