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

Optimal. Leaf size=120 \[ \frac{(A (n+3)+C (n+2)) \sin (c+d x) (b \sec (c+d x))^{n+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (-n-1),\frac{1-n}{2},\cos ^2(c+d x)\right )}{b d (n+1) (n+3) \sqrt{\sin ^2(c+d x)}}+\frac{C \tan (c+d x) (b \sec (c+d x))^{n+2}}{b^2 d (n+3)} \]

[Out]

((C*(2 + n) + A*(3 + n))*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)*(3 + n)*Sqrt[Sin[c + d*x]^2]) + (C*(b*Sec[c + d*x])^(2 + n)*Tan[c + d*x])/(b^2*d*
(3 + n))

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Rubi [A]  time = 0.111613, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {16, 4046, 3772, 2643} \[ \frac{(A (n+3)+C (n+2)) \sin (c+d x) (b \sec (c+d x))^{n+1} \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-n-1);\frac{1-n}{2};\cos ^2(c+d x)\right )}{b d (n+1) (n+3) \sqrt{\sin ^2(c+d x)}}+\frac{C \tan (c+d x) (b \sec (c+d x))^{n+2}}{b^2 d (n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((C*(2 + n) + A*(3 + n))*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)*(3 + n)*Sqrt[Sin[c + d*x]^2]) + (C*(b*Sec[c + d*x])^(2 + n)*Tan[c + d*x])/(b^2*d*
(3 + 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 ^2(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))^{2+n} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b^2}\\ &=\frac{C (b \sec (c+d x))^{2+n} \tan (c+d x)}{b^2 d (3+n)}+\frac{\left (A+\frac{C (2+n)}{3+n}\right ) \int (b \sec (c+d x))^{2+n} \, dx}{b^2}\\ &=\frac{C (b \sec (c+d x))^{2+n} \tan (c+d x)}{b^2 d (3+n)}+\frac{\left (\left (A+\frac{C (2+n)}{3+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 )^{-2-n} \, dx}{b^2}\\ &=\frac{\left (A+\frac{C (2+n)}{3+n}\right ) \, _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{C (b \sec (c+d x))^{2+n} \tan (c+d x)}{b^2 d (3+n)}\\ \end{align*}

Mathematica [C]  time = 6.89962, size = 274, normalized size = 2.28 \[ -\frac{i 2^{n+3} e^{-i n (c+d x)} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \sec ^{-n-2}(c+d x) \left (A+C \sec ^2(c+d x)\right ) (b \sec (c+d x))^n \left (\frac{2 (A+2 C) e^{i (n+4) (c+d x)} \text{Hypergeometric2F1}\left (1,-\frac{n}{2}-1,\frac{n+6}{2},-e^{2 i (c+d x)}\right )}{n+4}+\frac{A e^{i (n+2) (c+d x)} \text{Hypergeometric2F1}\left (1,-\frac{n}{2}-2,\frac{n+4}{2},-e^{2 i (c+d x)}\right )}{n+2}+\frac{A e^{i (n+6) (c+d x)} \text{Hypergeometric2F1}\left (1,-\frac{n}{2},\frac{n+8}{2},-e^{2 i (c+d x)}\right )}{n+6}\right )}{d \left (1+e^{2 i (c+d x)}\right )^3 (A \cos (2 c+2 d x)+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*2^(3 + n)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^n*((A*E^(I*(2 + n)*(c + d*x))*Hypergeometric2F1[1,
-2 - n/2, (4 + n)/2, -E^((2*I)*(c + d*x))])/(2 + n) + (2*(A + 2*C)*E^(I*(4 + n)*(c + d*x))*Hypergeometric2F1[
1, -1 - n/2, (6 + n)/2, -E^((2*I)*(c + d*x))])/(4 + n) + (A*E^(I*(6 + n)*(c + d*x))*Hypergeometric2F1[1, -n/2,
(8 + n)/2, -E^((2*I)*(c + d*x))])/(6 + 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)))^3*(A + 2*C + A*Cos[2*c + 2*d*x]))

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Maple [F]  time = 0.533, 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+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)^2*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x)

[Out]

int(sec(d*x+c)^2*(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 )^{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+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)^2, 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 )^{4} + 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+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^4 + 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 + C \sec ^{2}{\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+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)**2, 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 )^{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+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)^2, x)