3.557 \(\int \frac{\cos ^4(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=45 \[ \frac{\sin (c+d x)}{a^3 d}+\frac{\log (\sin (c+d x))}{a^3 d}-\frac{4 \log (\sin (c+d x)+1)}{a^3 d} \]

[Out]

Log[Sin[c + d*x]]/(a^3*d) - (4*Log[1 + Sin[c + d*x]])/(a^3*d) + Sin[c + d*x]/(a^3*d)

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Rubi [A]  time = 0.08554, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2836, 12, 72} \[ \frac{\sin (c+d x)}{a^3 d}+\frac{\log (\sin (c+d x))}{a^3 d}-\frac{4 \log (\sin (c+d x)+1)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Cot[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

Log[Sin[c + d*x]]/(a^3*d) - (4*Log[1 + Sin[c + d*x]])/(a^3*d) + Sin[c + d*x]/(a^3*d)

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a (a-x)^2}{x (a+x)} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^2}{x (a+x)} \, dx,x,a \sin (c+d x)\right )}{a^4 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+\frac{a}{x}-\frac{4 a}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^4 d}\\ &=\frac{\log (\sin (c+d x))}{a^3 d}-\frac{4 \log (1+\sin (c+d x))}{a^3 d}+\frac{\sin (c+d x)}{a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.0378115, size = 32, normalized size = 0.71 \[ \frac{\sin (c+d x)+\log (\sin (c+d x))-4 \log (\sin (c+d x)+1)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Cot[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

(Log[Sin[c + d*x]] - 4*Log[1 + Sin[c + d*x]] + Sin[c + d*x])/(a^3*d)

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Maple [A]  time = 0.124, size = 46, normalized size = 1. \begin{align*}{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{d{a}^{3}}}-4\,{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{d{a}^{3}}}+{\frac{\sin \left ( dx+c \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c))^3,x)

[Out]

ln(sin(d*x+c))/a^3/d-4*ln(1+sin(d*x+c))/a^3/d+sin(d*x+c)/a^3/d

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Maxima [A]  time = 0.982692, size = 58, normalized size = 1.29 \begin{align*} -\frac{\frac{4 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} - \frac{\log \left (\sin \left (d x + c\right )\right )}{a^{3}} - \frac{\sin \left (d x + c\right )}{a^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.09148, size = 100, normalized size = 2.22 \begin{align*} \frac{\log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 4 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + \sin \left (d x + c\right )}{a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.34512, size = 139, normalized size = 3.09 \begin{align*} \frac{\frac{3 \, \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}{a^{3}} - \frac{8 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} + \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} a^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

(3*log(tan(1/2*d*x + 1/2*c)^2 + 1)/a^3 - 8*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 + log(abs(tan(1/2*d*x + 1/2*
c)))/a^3 - (3*tan(1/2*d*x + 1/2*c)^2 - 2*tan(1/2*d*x + 1/2*c) + 3)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3))/d