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

Optimal. Leaf size=227 \[ -\frac {3 (A (3 m+5)+C (3 m+2)) \sin (c+d x) (b \sec (c+d x))^{2/3} \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (1-3 m);\frac {1}{6} (7-3 m);\cos ^2(c+d x)\right )}{d (1-3 m) (3 m+5) \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{2/3} \sec ^m(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (-3 m-2);\frac {1}{6} (4-3 m);\cos ^2(c+d x)\right )}{d (3 m+2) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) (b \sec (c+d x))^{2/3} \sec ^{m+1}(c+d x)}{d (3 m+5)} \]

[Out]

3*C*sec(d*x+c)^(1+m)*(b*sec(d*x+c))^(2/3)*sin(d*x+c)/d/(5+3*m)-3*(C*(2+3*m)+A*(5+3*m))*hypergeom([1/2, 1/6-1/2
*m],[7/6-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-1+m)*(b*sec(d*x+c))^(2/3)*sin(d*x+c)/d/(-9*m^2-12*m+5)/(sin(d*x+c)^
2)^(1/2)+3*B*hypergeom([1/2, -1/3-1/2*m],[2/3-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^m*(b*sec(d*x+c))^(2/3)*sin(d*x+c
)/d/(2+3*m)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 227, 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} \[ -\frac {3 (A (3 m+5)+C (3 m+2)) \sin (c+d x) (b \sec (c+d x))^{2/3} \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (1-3 m);\frac {1}{6} (7-3 m);\cos ^2(c+d x)\right )}{d (1-3 m) (3 m+5) \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{2/3} \sec ^m(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (-3 m-2);\frac {1}{6} (4-3 m);\cos ^2(c+d x)\right )}{d (3 m+2) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) (b \sec (c+d x))^{2/3} \sec ^{m+1}(c+d x)}{d (3 m+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*Sec[c + d*x]^(1 + m)*(b*Sec[c + d*x])^(2/3)*Sin[c + d*x])/(d*(5 + 3*m)) - (3*(C*(2 + 3*m) + A*(5 + 3*m))*
Hypergeometric2F1[1/2, (1 - 3*m)/6, (7 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*(b*Sec[c + d*x])^(2/3)*
Sin[c + d*x])/(d*(1 - 3*m)*(5 + 3*m)*Sqrt[Sin[c + d*x]^2]) + (3*B*Hypergeometric2F1[1/2, (-2 - 3*m)/6, (4 - 3*
m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^m*(b*Sec[c + d*x])^(2/3)*Sin[c + d*x])/(d*(2 + 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 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 \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {(b \sec (c+d x))^{2/3} \int \sec ^{\frac {2}{3}+m}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{\sec ^{\frac {2}{3}}(c+d x)}\\ &=\frac {(b \sec (c+d x))^{2/3} \int \sec ^{\frac {2}{3}+m}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx}{\sec ^{\frac {2}{3}}(c+d x)}+\frac {\left (B (b \sec (c+d x))^{2/3}\right ) \int \sec ^{\frac {5}{3}+m}(c+d x) \, dx}{\sec ^{\frac {2}{3}}(c+d x)}\\ &=\frac {3 C \sec ^{1+m}(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (5+3 m)}+\frac {\left (\left (C \left (\frac {2}{3}+m\right )+A \left (\frac {5}{3}+m\right )\right ) (b \sec (c+d x))^{2/3}\right ) \int \sec ^{\frac {2}{3}+m}(c+d x) \, dx}{\left (\frac {5}{3}+m\right ) \sec ^{\frac {2}{3}}(c+d x)}+\left (B \cos ^{\frac {2}{3}+m}(c+d x) \sec ^m(c+d x) (b \sec (c+d x))^{2/3}\right ) \int \cos ^{-\frac {5}{3}-m}(c+d x) \, dx\\ &=\frac {3 C \sec ^{1+m}(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (5+3 m)}+\frac {3 B \, _2F_1\left (\frac {1}{2},\frac {1}{6} (-2-3 m);\frac {1}{6} (4-3 m);\cos ^2(c+d x)\right ) \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (2+3 m) \sqrt {\sin ^2(c+d x)}}+\frac {\left (\left (C \left (\frac {2}{3}+m\right )+A \left (\frac {5}{3}+m\right )\right ) \cos ^{\frac {2}{3}+m}(c+d x) \sec ^m(c+d x) (b \sec (c+d x))^{2/3}\right ) \int \cos ^{-\frac {2}{3}-m}(c+d x) \, dx}{\frac {5}{3}+m}\\ &=\frac {3 C \sec ^{1+m}(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (5+3 m)}-\frac {3 (C (2+3 m)+A (5+3 m)) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (1-3 m);\frac {1}{6} (7-3 m);\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (1-3 m) (5+3 m) \sqrt {\sin ^2(c+d x)}}+\frac {3 B \, _2F_1\left (\frac {1}{2},\frac {1}{6} (-2-3 m);\frac {1}{6} (4-3 m);\cos ^2(c+d x)\right ) \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (2+3 m) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 7.08, size = 547, normalized size = 2.41 \[ -\frac {3 i 2^{m+\frac {5}{3}} e^{-\frac {1}{3} i d (3 m+2) x} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m+\frac {2}{3}} \left (1+e^{2 i (c+d x)}\right )^{m+\frac {2}{3}} (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {A e^{4 i c+\frac {1}{3} i d (3 m+14) x} \, _2F_1\left (\frac {m}{2}+\frac {7}{3},m+\frac {8}{3};\frac {1}{6} (3 m+20);-e^{2 i (c+d x)}\right )}{3 m+14}+\frac {A e^{\frac {1}{3} i d (3 m+2) x} \, _2F_1\left (m+\frac {8}{3},\frac {1}{6} (3 m+2);\frac {1}{6} (3 m+8);-e^{2 i (c+d x)}\right )}{3 m+2}+\frac {2 A e^{\frac {1}{3} i (6 c+d (3 m+8) x)} \, _2F_1\left (m+\frac {8}{3},\frac {1}{6} (3 m+8);\frac {m}{2}+\frac {7}{3};-e^{2 i (c+d x)}\right )}{3 m+8}+\frac {2 B e^{\frac {1}{3} i (3 c+d (3 m+5) x)} \, _2F_1\left (m+\frac {8}{3},\frac {1}{6} (3 m+5);\frac {1}{6} (3 m+11);-e^{2 i (c+d x)}\right )}{3 m+5}+\frac {2 B e^{\frac {1}{3} i (9 c+d (3 m+11) x)} \, _2F_1\left (m+\frac {8}{3},\frac {1}{6} (3 m+11);\frac {1}{6} (3 m+17);-e^{2 i (c+d x)}\right )}{3 m+11}+\frac {4 C e^{\frac {1}{3} i (6 c+d (3 m+8) x)} \, _2F_1\left (m+\frac {8}{3},\frac {1}{6} (3 m+8);\frac {m}{2}+\frac {7}{3};-e^{2 i (c+d x)}\right )}{3 m+8}\right )}{d \sec ^{\frac {8}{3}}(c+d x) (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-3*I)*2^(5/3 + 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 + (I/3)*d*(14 + 3*m)*x)*Hypergeometric2F1[7/3 + m/2, 8/3 + m, (20 + 3*m)/6, -E^((2*I)*(c + d*x)
)])/(14 + 3*m) + (A*E^((I/3)*d*(2 + 3*m)*x)*Hypergeometric2F1[8/3 + m, (2 + 3*m)/6, (8 + 3*m)/6, -E^((2*I)*(c
+ d*x))])/(2 + 3*m) + (2*B*E^((I/3)*(3*c + d*(5 + 3*m)*x))*Hypergeometric2F1[8/3 + m, (5 + 3*m)/6, (11 + 3*m)/
6, -E^((2*I)*(c + d*x))])/(5 + 3*m) + (2*A*E^((I/3)*(6*c + d*(8 + 3*m)*x))*Hypergeometric2F1[8/3 + m, (8 + 3*m
)/6, 7/3 + m/2, -E^((2*I)*(c + d*x))])/(8 + 3*m) + (4*C*E^((I/3)*(6*c + d*(8 + 3*m)*x))*Hypergeometric2F1[8/3
+ m, (8 + 3*m)/6, 7/3 + m/2, -E^((2*I)*(c + d*x))])/(8 + 3*m) + (2*B*E^((I/3)*(9*c + d*(11 + 3*m)*x))*Hypergeo
metric2F1[8/3 + m, (11 + 3*m)/6, (17 + 3*m)/6, -E^((2*I)*(c + d*x))])/(11 + 3*m))*(b*Sec[c + d*x])^(2/3)*(A +
B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*E^((I/3)*d*(2 + 3*m)*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x
])*Sec[c + d*x]^(8/3))

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.65, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{m}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}} \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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