3.10.12 \(\int \cos ^2(c+d x) (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx\) [912]

Optimal. Leaf size=141 \[ -\frac {A (b \cos (c+d x))^{3+n} \, _2F_1\left (\frac {1}{2},\frac {3+n}{2};\frac {5+n}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{b^3 d (3+n) \sqrt {\sin ^2(c+d x)}}-\frac {B (b \cos (c+d x))^{4+n} \, _2F_1\left (\frac {1}{2},\frac {4+n}{2};\frac {6+n}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{b^4 d (4+n) \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {16, 2827, 2722} \begin {gather*} -\frac {A \sin (c+d x) (b \cos (c+d x))^{n+3} \, _2F_1\left (\frac {1}{2},\frac {n+3}{2};\frac {n+5}{2};\cos ^2(c+d x)\right )}{b^3 d (n+3) \sqrt {\sin ^2(c+d x)}}-\frac {B \sin (c+d x) (b \cos (c+d x))^{n+4} \, _2F_1\left (\frac {1}{2},\frac {n+4}{2};\frac {n+6}{2};\cos ^2(c+d x)\right )}{b^4 d (n+4) \sqrt {\sin ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((A*(b*Cos[c + d*x])^(3 + n)*Hypergeometric2F1[1/2, (3 + n)/2, (5 + n)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(b^3*
d*(3 + n)*Sqrt[Sin[c + d*x]^2])) - (B*(b*Cos[c + d*x])^(4 + n)*Hypergeometric2F1[1/2, (4 + n)/2, (6 + n)/2, Co
s[c + d*x]^2]*Sin[c + d*x])/(b^4*d*(4 + 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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx &=\frac {\int (b \cos (c+d x))^{2+n} (A+B \cos (c+d x)) \, dx}{b^2}\\ &=\frac {A \int (b \cos (c+d x))^{2+n} \, dx}{b^2}+\frac {B \int (b \cos (c+d x))^{3+n} \, dx}{b^3}\\ &=-\frac {A (b \cos (c+d x))^{3+n} \, _2F_1\left (\frac {1}{2},\frac {3+n}{2};\frac {5+n}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{b^3 d (3+n) \sqrt {\sin ^2(c+d x)}}-\frac {B (b \cos (c+d x))^{4+n} \, _2F_1\left (\frac {1}{2},\frac {4+n}{2};\frac {6+n}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{b^4 d (4+n) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.31, size = 120, normalized size = 0.85 \begin {gather*} -\frac {\cos ^2(c+d x) (b \cos (c+d x))^n \cot (c+d x) \left (A (4+n) \, _2F_1\left (\frac {1}{2},\frac {3+n}{2};\frac {5+n}{2};\cos ^2(c+d x)\right )+B (3+n) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {4+n}{2};\frac {6+n}{2};\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (3+n) (4+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[c + d*x]^2*(b*Cos[c + d*x])^n*Cot[c + d*x]*(A*(4 + n)*Hypergeometric2F1[1/2, (3 + n)/2, (5 + n)/2, Cos[
c + d*x]^2] + B*(3 + n)*Cos[c + d*x]*Hypergeometric2F1[1/2, (4 + n)/2, (6 + n)/2, Cos[c + d*x]^2])*Sqrt[Sin[c
+ d*x]^2])/(d*(3 + n)*(4 + n)))

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Maple [F]
time = 0.44, size = 0, normalized size = 0.00 \[\int \left (\cos ^{2}\left (d x +c \right )\right ) \left (b \cos \left (d x +c \right )\right )^{n} \left (A +B \cos \left (d x +c \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(d*x+c)^2*(b*cos(d*x+c))^n*(A+B*cos(d*x+c)),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)^2*(b*cos(d*x+c))^n*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*cos(d*x + c)^2, 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)^2*(b*cos(d*x+c))^n*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*cos(d*x + c)^3 + A*cos(d*x + c)^2)*(b*cos(d*x + c))^n, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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)^2*(b*cos(d*x+c))^n*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*cos(d*x + c)^2, 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 )}^2\,{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (A+B\,\cos \left (c+d\,x\right )\right ) \,d x \end {gather*}

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)),x)

[Out]

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

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