\(\int \sec ^m(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [71]

Optimal result

Integrand size = 39, antiderivative size = 226 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {C \sec ^{1+m}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (1+m+n)}-\frac {(C (m+n)+A (1+m+n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1-m-n),\frac {1}{2} (3-m-n),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (1-m-n) (1+m+n) \sqrt {\sin ^2(c+d x)}}+\frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-n),\frac {1}{2} (2-m-n),\cos ^2(c+d x)\right ) \sec ^m(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (m+n) \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {20, 4132, 3857, 2722, 4131} \[ \int \sec ^m(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {(A (m+n+1)+C (m+n)) \sin (c+d x) \sec ^{m-1}(c+d x) (b \sec (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-n+1),\frac {1}{2} (-m-n+3),\cos ^2(c+d x)\right )}{d (-m-n+1) (m+n+1) \sqrt {\sin ^2(c+d x)}}+\frac {B \sin (c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-n),\frac {1}{2} (-m-n+2),\cos ^2(c+d x)\right )}{d (m+n) \sqrt {\sin ^2(c+d x)}}+\frac {C \sin (c+d x) \sec ^{m+1}(c+d x) (b \sec (c+d x))^n}{d (m+n+1)} \]

[In]

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

[Out]

(C*Sec[c + d*x]^(1 + m)*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(1 + m + n)) - ((C*(m + n) + A*(1 + m + n))*Hyperg
eometric2F1[1/2, (1 - m - n)/2, (3 - m - n)/2, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m)*(b*Sec[c + d*x])^n*Sin[c
+ d*x])/(d*(1 - m - n)*(1 + m + n)*Sqrt[Sin[c + d*x]^2]) + (B*Hypergeometric2F1[1/2, (-m - n)/2, (2 - m - n)/2
, Cos[c + d*x]^2]*Sec[c + d*x]^m*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(m + n)*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}& = \left (\sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{m+n}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx \\ & = \left (\sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{m+n}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx+\left (B \sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{1+m+n}(c+d x) \, dx \\ & = \frac {C \sec ^{1+m}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (1+m+n)}+\left (B \cos ^{m+n}(c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n\right ) \int \cos ^{-1-m-n}(c+d x) \, dx+\left (\left (A+\frac {C (m+n)}{1+m+n}\right ) \sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{m+n}(c+d x) \, dx \\ & = \frac {C \sec ^{1+m}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (1+m+n)}+\frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-n),\frac {1}{2} (2-m-n),\cos ^2(c+d x)\right ) \sec ^m(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (m+n) \sqrt {\sin ^2(c+d x)}}+\left (\left (A+\frac {C (m+n)}{1+m+n}\right ) \cos ^{m+n}(c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n\right ) \int \cos ^{-m-n}(c+d x) \, dx \\ & = \frac {C \sec ^{1+m}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (1+m+n)}-\frac {\left (A+\frac {C (m+n)}{1+m+n}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1-m-n),\frac {1}{2} (3-m-n),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (1-m-n) \sqrt {\sin ^2(c+d x)}}+\frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-n),\frac {1}{2} (2-m-n),\cos ^2(c+d x)\right ) \sec ^m(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (m+n) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.33 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.71 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\csc (c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n \left (\frac {A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+n}{2},\frac {1}{2} (2+m+n),\sec ^2(c+d x)\right )}{m+n}+\frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1+m+n),\frac {1}{2} (3+m+n),\sec ^2(c+d x)\right )}{1+m+n}+\frac {C \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (2+m+n),\frac {1}{2} (4+m+n),\sec ^2(c+d x)\right ) \sec (c+d x)}{2+m+n}\right ) \sqrt {-\tan ^2(c+d x)}}{d} \]

[In]

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

[Out]

(Csc[c + d*x]*Sec[c + d*x]^m*(b*Sec[c + d*x])^n*((A*Cos[c + d*x]*Hypergeometric2F1[1/2, (m + n)/2, (2 + m + n)
/2, Sec[c + d*x]^2])/(m + n) + (B*Hypergeometric2F1[1/2, (1 + m + n)/2, (3 + m + n)/2, Sec[c + d*x]^2])/(1 + m
+ n) + (C*Hypergeometric2F1[1/2, (2 + m + n)/2, (4 + m + n)/2, Sec[c + d*x]^2]*Sec[c + d*x])/(2 + m + n))*Sqr
t[-Tan[c + d*x]^2])/d

Maple [F]

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

[In]

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

Fricas [F]

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

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^n*(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))^n*sec(d*x + c)^m, x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^n*(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))^n*sec(d*x + c)^m, x)

Giac [F]

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

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^n*(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))^n*sec(d*x + c)^m, x)

Mupad [F(-1)]

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

[In]

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