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

Optimal. Leaf size=189 \[ \frac{(A (n+3)+C (n+2)) \sin (c+d x) (b \sec (c+d x))^{n+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (-n-1),\frac{1-n}{2},\cos ^2(c+d x)\right )}{b d (n+1) (n+3) \sqrt{\sin ^2(c+d x)}}+\frac{B \sin (c+d x) (b \sec (c+d x))^{n+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (-n-2),-\frac{n}{2},\cos ^2(c+d x)\right )}{b^2 d (n+2) \sqrt{\sin ^2(c+d x)}}+\frac{C \tan (c+d x) (b \sec (c+d x))^{n+2}}{b^2 d (n+3)} \]

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

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Rubi [A]  time = 0.195178, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {16, 4047, 3772, 2643, 4046} \[ \frac{(A (n+3)+C (n+2)) \sin (c+d x) (b \sec (c+d x))^{n+1} \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-n-1);\frac{1-n}{2};\cos ^2(c+d x)\right )}{b d (n+1) (n+3) \sqrt{\sin ^2(c+d x)}}+\frac{B \sin (c+d x) (b \sec (c+d x))^{n+2} \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-n-2);-\frac{n}{2};\cos ^2(c+d x)\right )}{b^2 d (n+2) \sqrt{\sin ^2(c+d x)}}+\frac{C \tan (c+d x) (b \sec (c+d x))^{n+2}}{b^2 d (n+3)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(b*Sec[c + d*x])^n*(A + B*Sec[c + d*x] + 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]) + (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]) + (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 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 \sec ^2(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{\int (b \sec (c+d x))^{2+n} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b^2}\\ &=\frac{\int (b \sec (c+d x))^{2+n} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b^2}+\frac{B \int (b \sec (c+d x))^{3+n} \, dx}{b^3}\\ &=\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{\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{C (b \sec (c+d x))^{2+n} \tan (c+d x)}{b^2 d (3+n)}+\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)}}+\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)}+\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 [C]  time = 5.51426, size = 462, normalized size = 2.44 \[ -\frac{i 2^{n+3} e^{2 i c-i d n x} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \left (1+e^{2 i (c+d x)}\right )^n \sec ^{-n-2}(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{A e^{i d (n+2) x} \text{Hypergeometric2F1}\left (\frac{n+2}{2},n+4,\frac{n+4}{2},-e^{2 i (c+d x)}\right )}{n+2}+\frac{2 A e^{i (2 c+d (n+4) x)} \text{Hypergeometric2F1}\left (\frac{n+4}{2},n+4,\frac{n+6}{2},-e^{2 i (c+d x)}\right )}{n+4}+\frac{A e^{i (4 c+d (n+6) x)} \text{Hypergeometric2F1}\left (n+4,\frac{n+6}{2},\frac{n+8}{2},-e^{2 i (c+d x)}\right )}{n+6}+\frac{2 B e^{i (c+d (n+3) x)} \text{Hypergeometric2F1}\left (\frac{n+3}{2},n+4,\frac{n+5}{2},-e^{2 i (c+d x)}\right )}{n+3}+\frac{2 B e^{i (3 c+d (n+5) x)} \text{Hypergeometric2F1}\left (n+4,\frac{n+5}{2},\frac{n+7}{2},-e^{2 i (c+d x)}\right )}{n+5}+\frac{4 C e^{i (2 c+d (n+4) x)} \text{Hypergeometric2F1}\left (\frac{n+4}{2},n+4,\frac{n+6}{2},-e^{2 i (c+d x)}\right )}{n+4}\right )}{d (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*2^(3 + n)*E^((2*I)*c - I*d*n*x)*(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*(2 + n)*x)*Hypergeometric2F1[(2 + n)/2, 4 + n, (4 + n)/2, -E^((2*I)*(c + d*x))])/(2 + n) + (2*B*E
^(I*(c + d*(3 + n)*x))*Hypergeometric2F1[(3 + n)/2, 4 + n, (5 + n)/2, -E^((2*I)*(c + d*x))])/(3 + n) + (2*A*E^
(I*(2*c + d*(4 + n)*x))*Hypergeometric2F1[(4 + n)/2, 4 + n, (6 + n)/2, -E^((2*I)*(c + d*x))])/(4 + n) + (4*C*E
^(I*(2*c + d*(4 + n)*x))*Hypergeometric2F1[(4 + n)/2, 4 + n, (6 + n)/2, -E^((2*I)*(c + d*x))])/(4 + n) + (2*B*
E^(I*(3*c + d*(5 + n)*x))*Hypergeometric2F1[4 + n, (5 + n)/2, (7 + n)/2, -E^((2*I)*(c + d*x))])/(5 + n) + (A*E
^(I*(4*c + d*(6 + n)*x))*Hypergeometric2F1[4 + n, (6 + 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 + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A
*Cos[2*c + 2*d*x]))

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Maple [F]  time = 0.679, 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 ) +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+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

[Out]

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