3.45 \(\int (d \cot (e+f x))^n \csc ^2(e+f x) \, dx\)

Optimal. Leaf size=25 \[ -\frac {(d \cot (e+f x))^{n+1}}{d f (n+1)} \]

[Out]

-(d*cot(f*x+e))^(1+n)/d/f/(1+n)

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Rubi [A]  time = 0.04, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2607, 32} \[ -\frac {(d \cot (e+f x))^{n+1}}{d f (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cot[e + f*x])^n*Csc[e + f*x]^2,x]

[Out]

-((d*Cot[e + f*x])^(1 + n)/(d*f*(1 + n)))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2607

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 26, normalized size = 1.04 \[ -\frac {\cot (e+f x) (d \cot (e+f x))^n}{f (n+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Cot[e + f*x])^n*Csc[e + f*x]^2,x]

[Out]

-((Cot[e + f*x]*(d*Cot[e + f*x])^n)/(f*(1 + n)))

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fricas [A]  time = 1.32, size = 41, normalized size = 1.64 \[ -\frac {\left (\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}\right )^{n} \cos \left (f x + e\right )}{{\left (f n + f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cot(f*x+e))^n*csc(f*x+e)^2,x, algorithm="fricas")

[Out]

-(d*cos(f*x + e)/sin(f*x + e))^n*cos(f*x + e)/((f*n + f)*sin(f*x + e))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \cot \left (f x + e\right )\right )^{n} \csc \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cot(f*x+e))^n*csc(f*x+e)^2,x, algorithm="giac")

[Out]

integrate((d*cot(f*x + e))^n*csc(f*x + e)^2, x)

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maple [A]  time = 0.26, size = 26, normalized size = 1.04 \[ -\frac {\left (d \cot \left (f x +e \right )\right )^{1+n}}{d f \left (1+n \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cot(f*x+e))^n*csc(f*x+e)^2,x)

[Out]

-(d*cot(f*x+e))^(1+n)/d/f/(1+n)

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maxima [A]  time = 0.32, size = 25, normalized size = 1.00 \[ -\frac {\left (d \cot \left (f x + e\right )\right )^{n + 1}}{d f {\left (n + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cot(f*x+e))^n*csc(f*x+e)^2,x, algorithm="maxima")

[Out]

-(d*cot(f*x + e))^(n + 1)/(d*f*(n + 1))

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mupad [B]  time = 0.34, size = 73, normalized size = 2.92 \[ \frac {\left (\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{2\,n+2}-\frac {1}{2\,n+2}\right )\,{\left (-\frac {d\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}{2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}^n}{f\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cot(e + f*x))^n/sin(e + f*x)^2,x)

[Out]

((tan(e/2 + (f*x)/2)^2/(2*n + 2) - 1/(2*n + 2))*(-(d*(tan(e/2 + (f*x)/2)^2 - 1))/(2*tan(e/2 + (f*x)/2)))^n)/(f
*tan(e/2 + (f*x)/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \cot {\left (e + f x \right )}\right )^{n} \csc ^{2}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cot(f*x+e))**n*csc(f*x+e)**2,x)

[Out]

Integral((d*cot(e + f*x))**n*csc(e + f*x)**2, x)

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