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

Optimal. Leaf size=84 \[ -\frac{\left (a^2-b^2\right ) \log (\sin (c+d x))}{a^3 d}+\frac{\left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{a^3 d}+\frac{b \csc (c+d x)}{a^2 d}-\frac{\csc ^2(c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.0888076, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2721, 894} \[ -\frac{\left (a^2-b^2\right ) \log (\sin (c+d x))}{a^3 d}+\frac{\left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{a^3 d}+\frac{b \csc (c+d x)}{a^2 d}-\frac{\csc ^2(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [A]  time = 0.1613, size = 65, normalized size = 0.77 \[ -\frac{2 \left (a^2-b^2\right ) (\log (\sin (c+d x))-\log (a+b \sin (c+d x)))+a^2 \csc ^2(c+d x)-2 a b \csc (c+d x)}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.063, size = 106, normalized size = 1.3 \begin{align*}{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{da}}-{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ){b}^{2}}{d{a}^{3}}}-{\frac{1}{2\,da \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{da}}+{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ){b}^{2}}{d{a}^{3}}}+{\frac{b}{d{a}^{2}\sin \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.61607, size = 104, normalized size = 1.24 \begin{align*} \frac{\frac{2 \,{\left (a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{3}} - \frac{2 \,{\left (a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{3}} + \frac{2 \, b \sin \left (d x + c\right ) - a}{a^{2} \sin \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.5958, size = 271, normalized size = 3.23 \begin{align*} -\frac{2 \, a b \sin \left (d x + c\right ) - a^{2} - 2 \,{\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 2 \,{\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + b^{2}\right )} \log \left (-\frac{1}{2} \, \sin \left (d x + c\right )\right )}{2 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.0115, size = 154, normalized size = 1.83 \begin{align*} -\frac{\frac{2 \,{\left (a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac{2 \,{\left (a^{2} b - b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{3} b} - \frac{3 \, a^{2} \sin \left (d x + c\right )^{2} - 3 \, b^{2} \sin \left (d x + c\right )^{2} + 2 \, a b \sin \left (d x + c\right ) - a^{2}}{a^{3} \sin \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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