3.34 \(\int \sec ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=143 \[ \frac{A \sin (c+d x) (b \sec (c+d x))^{n+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (-n-1),\frac{1-n}{2},\cos ^2(c+d x)\right )}{b d (n+1) \sqrt{\sin ^2(c+d x)}}+\frac{B \sin (c+d x) (b \sec (c+d x))^{n+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (-n-2),-\frac{n}{2},\cos ^2(c+d x)\right )}{b^2 d (n+2) \sqrt{\sin ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.126553, antiderivative size = 143, 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 = {16, 3787, 3772, 2643} \[ \frac{A \sin (c+d x) (b \sec (c+d x))^{n+1} \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-n-1);\frac{1-n}{2};\cos ^2(c+d x)\right )}{b d (n+1) \sqrt{\sin ^2(c+d x)}}+\frac{B \sin (c+d x) (b \sec (c+d x))^{n+2} \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-n-2);-\frac{n}{2};\cos ^2(c+d x)\right )}{b^2 d (n+2) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(b*Sec[c + d*x])^n*(A + B*Sec[c + d*x]),x]

[Out]

(A*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]) + (B*Hypergeometric2F1[1/2, (-2 - n)/2, -n/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(2
+ n)*Sin[c + d*x])/(b^2*d*(2 + 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 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 ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx &=\frac{\int (b \sec (c+d x))^{2+n} (A+B \sec (c+d x)) \, dx}{b^2}\\ &=\frac{A \int (b \sec (c+d x))^{2+n} \, dx}{b^2}+\frac{B \int (b \sec (c+d x))^{3+n} \, dx}{b^3}\\ &=\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 )^{-2-n} \, dx}{b^2}+\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 )^{-3-n} \, dx}{b^3}\\ &=\frac{A \, _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)}}+\frac{B \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-2-n);-\frac{n}{2};\cos ^2(c+d x)\right ) \sec (c+d x) (b \sec (c+d x))^n \tan (c+d x)}{d (2+n) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.220685, size = 119, normalized size = 0.83 \[ \frac{\sqrt{-\tan ^2(c+d x)} \csc (c+d x) \sec (c+d x) (b \sec (c+d x))^n \left (A (n+3) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n+2}{2},\frac{n+4}{2},\sec ^2(c+d x)\right )+B (n+2) \sec (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n+3}{2},\frac{n+5}{2},\sec ^2(c+d x)\right )\right )}{d (n+2) (n+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(b*Sec[c + d*x])^n*(A + B*Sec[c + d*x]),x]

[Out]

(Csc[c + d*x]*Sec[c + d*x]*(b*Sec[c + d*x])^n*(A*(3 + n)*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, Sec[c +
d*x]^2] + B*(2 + n)*Hypergeometric2F1[1/2, (3 + n)/2, (5 + n)/2, Sec[c + d*x]^2]*Sec[c + d*x])*Sqrt[-Tan[c + d
*x]^2])/(d*(2 + n)*(3 + n))

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Maple [F]  time = 0.924, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{2} \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(sec(d*x+c)^2*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x)

[Out]

int(sec(d*x+c)^2*(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} \sec \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(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)^2, 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 )^{3} + A \sec \left (d x + c\right )^{2}\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(sec(d*x+c)^2*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*sec(d*x + c)^3 + A*sec(d*x + c)^2)*(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 ) \sec ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(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)**2, 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} \sec \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(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)^2, x)