3.204 \(\int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \, dx\)

Optimal. Leaf size=81 \[ \frac{3 b \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^m(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{6} (-3 m-1),\frac{1}{6} (5-3 m),\cos ^2(c+d x)\right )}{d (3 m+1) \sqrt{\sin ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0402926, antiderivative size = 81, normalized size of antiderivative = 1., 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, 3772, 2643} \[ \frac{3 b \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \sec ^m(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (-3 m-1);\frac{1}{6} (5-3 m);\cos ^2(c+d x)\right )}{d (3 m+1) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^m*(b*Sec[c + d*x])^(4/3),x]

[Out]

(3*b*Hypergeometric2F1[1/2, (-1 - 3*m)/6, (5 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^m*(b*Sec[c + d*x])^(1/3)*S
in[c + d*x])/(d*(1 + 3*m)*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 3772

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]

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 \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \, dx &=\frac{\left (b \sqrt [3]{b \sec (c+d x)}\right ) \int \sec ^{\frac{4}{3}+m}(c+d x) \, dx}{\sqrt [3]{\sec (c+d x)}}\\ &=\left (b \cos ^{\frac{1}{3}+m}(c+d x) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)}\right ) \int \cos ^{-\frac{4}{3}-m}(c+d x) \, dx\\ &=\frac{3 b \, _2F_1\left (\frac{1}{2},\frac{1}{6} (-1-3 m);\frac{1}{6} (5-3 m);\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (1+3 m) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.092348, size = 83, normalized size = 1.02 \[ \frac{\sqrt{-\tan ^2(c+d x)} \csc (c+d x) (b \sec (c+d x))^{4/3} \sec ^{m-1}(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} \left (m+\frac{4}{3}\right ),\frac{1}{2} \left (m+\frac{10}{3}\right ),\sec ^2(c+d x)\right )}{d \left (m+\frac{4}{3}\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^m*(b*Sec[c + d*x])^(4/3),x]

[Out]

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

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Maple [F]  time = 0.102, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{m} \left ( b\sec \left ( dx+c \right ) \right ) ^{{\frac{4}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^m*(b*sec(d*x+c))^(4/3),x)

[Out]

int(sec(d*x+c)^m*(b*sec(d*x+c))^(4/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec \left (d x + c\right )\right )^{\frac{4}{3}} \sec \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c))^(4/3)*sec(d*x + c)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c))^(1/3)*b*sec(d*x + c)^m*sec(d*x + c), 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(sec(d*x+c)**m*(b*sec(d*x+c))**(4/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec \left (d x + c\right )\right )^{\frac{4}{3}} \sec \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c))^(4/3)*sec(d*x + c)^m, x)