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

Optimal. Leaf size=58 \[ -\frac{3 \sin (c+d x) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )}{5 b^2 d \sqrt{\sin ^2(c+d x)}} \]

[Out]

(-3*(b*Cos[c + d*x])^(5/3)*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*b^2*d*Sqrt[Sin[c
+ d*x]^2])

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Rubi [A]  time = 0.0195851, antiderivative size = 58, 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} \[ -\frac{3 \sin (c+d x) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )}{5 b^2 d \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b*Cos[c + d*x])^(5/3)*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*b^2*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 \frac{\cos (c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx &=\frac{\int (b \cos (c+d x))^{2/3} \, dx}{b}\\ &=-\frac{3 (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{5 b^2 d \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0049375, size = 58, normalized size = 1. \[ -\frac{3 \sqrt{\sin ^2(c+d x)} \cot (c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b*Cos[c + d*x])^(2/3)*Cot[c + d*x]*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2]
)/(5*b*d)

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Maple [F]  time = 0.177, size = 0, normalized size = 0. \begin{align*} \int{\cos \left ( dx+c \right ){\frac{1}{\sqrt [3]{b\cos \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (b \cos \left (d x + c\right )\right )^{\frac{2}{3}}}{b}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*cos(d*x + c))^(2/3)/b, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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