3.1280 \(\int \frac{\cos (c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=55 \[ \frac{a^2 \log (a+b \sin (c+d x))}{b^3 d}-\frac{a \sin (c+d x)}{b^2 d}+\frac{\sin ^2(c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.0804525, antiderivative size = 55, 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} \[ \frac{a^2 \log (a+b \sin (c+d x))}{b^3 d}-\frac{a \sin (c+d x)}{b^2 d}+\frac{\sin ^2(c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.103084, size = 49, normalized size = 0.89 \[ \frac{2 a^2 \log (a+b \sin (c+d x))-2 a b \sin (c+d x)+b^2 \sin ^2(c+d x)}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.027, size = 54, normalized size = 1. \begin{align*}{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ){a}^{2}}{d{b}^{3}}}-{\frac{a\sin \left ( dx+c \right ) }{{b}^{2}d}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,bd}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.987241, size = 66, normalized size = 1.2 \begin{align*} \frac{\frac{2 \, a^{2} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{3}} + \frac{b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{b^{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+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.51032, size = 119, normalized size = 2.16 \begin{align*} -\frac{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) + 2 \, a b \sin \left (d x + c\right )}{2 \, b^{3} 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+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.50444, size = 87, normalized size = 1.58 \begin{align*} \begin{cases} \frac{x \sin ^{2}{\left (c \right )} \cos{\left (c \right )}}{a} & \text{for}\: b = 0 \wedge d = 0 \\\frac{x \sin ^{2}{\left (c \right )} \cos{\left (c \right )}}{a + b \sin{\left (c \right )}} & \text{for}\: d = 0 \\\frac{\sin ^{3}{\left (c + d x \right )}}{3 a d} & \text{for}\: b = 0 \\\frac{a^{2} \log{\left (\frac{a}{b} + \sin{\left (c + d x \right )} \right )}}{b^{3} d} - \frac{a \sin{\left (c + d x \right )}}{b^{2} d} - \frac{\cos ^{2}{\left (c + d x \right )}}{2 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)**2/(a+b*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.20535, size = 68, normalized size = 1.24 \begin{align*} \frac{\frac{2 \, a^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{3}} + \frac{b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{b^{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+b*sin(d*x+c)),x, algorithm="giac")

[Out]

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