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

Optimal. Leaf size=145 \[ \frac{3 C \sin (c+d x) \sec ^{m+1}(c+d x)}{d (3 m+1) (b \sec (c+d x))^{2/3}}-\frac{3 (3 A m+A-C (2-3 m)) \sin (c+d x) \sec ^{m-1}(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{6} (5-3 m),\frac{1}{6} (11-3 m),\cos ^2(c+d x)\right )}{d (5-3 m) (3 m+1) \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}} \]

[Out]

(3*C*Sec[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(1 + 3*m)*(b*Sec[c + d*x])^(2/3)) - (3*(A - C*(2 - 3*m) + 3*A*m)*Hy
pergeometric2F1[1/2, (5 - 3*m)/6, (11 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*Sin[c + d*x])/(d*(5 - 3*
m)*(1 + 3*m)*(b*Sec[c + d*x])^(2/3)*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.126757, antiderivative size = 145, 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} \[ \frac{3 C \sin (c+d x) \sec ^{m+1}(c+d x)}{d (3 m+1) (b \sec (c+d x))^{2/3}}-\frac{3 (3 A m+A-C (2-3 m)) \sin (c+d x) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (5-3 m);\frac{1}{6} (11-3 m);\cos ^2(c+d x)\right )}{d (5-3 m) (3 m+1) \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*Sec[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(1 + 3*m)*(b*Sec[c + d*x])^(2/3)) - (3*(A - C*(2 - 3*m) + 3*A*m)*Hy
pergeometric2F1[1/2, (5 - 3*m)/6, (11 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*Sin[c + d*x])/(d*(5 - 3*
m)*(1 + 3*m)*(b*Sec[c + d*x])^(2/3)*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))^{2/3}} \, dx &=\frac{\sec ^{\frac{2}{3}}(c+d x) \int \sec ^{-\frac{2}{3}+m}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx}{(b \sec (c+d x))^{2/3}}\\ &=\frac{3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+3 m) (b \sec (c+d x))^{2/3}}+\frac{\left (\left (C \left (-\frac{2}{3}+m\right )+A \left (\frac{1}{3}+m\right )\right ) \sec ^{\frac{2}{3}}(c+d x)\right ) \int \sec ^{-\frac{2}{3}+m}(c+d x) \, dx}{\left (\frac{1}{3}+m\right ) (b \sec (c+d x))^{2/3}}\\ &=\frac{3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+3 m) (b \sec (c+d x))^{2/3}}+\frac{\left (\left (C \left (-\frac{2}{3}+m\right )+A \left (\frac{1}{3}+m\right )\right ) \cos ^{\frac{1}{3}+m}(c+d x) \sec ^{1+m}(c+d x)\right ) \int \cos ^{\frac{2}{3}-m}(c+d x) \, dx}{\left (\frac{1}{3}+m\right ) (b \sec (c+d x))^{2/3}}\\ &=\frac{3 C \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+3 m) (b \sec (c+d x))^{2/3}}-\frac{3 (A-C (2-3 m)+3 A m) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (5-3 m);\frac{1}{6} (11-3 m);\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (5-3 m) (1+3 m) (b \sec (c+d x))^{2/3} \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [C]  time = 8.24401, size = 311, normalized size = 2.14 \[ -\frac{3 i 2^{m+\frac{1}{3}} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m-\frac{2}{3}} \left (1+e^{2 i (c+d x)}\right )^{m-\frac{2}{3}} \left (A+C \sec ^2(c+d x)\right ) \left ((3 m+10) \left (2 (3 m-2) (A+2 C) e^{2 i (c+d x)} \text{Hypergeometric2F1}\left (m+\frac{4}{3},\frac{1}{6} (3 m+4),\frac{m}{2}+\frac{5}{3},-e^{2 i (c+d x)}\right )+A (3 m+4) \text{Hypergeometric2F1}\left (m+\frac{4}{3},\frac{1}{6} (3 m-2),\frac{1}{6} (3 m+4),-e^{2 i (c+d x)}\right )\right )+A \left (9 m^2+6 m-8\right ) e^{4 i (c+d x)} \text{Hypergeometric2F1}\left (\frac{m}{2}+\frac{5}{3},m+\frac{4}{3},\frac{m}{2}+\frac{8}{3},-e^{2 i (c+d x)}\right )\right )}{d (3 m-2) (3 m+4) (3 m+10) \sec ^{\frac{4}{3}}(c+d x) (b \sec (c+d x))^{2/3} (A \cos (2 c+2 d x)+A+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^m*(A + C*Sec[c + d*x]^2))/(b*Sec[c + d*x])^(2/3),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^((4*I)*(c + d*x))*(-8 + 6*m + 9*m^2)*Hypergeometric2F1[5/3 + m/2, 4/3 + m, 8/3 + m/2, -E^((2*I)*(c + d*
x))] + (10 + 3*m)*(A*(4 + 3*m)*Hypergeometric2F1[4/3 + m, (-2 + 3*m)/6, (4 + 3*m)/6, -E^((2*I)*(c + d*x))] + 2
*(A + 2*C)*E^((2*I)*(c + d*x))*(-2 + 3*m)*Hypergeometric2F1[4/3 + m, (4 + 3*m)/6, 5/3 + m/2, -E^((2*I)*(c + d*
x))]))*(A + C*Sec[c + d*x]^2))/(d*(-2 + 3*m)*(4 + 3*m)*(10 + 3*m)*(A + 2*C + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^
(4/3)*(b*Sec[c + d*x])^(2/3))

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Maple [F]  time = 0.149, 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{2}{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))^(2/3),x)

[Out]

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

[Out]

integrate((C*sec(d*x + c)^2 + A)*sec(d*x + c)^m/(b*sec(d*x + c))^(2/3), 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{1}{3}} \sec \left (d x + c\right )^{m}}{b \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*(A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^(1/3)*sec(d*x + c)^m/(b*sec(d*x + c)), x)

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

[Out]

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

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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{2}{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))^(2/3),x, algorithm="giac")

[Out]

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