\(\int \sec ^m(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx\) [33]

Optimal result

Integrand size = 29, antiderivative size = 172 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=-\frac {A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1-m-n),\frac {1}{2} (3-m-n),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (1-m-n) \sqrt {\sin ^2(c+d x)}}+\frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-n),\frac {1}{2} (2-m-n),\cos ^2(c+d x)\right ) \sec ^m(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (m+n) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-A*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/(1-m-n)/(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 = 0.13 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {20, 3872, 3857, 2722} \[ \int \sec ^m(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {B \sin (c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-n),\frac {1}{2} (-m-n+2),\cos ^2(c+d x)\right )}{d (m+n) \sqrt {\sin ^2(c+d x)}}-\frac {A \sin (c+d x) \sec ^{m-1}(c+d x) (b \sec (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-n+1),\frac {1}{2} (-m-n+3),\cos ^2(c+d x)\right )}{d (-m-n+1) \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

-((A*Hypergeometric2F1[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)*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 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, 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) (A+B \sec (c+d x)) \, dx \\ & = \left (A \sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{m+n}(c+d x) \, dx+\left (B \sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{1+m+n}(c+d x) \, dx \\ & = \left (A \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+\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 \\ & = -\frac {A \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 = 0.17 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.73 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {\csc (c+d x) \left (A (1+m+n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+n}{2},\frac {1}{2} (2+m+n),\sec ^2(c+d x)\right )+B (m+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1+m+n),\frac {1}{2} (3+m+n),\sec ^2(c+d x)\right )\right ) \sec ^m(c+d x) (b \sec (c+d x))^n \sqrt {-\tan ^2(c+d x)}}{d (m+n) (1+m+n)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*sec(d*x + c) + A)*(b*sec(d*x + c))^n*sec(d*x + c)^m, x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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