3.382 \(\int \cot ^4(e+f x) (b \csc (e+f x))^m \, dx\)

Optimal. Leaf size=63 \[ -\frac{\cot ^5(e+f x) \sin ^2(e+f x)^{\frac{m+5}{2}} (b \csc (e+f x))^m \, _2F_1\left (\frac{5}{2},\frac{m+5}{2};\frac{7}{2};\cos ^2(e+f x)\right )}{5 f} \]

[Out]

-(Cot[e + f*x]^5*(b*Csc[e + f*x])^m*Hypergeometric2F1[5/2, (5 + m)/2, 7/2, Cos[e + f*x]^2]*(Sin[e + f*x]^2)^((
5 + m)/2))/(5*f)

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Rubi [A]  time = 0.0364904, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {2617} \[ -\frac{\cot ^5(e+f x) \sin ^2(e+f x)^{\frac{m+5}{2}} (b \csc (e+f x))^m \, _2F_1\left (\frac{5}{2},\frac{m+5}{2};\frac{7}{2};\cos ^2(e+f x)\right )}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^4*(b*Csc[e + f*x])^m,x]

[Out]

-(Cot[e + f*x]^5*(b*Csc[e + f*x])^m*Hypergeometric2F1[5/2, (5 + m)/2, 7/2, Cos[e + f*x]^2]*(Sin[e + f*x]^2)^((
5 + m)/2))/(5*f)

Rule 2617

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1)*(Cos[e + f*x]^2)^((m + n + 1)/2)*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2,
(n + 3)/2, Sin[e + f*x]^2])/(b*f*(n + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rubi steps

\begin{align*} \int \cot ^4(e+f x) (b \csc (e+f x))^m \, dx &=-\frac{\cot ^5(e+f x) (b \csc (e+f x))^m \, _2F_1\left (\frac{5}{2},\frac{5+m}{2};\frac{7}{2};\cos ^2(e+f x)\right ) \sin ^2(e+f x)^{\frac{5+m}{2}}}{5 f}\\ \end{align*}

Mathematica [B]  time = 4.85394, size = 302, normalized size = 4.79 \[ \frac{\cot ^3\left (\frac{1}{2} (e+f x)\right ) \sec ^2\left (\frac{1}{2} (e+f x)\right )^{-m} (b \csc (e+f x))^m \left (\tan ^4\left (\frac{1}{2} (e+f x)\right ) \left (\frac{\tan ^2\left (\frac{1}{2} (e+f x)\right ) \, _2F_1\left (\frac{3}{2}-\frac{m}{2},-m;\frac{5}{2}-\frac{m}{2};-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{3-m}-\frac{16 \, _2F_1\left (1-m,\frac{1}{2}-\frac{m}{2};\frac{3}{2}-\frac{m}{2};-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{m-1}+\frac{5 \, _2F_1\left (\frac{1}{2}-\frac{m}{2},-m;\frac{3}{2}-\frac{m}{2};-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{m-1}\right )+\frac{5 \tan ^2\left (\frac{1}{2} (e+f x)\right ) \, _2F_1\left (-\frac{m}{2}-\frac{1}{2},-m;\frac{1}{2}-\frac{m}{2};-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{m+1}-\frac{\, _2F_1\left (-\frac{m}{2}-\frac{3}{2},-m;-\frac{m}{2}-\frac{1}{2};-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{m+3}\right )}{8 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^4*(b*Csc[e + f*x])^m,x]

[Out]

(Cot[(e + f*x)/2]^3*(b*Csc[e + f*x])^m*(-(Hypergeometric2F1[-3/2 - m/2, -m, -1/2 - m/2, -Tan[(e + f*x)/2]^2]/(
3 + m)) + (5*Hypergeometric2F1[-1/2 - m/2, -m, 1/2 - m/2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)/(1 + m) + T
an[(e + f*x)/2]^4*((-16*Hypergeometric2F1[1 - m, 1/2 - m/2, 3/2 - m/2, -Tan[(e + f*x)/2]^2])/(-1 + m) + (5*Hyp
ergeometric2F1[1/2 - m/2, -m, 3/2 - m/2, -Tan[(e + f*x)/2]^2])/(-1 + m) + (Hypergeometric2F1[3/2 - m/2, -m, 5/
2 - m/2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)/(3 - m))))/(8*f*(Sec[(e + f*x)/2]^2)^m)

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Maple [F]  time = 0.234, size = 0, normalized size = 0. \begin{align*} \int \left ( \cot \left ( fx+e \right ) \right ) ^{4} \left ( b\csc \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^4*(b*csc(f*x+e))^m,x)

[Out]

int(cot(f*x+e)^4*(b*csc(f*x+e))^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \csc \left (f x + e\right )\right )^{m} \cot \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4*(b*csc(f*x+e))^m,x, algorithm="maxima")

[Out]

integrate((b*csc(f*x + e))^m*cot(f*x + e)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (b \csc \left (f x + e\right )\right )^{m} \cot \left (f x + e\right )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4*(b*csc(f*x+e))^m,x, algorithm="fricas")

[Out]

integral((b*csc(f*x + e))^m*cot(f*x + e)^4, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \csc{\left (e + f x \right )}\right )^{m} \cot ^{4}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**4*(b*csc(f*x+e))**m,x)

[Out]

Integral((b*csc(e + f*x))**m*cot(e + f*x)**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \csc \left (f x + e\right )\right )^{m} \cot \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4*(b*csc(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((b*csc(f*x + e))^m*cot(f*x + e)^4, x)