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]

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

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Rubi [A]  time = 0.07, antiderivative size = 72, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.12, 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))

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fricas [A]  time = 0.45, size = 110, normalized size = 1.53 \[ \frac {4 \, a b \sin \left (d x + c\right ) + 3 \, a^{2} + b^{2} - 2 \, {\left (b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{2 \, {\left (b^{5} d \cos \left (d x + c\right )^{2} - 2 \, a b^{4} d \sin \left (d x + c\right ) - {\left (a^{2} b^{3} + b^{5}\right )} d\right )}} \]

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)

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giac [A]  time = 1.02, size = 62, normalized size = 0.86 \[ -\frac {\frac {2 \, \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{3}} + \frac {4 \, a \sin \left (d x + c\right ) + \frac {3 \, a^{2} + b^{2}}{b}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2} b^{2}}}{2 \, d} \]

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

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maple [A]  time = 0.26, size = 85, normalized size = 1.18 \[ -\frac {\ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} d}-\frac {2 a}{b^{3} d \left (a +b \sin \left (d x +c \right )\right )}+\frac {a^{2}}{2 d \,b^{3} \left (a +b \sin \left (d x +c \right )\right )^{2}}-\frac {1}{2 b d \left (a +b \sin \left (d x +c \right )\right )^{2}} \]

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

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maxima [A]  time = 0.32, size = 76, normalized size = 1.06 \[ -\frac {\frac {4 \, a b \sin \left (d x + c\right ) + 3 \, a^{2} + b^{2}}{b^{5} \sin \left (d x + c\right )^{2} + 2 \, a b^{4} \sin \left (d x + c\right ) + a^{2} b^{3}} + \frac {2 \, \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{3}}}{2 \, d} \]

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

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mupad [B]  time = 0.09, size = 80, normalized size = 1.11 \[ -\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{b^3\,d}-\frac {\frac {3\,a^2+b^2}{2\,b^3}+\frac {2\,a\,\sin \left (c+d\,x\right )}{b^2}}{d\,\left (a^2+2\,a\,b\,\sin \left (c+d\,x\right )+b^2\,{\sin \left (c+d\,x\right )}^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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sympy [A]  time = 2.39, size = 398, normalized size = 5.53 \[ \begin {cases} \frac {x \cos ^{3}{\relax (c )}}{a^{3}} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\frac {2 \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {\sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d}}{a^{3}} & \text {for}\: b = 0 \\\frac {x \cos ^{3}{\relax (c )}}{\left (a + b \sin {\relax (c )}\right )^{3}} & \text {for}\: d = 0 \\- \frac {2 a^{2} \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )}}{2 a^{2} b^{3} d + 4 a b^{4} d \sin {\left (c + d x \right )} + 2 b^{5} d \sin ^{2}{\left (c + d x \right )}} - \frac {2 a^{2}}{2 a^{2} b^{3} d + 4 a b^{4} d \sin {\left (c + d x \right )} + 2 b^{5} d \sin ^{2}{\left (c + d x \right )}} - \frac {4 a b \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )} \sin {\left (c + d x \right )}}{2 a^{2} b^{3} d + 4 a b^{4} d \sin {\left (c + d x \right )} + 2 b^{5} d \sin ^{2}{\left (c + d x \right )}} - \frac {2 a b \sin {\left (c + d x \right )}}{2 a^{2} b^{3} d + 4 a b^{4} d \sin {\left (c + d x \right )} + 2 b^{5} d \sin ^{2}{\left (c + d x \right )}} - \frac {2 b^{2} \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )} \sin ^{2}{\left (c + d x \right )}}{2 a^{2} b^{3} d + 4 a b^{4} d \sin {\left (c + d x \right )} + 2 b^{5} d \sin ^{2}{\left (c + d x \right )}} - \frac {b^{2} \cos ^{2}{\left (c + d x \right )}}{2 a^{2} b^{3} d + 4 a b^{4} d \sin {\left (c + d x \right )} + 2 b^{5} d \sin ^{2}{\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

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

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