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

Optimal. Leaf size=109 \[ \frac{(A (n+2)+C (n+1)) \sin (c+d x) (b \sec (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},-\frac{n}{2},\frac{2-n}{2},\cos ^2(c+d x)\right )}{d n (n+2) \sqrt{\sin ^2(c+d x)}}+\frac{C \tan (c+d x) (b \sec (c+d x))^{n+1}}{b d (n+2)} \]

[Out]

((C*(1 + n) + A*(2 + n))*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*(2 + n)*Sqrt[Sin[c + d*x]^2]) + (C*(b*Sec[c + d*x])^(1 + n)*Tan[c + d*x])/(b*d*(2 + n))

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Rubi [A]  time = 0.102033, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {16, 4046, 3772, 2643} \[ \frac{(A (n+2)+C (n+1)) \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac{1}{2},-\frac{n}{2};\frac{2-n}{2};\cos ^2(c+d x)\right )}{d n (n+2) \sqrt{\sin ^2(c+d x)}}+\frac{C \tan (c+d x) (b \sec (c+d x))^{n+1}}{b d (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((C*(1 + n) + A*(2 + n))*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*(2 + n)*Sqrt[Sin[c + d*x]^2]) + (C*(b*Sec[c + d*x])^(1 + n)*Tan[c + d*x])/(b*d*(2 + n))

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 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

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 (c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{\int (b \sec (c+d x))^{1+n} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b}\\ &=\frac{C (b \sec (c+d x))^{1+n} \tan (c+d x)}{b d (2+n)}+\frac{\left (A+\frac{C (1+n)}{2+n}\right ) \int (b \sec (c+d x))^{1+n} \, dx}{b}\\ &=\frac{C (b \sec (c+d x))^{1+n} \tan (c+d x)}{b d (2+n)}+\frac{\left (\left (A+\frac{C (1+n)}{2+n}\right ) \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 (A+\frac{C (1+n)}{2+n}\right ) \, _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{C (b \sec (c+d x))^{1+n} \tan (c+d x)}{b d (2+n)}\\ \end{align*}

Mathematica [C]  time = 6.58729, size = 282, normalized size = 2.59 \[ -\frac{i 2^{n+2} e^{-i n (c+d x)} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \sec ^{-n-2}(c+d x) \left (A+C \sec ^2(c+d x)\right ) (b \sec (c+d x))^n \left (\frac{2 (A+2 C) e^{i (n+3) (c+d x)} \text{Hypergeometric2F1}\left (1,\frac{1}{2} (-n-1),\frac{n+5}{2},-e^{2 i (c+d x)}\right )}{n+3}+\frac{A e^{i (n+1) (c+d x)} \text{Hypergeometric2F1}\left (1,\frac{1}{2} (-n-3),\frac{n+3}{2},-e^{2 i (c+d x)}\right )}{n+1}+\frac{A e^{i (n+5) (c+d x)} \text{Hypergeometric2F1}\left (1,\frac{1-n}{2},\frac{n+7}{2},-e^{2 i (c+d x)}\right )}{n+5}\right )}{d \left (1+e^{2 i (c+d x)}\right )^2 (A \cos (2 c+2 d x)+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*2^(2 + n)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^n*((A*E^(I*(1 + n)*(c + d*x))*Hypergeometric2F1[1,
(-3 - n)/2, (3 + n)/2, -E^((2*I)*(c + d*x))])/(1 + n) + (2*(A + 2*C)*E^(I*(3 + n)*(c + d*x))*Hypergeometric2F
1[1, (-1 - n)/2, (5 + n)/2, -E^((2*I)*(c + d*x))])/(3 + n) + (A*E^(I*(5 + n)*(c + d*x))*Hypergeometric2F1[1, (
1 - n)/2, (7 + n)/2, -E^((2*I)*(c + d*x))])/(5 + n))*Sec[c + d*x]^(-2 - n)*(b*Sec[c + d*x])^n*(A + C*Sec[c + d
*x]^2))/(d*E^(I*n*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^2*(A + 2*C + A*Cos[2*c + 2*d*x]))

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Maple [F]  time = 0.712, size = 0, normalized size = 0. \begin{align*} \int \sec \left ( dx+c \right ) \left ( b\sec \left ( dx+c \right ) \right ) ^{n} \left ( A+C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x)

[Out]

int(sec(d*x+c)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^n*sec(d*x + c), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \sec \left (d x + c\right )^{3} + A \sec \left (d x + c\right )\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)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(b*sec(d*x+c))**n*(A+C*sec(d*x+c)**2),x)

[Out]

Integral((b*sec(c + d*x))**n*(A + C*sec(c + d*x)**2)*sec(c + d*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^n*sec(d*x + c), x)