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

Optimal. Leaf size=151 \[ -\frac {A b^2 \sin (c+d x) (b \sec (c+d x))^{n-2} \, _2F_1\left (\frac {1}{2},\frac {2-n}{2};\frac {4-n}{2};\cos ^2(c+d x)\right )}{d (2-n) \sqrt {\sin ^2(c+d x)}}-\frac {b B \sin (c+d x) (b \sec (c+d x))^{n-1} \, _2F_1\left (\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(c+d x)\right )}{d (1-n) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-A*b^2*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*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)

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Rubi [A]  time = 0.13, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {16, 3787, 3772, 2643} \[ -\frac {A b^2 \sin (c+d x) (b \sec (c+d x))^{n-2} \, _2F_1\left (\frac {1}{2},\frac {2-n}{2};\frac {4-n}{2};\cos ^2(c+d x)\right )}{d (2-n) \sqrt {\sin ^2(c+d x)}}-\frac {b B \sin (c+d x) (b \sec (c+d x))^{n-1} \, _2F_1\left (\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(c+d x)\right )}{d (1-n) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A*b^2*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*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])

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 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 107, normalized size = 0.71 \[ \frac {\sqrt {-\tan ^2(c+d x)} \cot (c+d x) (b \sec (c+d x))^n \left (A n \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {n-1}{2};\frac {n+1}{2};\sec ^2(c+d x)\right )+B (n-1) \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {n+2}{2};\sec ^2(c+d x)\right )\right )}{d (n-1) n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cot[c + d*x]*(A*n*Cos[c + d*x]*Hypergeometric2F1[1/2, (-1 + n)/2, (1 + n)/2, Sec[c + d*x]^2] + B*(-1 + n)*Hyp
ergeometric2F1[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*(-1 + n)*n)

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \cos \left (d x + c\right ) \sec \left (d x + c\right ) + A \cos \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

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)),x, algorithm="giac")

[Out]

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

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maple [F]  time = 4.00, size = 0, normalized size = 0.00 \[ \int \cos \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )\right )\, dx \]

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)),x)

[Out]

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

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

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)),x, algorithm="maxima")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)),x)

[Out]

Integral((b*sec(c + d*x))**n*(A + B*sec(c + d*x))*cos(c + d*x), x)

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