3.379 \(\int \frac{(b \cos (c+d x))^n (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt{\cos (c+d x)}} \, dx\)

Optimal. Leaf size=221 \[ -\frac{2 (A (2 n+3)+2 C n+C) \sin (c+d x) \sqrt{\cos (c+d x)} (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n+1);\frac{1}{4} (2 n+5);\cos ^2(c+d x)\right )}{d (2 n+1) (2 n+3) \sqrt{\sin ^2(c+d x)}}-\frac{2 B \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n+3);\frac{1}{4} (2 n+7);\cos ^2(c+d x)\right )}{d (2 n+3) \sqrt{\sin ^2(c+d x)}}+\frac{2 C \sin (c+d x) \sqrt{\cos (c+d x)} (b \cos (c+d x))^n}{d (2 n+3)} \]

[Out]

(2*C*Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^n*Sin[c + d*x])/(d*(3 + 2*n)) - (2*(C + 2*C*n + A*(3 + 2*n))*Sqrt[Cos
[c + d*x]]*(b*Cos[c + d*x])^n*Hypergeometric2F1[1/2, (1 + 2*n)/4, (5 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(
d*(1 + 2*n)*(3 + 2*n)*Sqrt[Sin[c + d*x]^2]) - (2*B*Cos[c + d*x]^(3/2)*(b*Cos[c + d*x])^n*Hypergeometric2F1[1/2
, (3 + 2*n)/4, (7 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(3 + 2*n)*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.196848, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {20, 3023, 2748, 2643} \[ -\frac{2 (A (2 n+3)+2 C n+C) \sin (c+d x) \sqrt{\cos (c+d x)} (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n+1);\frac{1}{4} (2 n+5);\cos ^2(c+d x)\right )}{d (2 n+1) (2 n+3) \sqrt{\sin ^2(c+d x)}}-\frac{2 B \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n+3);\frac{1}{4} (2 n+7);\cos ^2(c+d x)\right )}{d (2 n+3) \sqrt{\sin ^2(c+d x)}}+\frac{2 C \sin (c+d x) \sqrt{\cos (c+d x)} (b \cos (c+d x))^n}{d (2 n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*C*Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^n*Sin[c + d*x])/(d*(3 + 2*n)) - (2*(C + 2*C*n + A*(3 + 2*n))*Sqrt[Cos
[c + d*x]]*(b*Cos[c + d*x])^n*Hypergeometric2F1[1/2, (1 + 2*n)/4, (5 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(
d*(1 + 2*n)*(3 + 2*n)*Sqrt[Sin[c + d*x]^2]) - (2*B*Cos[c + d*x]^(3/2)*(b*Cos[c + d*x])^n*Hypergeometric2F1[1/2
, (3 + 2*n)/4, (7 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(3 + 2*n)*Sqrt[Sin[c + d*x]^2])

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 3023

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

Rule 2748

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]

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]

Rubi steps

\begin{align*} \int \frac{(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx &=\left (\cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{-\frac{1}{2}+n}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\\ &=\frac{2 C \sqrt{\cos (c+d x)} (b \cos (c+d x))^n \sin (c+d x)}{d (3+2 n)}+\frac{\left (2 \cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{-\frac{1}{2}+n}(c+d x) \left (\frac{1}{2} \left (2 C \left (\frac{1}{2}+n\right )+2 A \left (\frac{3}{2}+n\right )\right )+\frac{1}{2} B (3+2 n) \cos (c+d x)\right ) \, dx}{3+2 n}\\ &=\frac{2 C \sqrt{\cos (c+d x)} (b \cos (c+d x))^n \sin (c+d x)}{d (3+2 n)}+\left (B \cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{\frac{1}{2}+n}(c+d x) \, dx+\frac{\left ((C+2 C n+A (3+2 n)) \cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{-\frac{1}{2}+n}(c+d x) \, dx}{3+2 n}\\ &=\frac{2 C \sqrt{\cos (c+d x)} (b \cos (c+d x))^n \sin (c+d x)}{d (3+2 n)}-\frac{2 (C+2 C n+A (3+2 n)) \sqrt{\cos (c+d x)} (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (1+2 n);\frac{1}{4} (5+2 n);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+2 n) (3+2 n) \sqrt{\sin ^2(c+d x)}}-\frac{2 B \cos ^{\frac{3}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (3+2 n);\frac{1}{4} (7+2 n);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (3+2 n) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.374842, size = 162, normalized size = 0.73 \[ -\frac{2 \sin (c+d x) \sqrt{\cos (c+d x)} (b \cos (c+d x))^n \left ((A (2 n+3)+2 C n+C) \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n+1);\frac{1}{4} (2 n+5);\cos ^2(c+d x)\right )+(2 n+1) \left (B \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n+3);\frac{1}{4} (2 n+7);\cos ^2(c+d x)\right )-C \sqrt{\sin ^2(c+d x)}\right )\right )}{d (2 n+1) (2 n+3) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^n*Sin[c + d*x]*((C + 2*C*n + A*(3 + 2*n))*Hypergeometric2F1[1/2, (1 +
2*n)/4, (5 + 2*n)/4, Cos[c + d*x]^2] + (1 + 2*n)*(B*Cos[c + d*x]*Hypergeometric2F1[1/2, (3 + 2*n)/4, (7 + 2*n)
/4, Cos[c + d*x]^2] - C*Sqrt[Sin[c + d*x]^2])))/(d*(1 + 2*n)*(3 + 2*n)*Sqrt[Sin[c + d*x]^2])

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Maple [F]  time = 0.787, size = 0, normalized size = 0. \begin{align*} \int{ \left ( b\cos \left ( dx+c \right ) \right ) ^{n} \left ( A+B\cos \left ( dx+c \right ) +C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{\cos \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x)

[Out]

int((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\sqrt{\cos \left (d x + c\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))**n*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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