3.377 \(\int \cot ^3(e+f x) (b \csc (e+f x))^m \, dx\)

Optimal. Leaf size=43 \[ \frac {(b \csc (e+f x))^m}{f m}-\frac {(b \csc (e+f x))^{m+2}}{b^2 f (m+2)} \]

[Out]

(b*csc(f*x+e))^m/f/m-(b*csc(f*x+e))^(2+m)/b^2/f/(2+m)

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Rubi [A]  time = 0.05, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2606, 14} \[ \frac {(b \csc (e+f x))^m}{f m}-\frac {(b \csc (e+f x))^{m+2}}{b^2 f (m+2)} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^3*(b*Csc[e + f*x])^m,x]

[Out]

(b*Csc[e + f*x])^m/(f*m) - (b*Csc[e + f*x])^(2 + m)/(b^2*f*(2 + m))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rubi steps

\begin {align*} \int \cot ^3(e+f x) (b \csc (e+f x))^m \, dx &=-\frac {b \operatorname {Subst}\left (\int (b x)^{-1+m} \left (-1+x^2\right ) \, dx,x,\csc (e+f x)\right )}{f}\\ &=-\frac {b \operatorname {Subst}\left (\int \left (-(b x)^{-1+m}+\frac {(b x)^{1+m}}{b^2}\right ) \, dx,x,\csc (e+f x)\right )}{f}\\ &=\frac {(b \csc (e+f x))^m}{f m}-\frac {(b \csc (e+f x))^{2+m}}{b^2 f (2+m)}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 36, normalized size = 0.84 \[ \frac {\left (-m \csc ^2(e+f x)+m+2\right ) (b \csc (e+f x))^m}{f m (m+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^3*(b*Csc[e + f*x])^m,x]

[Out]

((b*Csc[e + f*x])^m*(2 + m - m*Csc[e + f*x]^2))/(f*m*(2 + m))

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fricas [A]  time = 0.55, size = 60, normalized size = 1.40 \[ -\frac {{\left ({\left (m + 2\right )} \cos \left (f x + e\right )^{2} - 2\right )} \left (\frac {b}{\sin \left (f x + e\right )}\right )^{m}}{f m^{2} - {\left (f m^{2} + 2 \, f m\right )} \cos \left (f x + e\right )^{2} + 2 \, f m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(b*csc(f*x+e))^m,x, algorithm="fricas")

[Out]

-((m + 2)*cos(f*x + e)^2 - 2)*(b/sin(f*x + e))^m/(f*m^2 - (f*m^2 + 2*f*m)*cos(f*x + e)^2 + 2*f*m)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \csc \left (f x + e\right )\right )^{m} \cot \left (f x + e\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(b*csc(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((b*csc(f*x + e))^m*cot(f*x + e)^3, x)

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maple [C]  time = 1.26, size = 6612, normalized size = 153.77 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^3*(b*csc(f*x+e))^m,x)

[Out]

result too large to display

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maxima [A]  time = 0.67, size = 50, normalized size = 1.16 \[ \frac {\frac {b^{m} \sin \left (f x + e\right )^{-m}}{m} - \frac {b^{m} \sin \left (f x + e\right )^{-m}}{{\left (m + 2\right )} \sin \left (f x + e\right )^{2}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(b*csc(f*x+e))^m,x, algorithm="maxima")

[Out]

(b^m*sin(f*x + e)^(-m)/m - b^m*sin(f*x + e)^(-m)/((m + 2)*sin(f*x + e)^2))/f

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mupad [B]  time = 3.44, size = 92, normalized size = 2.14 \[ \frac {{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^m\,\left (m+4\,{\sin \left (2\,e+2\,f\,x\right )}^2+m\,\left (2\,{\sin \left (2\,e+2\,f\,x\right )}^2-1\right )-16\,{\sin \left (e+f\,x\right )}^2\right )}{f\,m\,\left (2\,{\sin \left (2\,e+2\,f\,x\right )}^2-8\,{\sin \left (e+f\,x\right )}^2\right )\,\left (m+2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^3*(b/sin(e + f*x))^m,x)

[Out]

((b/sin(e + f*x))^m*(m + 4*sin(2*e + 2*f*x)^2 + m*(2*sin(2*e + 2*f*x)^2 - 1) - 16*sin(e + f*x)^2))/(f*m*(2*sin
(2*e + 2*f*x)^2 - 8*sin(e + f*x)^2)*(m + 2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} x \left (b \csc {\relax (e )}\right )^{m} \cot ^{3}{\relax (e )} & \text {for}\: f = 0 \\\frac {\int \frac {\cot ^{3}{\left (e + f x \right )}}{\csc ^{2}{\left (e + f x \right )}}\, dx}{b^{2}} & \text {for}\: m = -2 \\\frac {\log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} - \frac {\log {\left (\tan {\left (e + f x \right )} \right )}}{f} - \frac {1}{2 f \tan ^{2}{\left (e + f x \right )}} & \text {for}\: m = 0 \\- \frac {b^{m} m \cot ^{2}{\left (e + f x \right )} \csc ^{m}{\left (e + f x \right )}}{f m^{2} + 2 f m} + \frac {2 b^{m} \csc ^{m}{\left (e + f x \right )}}{f m^{2} + 2 f m} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**3*(b*csc(f*x+e))**m,x)

[Out]

Piecewise((x*(b*csc(e))**m*cot(e)**3, Eq(f, 0)), (Integral(cot(e + f*x)**3/csc(e + f*x)**2, x)/b**2, Eq(m, -2)
), (log(tan(e + f*x)**2 + 1)/(2*f) - log(tan(e + f*x))/f - 1/(2*f*tan(e + f*x)**2), Eq(m, 0)), (-b**m*m*cot(e
+ f*x)**2*csc(e + f*x)**m/(f*m**2 + 2*f*m) + 2*b**m*csc(e + f*x)**m/(f*m**2 + 2*f*m), True))

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