3.150 \(\int \cot ^2(a+b x) \csc (a+b x) \, dx\)

Optimal. Leaf size=34 \[ \frac{\tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{\cot (a+b x) \csc (a+b x)}{2 b} \]

[Out]

ArcTanh[Cos[a + b*x]]/(2*b) - (Cot[a + b*x]*Csc[a + b*x])/(2*b)

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Rubi [A]  time = 0.0226236, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2611, 3770} \[ \frac{\tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{\cot (a+b x) \csc (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cot[a + b*x]^2*Csc[a + b*x],x]

[Out]

ArcTanh[Cos[a + b*x]]/(2*b) - (Cot[a + b*x]*Csc[a + b*x])/(2*b)

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \cot ^2(a+b x) \csc (a+b x) \, dx &=-\frac{\cot (a+b x) \csc (a+b x)}{2 b}-\frac{1}{2} \int \csc (a+b x) \, dx\\ &=\frac{\tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{\cot (a+b x) \csc (a+b x)}{2 b}\\ \end{align*}

Mathematica [B]  time = 0.0268574, size = 75, normalized size = 2.21 \[ -\frac{\csc ^2\left (\frac{1}{2} (a+b x)\right )}{8 b}+\frac{\sec ^2\left (\frac{1}{2} (a+b x)\right )}{8 b}-\frac{\log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{2 b}+\frac{\log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[a + b*x]^2*Csc[a + b*x],x]

[Out]

-Csc[(a + b*x)/2]^2/(8*b) + Log[Cos[(a + b*x)/2]]/(2*b) - Log[Sin[(a + b*x)/2]]/(2*b) + Sec[(a + b*x)/2]^2/(8*
b)

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Maple [A]  time = 0.01, size = 55, normalized size = 1.6 \begin{align*} -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{3}}{2\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2}}}-{\frac{\cos \left ( bx+a \right ) }{2\,b}}-{\frac{\ln \left ( \csc \left ( bx+a \right ) -\cot \left ( bx+a \right ) \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2/sin(b*x+a)^3,x)

[Out]

-1/2/b*cos(b*x+a)^3/sin(b*x+a)^2-1/2*cos(b*x+a)/b-1/2/b*ln(csc(b*x+a)-cot(b*x+a))

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Maxima [A]  time = 0.981926, size = 62, normalized size = 1.82 \begin{align*} \frac{\frac{2 \, \cos \left (b x + a\right )}{\cos \left (b x + a\right )^{2} - 1} + \log \left (\cos \left (b x + a\right ) + 1\right ) - \log \left (\cos \left (b x + a\right ) - 1\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/sin(b*x+a)^3,x, algorithm="maxima")

[Out]

1/4*(2*cos(b*x + a)/(cos(b*x + a)^2 - 1) + log(cos(b*x + a) + 1) - log(cos(b*x + a) - 1))/b

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Fricas [B]  time = 2.57544, size = 200, normalized size = 5.88 \begin{align*} \frac{{\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) -{\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) + 2 \, \cos \left (b x + a\right )}{4 \,{\left (b \cos \left (b x + a\right )^{2} - b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/sin(b*x+a)^3,x, algorithm="fricas")

[Out]

1/4*((cos(b*x + a)^2 - 1)*log(1/2*cos(b*x + a) + 1/2) - (cos(b*x + a)^2 - 1)*log(-1/2*cos(b*x + a) + 1/2) + 2*
cos(b*x + a))/(b*cos(b*x + a)^2 - b)

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Sympy [A]  time = 1.6412, size = 58, normalized size = 1.71 \begin{align*} \begin{cases} - \frac{\log{\left (\tan{\left (\frac{a}{2} + \frac{b x}{2} \right )} \right )}}{2 b} + \frac{\tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{8 b} - \frac{1}{8 b \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )}} & \text{for}\: b \neq 0 \\\frac{x \cos ^{2}{\left (a \right )}}{\sin ^{3}{\left (a \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**2/sin(b*x+a)**3,x)

[Out]

Piecewise((-log(tan(a/2 + b*x/2))/(2*b) + tan(a/2 + b*x/2)**2/(8*b) - 1/(8*b*tan(a/2 + b*x/2)**2), Ne(b, 0)),
(x*cos(a)**2/sin(a)**3, True))

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Giac [B]  time = 1.17846, size = 126, normalized size = 3.71 \begin{align*} \frac{\frac{{\left (\frac{2 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + 1\right )}{\left (\cos \left (b x + a\right ) + 1\right )}}{\cos \left (b x + a\right ) - 1} - \frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 2 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/sin(b*x+a)^3,x, algorithm="giac")

[Out]

1/8*((2*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)*(cos(b*x + a) + 1)/(cos(b*x + a) - 1) - (cos(b*x + a) - 1)/
(cos(b*x + a) + 1) - 2*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b