\(\int \frac {\cos ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx\) [314]

Optimal result

Integrand size = 21, antiderivative size = 58 \[ \int \frac {\cos ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\frac {3 \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{3},\frac {4}{3},\sin ^2(e+f x)\right ) (b \sin (e+f x))^{2/3}}{2 b f \sqrt {\cos ^2(e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2657} \[ \int \frac {\cos ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\frac {3 \cos (e+f x) (b \sin (e+f x))^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{3},\frac {4}{3},\sin ^2(e+f x)\right )}{2 b f \sqrt {\cos ^2(e+f x)}} \]

[In]

Int[Cos[e + f*x]^4/(b*Sin[e + f*x])^(1/3),x]

[Out]

(3*Cos[e + f*x]*Hypergeometric2F1[-3/2, 1/3, 4/3, Sin[e + f*x]^2]*(b*Sin[e + f*x])^(2/3))/(2*b*f*Sqrt[Cos[e +
f*x]^2])

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*} \text {integral}& = \frac {3 \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{3},\frac {4}{3},\sin ^2(e+f x)\right ) (b \sin (e+f x))^{2/3}}{2 b f \sqrt {\cos ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\frac {3 \sqrt {\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{3},\frac {4}{3},\sin ^2(e+f x)\right ) \tan (e+f x)}{2 f \sqrt [3]{b \sin (e+f x)}} \]

[In]

Integrate[Cos[e + f*x]^4/(b*Sin[e + f*x])^(1/3),x]

[Out]

(3*Sqrt[Cos[e + f*x]^2]*Hypergeometric2F1[-3/2, 1/3, 4/3, Sin[e + f*x]^2]*Tan[e + f*x])/(2*f*(b*Sin[e + f*x])^
(1/3))

Maple [F]

\[\int \frac {\cos ^{4}\left (f x +e \right )}{\left (b \sin \left (f x +e \right )\right )^{\frac {1}{3}}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\cos ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{\left (b \sin \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \]

[In]

integrate(cos(f*x+e)^4/(b*sin(f*x+e))^(1/3),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\text {Timed out} \]

[In]

integrate(cos(f*x+e)**4/(b*sin(f*x+e))**(1/3),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{\left (b \sin \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \]

[In]

integrate(cos(f*x+e)^4/(b*sin(f*x+e))^(1/3),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {\cos ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int { \frac {\cos \left (f x + e\right )^{4}}{\left (b \sin \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \]

[In]

integrate(cos(f*x+e)^4/(b*sin(f*x+e))^(1/3),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^4(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^4}{{\left (b\,\sin \left (e+f\,x\right )\right )}^{1/3}} \,d x \]

[In]

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

[Out]

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