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

Optimal. Leaf size=39 \[ -\frac {2}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {\log (\sin (c+d x)+1)}{a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 39, 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 = {2667, 43} \[ -\frac {2}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {\log (\sin (c+d x)+1)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[1 + Sin[c + d*x]]/(a^3*d)) - 2/(d*(a^3 + a^3*Sin[c + d*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])

Rule 2667

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 + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 58, normalized size = 1.49 \[ -\frac {\sin (c+d x) \log (\sin (c+d x)+1)+\log (\sin (c+d x)+1)+2}{a^3 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2 + Log[1 + Sin[c + d*x]] + Log[1 + Sin[c + d*x]]*Sin[c + d*x])/(a^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]
)^2))

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fricas [A]  time = 0.80, size = 41, normalized size = 1.05 \[ -\frac {{\left (\sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2}{a^{3} d \sin \left (d x + c\right ) + a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.25, size = 35, normalized size = 0.90 \[ -\frac {\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} + \frac {2}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.20, size = 37, normalized size = 0.95 \[ -\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{a^{3} d}-\frac {2}{a^{3} d \left (1+\sin \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a+a*sin(d*x+c))^3,x)

[Out]

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

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maxima [A]  time = 0.37, size = 37, normalized size = 0.95 \[ -\frac {\frac {2}{a^{3} \sin \left (d x + c\right ) + a^{3}} + \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.54, size = 36, normalized size = 0.92 \[ -\frac {2}{a^3\,d\,\left (\sin \left (c+d\,x\right )+1\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a + a*sin(c + d*x))^3,x)

[Out]

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

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sympy [A]  time = 1.95, size = 299, normalized size = 7.67 \[ \begin {cases} - \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {4 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {2 \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {\cos ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {2}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{3}{\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((-2*log(sin(c + d*x) + 1)*sin(c + d*x)**2/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3
*d) - 4*log(sin(c + d*x) + 1)*sin(c + d*x)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) - 2*l
og(sin(c + d*x) + 1)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) - 2*sin(c + d*x)/(2*a**3*d*
sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) - cos(c + d*x)**2/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin
(c + d*x) + 2*a**3*d) - 2/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d), Ne(d, 0)), (x*cos(c)*
*3/(a*sin(c) + a)**3, True))

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