3.536 \(\int \frac {\cos ^4(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=65 \[ \frac {\sin ^3(c+d x)}{3 a d}-\frac {\sin ^2(c+d x)}{2 a d}-\frac {\sin (c+d x)}{a d}+\frac {\log (\sin (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 65, 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, 75} \[ \frac {\sin ^3(c+d x)}{3 a d}-\frac {\sin ^2(c+d x)}{2 a d}-\frac {\sin (c+d x)}{a d}+\frac {\log (\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Cot[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

Log[Sin[c + d*x]]/(a*d) - Sin[c + d*x]/(a*d) - Sin[c + d*x]^2/(2*a*d) + Sin[c + d*x]^3/(3*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 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 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 ^4(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a (a-x)^2 (a+x)}{x} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(a-x)^2 (a+x)}{x} \, dx,x,a \sin (c+d x)\right )}{a^4 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-a^2+\frac {a^3}{x}-a x+x^2\right ) \, dx,x,a \sin (c+d x)\right )}{a^4 d}\\ &=\frac {\log (\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{a d}-\frac {\sin ^2(c+d x)}{2 a d}+\frac {\sin ^3(c+d x)}{3 a d}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 49, normalized size = 0.75 \[ \frac {2 \sin ^3(c+d x)-3 \sin ^2(c+d x)-6 \sin (c+d x)+6 \log (\sin (c+d x))-2}{6 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Cot[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

(-2 + 6*Log[Sin[c + d*x]] - 6*Sin[c + d*x] - 3*Sin[c + d*x]^2 + 2*Sin[c + d*x]^3)/(6*a*d)

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fricas [A]  time = 0.63, size = 48, normalized size = 0.74 \[ \frac {3 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) + 6 \, \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right )}{6 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*cos(d*x + c)^2 - 2*(cos(d*x + c)^2 + 2)*sin(d*x + c) + 6*log(1/2*sin(d*x + c)))/(a*d)

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giac [A]  time = 0.17, size = 61, normalized size = 0.94 \[ \frac {\frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{3} - 3 \, a^{2} \sin \left (d x + c\right )^{2} - 6 \, a^{2} \sin \left (d x + c\right )}{a^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.38, size = 62, normalized size = 0.95 \[ \frac {\ln \left (\sin \left (d x +c \right )\right )}{a d}-\frac {\sin \left (d x +c \right )}{a d}-\frac {\sin ^{2}\left (d x +c \right )}{2 a d}+\frac {\sin ^{3}\left (d x +c \right )}{3 d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.83, size = 51, normalized size = 0.78 \[ \frac {\frac {2 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} - 6 \, \sin \left (d x + c\right )}{a} + \frac {6 \, \log \left (\sin \left (d x + c\right )\right )}{a}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 9.05, size = 102, normalized size = 1.57 \[ \frac {\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d}-\frac {\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{a\,d}-\frac {2\,\sin \left (c+d\,x\right )}{3\,a\,d}+\frac {{\cos \left (c+d\,x\right )}^2}{2\,a\,d}-\frac {{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))/(a*d) - log(1/cos(c/2 + (d*x)/2)^2)/(a*d) - (2*sin(c + d*x))/(3*a*d
) + cos(c + d*x)^2/(2*a*d) - (cos(c + d*x)^2*sin(c + d*x))/(3*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)**5*csc(d*x+c)/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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