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

Optimal. Leaf size=72 \[ \frac{a^2-b^2}{2 b^3 d (a+b \sin (c+d x))^2}-\frac{2 a}{b^3 d (a+b \sin (c+d x))}-\frac{\log (a+b \sin (c+d x))}{b^3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.071842, antiderivative size = 72, 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} \[ \frac{a^2-b^2}{2 b^3 d (a+b \sin (c+d x))^2}-\frac{2 a}{b^3 d (a+b \sin (c+d x))}-\frac{\log (a+b \sin (c+d x))}{b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.127896, size = 55, normalized size = 0.76 \[ -\frac{\frac{3 a^2+4 a b \sin (c+d x)+b^2}{2 (a+b \sin (c+d x))^2}+\log (a+b \sin (c+d x))}{b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Log[a + b*Sin[c + d*x]] + (3*a^2 + b^2 + 4*a*b*Sin[c + d*x])/(2*(a + b*Sin[c + d*x])^2))/(b^3*d))

________________________________________________________________________________________

Maple [A]  time = 0.092, size = 85, normalized size = 1.2 \begin{align*} -{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{{b}^{3}d}}+{\frac{{a}^{2}}{2\,{b}^{3}d \left ( a+b\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{1}{2\,bd \left ( a+b\sin \left ( dx+c \right ) \right ) ^{2}}}-2\,{\frac{a}{{b}^{3}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)^3/(a+b*sin(d*x+c))^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.954816, size = 103, normalized size = 1.43 \begin{align*} -\frac{\frac{4 \, a b \sin \left (d x + c\right ) + 3 \, a^{2} + b^{2}}{b^{5} \sin \left (d x + c\right )^{2} + 2 \, a b^{4} \sin \left (d x + c\right ) + a^{2} b^{3}} + \frac{2 \, \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{3}}}{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))^3,x, algorithm="maxima")

[Out]

-1/2*((4*a*b*sin(d*x + c) + 3*a^2 + b^2)/(b^5*sin(d*x + c)^2 + 2*a*b^4*sin(d*x + c) + a^2*b^3) + 2*log(b*sin(d
*x + c) + a)/b^3)/d

________________________________________________________________________________________

Fricas [A]  time = 2.77337, size = 257, normalized size = 3.57 \begin{align*} \frac{4 \, a b \sin \left (d x + c\right ) + 3 \, a^{2} + b^{2} - 2 \,{\left (b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{2 \,{\left (b^{5} d \cos \left (d x + c\right )^{2} - 2 \, a b^{4} d \sin \left (d x + c\right ) -{\left (a^{2} b^{3} + b^{5}\right )} d\right )}} \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))^3,x, algorithm="fricas")

[Out]

1/2*(4*a*b*sin(d*x + c) + 3*a^2 + b^2 - 2*(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)*log(b*sin(d*x
+ c) + a))/(b^5*d*cos(d*x + c)^2 - 2*a*b^4*d*sin(d*x + c) - (a^2*b^3 + b^5)*d)

________________________________________________________________________________________

Sympy [A]  time = 2.74073, size = 670, normalized size = 9.31 \begin{align*} \text{result too large to display} \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))**3,x)

[Out]

Piecewise((zoo*x*cos(c)**3/sin(c)**3, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((-log(sin(c + d*x))/d - cos(c + d*x)**
2/(2*d*sin(c + d*x)**2))/b**3, Eq(a, 0)), ((2*sin(c + d*x)**3/(3*d) + sin(c + d*x)*cos(c + d*x)**2/d)/a**3, Eq
(b, 0)), (x*cos(c)**3/(a + b*sin(c))**3, Eq(d, 0)), (-2*a**4*log(a/b + sin(c + d*x))/(2*a**4*b**3*d + 4*a**3*b
**4*d*sin(c + d*x) + 2*a**2*b**5*d*sin(c + d*x)**2) - a**4/(2*a**4*b**3*d + 4*a**3*b**4*d*sin(c + d*x) + 2*a**
2*b**5*d*sin(c + d*x)**2) - 4*a**3*b*log(a/b + sin(c + d*x))*sin(c + d*x)/(2*a**4*b**3*d + 4*a**3*b**4*d*sin(c
+ d*x) + 2*a**2*b**5*d*sin(c + d*x)**2) - 2*a**2*b**2*log(a/b + sin(c + d*x))*sin(c + d*x)**2/(2*a**4*b**3*d
+ 4*a**3*b**4*d*sin(c + d*x) + 2*a**2*b**5*d*sin(c + d*x)**2) + 2*a**2*b**2*sin(c + d*x)**2/(2*a**4*b**3*d + 4
*a**3*b**4*d*sin(c + d*x) + 2*a**2*b**5*d*sin(c + d*x)**2) + 2*a*b**3*sin(c + d*x)**3/(2*a**4*b**3*d + 4*a**3*
b**4*d*sin(c + d*x) + 2*a**2*b**5*d*sin(c + d*x)**2) + 2*a*b**3*sin(c + d*x)*cos(c + d*x)**2/(2*a**4*b**3*d +
4*a**3*b**4*d*sin(c + d*x) + 2*a**2*b**5*d*sin(c + d*x)**2) + b**4*sin(c + d*x)**4/(2*a**4*b**3*d + 4*a**3*b**
4*d*sin(c + d*x) + 2*a**2*b**5*d*sin(c + d*x)**2) + b**4*sin(c + d*x)**2*cos(c + d*x)**2/(2*a**4*b**3*d + 4*a*
*3*b**4*d*sin(c + d*x) + 2*a**2*b**5*d*sin(c + d*x)**2), True))

________________________________________________________________________________________

Giac [A]  time = 1.12366, size = 84, normalized size = 1.17 \begin{align*} -\frac{\frac{2 \, \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{3}} + \frac{4 \, a \sin \left (d x + c\right ) + \frac{3 \, a^{2} + b^{2}}{b}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2} b^{2}}}{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))^3,x, algorithm="giac")

[Out]

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