3.375 \(\int (b \csc (e+f x))^m \tan (e+f x) \, dx\)

Optimal. Leaf size=39 \[ \frac{(b \csc (e+f x))^m \, _2F_1\left (1,\frac{m}{2};\frac{m+2}{2};\csc ^2(e+f x)\right )}{f m} \]

[Out]

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

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Rubi [A]  time = 0.0357686, antiderivative size = 39, normalized size of antiderivative = 1., 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} \[ \frac{(b \csc (e+f x))^m \, _2F_1\left (1,\frac{m}{2};\frac{m+2}{2};\csc ^2(e+f x)\right )}{f m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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])

Rubi steps

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

Mathematica [A]  time = 0.0491627, size = 52, normalized size = 1.33 \[ -\frac{\sin ^2(e+f x) (b \csc (e+f x))^m \, _2F_1\left (1,1-\frac{m}{2};2-\frac{m}{2};\sin ^2(e+f x)\right )}{f (m-2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.441, size = 0, normalized size = 0. \begin{align*} \int \left ( b\csc \left ( fx+e \right ) \right ) ^{m}\tan \left ( fx+e \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((b*csc(f*x+e))^m*tan(f*x+e),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} \tan \left (f x + e\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*csc(f*x + e))^m*tan(f*x + e), 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} \tan \left (f x + e\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*csc(f*x + e))^m*tan(f*x + e), 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} \tan{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*csc(e + f*x))**m*tan(e + f*x), 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} \tan \left (f x + e\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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