3.2.98 \(\int \frac {\cos ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx\) [198]

Optimal. Leaf size=58 \[ -\frac {3 b^3 \, _2F_1\left (\frac {1}{2},\frac {5}{3};\frac {8}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{10 d (b \sec (c+d x))^{10/3} \sqrt {\sin ^2(c+d x)}} \]

[Out]

-3/10*b^3*hypergeom([1/2, 5/3],[8/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(10/3)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {16, 3857, 2722} \begin {gather*} -\frac {3 b^3 \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{3};\frac {8}{3};\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{10/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(b*Sec[c + d*x])^(1/3),x]

[Out]

(-3*b^3*Hypergeometric2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2]*Sin[c + d*x])/(10*d*(b*Sec[c + d*x])^(10/3)*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 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 {\cos ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx &=b^2 \int \frac {1}{(b \sec (c+d x))^{7/3}} \, dx\\ &=\left (b^2 \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \left (\frac {\cos (c+d x)}{b}\right )^{7/3} \, dx\\ &=-\frac {3 \cos ^4(c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{3};\frac {8}{3};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{10 b d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 60, normalized size = 1.03 \begin {gather*} -\frac {3 b^2 \cot (c+d x) \, _2F_1\left (-\frac {7}{6},\frac {1}{2};-\frac {1}{6};\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{7 d (b \sec (c+d x))^{7/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(b*Sec[c + d*x])^(1/3),x]

[Out]

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

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Maple [F]
time = 0.73, size = 0, normalized size = 0.00 \[\int \frac {\cos ^{2}\left (d x +c \right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(b*sec(d*x+c))^(1/3),x)

[Out]

int(cos(d*x+c)^2/(b*sec(d*x+c))^(1/3),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(cos(d*x+c)^2/(b*sec(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^2/(b*sec(d*x + c))^(1/3), 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(cos(d*x+c)^2/(b*sec(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{2}{\left (c + d x \right )}}{\sqrt [3]{b \sec {\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2/(b*sec(d*x+c))**(1/3),x)

[Out]

Integral(cos(c + d*x)**2/(b*sec(c + d*x))**(1/3), 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(cos(d*x+c)^2/(b*sec(d*x+c))^(1/3),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^2/(b*sec(d*x + c))^(1/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^2}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2/(b/cos(c + d*x))^(1/3),x)

[Out]

int(cos(c + d*x)^2/(b/cos(c + d*x))^(1/3), x)

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