3.549 \(\int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=47 \[ -\frac {\csc ^2(c+d x)}{2 a^2 d}+\frac {2 \csc (c+d x)}{a^2 d}+\frac {\log (\sin (c+d x))}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 47, 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, 43} \[ -\frac {\csc ^2(c+d x)}{2 a^2 d}+\frac {2 \csc (c+d x)}{a^2 d}+\frac {\log (\sin (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

(2*Csc[c + d*x])/(a^2*d) - Csc[c + d*x]^2/(2*a^2*d) + Log[Sin[c + d*x]]/(a^2*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

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Mathematica [A]  time = 0.05, size = 38, normalized size = 0.81 \[ \frac {-\csc ^2(c+d x)+4 \csc (c+d x)+2 \log (\sin (c+d x))}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

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

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fricas [A]  time = 0.69, size = 55, normalized size = 1.17 \[ \frac {2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 4 \, \sin \left (d x + c\right ) + 1}{2 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 52, normalized size = 1.11 \[ \frac {\frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} - \frac {3 \, \sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) + 1}{a^{2} \sin \left (d x + c\right )^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.53, size = 48, normalized size = 1.02 \[ \frac {2}{a^{2} d \sin \left (d x +c \right )}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a^{2} d}-\frac {1}{2 d \,a^{2} \sin \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.85, size = 40, normalized size = 0.85 \[ \frac {\frac {2 \, \log \left (\sin \left (d x + c\right )\right )}{a^{2}} + \frac {4 \, \sin \left (d x + c\right ) - 1}{a^{2} \sin \left (d x + c\right )^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 8.89, size = 104, normalized size = 2.21 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{8}\right )}{a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^3*(a + a*sin(c + d*x))^2),x)

[Out]

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

[Out]

Timed out

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