\(\int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [66]

Optimal result

Integrand size = 41, antiderivative size = 227 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 C \sec ^{1+m}(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (5+3 m)}-\frac {3 (C (2+3 m)+A (5+3 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (1-3 m) (5+3 m) \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-2-3 m),\frac {1}{6} (4-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (2+3 m) \sqrt {\sin ^2(c+d x)}} \]

[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)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {20, 4132, 3857, 2722, 4131} \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {3 (A (3 m+5)+C (3 m+2)) \sin (c+d x) (b \sec (c+d x))^{2/3} \sec ^{m-1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right )}{d (1-3 m) (3 m+5) \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{2/3} \sec ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-3 m-2),\frac {1}{6} (4-3 m),\cos ^2(c+d x)\right )}{d (3 m+2) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) (b \sec (c+d x))^{2/3} \sec ^{m+1}(c+d x)}{d (3 m+5)} \]

[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 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3857

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 4131

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 4132

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*} \text {integral}& = \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 \operatorname {Hypergeometric2F1}\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)) \operatorname {Hypergeometric2F1}\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 \operatorname {Hypergeometric2F1}\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*}

Mathematica [A] (verified)

Time = 2.65 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.90 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b \csc (c+d x) \sec ^m(c+d x) \left (A \left (40+39 m+9 m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\sec ^2(c+d x)\right )+(2+3 m) \left (B (8+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5+3 m),\frac {1}{6} (11+3 m),\sec ^2(c+d x)\right )+C (5+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (8+3 m),\frac {7}{3}+\frac {m}{2},\sec ^2(c+d x)\right )\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (2+3 m) (5+3 m) (8+3 m) \sqrt [3]{b \sec (c+d x)}} \]

[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*b*Csc[c + d*x]*Sec[c + d*x]^m*(A*(40 + 39*m + 9*m^2)*Hypergeometric2F1[1/2, (2 + 3*m)/6, (8 + 3*m)/6, Sec[c
+ d*x]^2] + (2 + 3*m)*(B*(8 + 3*m)*Cos[c + d*x]*Hypergeometric2F1[1/2, (5 + 3*m)/6, (11 + 3*m)/6, Sec[c + d*x
]^2] + C*(5 + 3*m)*Hypergeometric2F1[1/2, (8 + 3*m)/6, 7/3 + m/2, Sec[c + d*x]^2])*Sec[c + d*x]^2)*Sqrt[-Tan[c
+ d*x]^2])/(d*(2 + 3*m)*(5 + 3*m)*(8 + 3*m)*(b*Sec[c + d*x])^(1/3))

Maple [F]

\[\int \sec \left (d x +c \right )^{m} \left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )d x\]

[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)

Fricas [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right )^{m} \,d x } \]

[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)

Sympy [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{\frac {2}{3}} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{m}{\left (c + d x \right )}\, dx \]

[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]

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

Maxima [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right )^{m} \,d x } \]

[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)

Giac [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right )^{m} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

[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)