3.32 \(\int \frac{\sec ^m(c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx\)

Optimal. Leaf size=173 \[ -\frac{3 A \sin (c+d x) \sec ^{m-2}(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{6} (7-3 m),\frac{1}{6} (13-3 m),\cos ^2(c+d x)\right )}{b d (7-3 m) \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac{3 B \sin (c+d x) \sec ^{m-1}(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{6} (4-3 m),\frac{1}{6} (10-3 m),\cos ^2(c+d x)\right )}{b d (4-3 m) \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}} \]

[Out]

(-3*A*Hypergeometric2F1[1/2, (7 - 3*m)/6, (13 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-2 + m)*Sin[c + d*x])/(b
*d*(7 - 3*m)*(b*Sec[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2]) - (3*B*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])/(b*d*(4 - 3*m)*(b*Sec[c + d*x])^(1/3)*Sqrt[Sin[c + d*
x]^2])

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Rubi [A]  time = 0.11961, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {20, 3787, 3772, 2643} \[ -\frac{3 A \sin (c+d x) \sec ^{m-2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (7-3 m);\frac{1}{6} (13-3 m);\cos ^2(c+d x)\right )}{b d (7-3 m) \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac{3 B \sin (c+d x) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (4-3 m);\frac{1}{6} (10-3 m);\cos ^2(c+d x)\right )}{b d (4-3 m) \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*A*Hypergeometric2F1[1/2, (7 - 3*m)/6, (13 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-2 + m)*Sin[c + d*x])/(b
*d*(7 - 3*m)*(b*Sec[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2]) - (3*B*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])/(b*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 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

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 \frac{\sec ^m(c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx &=\frac{\sqrt [3]{\sec (c+d x)} \int \sec ^{-\frac{4}{3}+m}(c+d x) (A+B \sec (c+d x)) \, dx}{b \sqrt [3]{b \sec (c+d x)}}\\ &=\frac{\left (A \sqrt [3]{\sec (c+d x)}\right ) \int \sec ^{-\frac{4}{3}+m}(c+d x) \, dx}{b \sqrt [3]{b \sec (c+d x)}}+\frac{\left (B \sqrt [3]{\sec (c+d x)}\right ) \int \sec ^{-\frac{1}{3}+m}(c+d x) \, dx}{b \sqrt [3]{b \sec (c+d x)}}\\ &=\frac{\left (A \cos ^{\frac{2}{3}+m}(c+d x) \sec ^{1+m}(c+d x)\right ) \int \cos ^{\frac{4}{3}-m}(c+d x) \, dx}{b \sqrt [3]{b \sec (c+d x)}}+\frac{\left (B \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}{b \sqrt [3]{b \sec (c+d x)}}\\ &=-\frac{3 A \, _2F_1\left (\frac{1}{2},\frac{1}{6} (7-3 m);\frac{1}{6} (13-3 m);\cos ^2(c+d x)\right ) \sec ^{-2+m}(c+d x) \sin (c+d x)}{b d (7-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt{\sin ^2(c+d x)}}-\frac{3 B \, _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)}{b d (4-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.33367, size = 140, normalized size = 0.81 \[ \frac{3 \sqrt{-\tan ^2(c+d x)} \csc (c+d x) \sec ^m(c+d x) \left (A (3 m-1) \cos (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{6} (3 m-4),\frac{1}{6} (3 m+2),\sec ^2(c+d x)\right )+B (3 m-4) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{6} (3 m-1),\frac{1}{6} (3 m+5),\sec ^2(c+d x)\right )\right )}{d (3 m-4) (3 m-1) (b \sec (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Csc[c + d*x]*(A*(-1 + 3*m)*Cos[c + d*x]*Hypergeometric2F1[1/2, (-4 + 3*m)/6, (2 + 3*m)/6, Sec[c + d*x]^2] +
B*(-4 + 3*m)*Hypergeometric2F1[1/2, (-1 + 3*m)/6, (5 + 3*m)/6, Sec[c + d*x]^2])*Sec[c + d*x]^m*Sqrt[-Tan[c +
d*x]^2])/(d*(-4 + 3*m)*(-1 + 3*m)*(b*Sec[c + d*x])^(4/3))

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Maple [F]  time = 0.131, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sec \left ( dx+c \right ) \right ) ^{m} \left ( A+B\sec \left ( dx+c \right ) \right ) \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*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*sec(d*x + c) + A)*(b*sec(d*x + c))^(2/3)*sec(d*x + c)^m/(b^2*sec(d*x + c)^2), 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*(A+B*sec(d*x+c))/(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 \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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