3.3.14 \(\int \cos (c+d x) (b \sec (c+d x))^n \, dx\) [214]

Optimal. Leaf size=75 \[ -\frac {b^2 \, _2F_1\left (\frac {1}{2},\frac {2-n}{2};\frac {4-n}{2};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-2+n} \sin (c+d x)}{d (2-n) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-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)

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Rubi [A]
time = 0.04, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {16, 3857, 2722} \begin {gather*} -\frac {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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((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]))

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]

Rubi steps

\begin {align*} \int \cos (c+d x) (b \sec (c+d x))^n \, dx &=b \int (b \sec (c+d x))^{-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 )^{1-n} \, dx\\ &=-\frac {\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 = 0.10, size = 68, normalized size = 0.91 \begin {gather*} \frac {b \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 ) (b \sec (c+d x))^{-1+n} \sqrt {-\tan ^2(c+d x)}}{d (-1+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.24, size = 0, normalized size = 0.00 \[\int \cos \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(d*x+c)*(b*sec(d*x+c))^n,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,x, algorithm="maxima")

[Out]

integrate((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,x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c))^n*cos(d*x + c), 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} \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,x)

[Out]

Integral((b*sec(c + d*x))**n*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,x, algorithm="giac")

[Out]

integrate((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 \,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,x)

[Out]

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

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