3.380 \(\int (b \csc (e+f x))^m \tan ^2(e+f x) \, dx\)

Optimal. Leaf size=58 \[ \frac{\tan (e+f x) \sin ^2(e+f x)^{\frac{m-1}{2}} (b \csc (e+f x))^m \, _2F_1\left (-\frac{1}{2},\frac{m-1}{2};\frac{1}{2};\cos ^2(e+f x)\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.0357518, antiderivative size = 58, 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} \[ \frac{\tan (e+f x) \sin ^2(e+f x)^{\frac{m-1}{2}} (b \csc (e+f x))^m \, _2F_1\left (-\frac{1}{2},\frac{m-1}{2};\frac{1}{2};\cos ^2(e+f x)\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[(b*Csc[e + f*x])^m*Tan[e + f*x]^2,x]

[Out]

((b*Csc[e + f*x])^m*Hypergeometric2F1[-1/2, (-1 + m)/2, 1/2, Cos[e + f*x]^2]*(Sin[e + f*x]^2)^((-1 + m)/2)*Tan
[e + f*x])/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 (b \csc (e+f x))^m \tan ^2(e+f x) \, dx &=\frac{(b \csc (e+f x))^m \, _2F_1\left (-\frac{1}{2},\frac{1}{2} (-1+m);\frac{1}{2};\cos ^2(e+f x)\right ) \sin ^2(e+f x)^{\frac{1}{2} (-1+m)} \tan (e+f x)}{f}\\ \end{align*}

Mathematica [A]  time = 0.56063, size = 79, normalized size = 1.36 \[ \frac{\tan ^3(e+f x) \sec ^2(e+f x)^{-m/2} (b \csc (e+f x))^m \, _2F_1\left (1-\frac{m}{2},\frac{3}{2}-\frac{m}{2};\frac{5}{2}-\frac{m}{2};-\tan ^2(e+f x)\right )}{f (3-m)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Csc[e + f*x])^m*Tan[e + f*x]^2,x]

[Out]

((b*Csc[e + f*x])^m*Hypergeometric2F1[1 - m/2, 3/2 - m/2, 5/2 - m/2, -Tan[e + f*x]^2]*Tan[e + f*x]^3)/(f*(3 -
m)*(Sec[e + f*x]^2)^(m/2))

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Maple [F]  time = 0.199, size = 0, normalized size = 0. \begin{align*} \int \left ( b\csc \left ( fx+e \right ) \right ) ^{m} \left ( \tan \left ( fx+e \right ) \right ) ^{2}\, 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)^2,x)

[Out]

int((b*csc(f*x+e))^m*tan(f*x+e)^2,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 )^{2}\,{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)^2,x, algorithm="maxima")

[Out]

integrate((b*csc(f*x + e))^m*tan(f*x + e)^2, 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 )^{2}, 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)^2,x, algorithm="fricas")

[Out]

integral((b*csc(f*x + e))^m*tan(f*x + e)^2, 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 ^{2}{\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)**2,x)

[Out]

Integral((b*csc(e + f*x))**m*tan(e + f*x)**2, 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 )^{2}\,{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)^2,x, algorithm="giac")

[Out]

integrate((b*csc(f*x + e))^m*tan(f*x + e)^2, x)