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

Optimal. Leaf size=47 \[ -\frac {\csc (c+d x)}{a^3 d}-\frac {3 \log (\sin (c+d x))}{a^3 d}+\frac {4 \log (\sin (c+d x)+1)}{a^3 d} \]

[Out]

-csc(d*x+c)/a^3/d-3*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 = 0.11, 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, 88} \[ -\frac {\csc (c+d x)}{a^3 d}-\frac {3 \log (\sin (c+d x))}{a^3 d}+\frac {4 \log (\sin (c+d x)+1)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[c + d*x]/(a^3*d)) - (3*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 {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a^2 (a-x)^2}{x^2 (a+x)} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(a-x)^2}{x^2 (a+x)} \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a}{x^2}-\frac {3}{x}+\frac {4}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=-\frac {\csc (c+d x)}{a^3 d}-\frac {3 \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 = 0.05, size = 35, normalized size = 0.74 \[ -\frac {\csc (c+d x)+3 \log (\sin (c+d x))-4 \log (\sin (c+d x)+1)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.85, size = 52, normalized size = 1.11 \[ -\frac {3 \, \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 4 \, \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 1}{a^{3} d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.25, size = 101, normalized size = 2.15 \[ -\frac {\frac {2 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}{a^{3}} - \frac {16 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} + \frac {6 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} - \frac {6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*log(tan(1/2*d*x + 1/2*c)^2 + 1)/a^3 - 16*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 + 6*log(abs(tan(1/2*d*
x + 1/2*c)))/a^3 + tan(1/2*d*x + 1/2*c)/a^3 - (6*tan(1/2*d*x + 1/2*c) - 1)/(a^3*tan(1/2*d*x + 1/2*c)))/d

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maple [A]  time = 0.58, size = 50, normalized size = 1.06 \[ -\frac {1}{a^{3} d \sin \left (d x +c \right )}-\frac {3 \ln \left (\sin \left (d x +c \right )\right )}{a^{3} d}+\frac {4 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a^3/d/sin(d*x+c)-3*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 = 44, normalized size = 0.94 \[ \frac {\frac {4 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} - \frac {3 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}} - \frac {1}{a^{3} \sin \left (d x + c\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.98, size = 71, normalized size = 1.51 \[ -\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )+\frac {\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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