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

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Rubi [A]  time = 0.12, antiderivative size = 117, normalized size of antiderivative = 1.00, 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 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]))

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]

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*}

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Mathematica [A]  time = 0.13, 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]

-1/60*(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]])/(a^3*d)

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fricas [A]  time = 0.58, size = 161, normalized size = 1.38 \[ -\frac {240 \, \cos \left (d x + c\right )^{4} + 240 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 240 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 560 \, \cos \left (d x + c\right )^{2} + 15 \, {\left (8 \, \cos \left (d x + c\right )^{2} - 11\right )} \sin \left (d x + c\right ) + 332}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )} \sin \left (d x + c\right )} \]

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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giac [A]  time = 0.27, size = 204, normalized size = 1.74 \[ \frac {\frac {7680 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {3840 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {8768 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2460 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 660 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 190 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} - \frac {6 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 45 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 190 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 660 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2460 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{960 \, d} \]

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

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maple [A]  time = 0.66, size = 114, normalized size = 0.97 \[ -\frac {1}{5 d \,a^{3} \sin \left (d x +c \right )^{5}}+\frac {3}{4 d \,a^{3} \sin \left (d x +c \right )^{4}}-\frac {4}{3 d \,a^{3} \sin \left (d x +c \right )^{3}}+\frac {2}{a^{3} d \sin \left (d x +c \right )^{2}}-\frac {4}{a^{3} d \sin \left (d x +c \right )}-\frac {4 \ln \left (\sin \left (d x +c \right )\right )}{a^{3} d}+\frac {4 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{3} d} \]

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]

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

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maxima [A]  time = 0.87, size = 85, normalized size = 0.73 \[ \frac {\frac {240 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} - \frac {240 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}} - \frac {240 \, \sin \left (d x + c\right )^{4} - 120 \, \sin \left (d x + c\right )^{3} + 80 \, \sin \left (d x + c\right )^{2} - 45 \, \sin \left (d x + c\right ) + 12}{a^{3} \sin \left (d x + c\right )^{5}}}{60 \, d} \]

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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mupad [B]  time = 8.96, size = 203, normalized size = 1.74 \[ \frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,a^3\,d}-\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,a^3\,d}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a^3\,d}-\frac {4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}+\frac {8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^3\,d}-\frac {41\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a^3\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (82\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {1}{5}\right )}{32\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^6*(a + a*sin(c + d*x))^3),x)

[Out]

(11*tan(c/2 + (d*x)/2)^2)/(16*a^3*d) - (19*tan(c/2 + (d*x)/2)^3)/(96*a^3*d) + (3*tan(c/2 + (d*x)/2)^4)/(64*a^3
*d) - tan(c/2 + (d*x)/2)^5/(160*a^3*d) - (4*log(tan(c/2 + (d*x)/2)))/(a^3*d) + (8*log(tan(c/2 + (d*x)/2) + 1))
/(a^3*d) - (41*tan(c/2 + (d*x)/2))/(16*a^3*d) - (cot(c/2 + (d*x)/2)^5*((19*tan(c/2 + (d*x)/2)^2)/3 - (3*tan(c/
2 + (d*x)/2))/2 - 22*tan(c/2 + (d*x)/2)^3 + 82*tan(c/2 + (d*x)/2)^4 + 1/5))/(32*a^3*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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