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

Optimal. Leaf size=117 \[ \frac {C \sin (c+d x) (b \cos (c+d x))^{n+3}}{b^3 d (n+4)}-\frac {(A (n+4)+C (n+3)) \sin (c+d x) (b \cos (c+d x))^{n+3} \, _2F_1\left (\frac {1}{2},\frac {n+3}{2};\frac {n+5}{2};\cos ^2(c+d x)\right )}{b^3 d (n+3) (n+4) \sqrt {\sin ^2(c+d x)}} \]

[Out]

C*(b*cos(d*x+c))^(3+n)*sin(d*x+c)/b^3/d/(4+n)-(C*(3+n)+A*(4+n))*(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)/(4+n)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {16, 3014, 2643} \[ \frac {C \sin (c+d x) (b \cos (c+d x))^{n+3}}{b^3 d (n+4)}-\frac {(A (n+4)+C (n+3)) \sin (c+d x) (b \cos (c+d x))^{n+3} \, _2F_1\left (\frac {1}{2},\frac {n+3}{2};\frac {n+5}{2};\cos ^2(c+d x)\right )}{b^3 d (n+3) (n+4) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*(b*Cos[c + d*x])^(3 + n)*Sin[c + d*x])/(b^3*d*(4 + n)) - ((C*(3 + n) + A*(4 + n))*(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)*(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 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 3014

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[
e + f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e + f*
x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac {\int (b \cos (c+d x))^{2+n} \left (A+C \cos ^2(c+d x)\right ) \, dx}{b^2}\\ &=\frac {C (b \cos (c+d x))^{3+n} \sin (c+d x)}{b^3 d (4+n)}+\frac {\left (A+\frac {C (3+n)}{4+n}\right ) \int (b \cos (c+d x))^{2+n} \, dx}{b^2}\\ &=\frac {C (b \cos (c+d x))^{3+n} \sin (c+d x)}{b^3 d (4+n)}-\frac {\left (A+\frac {C (3+n)}{4+n}\right ) (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)}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 122, normalized size = 1.04 \[ -\frac {\sqrt {\sin ^2(c+d x)} \cos ^2(c+d x) \cot (c+d x) (b \cos (c+d x))^n \left (A (n+5) \, _2F_1\left (\frac {1}{2},\frac {n+3}{2};\frac {n+5}{2};\cos ^2(c+d x)\right )+C (n+3) \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {n+5}{2};\frac {n+7}{2};\cos ^2(c+d x)\right )\right )}{d (n+3) (n+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[c + d*x]^2*(b*Cos[c + d*x])^n*Cot[c + d*x]*(A*(5 + n)*Hypergeometric2F1[1/2, (3 + n)/2, (5 + n)/2, Cos[
c + d*x]^2] + C*(3 + n)*Cos[c + d*x]^2*Hypergeometric2F1[1/2, (5 + n)/2, (7 + n)/2, Cos[c + d*x]^2])*Sqrt[Sin[
c + d*x]^2])/(d*(3 + n)*(5 + n)))

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{4} + A \cos \left (d x + c\right )^{2}\right )} \left (b \cos \left (d x + c\right )\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)^4 + A*cos(d*x + c)^2)*(b*cos(d*x + c))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.78, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (d x +c \right )\right ) \left (b \cos \left (d x +c \right )\right )^{n} \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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