3.1555 \(\int \frac{\cos (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=48 \[ \frac{B \log (a+b \sin (c+d x))}{b^2 d}-\frac{A b-a B}{b^2 d (a+b \sin (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.0767375, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {2833, 43} \[ \frac{B \log (a+b \sin (c+d x))}{b^2 d}-\frac{A b-a B}{b^2 d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0868795, size = 42, normalized size = 0.88 \[ \frac{\frac{a B-A b}{a+b \sin (c+d x)}+B \log (a+b \sin (c+d x))}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.056, size = 63, normalized size = 1.3 \begin{align*}{\frac{B\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{{b}^{2}d}}-{\frac{A}{bd \left ( a+b\sin \left ( dx+c \right ) \right ) }}+{\frac{aB}{{b}^{2}d \left ( a+b\sin \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.992207, size = 65, normalized size = 1.35 \begin{align*} \frac{\frac{B a - A b}{b^{3} \sin \left (d x + c\right ) + a b^{2}} + \frac{B \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.42007, size = 128, normalized size = 2.67 \begin{align*} \frac{B a - A b +{\left (B b \sin \left (d x + c\right ) + B a\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{3} d \sin \left (d x + c\right ) + a b^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.27253, size = 178, normalized size = 3.71 \begin{align*} \begin{cases} \frac{x \left (A + B \sin{\left (c \right )}\right ) \cos{\left (c \right )}}{a^{2}} & \text{for}\: b = 0 \wedge d = 0 \\\frac{\frac{A \sin{\left (c + d x \right )}}{d} - \frac{B \cos ^{2}{\left (c + d x \right )}}{2 d}}{a^{2}} & \text{for}\: b = 0 \\\frac{x \left (A + B \sin{\left (c \right )}\right ) \cos{\left (c \right )}}{\left (a + b \sin{\left (c \right )}\right )^{2}} & \text{for}\: d = 0 \\- \frac{A b}{a b^{2} d + b^{3} d \sin{\left (c + d x \right )}} + \frac{B a \log{\left (\frac{a}{b} + \sin{\left (c + d x \right )} \right )}}{a b^{2} d + b^{3} d \sin{\left (c + d x \right )}} + \frac{B a}{a b^{2} d + b^{3} d \sin{\left (c + d x \right )}} + \frac{B b \log{\left (\frac{a}{b} + \sin{\left (c + d x \right )} \right )} \sin{\left (c + d x \right )}}{a b^{2} d + b^{3} d \sin{\left (c + d x \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))**2,x)

[Out]

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

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Giac [A]  time = 1.35983, size = 108, normalized size = 2.25 \begin{align*} -\frac{\frac{B{\left (\frac{\log \left (\frac{{\left | b \sin \left (d x + c\right ) + a \right |}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}{\left | b \right |}}\right )}{b} - \frac{a}{{\left (b \sin \left (d x + c\right ) + a\right )} b}\right )}}{b} + \frac{A}{{\left (b \sin \left (d x + c\right ) + a\right )} b}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*(log(abs(b*sin(d*x + c) + a)/((b*sin(d*x + c) + a)^2*abs(b)))/b - a/((b*sin(d*x + c) + a)*b))/b + A/((b*si
n(d*x + c) + a)*b))/d