3.354 \(\int \cot (e+f x) (b \sec (e+f x))^m \, dx\)

Optimal. Leaf size=40 \[ -\frac {(b \sec (e+f x))^m \, _2F_1\left (1,\frac {m}{2};\frac {m+2}{2};\sec ^2(e+f x)\right )}{f m} \]

[Out]

-hypergeom([1, 1/2*m],[1+1/2*m],sec(f*x+e)^2)*(b*sec(f*x+e))^m/f/m

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Rubi [A]  time = 0.04, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2606, 364} \[ -\frac {(b \sec (e+f x))^m \, _2F_1\left (1,\frac {m}{2};\frac {m+2}{2};\sec ^2(e+f x)\right )}{f m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Hypergeometric2F1[1, m/2, (2 + m)/2, Sec[e + f*x]^2]*(b*Sec[e + f*x])^m)/(f*m))

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

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 (e+f x) (b \sec (e+f x))^m \, dx &=\frac {b \operatorname {Subst}\left (\int \frac {(b x)^{-1+m}}{-1+x^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\, _2F_1\left (1,\frac {m}{2};\frac {2+m}{2};\sec ^2(e+f x)\right ) (b \sec (e+f x))^m}{f m}\\ \end {align*}

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Mathematica [B]  time = 0.82, size = 124, normalized size = 3.10 \[ \frac {b \sec ^2\left (\frac {1}{2} (e+f x)\right ) (b \sec (e+f x))^{m-1} \left ((\cos (e+f x)+1) \, _2F_1(1,1-m;2-m;\cos (e+f x))-2^m \sec ^2\left (\frac {1}{2} (e+f x)\right )^{-m} \, _2F_1\left (1-m,1-m;2-m;\frac {1}{2} \cos (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )}{4 f (m-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sec[(e + f*x)/2]^2*((1 + Cos[e + f*x])*Hypergeometric2F1[1, 1 - m, 2 - m, Cos[e + f*x]] - (2^m*Hypergeometr
ic2F1[1 - m, 1 - m, 2 - m, (Cos[e + f*x]*Sec[(e + f*x)/2]^2)/2])/(Sec[(e + f*x)/2]^2)^m)*(b*Sec[e + f*x])^(-1
+ m))/(4*f*(-1 + m))

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (b \sec \left (f x + e\right )\right )^{m} \cot \left (f x + e\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*sec(f*x + e))^m*cot(f*x + e), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.11, size = 0, normalized size = 0.00 \[ \int \cot \left (f x +e \right ) \left (b \sec \left (f x +e \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \mathrm {cot}\left (e+f\,x\right )\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec {\left (e + f x \right )}\right )^{m} \cot {\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*sec(e + f*x))**m*cot(e + f*x), x)

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