3.379 \(\int (b \csc (e+f x))^m \tan ^4(e+f x) \, dx\)

Optimal. Leaf size=63 \[ \frac{\tan ^3(e+f x) \sin ^2(e+f x)^{\frac{m-3}{2}} (b \csc (e+f x))^m \, _2F_1\left (-\frac{3}{2},\frac{m-3}{2};-\frac{1}{2};\cos ^2(e+f x)\right )}{3 f} \]

[Out]

((b*Csc[e + f*x])^m*Hypergeometric2F1[-3/2, (-3 + m)/2, -1/2, Cos[e + f*x]^2]*(Sin[e + f*x]^2)^((-3 + m)/2)*Ta
n[e + f*x]^3)/(3*f)

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Rubi [A]  time = 0.0361783, 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} \[ \frac{\tan ^3(e+f x) \sin ^2(e+f x)^{\frac{m-3}{2}} (b \csc (e+f x))^m \, _2F_1\left (-\frac{3}{2},\frac{m-3}{2};-\frac{1}{2};\cos ^2(e+f x)\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Csc[e + f*x])^m*Hypergeometric2F1[-3/2, (-3 + m)/2, -1/2, Cos[e + f*x]^2]*(Sin[e + f*x]^2)^((-3 + m)/2)*Ta
n[e + f*x]^3)/(3*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 ^4(e+f x) \, dx &=\frac{(b \csc (e+f x))^m \, _2F_1\left (-\frac{3}{2},\frac{1}{2} (-3+m);-\frac{1}{2};\cos ^2(e+f x)\right ) \sin ^2(e+f x)^{\frac{1}{2} (-3+m)} \tan ^3(e+f x)}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.699332, size = 79, normalized size = 1.25 \[ \frac{\tan ^5(e+f x) \sec ^2(e+f x)^{-m/2} (b \csc (e+f x))^m \, _2F_1\left (1-\frac{m}{2},\frac{5}{2}-\frac{m}{2};\frac{7}{2}-\frac{m}{2};-\tan ^2(e+f x)\right )}{f (5-m)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Csc[e + f*x])^m*Hypergeometric2F1[1 - m/2, 5/2 - m/2, 7/2 - m/2, -Tan[e + f*x]^2]*Tan[e + f*x]^5)/(f*(5 -
m)*(Sec[e + f*x]^2)^(m/2))

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Maple [F]  time = 0.221, size = 0, normalized size = 0. \begin{align*} \int \left ( b\csc \left ( fx+e \right ) \right ) ^{m} \left ( \tan \left ( fx+e \right ) \right ) ^{4}\, 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)^4,x)

[Out]

int((b*csc(f*x+e))^m*tan(f*x+e)^4,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 )^{4}\,{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)^4,x, algorithm="maxima")

[Out]

integrate((b*csc(f*x + e))^m*tan(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} \tan \left (f x + e\right )^{4}, 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)^4,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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 )^{4}\,{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)^4,x, algorithm="giac")

[Out]

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