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

Optimal. Leaf size=63 \[ \frac{(a+b) \log \left (a+b \sin ^2(c+d x)\right )}{2 a^2 d}-\frac{(a+b) \log (\sin (c+d x))}{a^2 d}-\frac{\csc ^2(c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.0749722, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3194, 77} \[ \frac{(a+b) \log \left (a+b \sin ^2(c+d x)\right )}{2 a^2 d}-\frac{(a+b) \log (\sin (c+d x))}{a^2 d}-\frac{\csc ^2(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3194

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.141051, size = 50, normalized size = 0.79 \[ -\frac{(a+b) \left (2 \log (\sin (c+d x))-\log \left (a+b \sin ^2(c+d x)\right )\right )+a \csc ^2(c+d x)}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.094, size = 161, normalized size = 2.6 \begin{align*}{\frac{1}{4\,da \left ( -1+\cos \left ( dx+c \right ) \right ) }}-{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{2\,da}}-{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) b}{2\,{a}^{2}d}}-{\frac{1}{4\,da \left ( 1+\cos \left ( dx+c \right ) \right ) }}-{\frac{\ln \left ( 1+\cos \left ( dx+c \right ) \right ) }{2\,da}}-{\frac{\ln \left ( 1+\cos \left ( dx+c \right ) \right ) b}{2\,{a}^{2}d}}+{\frac{\ln \left ( b \left ( \cos \left ( dx+c \right ) \right ) ^{2}-a-b \right ) }{2\,da}}+{\frac{\ln \left ( b \left ( \cos \left ( dx+c \right ) \right ) ^{2}-a-b \right ) b}{2\,{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3/(a+sin(d*x+c)^2*b),x)

[Out]

1/4/d/a/(-1+cos(d*x+c))-1/2/d/a*ln(-1+cos(d*x+c))-1/2/d/a^2*ln(-1+cos(d*x+c))*b-1/4/a/d/(1+cos(d*x+c))-1/2/d/a
*ln(1+cos(d*x+c))-1/2/d/a^2*ln(1+cos(d*x+c))*b+1/2/d/a*ln(b*cos(d*x+c)^2-a-b)+1/2/d/a^2*ln(b*cos(d*x+c)^2-a-b)
*b

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Maxima [A]  time = 0.986388, size = 76, normalized size = 1.21 \begin{align*} \frac{\frac{{\left (a + b\right )} \log \left (b \sin \left (d x + c\right )^{2} + a\right )}{a^{2}} - \frac{{\left (a + b\right )} \log \left (\sin \left (d x + c\right )^{2}\right )}{a^{2}} - \frac{1}{a \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)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.0216, size = 223, normalized size = 3.54 \begin{align*} \frac{{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \log \left (-b \cos \left (d x + c\right )^{2} + a + b\right ) - 2 \,{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) + a}{2 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} 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)^2),x, algorithm="fricas")

[Out]

1/2*(((a + b)*cos(d*x + c)^2 - a - b)*log(-b*cos(d*x + c)^2 + a + b) - 2*((a + b)*cos(d*x + c)^2 - a - b)*log(
1/2*sin(d*x + c)) + a)/(a^2*d*cos(d*x + c)^2 - a^2*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 ^{2}{\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)**2),x)

[Out]

Integral(cot(c + d*x)**3/(a + b*sin(c + d*x)**2), x)

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Giac [A]  time = 1.18514, size = 146, normalized size = 2.32 \begin{align*} \frac{\frac{\frac{\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}}{a} + \frac{4 \,{\left (a + b\right )} \log \left ({\left | -a{\left (\frac{\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} + \frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1}\right )} + 2 \, a + 4 \, b \right |}\right )}{a^{2}}}{8 \, 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)^2),x, algorithm="giac")

[Out]

1/8*(((cos(d*x + c) + 1)/(cos(d*x + c) - 1) + (cos(d*x + c) - 1)/(cos(d*x + c) + 1))/a + 4*(a + b)*log(abs(-a*
((cos(d*x + c) + 1)/(cos(d*x + c) - 1) + (cos(d*x + c) - 1)/(cos(d*x + c) + 1)) + 2*a + 4*b))/a^2)/d