3.689 \(\int \frac {\cos (c+d x) \cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=100 \[ -\frac {\csc ^5(c+d x)}{5 a d}+\frac {\csc ^4(c+d x)}{4 a d}+\frac {2 \csc ^3(c+d x)}{3 a d}-\frac {\csc ^2(c+d x)}{a d}-\frac {\csc (c+d x)}{a d}-\frac {\log (\sin (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 100, 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} \[ -\frac {\csc ^5(c+d x)}{5 a d}+\frac {\csc ^4(c+d x)}{4 a d}+\frac {2 \csc ^3(c+d x)}{3 a d}-\frac {\csc ^2(c+d x)}{a d}-\frac {\csc (c+d x)}{a d}-\frac {\log (\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*Cot[c + d*x]^6)/(a + a*Sin[c + d*x]),x]

[Out]

-(Csc[c + d*x]/(a*d)) - Csc[c + d*x]^2/(a*d) + (2*Csc[c + d*x]^3)/(3*a*d) + Csc[c + d*x]^4/(4*a*d) - Csc[c + d
*x]^5/(5*a*d) - Log[Sin[c + d*x]]/(a*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 {\cos (c+d x) \cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a^6 (a-x)^3 (a+x)^2}{x^6} \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(a-x)^3 (a+x)^2}{x^6} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a^5}{x^6}-\frac {a^4}{x^5}-\frac {2 a^3}{x^4}+\frac {2 a^2}{x^3}+\frac {a}{x^2}-\frac {1}{x}\right ) \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=-\frac {\csc (c+d x)}{a d}-\frac {\csc ^2(c+d x)}{a d}+\frac {2 \csc ^3(c+d x)}{3 a d}+\frac {\csc ^4(c+d x)}{4 a d}-\frac {\csc ^5(c+d x)}{5 a d}-\frac {\log (\sin (c+d x))}{a d}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 68, normalized size = 0.68 \[ -\frac {12 \csc ^5(c+d x)-15 \csc ^4(c+d x)-40 \csc ^3(c+d x)+60 \csc ^2(c+d x)+60 \csc (c+d x)+60 \log (\sin (c+d x))}{60 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*Cot[c + d*x]^6)/(a + a*Sin[c + d*x]),x]

[Out]

-1/60*(60*Csc[c + d*x] + 60*Csc[c + d*x]^2 - 40*Csc[c + d*x]^3 - 15*Csc[c + d*x]^4 + 12*Csc[c + d*x]^5 + 60*Lo
g[Sin[c + d*x]])/(a*d)

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fricas [A]  time = 0.47, size = 118, normalized size = 1.18 \[ -\frac {60 \, \cos \left (d x + c\right )^{4} + 60 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 80 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (4 \, \cos \left (d x + c\right )^{2} - 3\right )} \sin \left (d x + c\right ) + 32}{60 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/60*(60*cos(d*x + c)^4 + 60*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*sin(d*x + c))*sin(d*x + c) - 80*
cos(d*x + c)^2 - 15*(4*cos(d*x + c)^2 - 3)*sin(d*x + c) + 32)/((a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*
d)*sin(d*x + c))

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giac [A]  time = 0.22, size = 82, normalized size = 0.82 \[ -\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} - \frac {137 \, \sin \left (d x + c\right )^{5} - 60 \, \sin \left (d x + c\right )^{4} - 60 \, \sin \left (d x + c\right )^{3} + 40 \, \sin \left (d x + c\right )^{2} + 15 \, \sin \left (d x + c\right ) - 12}{a \sin \left (d x + c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/60*(60*log(abs(sin(d*x + c)))/a - (137*sin(d*x + c)^5 - 60*sin(d*x + c)^4 - 60*sin(d*x + c)^3 + 40*sin(d*x
+ c)^2 + 15*sin(d*x + c) - 12)/(a*sin(d*x + c)^5))/d

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maple [A]  time = 0.48, size = 97, normalized size = 0.97 \[ -\frac {1}{d a \sin \left (d x +c \right )}-\frac {\ln \left (\sin \left (d x +c \right )\right )}{a d}-\frac {1}{5 a d \sin \left (d x +c \right )^{5}}-\frac {1}{a d \sin \left (d x +c \right )^{2}}+\frac {1}{4 a d \sin \left (d x +c \right )^{4}}+\frac {2}{3 a d \sin \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*csc(d*x+c)^6/(a+a*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.36, size = 70, normalized size = 0.70 \[ -\frac {\frac {60 \, \log \left (\sin \left (d x + c\right )\right )}{a} + \frac {60 \, \sin \left (d x + c\right )^{4} + 60 \, \sin \left (d x + c\right )^{3} - 40 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) + 12}{a \sin \left (d x + c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(60*log(sin(d*x + c))/a + (60*sin(d*x + c)^4 + 60*sin(d*x + c)^3 - 40*sin(d*x + c)^2 - 15*sin(d*x + c) +
12)/(a*sin(d*x + c)^5))/d

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mupad [B]  time = 9.24, size = 204, normalized size = 2.04 \[ \frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,a\,d}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,a\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a\,d}+\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {1}{5}\right )}{32\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7/(sin(c + d*x)^6*(a + a*sin(c + d*x))),x)

[Out]

(5*tan(c/2 + (d*x)/2)^3)/(96*a*d) - (3*tan(c/2 + (d*x)/2)^2)/(16*a*d) + tan(c/2 + (d*x)/2)^4/(64*a*d) - tan(c/
2 + (d*x)/2)^5/(160*a*d) - log(tan(c/2 + (d*x)/2))/(a*d) - (5*tan(c/2 + (d*x)/2))/(16*a*d) + log(tan(c/2 + (d*
x)/2)^2 + 1)/(a*d) - (cot(c/2 + (d*x)/2)^5*(6*tan(c/2 + (d*x)/2)^3 - (5*tan(c/2 + (d*x)/2)^2)/3 - tan(c/2 + (d
*x)/2)/2 + 10*tan(c/2 + (d*x)/2)^4 + 1/5))/(32*a*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)**7*csc(d*x+c)**6/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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