3.178 \(\int \cot ^2(a+b x) \csc ^3(a+b x) \, dx\)

Optimal. Leaf size=55 \[ \frac{\tanh ^{-1}(\cos (a+b x))}{8 b}-\frac{\cot (a+b x) \csc ^3(a+b x)}{4 b}+\frac{\cot (a+b x) \csc (a+b x)}{8 b} \]

[Out]

ArcTanh[Cos[a + b*x]]/(8*b) + (Cot[a + b*x]*Csc[a + b*x])/(8*b) - (Cot[a + b*x]*Csc[a + b*x]^3)/(4*b)

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Rubi [A]  time = 0.0451343, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2611, 3768, 3770} \[ \frac{\tanh ^{-1}(\cos (a+b x))}{8 b}-\frac{\cot (a+b x) \csc ^3(a+b x)}{4 b}+\frac{\cot (a+b x) \csc (a+b x)}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Cot[a + b*x]^2*Csc[a + b*x]^3,x]

[Out]

ArcTanh[Cos[a + b*x]]/(8*b) + (Cot[a + b*x]*Csc[a + b*x])/(8*b) - (Cot[a + b*x]*Csc[a + b*x]^3)/(4*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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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 ^3(a+b x) \, dx &=-\frac{\cot (a+b x) \csc ^3(a+b x)}{4 b}-\frac{1}{4} \int \csc ^3(a+b x) \, dx\\ &=\frac{\cot (a+b x) \csc (a+b x)}{8 b}-\frac{\cot (a+b x) \csc ^3(a+b x)}{4 b}-\frac{1}{8} \int \csc (a+b x) \, dx\\ &=\frac{\tanh ^{-1}(\cos (a+b x))}{8 b}+\frac{\cot (a+b x) \csc (a+b x)}{8 b}-\frac{\cot (a+b x) \csc ^3(a+b x)}{4 b}\\ \end{align*}

Mathematica [B]  time = 0.0342107, size = 113, normalized size = 2.05 \[ -\frac{\csc ^4\left (\frac{1}{2} (a+b x)\right )}{64 b}+\frac{\csc ^2\left (\frac{1}{2} (a+b x)\right )}{32 b}+\frac{\sec ^4\left (\frac{1}{2} (a+b x)\right )}{64 b}-\frac{\sec ^2\left (\frac{1}{2} (a+b x)\right )}{32 b}-\frac{\log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{8 b}+\frac{\log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[a + b*x]^2*Csc[a + b*x]^3,x]

[Out]

Csc[(a + b*x)/2]^2/(32*b) - Csc[(a + b*x)/2]^4/(64*b) + Log[Cos[(a + b*x)/2]]/(8*b) - Log[Sin[(a + b*x)/2]]/(8
*b) - Sec[(a + b*x)/2]^2/(32*b) + Sec[(a + b*x)/2]^4/(64*b)

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Maple [A]  time = 0.011, size = 76, normalized size = 1.4 \begin{align*} -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{3}}{4\,b \left ( \sin \left ( bx+a \right ) \right ) ^{4}}}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{3}}{8\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2}}}-{\frac{\cos \left ( bx+a \right ) }{8\,b}}-{\frac{\ln \left ( \csc \left ( bx+a \right ) -\cot \left ( bx+a \right ) \right ) }{8\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2/sin(b*x+a)^5,x)

[Out]

-1/4/b*cos(b*x+a)^3/sin(b*x+a)^4-1/8/b*cos(b*x+a)^3/sin(b*x+a)^2-1/8*cos(b*x+a)/b-1/8/b*ln(csc(b*x+a)-cot(b*x+
a))

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Maxima [A]  time = 0.970983, size = 88, normalized size = 1.6 \begin{align*} -\frac{\frac{2 \,{\left (\cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )\right )}}{\cos \left (b x + a\right )^{4} - 2 \, \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 )}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/sin(b*x+a)^5,x, algorithm="maxima")

[Out]

-1/16*(2*(cos(b*x + a)^3 + cos(b*x + a))/(cos(b*x + a)^4 - 2*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 = 1.95067, size = 308, normalized size = 5.6 \begin{align*} -\frac{2 \, \cos \left (b x + a\right )^{3} -{\left (\cos \left (b x + a\right )^{4} - 2 \, \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 )^{4} - 2 \, \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 )}{16 \,{\left (b \cos \left (b x + a\right )^{4} - 2 \, 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)^5,x, algorithm="fricas")

[Out]

-1/16*(2*cos(b*x + a)^3 - (cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*log(1/2*cos(b*x + a) + 1/2) + (cos(b*x + a)^
4 - 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)^4 - 2*b*cos(b*x + a)^
2 + b)

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Sympy [A]  time = 4.73168, size = 58, normalized size = 1.05 \begin{align*} \begin{cases} - \frac{\log{\left (\tan{\left (\frac{a}{2} + \frac{b x}{2} \right )} \right )}}{8 b} + \frac{\tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{64 b} - \frac{1}{64 b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )}} & \text{for}\: b \neq 0 \\\frac{x \cos ^{2}{\left (a \right )}}{\sin ^{5}{\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)**5,x)

[Out]

Piecewise((-log(tan(a/2 + b*x/2))/(8*b) + tan(a/2 + b*x/2)**4/(64*b) - 1/(64*b*tan(a/2 + b*x/2)**4), Ne(b, 0))
, (x*cos(a)**2/sin(a)**5, True))

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Giac [A]  time = 1.16238, size = 132, normalized size = 2.4 \begin{align*} \frac{\frac{{\left (\frac{2 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 1\right )}{\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} + \frac{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 4 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{64 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/sin(b*x+a)^5,x, algorithm="giac")

[Out]

1/64*((2*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 - 1)*(cos(b*x + a) + 1)^2/(cos(b*x + a) - 1)^2 + (cos(b*x +
a) - 1)^2/(cos(b*x + a) + 1)^2 - 4*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b