3.30 \(\int \frac {1}{\cos ^{\frac {4}{3}}(a+b x)} \, dx\)

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

[Out]

3*hypergeom([-1/6, 1/2],[5/6],cos(b*x+a)^2)*sin(b*x+a)/b/cos(b*x+a)^(1/3)/(sin(b*x+a)^2)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2643} \[ \frac {3 \sin (a+b x) \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(a+b x)\right )}{b \sqrt {\sin ^2(a+b x)} \sqrt [3]{\cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^(-4/3),x]

[Out]

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

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin {align*} \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 ) \sin (a+b x)}{b \sqrt [3]{\cos (a+b x)} \sqrt {\sin ^2(a+b x)}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 51, normalized size = 1.00 \[ \frac {3 \sqrt {\sin ^2(a+b x)} \csc (a+b x) \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(a+b x)\right )}{b \sqrt [3]{\cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^(-4/3),x]

[Out]

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

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\cos \left (b x + a\right )^{\frac {4}{3}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(b*x+a)^(4/3),x, algorithm="fricas")

[Out]

integral(cos(b*x + a)^(-4/3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\cos \left (b x + a\right )^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(b*x+a)^(4/3),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)^(-4/3), x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {1}{\cos \left (b x +a \right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(b*x+a)^(4/3),x)

[Out]

int(1/cos(b*x+a)^(4/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\cos \left (b x + a\right )^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(b*x+a)^(4/3),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^(-4/3), x)

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mupad [B]  time = 0.23, size = 42, normalized size = 0.82 \[ \frac {3\,\sin \left (a+b\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{6},\frac {1}{2};\ \frac {5}{6};\ {\cos \left (a+b\,x\right )}^2\right )}{b\,{\cos \left (a+b\,x\right )}^{1/3}\,\sqrt {{\sin \left (a+b\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(a + b*x)^(4/3),x)

[Out]

(3*sin(a + b*x)*hypergeom([-1/6, 1/2], 5/6, cos(a + b*x)^2))/(b*cos(a + b*x)^(1/3)*(sin(a + b*x)^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\cos ^{\frac {4}{3}}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(b*x+a)**(4/3),x)

[Out]

Integral(cos(a + b*x)**(-4/3), x)

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