3.35 \(\int \sec (c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=136 \[ \frac {A \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},-\frac {n}{2};\frac {2-n}{2};\cos ^2(c+d x)\right )}{d n \sqrt {\sin ^2(c+d x)}}+\frac {B \sin (c+d x) (b \sec (c+d x))^{n+1} \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-n-1);\frac {1-n}{2};\cos ^2(c+d x)\right )}{b d (n+1) \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {16, 3787, 3772, 2643} \[ \frac {A \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},-\frac {n}{2};\frac {2-n}{2};\cos ^2(c+d x)\right )}{d n \sqrt {\sin ^2(c+d x)}}+\frac {B \sin (c+d x) (b \sec (c+d x))^{n+1} \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-n-1);\frac {1-n}{2};\cos ^2(c+d x)\right )}{b d (n+1) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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]

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 119, normalized size = 0.88 \[ \frac {\sqrt {-\tan ^2(c+d x)} \csc (c+d x) \sec (c+d x) (b \sec (c+d x))^n \left (A (n+2) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\sec ^2(c+d x)\right )+B (n+1) \, _2F_1\left (\frac {1}{2},\frac {n+2}{2};\frac {n+4}{2};\sec ^2(c+d x)\right )\right )}{d (n+1) (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \sec \left (d x + c\right )^{2} + A \sec \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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), x)

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maple [F]  time = 2.95, size = 0, normalized size = 0.00 \[ \int \sec \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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), x)

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