3.1.75 \(\int \cos (c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [75]

Optimal. Leaf size=191 \[ \frac {b^2 (C (1-n)-A n) \, _2F_1\left (\frac {1}{2},\frac {2-n}{2};\frac {4-n}{2};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-2+n} \sin (c+d x)}{d (2-n) n \sqrt {\sin ^2(c+d x)}}-\frac {b B \, _2F_1\left (\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-1+n} \sin (c+d x)}{d (1-n) \sqrt {\sin ^2(c+d x)}}+\frac {b C (b \sec (c+d x))^{-1+n} \tan (c+d x)}{d n} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {16, 4132, 3857, 2722, 4131} \begin {gather*} \frac {b^2 (C (1-n)-A n) \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) n \sqrt {\sin ^2(c+d x)}}-\frac {b B \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) \sqrt {\sin ^2(c+d x)}}+\frac {b C \tan (c+d x) (b \sec (c+d x))^{n-1}}{d n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*(C*(1 - n) - A*n)*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)*n*Sqrt[Sin[c + d*x]^2]) - (b*B*Hypergeometric2F1[1/2, (1 - n)/2, (3 - n)/2, Cos[c + d
*x]^2]*(b*Sec[c + d*x])^(-1 + n)*Sin[c + d*x])/(d*(1 - n)*Sqrt[Sin[c + d*x]^2]) + (b*C*(b*Sec[c + d*x])^(-1 +
n)*Tan[c + d*x])/(d*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 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3857

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 4131

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 4132

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

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Mathematica [A]
time = 0.40, size = 161, normalized size = 0.84 \begin {gather*} \frac {\left (A n (1+n) \cos (c+d x) \cot (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-1+n);\frac {1+n}{2};\sec ^2(c+d x)\right )+(-1+n) \csc (c+d x) \left (B (1+n) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {2+n}{2};\sec ^2(c+d x)\right )+C n \, _2F_1\left (\frac {1}{2},\frac {1+n}{2};\frac {3+n}{2};\sec ^2(c+d x)\right )\right )\right ) (b \sec (c+d x))^n \sqrt {-\tan ^2(c+d x)}}{d (-1+n) n (1+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((A*n*(1 + n)*Cos[c + d*x]*Cot[c + d*x]*Hypergeometric2F1[1/2, (-1 + n)/2, (1 + n)/2, Sec[c + d*x]^2] + (-1 +
n)*Csc[c + d*x]*(B*(1 + n)*Cos[c + d*x]*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Sec[c + d*x]^2] + C*n*Hypergeom
etric2F1[1/2, (1 + n)/2, (3 + n)/2, Sec[c + d*x]^2]))*(b*Sec[c + d*x])^n*Sqrt[-Tan[c + d*x]^2])/(d*(-1 + n)*n*
(1 + n))

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Maple [F]
time = 0.39, size = 0, normalized size = 0.00 \[\int \cos \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(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)*(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \cos \left (c+d\,x\right )\,{\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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