3.482 \(\int \sec ^m(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=177 \[ -\frac{\sin (c+d x) (a A (m+1)+b B m) \sec ^{m-1}(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1-m}{2},\frac{3-m}{2},\cos ^2(c+d x)\right )}{d \left (1-m^2\right ) \sqrt{\sin ^2(c+d x)}}+\frac{(a B+A b) \sin (c+d x) \sec ^m(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},-\frac{m}{2},\frac{2-m}{2},\cos ^2(c+d x)\right )}{d m \sqrt{\sin ^2(c+d x)}}+\frac{b B \sin (c+d x) \sec ^{m+1}(c+d x)}{d (m+1)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.201102, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {3997, 3787, 3772, 2643} \[ -\frac{\sin (c+d x) (a A (m+1)+b B m) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1-m}{2};\frac{3-m}{2};\cos ^2(c+d x)\right )}{d \left (1-m^2\right ) \sqrt{\sin ^2(c+d x)}}+\frac{(a B+A b) \sin (c+d x) \sec ^m(c+d x) \, _2F_1\left (\frac{1}{2},-\frac{m}{2};\frac{2-m}{2};\cos ^2(c+d x)\right )}{d m \sqrt{\sin ^2(c+d x)}}+\frac{b B \sin (c+d x) \sec ^{m+1}(c+d x)}{d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f
, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 3787

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]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.387213, size = 168, normalized size = 0.95 \[ \frac{\sqrt{-\tan ^2(c+d x)} \csc (c+d x) \sec ^{m+1}(c+d x) \left (m (m+2) (a B+A b) \cos (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},\sec ^2(c+d x)\right )+a A \left (m^2+3 m+2\right ) \cos ^2(c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m}{2},\frac{m+2}{2},\sec ^2(c+d x)\right )+b B m (m+1) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},\sec ^2(c+d x)\right )\right )}{d m (m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]*(a*A*(2 + 3*m + m^2)*Cos[c + d*x]^2*Hypergeometric2F1[1/2, m/2, (2 + m)/2, Sec[c + d*x]^2] + (A*
b + a*B)*m*(2 + m)*Cos[c + d*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sec[c + d*x]^2] + b*B*m*(1 + m)*H
ypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Sec[c + d*x]^2])*Sec[c + d*x]^(1 + m)*Sqrt[-Tan[c + d*x]^2])/(d*m*
(1 + m)*(2 + m))

________________________________________________________________________________________

Maple [F]  time = 0.559, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{m} \left ( a+b\sec \left ( dx+c \right ) \right ) \left ( A+B\sec \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B b \sec \left (d x + c\right )^{2} + A a +{\left (B a + A b\right )} \sec \left (d x + c\right )\right )} \sec \left (d x + c\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B \sec{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right ) \sec ^{m}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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