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

Optimal. Leaf size=51 \[ \frac {3 \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(a+b x)\right ) \sqrt [3]{\sec (a+b x)} \sin (a+b x)}{b \sqrt {\sin ^2(a+b x)}} \]

[Out]

3*hypergeom([-1/6, 1/2],[5/6],cos(b*x+a)^2)*sec(b*x+a)^(1/3)*sin(b*x+a)/b/(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) \sqrt [3]{\sec (a+b x)} \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(a+b x)\right )}{b \sqrt {\sin ^2(a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^(4/3),x]

[Out]

(3*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[a + b*x]^2]*Sec[a + b*x]^(1/3)*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]

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 {4}{3}}(a+b x) \, dx &=\sqrt [3]{\cos (a+b x)} \sqrt [3]{\sec (a+b x)} \int \frac {1}{\cos ^{\frac {4}{3}}(a+b x)} \, dx\\ &=\frac {3 \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(a+b x)\right ) \sqrt [3]{\sec (a+b x)} \sin (a+b x)}{b \sqrt {\sin ^2(a+b x)}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 55, normalized size = 1.08 \begin {gather*} \frac {3 \csc (a+b x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\sec ^2(a+b x)\right ) \sqrt [3]{\sec (a+b x)} \sqrt {-\tan ^2(a+b x)}}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^(4/3),x]

[Out]

(3*Csc[a + b*x]*Hypergeometric2F1[1/2, 2/3, 5/3, Sec[a + b*x]^2]*Sec[a + b*x]^(1/3)*Sqrt[-Tan[a + b*x]^2])/(4*
b)

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Maple [F]
time = 0.22, size = 0, normalized size = 0.00 \[\int \sec ^{\frac {4}{3}}\left (b x +a \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral(sec(b*x + a)^(4/3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sec ^{\frac {4}{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)**(4/3),x)

[Out]

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

[Out]

integrate(sec(b*x + a)^(4/3), 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 )}^{4/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(a + b*x))^(4/3),x)

[Out]

int((1/cos(a + b*x))^(4/3), x)

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