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 \[ \frac {(A (n+3)+C (n+2)) \sin (c+d x) (b \sec (c+d x))^{n+1} \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-n-1);\frac {1-n}{2};\cos ^2(c+d x)\right )}{b d (n+1) (n+3) \sqrt {\sin ^2(c+d x)}}+\frac {B \sin (c+d x) (b \sec (c+d x))^{n+2} \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-n-2);-\frac {n}{2};\cos ^2(c+d x)\right )}{b^2 d (n+2) \sqrt {\sin ^2(c+d x)}}+\frac {C \tan (c+d x) (b \sec (c+d x))^{n+2}}{b^2 d (n+3)} \]

[Out]

(C*(2+n)+A*(3+n))*hypergeom([1/2, -1/2-1/2*n],[1/2-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^(1+n)*sin(d*x+c)/b/d/(1
+n)/(3+n)/(sin(d*x+c)^2)^(1/2)+B*hypergeom([1/2, -1-1/2*n],[-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^(2+n)*sin(d*x
+c)/b^2/d/(2+n)/(sin(d*x+c)^2)^(1/2)+C*(b*sec(d*x+c))^(2+n)*tan(d*x+c)/b^2/d/(3+n)

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Rubi [A]  time = 0.20, antiderivative size = 189, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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*}

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Mathematica [C]  time = 2.77, size = 296, normalized size = 1.57 \[ -\frac {i 2^{n+3} e^{2 i (c+d x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \sec ^{-n-2}(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {A \left (1+e^{2 i (c+d x)}\right )^2 \, _2F_1\left (1,-\frac {n}{2};\frac {n+4}{2};-e^{2 i (c+d x)}\right )}{n+2}+\frac {2 B e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \, _2F_1\left (1,\frac {1}{2} (-n-1);\frac {n+5}{2};-e^{2 i (c+d x)}\right )}{n+3}+\frac {4 C e^{2 i (c+d x)} \, _2F_1\left (1,-\frac {n}{2}-1;\frac {n+6}{2};-e^{2 i (c+d x)}\right )}{n+4}\right )}{d \left (1+e^{2 i (c+d x)}\right )^3 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

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 + d*x))*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^n*((2*B*E^(I*(c + d*x))*(1 + E
^((2*I)*(c + d*x)))*Hypergeometric2F1[1, (-1 - n)/2, (5 + n)/2, -E^((2*I)*(c + d*x))])/(3 + n) + (4*C*E^((2*I)
*(c + d*x))*Hypergeometric2F1[1, -1 - n/2, (6 + n)/2, -E^((2*I)*(c + d*x))])/(4 + n) + (A*(1 + E^((2*I)*(c + d
*x)))^2*Hypergeometric2F1[1, -1/2*n, (4 + n)/2, -E^((2*I)*(c + d*x))])/(2 + 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*(1 + E^((2*I)*(c + d*x)))^3*(A + 2*C + 2*B*Cos[c + d*x]
+ A*Cos[2*(c + d*x)]))

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \sec \left (d x + c\right )^{4} + B \sec \left (d x + c\right )^{3} + A \sec \left (d x + c\right )^{2}\right )} \left (b \sec \left (d x + c\right )\right )^{n}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{2}\,{d x} \]

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)

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maple [F]  time = 2.59, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{2}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{2}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\cos \left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b/cos(c + d*x))^n*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x)^2,x)

[Out]

int(((b/cos(c + d*x))^n*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \]

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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