3.211 \(\int \cos ^m(c+d x) (b \cos (c+d x))^{2/3} (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=167 \[ -\frac{3 B \sin (c+d x) (b \cos (c+d x))^{2/3} \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+8);\frac{1}{6} (3 m+14);\cos ^2(c+d x)\right )}{d (3 m+8) \sqrt{\sin ^2(c+d x)}}-\frac{3 C \sin (c+d x) (b \cos (c+d x))^{2/3} \cos ^{m+3}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+11);\frac{1}{6} (3 m+17);\cos ^2(c+d x)\right )}{d (3 m+11) \sqrt{\sin ^2(c+d x)}} \]

[Out]

(-3*B*Cos[c + d*x]^(2 + m)*(b*Cos[c + d*x])^(2/3)*Hypergeometric2F1[1/2, (8 + 3*m)/6, (14 + 3*m)/6, Cos[c + d*
x]^2]*Sin[c + d*x])/(d*(8 + 3*m)*Sqrt[Sin[c + d*x]^2]) - (3*C*Cos[c + d*x]^(3 + m)*(b*Cos[c + d*x])^(2/3)*Hype
rgeometric2F1[1/2, (11 + 3*m)/6, (17 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(11 + 3*m)*Sqrt[Sin[c + d*x]^2
])

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Rubi [A]  time = 0.132205, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {20, 3010, 2748, 2643} \[ -\frac{3 B \sin (c+d x) (b \cos (c+d x))^{2/3} \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+8);\frac{1}{6} (3 m+14);\cos ^2(c+d x)\right )}{d (3 m+8) \sqrt{\sin ^2(c+d x)}}-\frac{3 C \sin (c+d x) (b \cos (c+d x))^{2/3} \cos ^{m+3}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+11);\frac{1}{6} (3 m+17);\cos ^2(c+d x)\right )}{d (3 m+11) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^m*(b*Cos[c + d*x])^(2/3)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(-3*B*Cos[c + d*x]^(2 + m)*(b*Cos[c + d*x])^(2/3)*Hypergeometric2F1[1/2, (8 + 3*m)/6, (14 + 3*m)/6, Cos[c + d*
x]^2]*Sin[c + d*x])/(d*(8 + 3*m)*Sqrt[Sin[c + d*x]^2]) - (3*C*Cos[c + d*x]^(3 + m)*(b*Cos[c + d*x])^(2/3)*Hype
rgeometric2F1[1/2, (11 + 3*m)/6, (17 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(11 + 3*m)*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 3010

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x
_Symbol] :> Dist[1/b, Int[(b*Sin[e + f*x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x
]

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 \cos ^m(c+d x) (b \cos (c+d x))^{2/3} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{(b \cos (c+d x))^{2/3} \int \cos ^{\frac{2}{3}+m}(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx}{\cos ^{\frac{2}{3}}(c+d x)}\\ &=\frac{(b \cos (c+d x))^{2/3} \int \cos ^{\frac{5}{3}+m}(c+d x) (B+C \cos (c+d x)) \, dx}{\cos ^{\frac{2}{3}}(c+d x)}\\ &=\frac{\left (B (b \cos (c+d x))^{2/3}\right ) \int \cos ^{\frac{5}{3}+m}(c+d x) \, dx}{\cos ^{\frac{2}{3}}(c+d x)}+\frac{\left (C (b \cos (c+d x))^{2/3}\right ) \int \cos ^{\frac{8}{3}+m}(c+d x) \, dx}{\cos ^{\frac{2}{3}}(c+d x)}\\ &=-\frac{3 B \cos ^{2+m}(c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac{1}{2},\frac{1}{6} (8+3 m);\frac{1}{6} (14+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (8+3 m) \sqrt{\sin ^2(c+d x)}}-\frac{3 C \cos ^{3+m}(c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac{1}{2},\frac{1}{6} (11+3 m);\frac{1}{6} (17+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (11+3 m) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.367617, size = 140, normalized size = 0.84 \[ -\frac{3 \sqrt{\sin ^2(c+d x)} \csc (c+d x) (b \cos (c+d x))^{2/3} \cos ^{m+2}(c+d x) \left (B (3 m+11) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+8);\frac{m}{2}+\frac{7}{3};\cos ^2(c+d x)\right )+C (3 m+8) \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+11);\frac{1}{6} (3 m+17);\cos ^2(c+d x)\right )\right )}{d (3 m+8) (3 m+11)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^m*(b*Cos[c + d*x])^(2/3)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(-3*Cos[c + d*x]^(2 + m)*(b*Cos[c + d*x])^(2/3)*Csc[c + d*x]*(B*(11 + 3*m)*Hypergeometric2F1[1/2, (8 + 3*m)/6,
7/3 + m/2, Cos[c + d*x]^2] + C*(8 + 3*m)*Cos[c + d*x]*Hypergeometric2F1[1/2, (11 + 3*m)/6, (17 + 3*m)/6, Cos[
c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(8 + 3*m)*(11 + 3*m))

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Maple [F]  time = 0.312, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{m} \left ( b\cos \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}} \left ( B\cos \left ( dx+c \right ) +C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^m*(b*cos(d*x+c))^(2/3)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

int(cos(d*x+c)^m*(b*cos(d*x+c))^(2/3)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{2}{3}} \cos \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(2/3)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c))^(2/3)*cos(d*x + c)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{2}{3}} \cos \left (d x + c\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(2/3)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c))^(2/3)*cos(d*x + c)^m, 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(cos(d*x+c)**m*(b*cos(d*x+c))**(2/3)*(B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{2}{3}} \cos \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(2/3)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c))^(2/3)*cos(d*x + c)^m, x)