3.26 \(\int \frac{\sec ^m(c+d x) (A+C \sec ^2(c+d x))}{(b \sec (c+d x))^{4/3}} \, dx\)

Optimal. Leaf size=148 \[ -\frac{3 (-3 A m+A+C (4-3 m)) \sin (c+d x) \sec ^{m-2}(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{6} (7-3 m),\frac{1}{6} (13-3 m),\cos ^2(c+d x)\right )}{b d (1-3 m) (7-3 m) \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac{3 C \sin (c+d x) \sec ^m(c+d x)}{b d (1-3 m) \sqrt [3]{b \sec (c+d x)}} \]

[Out]

(-3*C*Sec[c + d*x]^m*Sin[c + d*x])/(b*d*(1 - 3*m)*(b*Sec[c + d*x])^(1/3)) - (3*(A + C*(4 - 3*m) - 3*A*m)*Hyper
geometric2F1[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*(1 - 3*m
)*(7 - 3*m)*(b*Sec[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.134481, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {20, 4046, 3772, 2643} \[ -\frac{3 (-3 A m+A+C (4-3 m)) \sin (c+d x) \sec ^{m-2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (7-3 m);\frac{1}{6} (13-3 m);\cos ^2(c+d x)\right )}{b d (1-3 m) (7-3 m) \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}-\frac{3 C \sin (c+d x) \sec ^m(c+d x)}{b d (1-3 m) \sqrt [3]{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^m*(A + C*Sec[c + d*x]^2))/(b*Sec[c + d*x])^(4/3),x]

[Out]

(-3*C*Sec[c + d*x]^m*Sin[c + d*x])/(b*d*(1 - 3*m)*(b*Sec[c + d*x])^(1/3)) - (3*(A + C*(4 - 3*m) - 3*A*m)*Hyper
geometric2F1[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*(1 - 3*m
)*(7 - 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 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

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

Mathematica [C]  time = 3.51081, size = 340, normalized size = 2.3 \[ -\frac{3 i 2^{m-\frac{1}{3}} e^{-\frac{1}{3} i (6 c+d (3 m+2) x)} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m+\frac{2}{3}} \left (1+e^{2 i (c+d x)}\right )^{m+\frac{2}{3}} \left (A+C \sec ^2(c+d x)\right ) \left (\frac{e^{\frac{1}{3} i (6 c+d (3 m+2) x)} \left (2 (3 m+8) (A+2 C) \text{Hypergeometric2F1}\left (m+\frac{2}{3},\frac{1}{6} (3 m+2),\frac{1}{6} (3 m+8),-e^{2 i (c+d x)}\right )+A (3 m+2) e^{2 i (c+d x)} \text{Hypergeometric2F1}\left (m+\frac{2}{3},\frac{1}{6} (3 m+8),\frac{m}{2}+\frac{7}{3},-e^{2 i (c+d x)}\right )\right )}{(3 m+2) (3 m+8)}+\frac{A e^{\frac{1}{3} i d (3 m-4) x} \text{Hypergeometric2F1}\left (m+\frac{2}{3},\frac{1}{6} (3 m-4),\frac{1}{6} (3 m+2),-e^{2 i (c+d x)}\right )}{3 m-4}\right )}{d \sec ^{\frac{2}{3}}(c+d x) (b \sec (c+d x))^{4/3} (A \cos (2 c+2 d x)+A+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^m*(A + C*Sec[c + d*x]^2))/(b*Sec[c + d*x])^(4/3),x]

[Out]

((-3*I)*2^(-1/3 + m)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(2/3 + m)*(1 + E^((2*I)*(c + d*x)))^(2/3 + m)
*((A*E^((I/3)*d*(-4 + 3*m)*x)*Hypergeometric2F1[2/3 + m, (-4 + 3*m)/6, (2 + 3*m)/6, -E^((2*I)*(c + d*x))])/(-4
+ 3*m) + (E^((I/3)*(6*c + d*(2 + 3*m)*x))*(2*(A + 2*C)*(8 + 3*m)*Hypergeometric2F1[2/3 + m, (2 + 3*m)/6, (8 +
3*m)/6, -E^((2*I)*(c + d*x))] + A*E^((2*I)*(c + d*x))*(2 + 3*m)*Hypergeometric2F1[2/3 + m, (8 + 3*m)/6, 7/3 +
m/2, -E^((2*I)*(c + d*x))]))/((2 + 3*m)*(8 + 3*m)))*(A + C*Sec[c + d*x]^2))/(d*E^((I/3)*(6*c + d*(2 + 3*m)*x)
)*(A + 2*C + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(2/3)*(b*Sec[c + d*x])^(4/3))

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Maple [F]  time = 0.155, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sec \left ( dx+c \right ) \right ) ^{m} \left ( A+C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \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+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(4/3),x)

[Out]

int(sec(d*x+c)^m*(A+C*sec(d*x+c)^2)/(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 (C \sec \left (d x + c\right )^{2} + 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+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + 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 (C \sec \left (d x + c\right )^{2} + 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+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + 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+C*sec(d*x+c)**2)/(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 (C \sec \left (d x + c\right )^{2} + 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+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*sec(d*x + c)^m/(b*sec(d*x + c))^(4/3), x)