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

Optimal. Leaf size=137 \[ \frac{B \sin (c+d x) (b \sec (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},-\frac{n}{2},\frac{2-n}{2},\cos ^2(c+d x)\right )}{d n \sqrt{\sin ^2(c+d x)}}-\frac{A b \sin (c+d x) (b \sec (c+d x))^{n-1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1-n}{2},\frac{3-n}{2},\cos ^2(c+d x)\right )}{d (1-n) \sqrt{\sin ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0962599, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3787, 3772, 2643} \[ \frac{B \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac{1}{2},-\frac{n}{2};\frac{2-n}{2};\cos ^2(c+d x)\right )}{d n \sqrt{\sin ^2(c+d x)}}-\frac{A b \sin (c+d x) (b \sec (c+d x))^{n-1} \, _2F_1\left (\frac{1}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(c+d x)\right )}{d (1-n) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.14961, size = 107, normalized size = 0.78 \[ \frac{\sqrt{-\tan ^2(c+d x)} \csc (c+d x) (b \sec (c+d x))^n \left (A (n+1) \cos (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n}{2},\frac{n+2}{2},\sec ^2(c+d x)\right )+B n \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n+1}{2},\frac{n+3}{2},\sec ^2(c+d x)\right )\right )}{d n (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]*(A*(1 + n)*Cos[c + d*x]*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Sec[c + d*x]^2] + B*n*Hypergeomet
ric2F1[1/2, (1 + n)/2, (3 + n)/2, Sec[c + d*x]^2])*(b*Sec[c + d*x])^n*Sqrt[-Tan[c + d*x]^2])/(d*n*(1 + n))

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Maple [F]  time = 0.605, size = 0, normalized size = 0. \begin{align*} \int \left ( b\sec \left ( dx+c \right ) \right ) ^{n} \left ( A+B\sec \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)), x)

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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 )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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, x)