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]

1/3*(b*csc(f*x+e))^m*hypergeom([-3/2, -3/2+1/2*m],[-1/2],cos(f*x+e)^2)*(sin(f*x+e)^2)^(-3/2+1/2*m)*tan(f*x+e)^
3/f

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Rubi [A]  time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, 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*}

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Mathematica [B]  time = 1.38, size = 212, normalized size = 3.37 \[ -\frac {\tan (e+f x) \sec ^2(e+f x)^{-m/2} (b \csc (e+f x))^m \left (\sqrt {\sin ^2(e+f x)} \, _2F_1\left (\frac {1-m}{2},-\frac {m}{2}-1;\frac {3-m}{2};-\tan ^2(e+f x)\right )-2 \sqrt {\sin ^2(e+f x)} \, _2F_1\left (\frac {1-m}{2},-\frac {m}{2};\frac {3-m}{2};-\tan ^2(e+f x)\right )+(m-1) \cos ^2(e+f x) \sin ^2(e+f x)^{m/2} \sec ^2(e+f x)^{m/2} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {3}{2};\cos ^2(e+f x)\right )\right )}{f (m-1) \sqrt {\sin ^2(e+f x)}} \]

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, -1 - m/2, (3 - m)/2, -Tan[e + f*x]^2]*Sqrt[Sin[e + f*x]^2]
- 2*Hypergeometric2F1[(1 - m)/2, -1/2*m, (3 - m)/2, -Tan[e + f*x]^2]*Sqrt[Sin[e + f*x]^2] + (-1 + m)*Cos[e +
f*x]^2*Hypergeometric2F1[1/2, (1 + m)/2, 3/2, Cos[e + f*x]^2]*(Sec[e + f*x]^2)^(m/2)*(Sin[e + f*x]^2)^(m/2))*T
an[e + f*x])/(f*(-1 + m)*(Sec[e + f*x]^2)^(m/2)*Sqrt[Sin[e + f*x]^2]))

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (b \csc \left (f x + e\right )\right )^{m} \tan \left (f x + e\right )^{4}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \csc \left (f x + e\right )\right )^{m} \tan \left (f x + e\right )^{4}\,{d x} \]

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)

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maple [F]  time = 0.48, size = 0, normalized size = 0.00 \[ \int \left (b \csc \left (f x +e \right )\right )^{m} \left (\tan ^{4}\left (f x +e \right )\right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \csc \left (f x + e\right )\right )^{m} \tan \left (f x + e\right )^{4}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^4\,{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^4*(b/sin(e + f*x))^m,x)

[Out]

int(tan(e + f*x)^4*(b/sin(e + f*x))^m, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \csc {\left (e + f x \right )}\right )^{m} \tan ^{4}{\left (e + f x \right )}\, dx \]

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]

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

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