\(\int \frac {1}{\cos ^{\frac {2}{3}}(a+b x)} \, dx\) [29]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 10, antiderivative size = 51 \[ \int \frac {1}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=-\frac {3 \sqrt [3]{\cos (a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(a+b x)\right ) \sin (a+b x)}{b \sqrt {\sin ^2(a+b x)}} \]

[Out]

-3*cos(b*x+a)^(1/3)*hypergeom([1/6, 1/2],[7/6],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 = 51, 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} \[ \int \frac {1}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=-\frac {3 \sin (a+b x) \sqrt [3]{\cos (a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(a+b x)\right )}{b \sqrt {\sin ^2(a+b x)}} \]

[In]

Int[Cos[a + b*x]^(-2/3),x]

[Out]

(-3*Cos[a + b*x]^(1/3)*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[a + b*x]^2]*Sin[a + b*x])/(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 \sqrt [3]{\cos (a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(a+b x)\right ) \sin (a+b x)}{b \sqrt {\sin ^2(a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=-\frac {3 \sqrt [3]{\cos (a+b x)} \csc (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(a+b x)\right ) \sqrt {\sin ^2(a+b x)}}{b} \]

[In]

Integrate[Cos[a + b*x]^(-2/3),x]

[Out]

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

Maple [F]

\[\int \frac {1}{\cos \left (b x +a \right )^{\frac {2}{3}}}d x\]

[In]

int(1/cos(b*x+a)^(2/3),x)

[Out]

int(1/cos(b*x+a)^(2/3),x)

Fricas [F]

\[ \int \frac {1}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=\int { \frac {1}{\cos \left (b x + a\right )^{\frac {2}{3}}} \,d x } \]

[In]

integrate(1/cos(b*x+a)^(2/3),x, algorithm="fricas")

[Out]

integral(cos(b*x + a)^(-2/3), x)

Sympy [F]

\[ \int \frac {1}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=\int \frac {1}{\cos ^{\frac {2}{3}}{\left (a + b x \right )}}\, dx \]

[In]

integrate(1/cos(b*x+a)**(2/3),x)

[Out]

Integral(cos(a + b*x)**(-2/3), x)

Maxima [F]

\[ \int \frac {1}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=\int { \frac {1}{\cos \left (b x + a\right )^{\frac {2}{3}}} \,d x } \]

[In]

integrate(1/cos(b*x+a)^(2/3),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^(-2/3), x)

Giac [F]

\[ \int \frac {1}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=\int { \frac {1}{\cos \left (b x + a\right )^{\frac {2}{3}}} \,d x } \]

[In]

integrate(1/cos(b*x+a)^(2/3),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)^(-2/3), x)

Mupad [B] (verification not implemented)

Time = 15.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=-\frac {3\,{\cos \left (a+b\,x\right )}^{1/3}\,\sin \left (a+b\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{6},\frac {1}{2};\ \frac {7}{6};\ {\cos \left (a+b\,x\right )}^2\right )}{b\,\sqrt {{\sin \left (a+b\,x\right )}^2}} \]

[In]

int(1/cos(a + b*x)^(2/3),x)

[Out]

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