\(\int \cos ^{\frac {4}{3}}(a+b x) \, dx\) [25]

Optimal result

Integrand size = 10, antiderivative size = 53 \[ \int \cos ^{\frac {4}{3}}(a+b x) \, dx=-\frac {3 \cos ^{\frac {7}{3}}(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(a+b x)\right ) \sin (a+b x)}{7 b \sqrt {\sin ^2(a+b x)}} \]

[Out]

-3/7*cos(b*x+a)^(7/3)*hypergeom([1/2, 7/6],[13/6],cos(b*x+a)^2)*sin(b*x+a)/b/(sin(b*x+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2722} \[ \int \cos ^{\frac {4}{3}}(a+b x) \, dx=-\frac {3 \sin (a+b x) \cos ^{\frac {7}{3}}(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(a+b x)\right )}{7 b \sqrt {\sin ^2(a+b x)}} \]

[In]

Int[Cos[a + b*x]^(4/3),x]

[Out]

(-3*Cos[a + b*x]^(7/3)*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[a + b*x]^2]*Sin[a + b*x])/(7*b*Sqrt[Sin[a + b*x]^
2])

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 \cos ^{\frac {7}{3}}(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(a+b x)\right ) \sin (a+b x)}{7 b \sqrt {\sin ^2(a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int \cos ^{\frac {4}{3}}(a+b x) \, dx=-\frac {3 \cos ^{\frac {7}{3}}(a+b x) \csc (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(a+b x)\right ) \sqrt {\sin ^2(a+b x)}}{7 b} \]

[In]

Integrate[Cos[a + b*x]^(4/3),x]

[Out]

(-3*Cos[a + b*x]^(7/3)*Csc[a + b*x]*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[a + b*x]^2]*Sqrt[Sin[a + b*x]^2])/(7
*b)

Maple [F]

\[\int \left (\cos ^{\frac {4}{3}}\left (b x +a \right )\right )d x\]

[In]

int(cos(b*x+a)^(4/3),x)

[Out]

int(cos(b*x+a)^(4/3),x)

Fricas [F]

\[ \int \cos ^{\frac {4}{3}}(a+b x) \, dx=\int { \cos \left (b x + a\right )^{\frac {4}{3}} \,d x } \]

[In]

integrate(cos(b*x+a)^(4/3),x, algorithm="fricas")

[Out]

integral(cos(b*x + a)^(4/3), x)

Sympy [F(-1)]

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

[In]

integrate(cos(b*x+a)**(4/3),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos ^{\frac {4}{3}}(a+b x) \, dx=\int { \cos \left (b x + a\right )^{\frac {4}{3}} \,d x } \]

[In]

integrate(cos(b*x+a)^(4/3),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^(4/3), x)

Giac [F]

\[ \int \cos ^{\frac {4}{3}}(a+b x) \, dx=\int { \cos \left (b x + a\right )^{\frac {4}{3}} \,d x } \]

[In]

integrate(cos(b*x+a)^(4/3),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)^(4/3), x)

Mupad [B] (verification not implemented)

Time = 13.89 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.79 \[ \int \cos ^{\frac {4}{3}}(a+b x) \, dx=-\frac {3\,{\cos \left (a+b\,x\right )}^{7/3}\,\sin \left (a+b\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{6};\ \frac {13}{6};\ {\cos \left (a+b\,x\right )}^2\right )}{7\,b\,\sqrt {{\sin \left (a+b\,x\right )}^2}} \]

[In]

int(cos(a + b*x)^(4/3),x)

[Out]

-(3*cos(a + b*x)^(7/3)*sin(a + b*x)*hypergeom([1/2, 7/6], 13/6, cos(a + b*x)^2))/(7*b*(sin(a + b*x)^2)^(1/2))