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

Optimal. Leaf size=100 \[ \frac {C \sin (c+d x) (b \cos (c+d x))^n}{d (n+1)}-\frac {(A n+A+C n) \sin (c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {n+2}{2};\cos ^2(c+d x)\right )}{d n (n+1) \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {16, 3014, 2643} \[ \frac {C \sin (c+d x) (b \cos (c+d x))^n}{d (n+1)}-\frac {(A n+A+C n) \sin (c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {n+2}{2};\cos ^2(c+d x)\right )}{d n (n+1) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*(b*Cos[c + d*x])^n*Sin[c + d*x])/(d*(1 + n)) - ((A + A*n + C*n)*(b*Cos[c + d*x])^n*Hypergeometric2F1[1/2, n
/2, (2 + n)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*n*(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 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 (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx &=b \int (b \cos (c+d x))^{-1+n} \left (A+C \cos ^2(c+d x)\right ) \, dx\\ &=\frac {C (b \cos (c+d x))^n \sin (c+d x)}{d (1+n)}+\frac {(b (A+A n+C n)) \int (b \cos (c+d x))^{-1+n} \, dx}{1+n}\\ &=\frac {C (b \cos (c+d x))^n \sin (c+d x)}{d (1+n)}-\frac {(A+A n+C n) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {2+n}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{d n (1+n) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 111, normalized size = 1.11 \[ -\frac {b \sqrt {\sin ^2(c+d x)} \cot (c+d x) (b \cos (c+d x))^{n-1} \left (A (n+2) \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {n+2}{2};\cos ^2(c+d x)\right )+C n \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {n+2}{2};\frac {n+4}{2};\cos ^2(c+d x)\right )\right )}{d n (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 3.40, size = 0, normalized size = 0.00 \[ \int \left (b \cos \left (d x +c \right )\right )^{n} \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \sec \left (d x +c \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^n}{\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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