3.206 \(\int \sqrt [3]{b \cos (c+d x)} \sec (c+d x) \, dx\)

Optimal. Leaf size=53 \[ -\frac{3 \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};\cos ^2(c+d x)\right )}{d \sqrt{\sin ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0244257, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {16, 2643} \[ -\frac{3 \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};\cos ^2(c+d x)\right )}{d \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Cos[c + d*x])^(1/3)*Sec[c + d*x],x]

[Out]

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

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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

Mathematica [A]  time = 0.0484639, size = 54, normalized size = 1.02 \[ -\frac{3 b \sqrt{\sin ^2(c+d x)} \cot (c+d x) \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};\cos ^2(c+d x)\right )}{d (b \cos (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Cos[c + d*x])^(1/3)*Sec[c + d*x],x]

[Out]

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

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Maple [F]  time = 0.181, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{b\cos \left ( dx+c \right ) }\sec \left ( dx+c \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \cos \left (d x + c\right )\right )^{\frac{1}{3}} \sec \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (b \cos \left (d x + c\right )\right )^{\frac{1}{3}} \sec \left (d x + c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*cos(d*x + c))^(1/3)*sec(d*x + c), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{b \cos{\left (c + d x \right )}} \sec{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*cos(c + d*x))**(1/3)*sec(c + d*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \cos \left (d x + c\right )\right )^{\frac{1}{3}} \sec \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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