3.3.80 \(\int \sec ^{\frac {10}{3}}(e+f x) \sin ^4(e+f x) \, dx\) [280]

Optimal. Leaf size=53 \[ \frac {3 \, _2F_1\left (-\frac {3}{2},-\frac {7}{6};-\frac {1}{6};\cos ^2(e+f x)\right ) \sec ^{\frac {7}{3}}(e+f x) \sin (e+f x)}{7 f \sqrt {\sin ^2(e+f x)}} \]

[Out]

3/7*hypergeom([-3/2, -7/6],[-1/6],cos(f*x+e)^2)*sec(f*x+e)^(7/3)*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 {7}{3}}(e+f x) \, _2F_1\left (-\frac {3}{2},-\frac {7}{6};-\frac {1}{6};\cos ^2(e+f x)\right )}{7 f \sqrt {\sin ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^(10/3)*Sin[e + f*x]^4,x]

[Out]

(3*Hypergeometric2F1[-3/2, -7/6, -1/6, Cos[e + f*x]^2]*Sec[e + f*x]^(7/3)*Sin[e + f*x])/(7*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 {10}{3}}(e+f x) \sin ^4(e+f x) \, dx &=\left (\sqrt [3]{\cos (e+f x)} \sqrt [3]{\sec (e+f x)}\right ) \int \frac {\sin ^4(e+f x)}{\cos ^{\frac {10}{3}}(e+f x)} \, dx\\ &=\frac {3 \, _2F_1\left (-\frac {3}{2},-\frac {7}{6};-\frac {1}{6};\cos ^2(e+f x)\right ) \sec ^{\frac {7}{3}}(e+f x) \sin (e+f x)}{7 f \sqrt {\sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 77, normalized size = 1.45 \begin {gather*} \frac {3 \sqrt [3]{\sec (e+f x)} \left (-10 \sin (e+f x)+9 \sqrt [6]{\cos ^2(e+f x)} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {3}{2};\sin ^2(e+f x)\right ) \sin (e+f x)+\sec (e+f x) \tan (e+f x)\right )}{7 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^(10/3)*Sin[e + f*x]^4,x]

[Out]

(3*Sec[e + f*x]^(1/3)*(-10*Sin[e + f*x] + 9*(Cos[e + f*x]^2)^(1/6)*Hypergeometric2F1[1/6, 1/2, 3/2, Sin[e + f*
x]^2]*Sin[e + f*x] + Sec[e + f*x]*Tan[e + f*x]))/(7*f)

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\tan ^{4}\left (f x +e \right )}{\sec \left (f x +e \right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^4/sec(f*x+e)^(2/3),x)

[Out]

int(tan(f*x+e)^4/sec(f*x+e)^(2/3),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(tan(f*x+e)^4/sec(f*x+e)^(2/3),x, algorithm="maxima")

[Out]

integrate(tan(f*x + e)^4/sec(f*x + e)^(2/3), 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(tan(f*x+e)^4/sec(f*x+e)^(2/3),x, algorithm="fricas")

[Out]

integral(tan(f*x + e)^4/sec(f*x + e)^(2/3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{4}{\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(tan(f*x+e)**4/sec(f*x+e)**(2/3),x)

[Out]

Integral(tan(e + f*x)**4/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(tan(f*x+e)^4/sec(f*x+e)^(2/3),x, algorithm="giac")

[Out]

integrate(tan(f*x + e)^4/sec(f*x + e)^(2/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^4}{{\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)^4/(1/cos(e + f*x))^(2/3),x)

[Out]

int(tan(e + f*x)^4/(1/cos(e + f*x))^(2/3), x)

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