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

Optimal. Leaf size=208 \[ -\frac {b^3 (A (1-n)+C (2-n)) \sin (c+d x) (b \sec (c+d x))^{n-3} \, _2F_1\left (\frac {1}{2},\frac {3-n}{2};\frac {5-n}{2};\cos ^2(c+d x)\right )}{d (1-n) (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^2 B \sin (c+d x) (b \sec (c+d x))^{n-2} \, _2F_1\left (\frac {1}{2},\frac {2-n}{2};\frac {4-n}{2};\cos ^2(c+d x)\right )}{d (2-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^2 C \tan (c+d x) (b \sec (c+d x))^{n-2}}{d (1-n)} \]

[Out]

-b^3*(A*(1-n)+C*(2-n))*hypergeom([1/2, 3/2-1/2*n],[5/2-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^(-3+n)*sin(d*x+c)/d
/(n^2-4*n+3)/(sin(d*x+c)^2)^(1/2)-b^2*B*hypergeom([1/2, 1-1/2*n],[2-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^(-2+n)
*sin(d*x+c)/d/(2-n)/(sin(d*x+c)^2)^(1/2)-b^2*C*(b*sec(d*x+c))^(-2+n)*tan(d*x+c)/d/(1-n)

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Rubi [A]  time = 0.22, antiderivative size = 208, 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} \[ -\frac {b^3 (A (1-n)+C (2-n)) \sin (c+d x) (b \sec (c+d x))^{n-3} \, _2F_1\left (\frac {1}{2},\frac {3-n}{2};\frac {5-n}{2};\cos ^2(c+d x)\right )}{d (1-n) (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^2 B \sin (c+d x) (b \sec (c+d x))^{n-2} \, _2F_1\left (\frac {1}{2},\frac {2-n}{2};\frac {4-n}{2};\cos ^2(c+d x)\right )}{d (2-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^2 C \tan (c+d x) (b \sec (c+d x))^{n-2}}{d (1-n)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^3*(A*(1 - n) + C*(2 - n))*Hypergeometric2F1[1/2, (3 - n)/2, (5 - n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(
-3 + n)*Sin[c + d*x])/(d*(1 - n)*(3 - n)*Sqrt[Sin[c + d*x]^2])) - (b^2*B*Hypergeometric2F1[1/2, (2 - n)/2, (4
- n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(-2 + n)*Sin[c + d*x])/(d*(2 - n)*Sqrt[Sin[c + d*x]^2]) - (b^2*C*(b*S
ec[c + d*x])^(-2 + n)*Tan[c + d*x])/(d*(1 - 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 \cos ^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 &=b^2 \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 \int (b \sec (c+d x))^{-2+n} \left (A+C \sec ^2(c+d x)\right ) \, dx+(b B) \int (b \sec (c+d x))^{-1+n} \, dx\\ &=-\frac {b^2 C (b \sec (c+d x))^{-2+n} \tan (c+d x)}{d (1-n)}+\left (b^2 \left (A+\frac {C (2-n)}{1-n}\right )\right ) \int (b \sec (c+d x))^{-2+n} \, dx+\left (b 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\\ &=-\frac {B \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {2-n}{2};\frac {4-n}{2};\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (2-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^2 C (b \sec (c+d x))^{-2+n} \tan (c+d x)}{d (1-n)}+\left (b^2 \left (A+\frac {C (2-n)}{1-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\\ &=-\frac {B \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {2-n}{2};\frac {4-n}{2};\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (2-n) \sqrt {\sin ^2(c+d x)}}-\frac {\left (A+\frac {C (2-n)}{1-n}\right ) \cos ^3(c+d x) \, _2F_1\left (\frac {1}{2},\frac {3-n}{2};\frac {5-n}{2};\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^2 C (b \sec (c+d x))^{-2+n} \tan (c+d x)}{d (1-n)}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 155, normalized size = 0.75 \[ \frac {\sqrt {-\tan ^2(c+d x)} \cot (c+d x) (b \sec (c+d x))^n \left (A (n-1) n \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {n-2}{2};\frac {n}{2};\sec ^2(c+d x)\right )+(n-2) \left (B n \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {n-1}{2};\frac {n+1}{2};\sec ^2(c+d x)\right )+C (n-1) \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {n+2}{2};\sec ^2(c+d x)\right )\right )\right )}{d (n-2) (n-1) n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + B \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + A \cos \left (d x + c\right )^{2}\right )} \left (b \sec \left (d x + c\right )\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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*cos(d*x + c)^2*sec(d*x + c)^2 + B*cos(d*x + c)^2*sec(d*x + c) + A*cos(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 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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*cos(d*x + c)^2, x)

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maple [F]  time = 5.52, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(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(cos(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 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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*cos(d*x + c)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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)*cos(c + d*x)**2, x)

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