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

Optimal. Leaf size=226 \[ -\frac {3 (3 A m+A-C (2-3 m)) \sin (c+d x) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (5-3 m);\frac {1}{6} (11-3 m);\cos ^2(c+d x)\right )}{d (5-3 m) (3 m+1) \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}-\frac {3 B \sin (c+d x) \sec ^m(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (2-3 m);\frac {1}{6} (8-3 m);\cos ^2(c+d x)\right )}{d (2-3 m) \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac {3 C \sin (c+d x) \sec ^{m+1}(c+d x)}{d (3 m+1) (b \sec (c+d x))^{2/3}} \]

[Out]

3*C*sec(d*x+c)^(1+m)*sin(d*x+c)/d/(1+3*m)/(b*sec(d*x+c))^(2/3)-3*(A-C*(2-3*m)+3*A*m)*hypergeom([1/2, 5/6-1/2*m
],[11/6-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-1+m)*sin(d*x+c)/d/(-9*m^2+12*m+5)/(b*sec(d*x+c))^(2/3)/(sin(d*x+c)^2
)^(1/2)-3*B*hypergeom([1/2, 1/3-1/2*m],[4/3-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^m*sin(d*x+c)/d/(2-3*m)/(b*sec(d*x+
c))^(2/3)/(sin(d*x+c)^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {20, 4047, 3772, 2643, 4046} \[ -\frac {3 (3 A m+A-C (2-3 m)) \sin (c+d x) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (5-3 m);\frac {1}{6} (11-3 m);\cos ^2(c+d x)\right )}{d (5-3 m) (3 m+1) \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}-\frac {3 B \sin (c+d x) \sec ^m(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (2-3 m);\frac {1}{6} (8-3 m);\cos ^2(c+d x)\right )}{d (2-3 m) \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac {3 C \sin (c+d x) \sec ^{m+1}(c+d x)}{d (3 m+1) (b \sec (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^m*(A + B*Sec[c + d*x] + 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]) - (3*B*Hypergeometric2F1[1/2, (2 - 3*m)/6, (8 - 3*m)
/6, Cos[c + d*x]^2]*Sec[c + d*x]^m*Sin[c + d*x])/(d*(2 - 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 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]

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 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 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rubi steps

\begin {align*} \int \frac {\sec ^m(c+d x) \left (A+B \sec (c+d x)+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+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{(b \sec (c+d x))^{2/3}}\\ &=\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 {\left (B \sec ^{\frac {2}{3}}(c+d x)\right ) \int \sec ^{\frac {1}{3}+m}(c+d x) \, 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 {\left (B \cos ^{\frac {1}{3}+m}(c+d x) \sec ^{1+m}(c+d x)\right ) \int \cos ^{-\frac {1}{3}-m}(c+d x) \, 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 {3 B \, _2F_1\left (\frac {1}{2},\frac {1}{6} (2-3 m);\frac {1}{6} (8-3 m);\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d (2-3 m) (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}+\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)}}-\frac {3 B \, _2F_1\left (\frac {1}{2},\frac {1}{6} (2-3 m);\frac {1}{6} (8-3 m);\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d (2-3 m) (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 10.81, size = 545, normalized size = 2.41 \[ -\frac {3 i 2^{m+\frac {1}{3}} e^{-\frac {1}{3} i (3 c+d (3 m+1) x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m+\frac {1}{3}} \left (1+e^{2 i (c+d x)}\right )^{m+\frac {1}{3}} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left ((3 m+10) \left (2 (3 m-2) e^{\frac {1}{3} i (3 c+d (3 m+1) x)} \left ((3 m+1) e^{i (c+d x)} \left ((3 m+7) (A+2 C) \, _2F_1\left (m+\frac {4}{3},\frac {1}{6} (3 m+4);\frac {m}{2}+\frac {5}{3};-e^{2 i (c+d x)}\right )+B (3 m+4) e^{i (c+d x)} \, _2F_1\left (m+\frac {4}{3},\frac {1}{6} (3 m+7);\frac {1}{6} (3 m+13);-e^{2 i (c+d x)}\right )\right )+B \left (9 m^2+33 m+28\right ) \, _2F_1\left (m+\frac {4}{3},\frac {1}{6} (3 m+1);\frac {1}{6} (3 m+7);-e^{2 i (c+d x)}\right )\right )+A \left (27 m^3+108 m^2+117 m+28\right ) e^{\frac {1}{3} i d (3 m-2) x} \, _2F_1\left (m+\frac {4}{3},\frac {1}{6} (3 m-2);\frac {1}{6} (3 m+4);-e^{2 i (c+d x)}\right )\right )+A \left (81 m^4+270 m^3+135 m^2-150 m-56\right ) e^{4 i c+\frac {1}{3} i d (3 m+10) x} \, _2F_1\left (\frac {m}{2}+\frac {5}{3},m+\frac {4}{3};\frac {m}{2}+\frac {8}{3};-e^{2 i (c+d x)}\right )\right )}{d (3 m-2) (3 m+1) (3 m+4) (3 m+7) (3 m+10) \sec ^{\frac {4}{3}}(c+d x) (b \sec (c+d x))^{2/3} (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]^m*(A + B*Sec[c + d*x] + 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))))^(1/3 + m)*(1 + E^((2*I)*(c + d*x)))^(1/3 + m)*
(A*E^((4*I)*c + (I/3)*d*(10 + 3*m)*x)*(-56 - 150*m + 135*m^2 + 270*m^3 + 81*m^4)*Hypergeometric2F1[5/3 + m/2,
4/3 + m, 8/3 + m/2, -E^((2*I)*(c + d*x))] + (10 + 3*m)*(A*E^((I/3)*d*(-2 + 3*m)*x)*(28 + 117*m + 108*m^2 + 27*
m^3)*Hypergeometric2F1[4/3 + m, (-2 + 3*m)/6, (4 + 3*m)/6, -E^((2*I)*(c + d*x))] + 2*E^((I/3)*(3*c + d*(1 + 3*
m)*x))*(-2 + 3*m)*(B*(28 + 33*m + 9*m^2)*Hypergeometric2F1[4/3 + m, (1 + 3*m)/6, (7 + 3*m)/6, -E^((2*I)*(c + d
*x))] + E^(I*(c + d*x))*(1 + 3*m)*((A + 2*C)*(7 + 3*m)*Hypergeometric2F1[4/3 + m, (4 + 3*m)/6, 5/3 + m/2, -E^(
(2*I)*(c + d*x))] + B*E^(I*(c + d*x))*(4 + 3*m)*Hypergeometric2F1[4/3 + m, (7 + 3*m)/6, (13 + 3*m)/6, -E^((2*I
)*(c + d*x))]))))*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*E^((I/3)*(3*c + d*(1 + 3*m)*x))*(-2 + 3*m)*(1 +
3*m)*(4 + 3*m)*(7 + 3*m)*(10 + 3*m)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(4/3)*(b*Se
c[c + d*x])^(2/3))

________________________________________________________________________________________

fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{m}}{b \sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^m*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^m*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(2/3),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [F]  time = 1.18, size = 0, normalized size = 0.00 \[ \int \frac {\left (\sec ^{m}\left (d x +c \right )\right ) \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{m}}{\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1/cos(c + d*x))^m*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(b/cos(c + d*x))^(2/3),x)

[Out]

int(((1/cos(c + d*x))^m*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(b/cos(c + d*x))^(2/3), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{m}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________