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

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

[Out]

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

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Rubi [A]  time = 0.0316683, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {16, 2643} \[ \frac{3 b^2 \sin (c+d x) \, _2F_1\left (-\frac{5}{6},\frac{1}{2};\frac{1}{6};\cos ^2(c+d x)\right )}{5 d \sqrt{\sin ^2(c+d x)} (b \cos (c+d x))^{5/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^2*Hypergeometric2F1[-5/6, 1/2, 1/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*d*(b*Cos[c + d*x])^(5/3)*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 ^3(c+d x) \, dx &=b^3 \int \frac{1}{(b \cos (c+d x))^{8/3}} \, dx\\ &=\frac{3 b^2 \, _2F_1\left (-\frac{5}{6},\frac{1}{2};\frac{1}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d (b \cos (c+d x))^{5/3} \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0482712, size = 63, normalized size = 1.09 \[ \frac{3 \sqrt{\sin ^2(c+d x)} \csc (c+d x) \sec ^2(c+d x) \sqrt [3]{b \cos (c+d x)} \, _2F_1\left (-\frac{5}{6},\frac{1}{2};\frac{1}{6};\cos ^2(c+d x)\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b*Cos[c + d*x])^(1/3)*Csc[c + d*x]*Hypergeometric2F1[-5/6, 1/2, 1/6, Cos[c + d*x]^2]*Sec[c + d*x]^2*Sqrt[S
in[c + d*x]^2])/(5*d)

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Maple [F]  time = 0.161, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{b\cos \left ( dx+c \right ) } \left ( \sec \left ( dx+c \right ) \right ) ^{3}\, 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)^3,x)

[Out]

int((b*cos(d*x+c))^(1/3)*sec(d*x+c)^3,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 )^{3}\,{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)^3,x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

Timed out

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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 )^{3}\,{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)^3,x, algorithm="giac")

[Out]

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