3.4.19 \(\int \frac {\cos ^4(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx\) [319]

Optimal. Leaf size=58 \[ -\frac {3 \cos (e+f x) \, _2F_1\left (-\frac {3}{2},-\frac {1}{3};\frac {2}{3};\sin ^2(e+f x)\right )}{2 b f \sqrt {\cos ^2(e+f x)} (b \sin (e+f x))^{2/3}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2657} \begin {gather*} -\frac {3 \cos (e+f x) \, _2F_1\left (-\frac {3}{2},-\frac {1}{3};\frac {2}{3};\sin ^2(e+f x)\right )}{2 b f \sqrt {\cos ^2(e+f x)} (b \sin (e+f x))^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^4/(b*Sin[e + f*x])^(5/3),x]

[Out]

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

Rule 2657

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

Rubi steps

\begin {align*} \int \frac {\cos ^4(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx &=-\frac {3 \cos (e+f x) \, _2F_1\left (-\frac {3}{2},-\frac {1}{3};\frac {2}{3};\sin ^2(e+f x)\right )}{2 b f \sqrt {\cos ^2(e+f x)} (b \sin (e+f x))^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 55, normalized size = 0.95 \begin {gather*} -\frac {3 \sqrt {\cos ^2(e+f x)} \, _2F_1\left (-\frac {3}{2},-\frac {1}{3};\frac {2}{3};\sin ^2(e+f x)\right ) \tan (e+f x)}{2 f (b \sin (e+f x))^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^4/(b*Sin[e + f*x])^(5/3),x]

[Out]

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

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\cos ^{4}\left (f x +e \right )}{\left (b \sin \left (f x +e \right )\right )^{\frac {5}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^4/(b*sin(f*x+e))^(5/3),x)

[Out]

int(cos(f*x+e)^4/(b*sin(f*x+e))^(5/3),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(cos(f*x+e)^4/(b*sin(f*x+e))^(5/3),x, algorithm="maxima")

[Out]

integrate(cos(f*x + e)^4/(b*sin(f*x + e))^(5/3), 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(cos(f*x+e)^4/(b*sin(f*x+e))^(5/3),x, algorithm="fricas")

[Out]

integral(-(b*sin(f*x + e))^(1/3)*cos(f*x + e)^4/(b^2*cos(f*x + e)^2 - b^2), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**4/(b*sin(f*x+e))**(5/3),x)

[Out]

Timed out

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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(cos(f*x+e)^4/(b*sin(f*x+e))^(5/3),x, algorithm="giac")

[Out]

integrate(cos(f*x + e)^4/(b*sin(f*x + e))^(5/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\cos \left (e+f\,x\right )}^4}{{\left (b\,\sin \left (e+f\,x\right )\right )}^{5/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(e + f*x)^4/(b*sin(e + f*x))^(5/3),x)

[Out]

int(cos(e + f*x)^4/(b*sin(e + f*x))^(5/3), x)

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