3.3.1 \(\int \frac {1}{(b \sec (c+d x))^{4/3}} \, dx\) [201]

Optimal. Leaf size=56 \[ -\frac {3 b \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{7 d (b \sec (c+d x))^{7/3} \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3857, 2722} \begin {gather*} -\frac {3 b \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x])^(-4/3),x]

[Out]

(-3*b*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x])/(7*d*(b*Sec[c + d*x])^(7/3)*Sqrt[Sin[c +
d*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 \frac {1}{(b \sec (c+d x))^{4/3}} \, dx &=\left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \left (\frac {\cos (c+d x)}{b}\right )^{4/3} \, dx\\ &=-\frac {3 \cos ^3(c+d x) \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{7 b^2 d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 57, normalized size = 1.02 \begin {gather*} -\frac {3 \cot (c+d x) \, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{4 d (b \sec (c+d x))^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[c + d*x])^(-4/3),x]

[Out]

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

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Maple [F]
time = 0.28, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sec(d*x+c))^(4/3),x)

[Out]

int(1/(b*sec(d*x+c))^(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(1/(b*sec(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c))^(-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(1/(b*sec(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c))^(2/3)/(b^2*sec(d*x + c)^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(d*x+c))**(4/3),x)

[Out]

Integral((b*sec(c + d*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(1/(b*sec(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c))^(-4/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/cos(c + d*x))^(4/3),x)

[Out]

int(1/(b/cos(c + d*x))^(4/3), x)

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