3.3.7 \(\int \frac {\sec ^m(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx\) [207]

Optimal. Leaf size=82 \[ -\frac {3 \, _2F_1\left (\frac {1}{2},\frac {1}{6} (4-3 m);\frac {1}{6} (10-3 m);\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (4-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \]

[Out]

-3*hypergeom([1/2, 2/3-1/2*m],[5/3-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-1+m)*sin(d*x+c)/d/(4-3*m)/(b*sec(d*x+c))^
(1/3)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 82, 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 = {20, 3857, 2722} \begin {gather*} -\frac {3 \sin (c+d x) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (4-3 m);\frac {1}{6} (10-3 m);\cos ^2(c+d x)\right )}{d (4-3 m) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Hypergeometric2F1[1/2, (4 - 3*m)/6, (10 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*Sin[c + d*x])/(d*(
4 - 3*m)*(b*Sec[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2])

Rule 20

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

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 {\sec ^m(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx &=\frac {\sqrt [3]{\sec (c+d x)} \int \sec ^{-\frac {1}{3}+m}(c+d x) \, dx}{\sqrt [3]{b \sec (c+d x)}}\\ &=\frac {\left (\cos ^{\frac {2}{3}+m}(c+d x) \sec ^{1+m}(c+d x)\right ) \int \cos ^{\frac {1}{3}-m}(c+d x) \, dx}{\sqrt [3]{b \sec (c+d x)}}\\ &=-\frac {3 \, _2F_1\left (\frac {1}{2},\frac {1}{6} (4-3 m);\frac {1}{6} (10-3 m);\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (4-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 83, normalized size = 1.01 \begin {gather*} \frac {\csc (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} \left (-\frac {1}{3}+m\right );\frac {1}{2} \left (\frac {5}{3}+m\right );\sec ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sqrt {-\tan ^2(c+d x)}}{d \left (-\frac {1}{3}+m\right ) \sqrt [3]{b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]*Hypergeometric2F1[1/2, (-1/3 + m)/2, (5/3 + m)/2, Sec[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*Sqrt[-Ta
n[c + d*x]^2])/(d*(-1/3 + m)*(b*Sec[c + d*x])^(1/3))

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Maple [F]
time = 0.42, size = 0, normalized size = 0.00 \[\int \frac {\sec ^{m}\left (d x +c \right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((b*sec(d*x + c))^(2/3)*sec(d*x + c)^m/(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 {\sec ^{m}{\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(sec(d*x+c)**m/(b*sec(d*x+c))**(1/3),x)

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m}{{\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((1/cos(c + d*x))^m/(b/cos(c + d*x))^(1/3),x)

[Out]

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

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