3.3.72 \(\int \sec ^{\frac {8}{3}}(e+f x) \sin ^2(e+f x) \, dx\) [272]

Optimal. Leaf size=53 \[ \frac {3 \, _2F_1\left (-\frac {5}{6},-\frac {1}{2};\frac {1}{6};\cos ^2(e+f x)\right ) \sec ^{\frac {5}{3}}(e+f x) \sin (e+f x)}{5 f \sqrt {\sin ^2(e+f x)}} \]

[Out]

3/5*hypergeom([-5/6, -1/2],[1/6],cos(f*x+e)^2)*sec(f*x+e)^(5/3)*sin(f*x+e)/f/(sin(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2712, 2656} \begin {gather*} \frac {3 \sin (e+f x) \sec ^{\frac {5}{3}}(e+f x) \, _2F_1\left (-\frac {5}{6},-\frac {1}{2};\frac {1}{6};\cos ^2(e+f x)\right )}{5 f \sqrt {\sin ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^(8/3)*Sin[e + f*x]^2,x]

[Out]

(3*Hypergeometric2F1[-5/6, -1/2, 1/6, Cos[e + f*x]^2]*Sec[e + f*x]^(5/3)*Sin[e + f*x])/(5*f*Sqrt[Sin[e + f*x]^
2])

Rule 2656

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^(2*IntPar
t[(n - 1)/2] + 1))*(b*Sin[e + f*x])^(2*FracPart[(n - 1)/2])*((a*Cos[e + f*x])^(m + 1)/(a*f*(m + 1)*(Sin[e + f*
x]^2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Cos[e + f*x]^2], x] /; FreeQ[{a
, b, e, f, m, n}, x] && SimplerQ[n, m]

Rule 2712

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a^2/b^2)*(a*
Sec[e + f*x])^(m - 1)*(b*Csc[e + f*x])^(n + 1)*(a*Cos[e + f*x])^(m - 1)*(b*Sin[e + f*x])^(n + 1), Int[1/((a*Co
s[e + f*x])^m*(b*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, e, f, m, n}, x]

Rubi steps

\begin {align*} \int \sec ^{\frac {8}{3}}(e+f x) \sin ^2(e+f x) \, dx &=\left (\cos ^{\frac {2}{3}}(e+f x) \sec ^{\frac {2}{3}}(e+f x)\right ) \int \frac {\sin ^2(e+f x)}{\cos ^{\frac {8}{3}}(e+f x)} \, dx\\ &=\frac {3 \, _2F_1\left (-\frac {5}{6},-\frac {1}{2};\frac {1}{6};\cos ^2(e+f x)\right ) \sec ^{\frac {5}{3}}(e+f x) \sin (e+f x)}{5 f \sqrt {\sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 56, normalized size = 1.06 \begin {gather*} -\frac {3 \left (-1+\cos ^2(e+f x)^{5/6} \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\sin ^2(e+f x)\right )\right ) \sec ^{\frac {5}{3}}(e+f x) \sin (e+f x)}{5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^(8/3)*Sin[e + f*x]^2,x]

[Out]

(-3*(-1 + (Cos[e + f*x]^2)^(5/6)*Hypergeometric2F1[1/2, 5/6, 3/2, Sin[e + f*x]^2])*Sec[e + f*x]^(5/3)*Sin[e +
f*x])/(5*f)

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Maple [F]
time = 0.09, size = 0, normalized size = 0.00 \[\int \left (\sec ^{\frac {2}{3}}\left (f x +e \right )\right ) \left (\tan ^{2}\left (f x +e \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^(2/3)*tan(f*x+e)^2,x)

[Out]

int(sec(f*x+e)^(2/3)*tan(f*x+e)^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^(2/3)*tan(f*x+e)^2,x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)^(2/3)*tan(f*x + e)^2, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^(2/3)*tan(f*x+e)^2,x, algorithm="fricas")

[Out]

integral(sec(f*x + e)^(2/3)*tan(f*x + e)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \tan ^{2}{\left (e + f x \right )} \sec ^{\frac {2}{3}}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**(2/3)*tan(f*x+e)**2,x)

[Out]

Integral(tan(e + f*x)**2*sec(e + f*x)**(2/3), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^(2/3)*tan(f*x+e)^2,x, algorithm="giac")

[Out]

integrate(sec(f*x + e)^(2/3)*tan(f*x + e)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\mathrm {tan}\left (e+f\,x\right )}^2\,{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^{2/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^2*(1/cos(e + f*x))^(2/3),x)

[Out]

int(tan(e + f*x)^2*(1/cos(e + f*x))^(2/3), x)

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