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

Optimal. Leaf size=175 \[ -\frac{b (A n+A+C n) \sin (c+d x) (b \sec (c+d x))^{n-1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1-n}{2},\frac{3-n}{2},\cos ^2(c+d x)\right )}{d (1-n) (n+1) \sqrt{\sin ^2(c+d x)}}+\frac{B \sin (c+d x) (b \sec (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},-\frac{n}{2},\frac{2-n}{2},\cos ^2(c+d x)\right )}{d n \sqrt{\sin ^2(c+d x)}}+\frac{C \tan (c+d x) (b \sec (c+d x))^n}{d (n+1)} \]

[Out]

-((b*(A + A*n + C*n)*Hypergeometric2F1[1/2, (1 - n)/2, (3 - n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(-1 + n)*Si
n[c + d*x])/(d*(1 - n)*(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]) + (C*(b*Sec[c + d*x])^n*Tan[c + d*x])/(d*(1 +
n))

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Rubi [A]  time = 0.142766, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {4047, 3772, 2643, 4046} \[ -\frac{b (A n+A+C n) \sin (c+d x) (b \sec (c+d x))^{n-1} \, _2F_1\left (\frac{1}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(c+d x)\right )}{d (1-n) (n+1) \sqrt{\sin ^2(c+d x)}}+\frac{B \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac{1}{2},-\frac{n}{2};\frac{2-n}{2};\cos ^2(c+d x)\right )}{d n \sqrt{\sin ^2(c+d x)}}+\frac{C \tan (c+d x) (b \sec (c+d x))^n}{d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(A + A*n + C*n)*Hypergeometric2F1[1/2, (1 - n)/2, (3 - n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(-1 + n)*Si
n[c + d*x])/(d*(1 - n)*(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]) + (C*(b*Sec[c + d*x])^n*Tan[c + d*x])/(d*(1 +
n))

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, 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]

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]

Rubi steps

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

Mathematica [C]  time = 4.8301, size = 401, normalized size = 2.29 \[ -\frac{i 2^{n+1} e^{-i d n x} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \left (1+e^{2 i (c+d x)}\right )^n \sec ^{-n-2}(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (e^{2 i c} \left (\frac{2 (A+2 C) e^{i d (n+2) x} \text{Hypergeometric2F1}\left (\frac{n+2}{2},n+2,\frac{n+4}{2},-e^{2 i (c+d x)}\right )}{n+2}+\frac{A e^{i (2 c+d (n+4) x)} \text{Hypergeometric2F1}\left (n+2,\frac{n+4}{2},\frac{n+6}{2},-e^{2 i (c+d x)}\right )}{n+4}+\frac{2 B e^{i (c+d (n+3) x)} \text{Hypergeometric2F1}\left (n+2,\frac{n+3}{2},\frac{n+5}{2},-e^{2 i (c+d x)}\right )}{n+3}\right )+\frac{A e^{i d n x} \text{Hypergeometric2F1}\left (\frac{n}{2},n+2,\frac{n+2}{2},-e^{2 i (c+d x)}\right )}{n}+\frac{2 B e^{i (c+d (n+1) x)} \text{Hypergeometric2F1}\left (\frac{n+1}{2},n+2,\frac{n+3}{2},-e^{2 i (c+d x)}\right )}{n+1}\right )}{d (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*2^(1 + n)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^n*(1 + E^((2*I)*(c + d*x)))^n*((A*E^(I*d*n*x)*Hype
rgeometric2F1[n/2, 2 + n, (2 + n)/2, -E^((2*I)*(c + d*x))])/n + (2*B*E^(I*(c + d*(1 + n)*x))*Hypergeometric2F1
[(1 + n)/2, 2 + n, (3 + n)/2, -E^((2*I)*(c + d*x))])/(1 + n) + E^((2*I)*c)*((2*(A + 2*C)*E^(I*d*(2 + n)*x)*Hyp
ergeometric2F1[(2 + n)/2, 2 + n, (4 + n)/2, -E^((2*I)*(c + d*x))])/(2 + n) + (2*B*E^(I*(c + d*(3 + n)*x))*Hype
rgeometric2F1[2 + n, (3 + n)/2, (5 + n)/2, -E^((2*I)*(c + d*x))])/(3 + n) + (A*E^(I*(2*c + d*(4 + n)*x))*Hyper
geometric2F1[2 + n, (4 + n)/2, (6 + n)/2, -E^((2*I)*(c + d*x))])/(4 + n)))*Sec[c + d*x]^(-2 - n)*(b*Sec[c + d*
x])^n*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*E^(I*d*n*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]
))

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Maple [F]  time = 0.612, 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 ) +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((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

[Out]

integrate((C*sec(d*x + c)^2 + 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 (C \sec \left (d x + c\right )^{2} + 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)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

[Out]

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

[Out]

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