3.562 \(\int \frac{\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=117 \[ -\frac{\csc ^5(c+d x)}{5 a^3 d}+\frac{3 \csc ^4(c+d x)}{4 a^3 d}-\frac{4 \csc ^3(c+d x)}{3 a^3 d}+\frac{2 \csc ^2(c+d x)}{a^3 d}-\frac{4 \csc (c+d x)}{a^3 d}-\frac{4 \log (\sin (c+d x))}{a^3 d}+\frac{4 \log (\sin (c+d x)+1)}{a^3 d} \]

[Out]

(-4*Csc[c + d*x])/(a^3*d) + (2*Csc[c + d*x]^2)/(a^3*d) - (4*Csc[c + d*x]^3)/(3*a^3*d) + (3*Csc[c + d*x]^4)/(4*
a^3*d) - Csc[c + d*x]^5/(5*a^3*d) - (4*Log[Sin[c + d*x]])/(a^3*d) + (4*Log[1 + Sin[c + d*x]])/(a^3*d)

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Rubi [A]  time = 0.119461, antiderivative size = 117, 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, 88} \[ -\frac{\csc ^5(c+d x)}{5 a^3 d}+\frac{3 \csc ^4(c+d x)}{4 a^3 d}-\frac{4 \csc ^3(c+d x)}{3 a^3 d}+\frac{2 \csc ^2(c+d x)}{a^3 d}-\frac{4 \csc (c+d x)}{a^3 d}-\frac{4 \log (\sin (c+d x))}{a^3 d}+\frac{4 \log (\sin (c+d x)+1)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^5*Csc[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

(-4*Csc[c + d*x])/(a^3*d) + (2*Csc[c + d*x]^2)/(a^3*d) - (4*Csc[c + d*x]^3)/(3*a^3*d) + (3*Csc[c + d*x]^4)/(4*
a^3*d) - Csc[c + d*x]^5/(5*a^3*d) - (4*Log[Sin[c + d*x]])/(a^3*d) + (4*Log[1 + 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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a^6 (a-x)^2}{x^6 (a+x)} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{a \operatorname{Subst}\left (\int \frac{(a-x)^2}{x^6 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a \operatorname{Subst}\left (\int \left (\frac{a}{x^6}-\frac{3}{x^5}+\frac{4}{a x^4}-\frac{4}{a^2 x^3}+\frac{4}{a^3 x^2}-\frac{4}{a^4 x}+\frac{4}{a^4 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{4 \csc (c+d x)}{a^3 d}+\frac{2 \csc ^2(c+d x)}{a^3 d}-\frac{4 \csc ^3(c+d x)}{3 a^3 d}+\frac{3 \csc ^4(c+d x)}{4 a^3 d}-\frac{\csc ^5(c+d x)}{5 a^3 d}-\frac{4 \log (\sin (c+d x))}{a^3 d}+\frac{4 \log (1+\sin (c+d x))}{a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.126769, size = 79, normalized size = 0.68 \[ -\frac{12 \csc ^5(c+d x)-45 \csc ^4(c+d x)+80 \csc ^3(c+d x)-120 \csc ^2(c+d x)+240 \csc (c+d x)+240 \log (\sin (c+d x))-240 \log (\sin (c+d x)+1)}{60 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^5*Csc[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

-(240*Csc[c + d*x] - 120*Csc[c + d*x]^2 + 80*Csc[c + d*x]^3 - 45*Csc[c + d*x]^4 + 12*Csc[c + d*x]^5 + 240*Log[
Sin[c + d*x]] - 240*Log[1 + Sin[c + d*x]])/(60*a^3*d)

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Maple [A]  time = 0.184, size = 114, normalized size = 1. \begin{align*} 4\,{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{d{a}^{3}}}-{\frac{1}{5\,d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}+{\frac{3}{4\,d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{4}{3\,d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+2\,{\frac{1}{d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-4\,{\frac{1}{d{a}^{3}\sin \left ( dx+c \right ) }}-4\,{\frac{\ln \left ( \sin \left ( dx+c \right ) \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)^6/(a+a*sin(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.07531, size = 115, normalized size = 0.98 \begin{align*} \frac{\frac{240 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} - \frac{240 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}} - \frac{240 \, \sin \left (d x + c\right )^{4} - 120 \, \sin \left (d x + c\right )^{3} + 80 \, \sin \left (d x + c\right )^{2} - 45 \, \sin \left (d x + c\right ) + 12}{a^{3} \sin \left (d x + c\right )^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(240*log(sin(d*x + c) + 1)/a^3 - 240*log(sin(d*x + c))/a^3 - (240*sin(d*x + c)^4 - 120*sin(d*x + c)^3 + 8
0*sin(d*x + c)^2 - 45*sin(d*x + c) + 12)/(a^3*sin(d*x + c)^5))/d

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Fricas [A]  time = 1.19092, size = 446, normalized size = 3.81 \begin{align*} -\frac{240 \, \cos \left (d x + c\right )^{4} + 240 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 240 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 560 \, \cos \left (d x + c\right )^{2} + 15 \,{\left (8 \, \cos \left (d x + c\right )^{2} - 11\right )} \sin \left (d x + c\right ) + 332}{60 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(240*cos(d*x + c)^4 + 240*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*sin(d*x + c))*sin(d*x + c) - 2
40*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(sin(d*x + c) + 1)*sin(d*x + c) - 560*cos(d*x + c)^2 + 15*(8*cos
(d*x + c)^2 - 11)*sin(d*x + c) + 332)/((a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d)*sin(d*x + c))

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

[Out]

Timed out

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Giac [A]  time = 1.33299, size = 275, normalized size = 2.35 \begin{align*} \frac{\frac{7680 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{3840 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac{8768 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 2460 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 660 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 190 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 45 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}} - \frac{6 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 45 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 190 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 660 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2460 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/960*(7680*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 3840*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + (8768*tan(1/2*d
*x + 1/2*c)^5 - 2460*tan(1/2*d*x + 1/2*c)^4 + 660*tan(1/2*d*x + 1/2*c)^3 - 190*tan(1/2*d*x + 1/2*c)^2 + 45*tan
(1/2*d*x + 1/2*c) - 6)/(a^3*tan(1/2*d*x + 1/2*c)^5) - (6*a^12*tan(1/2*d*x + 1/2*c)^5 - 45*a^12*tan(1/2*d*x + 1
/2*c)^4 + 190*a^12*tan(1/2*d*x + 1/2*c)^3 - 660*a^12*tan(1/2*d*x + 1/2*c)^2 + 2460*a^12*tan(1/2*d*x + 1/2*c))/
a^15)/d