3.1.26 \(\int \sec ^{\frac {2}{3}}(a+b x) \, dx\) [26]

Optimal. Leaf size=51 \[ -\frac {3 \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(a+b x)\right ) \sin (a+b x)}{b \sqrt [3]{\sec (a+b x)} \sqrt {\sin ^2(a+b x)}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3857, 2722} \begin {gather*} -\frac {3 \sin (a+b x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(a+b x)\right )}{b \sqrt {\sin ^2(a+b x)} \sqrt [3]{\sec (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^(2/3),x]

[Out]

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

Rule 3857

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \sec ^{\frac {2}{3}}(a+b x) \, dx &=\cos ^{\frac {2}{3}}(a+b x) \sec ^{\frac {2}{3}}(a+b x) \int \frac {1}{\cos ^{\frac {2}{3}}(a+b x)} \, dx\\ &=-\frac {3 \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(a+b x)\right ) \sin (a+b x)}{b \sqrt [3]{\sec (a+b x)} \sqrt {\sin ^2(a+b x)}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 55, normalized size = 1.08 \begin {gather*} \frac {3 \csc (a+b x) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\sec ^2(a+b x)\right ) \sqrt {-\tan ^2(a+b x)}}{2 b \sqrt [3]{\sec (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^(2/3),x]

[Out]

(3*Csc[a + b*x]*Hypergeometric2F1[1/3, 1/2, 4/3, Sec[a + b*x]^2]*Sqrt[-Tan[a + b*x]^2])/(2*b*Sec[a + b*x]^(1/3
))

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Maple [F]
time = 0.21, size = 0, normalized size = 0.00 \[\int \sec ^{\frac {2}{3}}\left (b x +a \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(sec(b*x+a)^(2/3),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(sec(b*x+a)^(2/3),x, algorithm="maxima")

[Out]

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

[Out]

integral(sec(b*x + a)^(2/3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sec ^{\frac {2}{3}}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(a + b*x)**(2/3), x)

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

[Out]

integrate(sec(b*x + a)^(2/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\left (\frac {1}{\cos \left (a+b\,x\right )}\right )}^{2/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(a + b*x))^(2/3),x)

[Out]

int((1/cos(a + b*x))^(2/3), x)

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