3.688 \(\int \frac {\cos ^2(c+d x) \cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=94 \[ -\frac {\sin (c+d x)}{a d}-\frac {\csc ^4(c+d x)}{4 a d}+\frac {\csc ^3(c+d x)}{3 a d}+\frac {\csc ^2(c+d x)}{a d}-\frac {2 \csc (c+d x)}{a d}+\frac {\log (\sin (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2836, 12, 88} \[ -\frac {\sin (c+d x)}{a d}-\frac {\csc ^4(c+d x)}{4 a d}+\frac {\csc ^3(c+d x)}{3 a d}+\frac {\csc ^2(c+d x)}{a d}-\frac {2 \csc (c+d x)}{a d}+\frac {\log (\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*Cot[c + d*x]^5)/(a + a*Sin[c + d*x]),x]

[Out]

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

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Mathematica [A]  time = 0.30, size = 66, normalized size = 0.70 \[ -\frac {12 \sin (c+d x)+3 \csc ^4(c+d x)-4 \csc ^3(c+d x)-12 \csc ^2(c+d x)+24 \csc (c+d x)-12 \log (\sin (c+d x))}{12 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^5)/(a + a*Sin[c + d*x]),x]

[Out]

-1/12*(24*Csc[c + d*x] - 12*Csc[c + d*x]^2 - 4*Csc[c + d*x]^3 + 3*Csc[c + d*x]^4 - 12*Log[Sin[c + d*x]] + 12*S
in[c + d*x])/(a*d)

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fricas [A]  time = 0.48, size = 104, normalized size = 1.11 \[ -\frac {12 \, \cos \left (d x + c\right )^{2} - 12 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 12 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) - 9}{12 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(12*cos(d*x + c)^2 - 12*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*sin(d*x + c)) + 4*(3*cos(d*x + c
)^4 - 12*cos(d*x + c)^2 + 8)*sin(d*x + c) - 9)/(a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d)

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giac [A]  time = 0.24, size = 83, normalized size = 0.88 \[ \frac {\frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} - \frac {12 \, \sin \left (d x + c\right )}{a} - \frac {25 \, \sin \left (d x + c\right )^{4} + 24 \, \sin \left (d x + c\right )^{3} - 12 \, \sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) + 3}{a \sin \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/12*(12*log(abs(sin(d*x + c)))/a - 12*sin(d*x + c)/a - (25*sin(d*x + c)^4 + 24*sin(d*x + c)^3 - 12*sin(d*x +
c)^2 - 4*sin(d*x + c) + 3)/(a*sin(d*x + c)^4))/d

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maple [A]  time = 0.46, size = 93, normalized size = 0.99 \[ -\frac {\sin \left (d x +c \right )}{a d}-\frac {2}{d a \sin \left (d x +c \right )}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a d}+\frac {1}{a d \sin \left (d x +c \right )^{2}}-\frac {1}{4 a d \sin \left (d x +c \right )^{4}}+\frac {1}{3 a d \sin \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*csc(d*x+c)^5/(a+a*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.33, size = 72, normalized size = 0.77 \[ \frac {\frac {12 \, \log \left (\sin \left (d x + c\right )\right )}{a} - \frac {12 \, \sin \left (d x + c\right )}{a} - \frac {24 \, \sin \left (d x + c\right )^{3} - 12 \, \sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) + 3}{a \sin \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(12*log(sin(d*x + c))/a - 12*sin(d*x + c)/a - (24*sin(d*x + c)^3 - 12*sin(d*x + c)^2 - 4*sin(d*x + c) + 3
)/(a*sin(d*x + c)^4))/d

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mupad [B]  time = 9.34, size = 214, normalized size = 2.28 \[ \frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,a\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}+\frac {-46\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-\frac {1}{4}}{d\,\left (16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7/(sin(c + d*x)^5*(a + a*sin(c + d*x))),x)

[Out]

(3*tan(c/2 + (d*x)/2)^2)/(16*a*d) + tan(c/2 + (d*x)/2)^3/(24*a*d) - tan(c/2 + (d*x)/2)^4/(64*a*d) + log(tan(c/
2 + (d*x)/2))/(a*d) + ((2*tan(c/2 + (d*x)/2))/3 + (11*tan(c/2 + (d*x)/2)^2)/4 - (40*tan(c/2 + (d*x)/2)^3)/3 +
3*tan(c/2 + (d*x)/2)^4 - 46*tan(c/2 + (d*x)/2)^5 - 1/4)/(d*(16*a*tan(c/2 + (d*x)/2)^4 + 16*a*tan(c/2 + (d*x)/2
)^6)) - (7*tan(c/2 + (d*x)/2))/(8*a*d) - log(tan(c/2 + (d*x)/2)^2 + 1)/(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)**5/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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