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

Optimal. Leaf size=32 \[ \frac{\csc (c+d x)}{a d}-\frac{\csc ^2(c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.0677649, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2706, 2606, 30, 8} \[ \frac{\csc (c+d x)}{a d}-\frac{\csc ^2(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.032614, size = 24, normalized size = 0.75 \[ -\frac{(\csc (c+d x)-2) \csc (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((-2 + Csc[c + d*x])*Csc[c + d*x])/(2*a*d)

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Maple [A]  time = 0.032, size = 30, normalized size = 0.9 \begin{align*} -{\frac{1}{da} \left ( - \left ( \sin \left ( dx+c \right ) \right ) ^{-1}+{\frac{1}{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3/(a+a*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.11797, size = 35, normalized size = 1.09 \begin{align*} \frac{2 \, \sin \left (d x + c\right ) - 1}{2 \, a d \sin \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.37048, size = 73, normalized size = 2.28 \begin{align*} -\frac{2 \, \sin \left (d x + c\right ) - 1}{2 \,{\left (a d \cos \left (d x + c\right )^{2} - a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\cot ^{3}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3/(a+a*sin(d*x+c)),x)

[Out]

Integral(cot(c + d*x)**3/(sin(c + d*x) + 1), x)/a

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Giac [A]  time = 1.32292, size = 35, normalized size = 1.09 \begin{align*} \frac{2 \, \sin \left (d x + c\right ) - 1}{2 \, a d \sin \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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