3.5 \(\int \csc ^5(a+b x) \, dx\)

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

[Out]

(-3*ArcTanh[Cos[a + b*x]])/(8*b) - (3*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 = 0.0259667, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3768, 3770} \[ -\frac{3 \tanh ^{-1}(\cos (a+b x))}{8 b}-\frac{\cot (a+b x) \csc ^3(a+b x)}{4 b}-\frac{3 \cot (a+b x) \csc (a+b x)}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^5,x]

[Out]

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

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 \csc ^5(a+b x) \, dx &=-\frac{\cot (a+b x) \csc ^3(a+b x)}{4 b}+\frac{3}{4} \int \csc ^3(a+b x) \, dx\\ &=-\frac{3 \cot (a+b x) \csc (a+b x)}{8 b}-\frac{\cot (a+b x) \csc ^3(a+b x)}{4 b}+\frac{3}{8} \int \csc (a+b x) \, dx\\ &=-\frac{3 \tanh ^{-1}(\cos (a+b x))}{8 b}-\frac{3 \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 = 0.0164213, size = 113, normalized size = 2.05 \[ -\frac{\csc ^4\left (\frac{1}{2} (a+b x)\right )}{64 b}-\frac{3 \csc ^2\left (\frac{1}{2} (a+b x)\right )}{32 b}+\frac{\sec ^4\left (\frac{1}{2} (a+b x)\right )}{64 b}+\frac{3 \sec ^2\left (\frac{1}{2} (a+b x)\right )}{32 b}+\frac{3 \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{8 b}-\frac{3 \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^5,x]

[Out]

(-3*Csc[(a + b*x)/2]^2)/(32*b) - Csc[(a + b*x)/2]^4/(64*b) - (3*Log[Cos[(a + b*x)/2]])/(8*b) + (3*Log[Sin[(a +
b*x)/2]])/(8*b) + (3*Sec[(a + b*x)/2]^2)/(32*b) + Sec[(a + b*x)/2]^4/(64*b)

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Maple [A]  time = 0.021, size = 59, normalized size = 1.1 \begin{align*} -{\frac{\cot \left ( bx+a \right ) \left ( \csc \left ( bx+a \right ) \right ) ^{3}}{4\,b}}-{\frac{3\,\csc \left ( bx+a \right ) \cot \left ( bx+a \right ) }{8\,b}}+{\frac{3\,\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(csc(b*x+a)^5,x)

[Out]

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

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Maxima [A]  time = 1.06597, size = 96, normalized size = 1.75 \begin{align*} \frac{\frac{2 \,{\left (3 \, \cos \left (b x + a\right )^{3} - 5 \, \cos \left (b x + a\right )\right )}}{\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1} - 3 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 3 \, \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(csc(b*x+a)^5,x, algorithm="maxima")

[Out]

1/16*(2*(3*cos(b*x + a)^3 - 5*cos(b*x + a))/(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1) - 3*log(cos(b*x + a) + 1)
+ 3*log(cos(b*x + a) - 1))/b

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Fricas [B]  time = 0.492159, size = 313, normalized size = 5.69 \begin{align*} \frac{6 \, \cos \left (b x + a\right )^{3} - 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 ) + 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 ) - 10 \, \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(csc(b*x+a)^5,x, algorithm="fricas")

[Out]

1/16*(6*cos(b*x + a)^3 - 3*(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*log(1/2*cos(b*x + a) + 1/2) + 3*(cos(b*x +
a)^4 - 2*cos(b*x + a)^2 + 1)*log(-1/2*cos(b*x + a) + 1/2) - 10*cos(b*x + a))/(b*cos(b*x + a)^4 - 2*b*cos(b*x +
a)^2 + b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \csc ^{5}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**5,x)

[Out]

Integral(csc(a + b*x)**5, x)

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Giac [B]  time = 1.28822, size = 186, normalized size = 3.38 \begin{align*} \frac{\frac{{\left (\frac{8 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac{18 \,{\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{8 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 12 \, \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(csc(b*x+a)^5,x, algorithm="giac")

[Out]

1/64*((8*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 18*(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 - 8*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + (cos(b*x + a) - 1)^2/(cos(b*x + a)
+ 1)^2 + 12*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b