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 \[ -\frac {\sin (c+d x) (a A (m+1)+b B m) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3-m}{2};\cos ^2(c+d x)\right )}{d \left (1-m^2\right ) \sqrt {\sin ^2(c+d x)}}+\frac {(a B+A b) \sin (c+d x) \sec ^m(c+d x) \, _2F_1\left (\frac {1}{2},-\frac {m}{2};\frac {2-m}{2};\cos ^2(c+d x)\right )}{d m \sqrt {\sin ^2(c+d x)}}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x)}{d (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 177, normalized size of antiderivative = 1.00, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.46, size = 168, normalized size = 0.95 \[ \frac {\sqrt {-\tan ^2(c+d x)} \csc (c+d x) \sec ^{m+1}(c+d x) \left (m (m+2) (a B+A b) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sec ^2(c+d x)\right )+a A \left (m^2+3 m+2\right ) \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m}{2};\frac {m+2}{2};\sec ^2(c+d x)\right )+b B m (m+1) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\sec ^2(c+d x)\right )\right )}{d m (m+1) (m+2)} \]

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))

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B b \sec \left (d x + c\right )^{2} + A a + {\left (B a + A b\right )} \sec \left (d x + c\right )\right )} \sec \left (d x + c\right )^{m}, x\right ) \]

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)

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

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)

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maple [F]  time = 2.18, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{m}\left (d x +c \right )\right ) \left (a +b \sec \left (d x +c \right )\right ) \left (A +B \sec \left (d x +c \right )\right )\, dx \]

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)

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

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \sec ^{m}{\left (c + d x \right )}\, dx \]

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)

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