3.320 \(\int \frac {\cos ^2(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx\)

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

[Out]

-3/2*cos(f*x+e)*hypergeom([-1/2, -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)

________________________________________________________________________________________

Rubi [A]  time = 0.05, 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 = {2577} \[ -\frac {3 \cos (e+f x) \, _2F_1\left (-\frac {1}{2},-\frac {1}{3};\frac {2}{3};\sin ^2(e+f x)\right )}{2 b f \sqrt {\cos ^2(e+f x)} (b \sin (e+f x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^2/(b*Sin[e + f*x])^(5/3),x]

[Out]

(-3*Cos[e + f*x]*Hypergeometric2F1[-1/2, -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 2577

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)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 55, normalized size = 0.95 \[ -\frac {3 \sqrt {\cos ^2(e+f x)} \tan (e+f x) \, _2F_1\left (-\frac {1}{2},-\frac {1}{3};\frac {2}{3};\sin ^2(e+f x)\right )}{2 f (b \sin (e+f x))^{5/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^2/(b*Sin[e + f*x])^(5/3),x]

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\left (b \sin \left (f x + e\right )\right )^{\frac {1}{3}} \cos \left (f x + e\right )^{2}}{b^{2} \cos \left (f x + e\right )^{2} - b^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2/(b*sin(f*x+e))^(5/3),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (f x + e\right )^{2}}{\left (b \sin \left (f x + e\right )\right )^{\frac {5}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2/(b*sin(f*x+e))^(5/3),x, algorithm="giac")

[Out]

integrate(cos(f*x + e)^2/(b*sin(f*x + e))^(5/3), x)

________________________________________________________________________________________

maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{2}\left (f x +e \right )}{\left (b \sin \left (f x +e \right )\right )^{\frac {5}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2/(b*sin(f*x+e))^(5/3),x)

[Out]

int(cos(f*x+e)^2/(b*sin(f*x+e))^(5/3),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (f x + e\right )^{2}}{\left (b \sin \left (f x + e\right )\right )^{\frac {5}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2/(b*sin(f*x+e))^(5/3),x, algorithm="maxima")

[Out]

integrate(cos(f*x + e)^2/(b*sin(f*x + e))^(5/3), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\cos \left (e+f\,x\right )}^2}{{\left (b\,\sin \left (e+f\,x\right )\right )}^{5/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(e + f*x)^2/(b*sin(e + f*x))^(5/3),x)

[Out]

int(cos(e + f*x)^2/(b*sin(e + f*x))^(5/3), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**2/(b*sin(f*x+e))**(5/3),x)

[Out]

Timed out

________________________________________________________________________________________