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

Optimal. Leaf size=63 \[ \frac{a^2-b^2}{b^3 d (a+b \sin (c+d x))}+\frac{2 a \log (a+b \sin (c+d x))}{b^3 d}-\frac{\sin (c+d x)}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.0651441, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2668, 697} \[ \frac{a^2-b^2}{b^3 d (a+b \sin (c+d x))}+\frac{2 a \log (a+b \sin (c+d x))}{b^3 d}-\frac{\sin (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.0376845, size = 52, normalized size = 0.83 \[ \frac{\frac{(a-b) (a+b)}{a+b \sin (c+d x)}+2 a \log (a+b \sin (c+d x))-b \sin (c+d x)}{b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.075, size = 78, normalized size = 1.2 \begin{align*} -{\frac{\sin \left ( dx+c \right ) }{{b}^{2}d}}+2\,{\frac{a\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{{b}^{3}d}}+{\frac{{a}^{2}}{{b}^{3}d \left ( a+b\sin \left ( dx+c \right ) \right ) }}-{\frac{1}{bd \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)^3/(a+b*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 0.943103, size = 82, normalized size = 1.3 \begin{align*} \frac{\frac{a^{2} - b^{2}}{b^{4} \sin \left (d x + c\right ) + a b^{3}} + \frac{2 \, a \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{3}} - \frac{\sin \left (d x + c\right )}{b^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.805, size = 188, normalized size = 2.98 \begin{align*} \frac{b^{2} \cos \left (d x + c\right )^{2} - a b \sin \left (d x + c\right ) + a^{2} - 2 \, b^{2} + 2 \,{\left (a b \sin \left (d x + c\right ) + a^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{4} d \sin \left (d x + c\right ) + a b^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.13483, size = 316, normalized size = 5.02 \begin{align*} \begin{cases} \frac{\tilde{\infty } x \cos ^{3}{\left (c \right )}}{\sin ^{2}{\left (c \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac{- \frac{2 \sin{\left (c + d x \right )}}{d} - \frac{\cos ^{2}{\left (c + d x \right )}}{d \sin{\left (c + d x \right )}}}{b^{2}} & \text{for}\: a = 0 \\\frac{\frac{2 \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{\sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d}}{a^{2}} & \text{for}\: b = 0 \\\frac{x \cos ^{3}{\left (c \right )}}{\left (a + b \sin{\left (c \right )}\right )^{2}} & \text{for}\: d = 0 \\\frac{2 a^{3} \log{\left (\frac{a}{b} + \sin{\left (c + d x \right )} \right )}}{a^{2} b^{3} d + a b^{4} d \sin{\left (c + d x \right )}} + \frac{2 a^{3}}{a^{2} b^{3} d + a b^{4} d \sin{\left (c + d x \right )}} + \frac{2 a^{2} b \log{\left (\frac{a}{b} + \sin{\left (c + d x \right )} \right )} \sin{\left (c + d x \right )}}{a^{2} b^{3} d + a b^{4} d \sin{\left (c + d x \right )}} - \frac{a b^{2} \sin ^{2}{\left (c + d x \right )}}{a^{2} b^{3} d + a b^{4} d \sin{\left (c + d x \right )}} + \frac{b^{3} \sin ^{3}{\left (c + d x \right )}}{a^{2} b^{3} d + a b^{4} d \sin{\left (c + d x \right )}} + \frac{b^{3} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{a^{2} b^{3} d + a b^{4} 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)**3/(a+b*sin(d*x+c))**2,x)

[Out]

Piecewise((zoo*x*cos(c)**3/sin(c)**2, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((-2*sin(c + d*x)/d - cos(c + d*x)**2/(
d*sin(c + d*x)))/b**2, Eq(a, 0)), ((2*sin(c + d*x)**3/(3*d) + sin(c + d*x)*cos(c + d*x)**2/d)/a**2, Eq(b, 0)),
(x*cos(c)**3/(a + b*sin(c))**2, Eq(d, 0)), (2*a**3*log(a/b + sin(c + d*x))/(a**2*b**3*d + a*b**4*d*sin(c + d*
x)) + 2*a**3/(a**2*b**3*d + a*b**4*d*sin(c + d*x)) + 2*a**2*b*log(a/b + sin(c + d*x))*sin(c + d*x)/(a**2*b**3*
d + a*b**4*d*sin(c + d*x)) - a*b**2*sin(c + d*x)**2/(a**2*b**3*d + a*b**4*d*sin(c + d*x)) + b**3*sin(c + d*x)*
*3/(a**2*b**3*d + a*b**4*d*sin(c + d*x)) + b**3*sin(c + d*x)*cos(c + d*x)**2/(a**2*b**3*d + a*b**4*d*sin(c + d
*x)), True))

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Giac [A]  time = 1.10733, size = 123, normalized size = 1.95 \begin{align*} -\frac{\frac{2 \, a \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^{3}} + \frac{b \sin \left (d x + c\right ) + a}{b^{3}} - \frac{a^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{3}} + \frac{1}{{\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)^3/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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