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

Optimal. Leaf size=227 \[ -\frac{3 (A (3 m+4)+3 C m+C) \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+1);\frac{1}{6} (3 m+7);\cos ^2(c+d x)\right )}{d (3 m+1) (3 m+4) \sqrt{\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}}-\frac{3 B \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+4);\frac{1}{6} (3 m+10);\cos ^2(c+d x)\right )}{d (3 m+4) \sqrt{\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}}+\frac{3 C \sin (c+d x) \cos ^{m+1}(c+d x)}{d (3 m+4) (b \cos (c+d x))^{2/3}} \]

[Out]

(3*C*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(4 + 3*m)*(b*Cos[c + d*x])^(2/3)) - (3*(C + 3*C*m + A*(4 + 3*m))*Co
s[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1 + 3*m)/6, (7 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 + 3*m)
*(4 + 3*m)*(b*Cos[c + d*x])^(2/3)*Sqrt[Sin[c + d*x]^2]) - (3*B*Cos[c + d*x]^(2 + m)*Hypergeometric2F1[1/2, (4
+ 3*m)/6, (10 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(4 + 3*m)*(b*Cos[c + d*x])^(2/3)*Sqrt[Sin[c + d*x]^2]
)

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Rubi [A]  time = 0.233094, antiderivative size = 227, 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} \[ -\frac{3 (A (3 m+4)+3 C m+C) \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+1);\frac{1}{6} (3 m+7);\cos ^2(c+d x)\right )}{d (3 m+1) (3 m+4) \sqrt{\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}}-\frac{3 B \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+4);\frac{1}{6} (3 m+10);\cos ^2(c+d x)\right )}{d (3 m+4) \sqrt{\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}}+\frac{3 C \sin (c+d x) \cos ^{m+1}(c+d x)}{d (3 m+4) (b \cos (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(4 + 3*m)*(b*Cos[c + d*x])^(2/3)) - (3*(C + 3*C*m + A*(4 + 3*m))*Co
s[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1 + 3*m)/6, (7 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 + 3*m)
*(4 + 3*m)*(b*Cos[c + d*x])^(2/3)*Sqrt[Sin[c + d*x]^2]) - (3*B*Cos[c + d*x]^(2 + m)*Hypergeometric2F1[1/2, (4
+ 3*m)/6, (10 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(4 + 3*m)*(b*Cos[c + d*x])^(2/3)*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{\cos ^m(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{2/3}} \, dx &=\frac{\cos ^{\frac{2}{3}}(c+d x) \int \cos ^{-\frac{2}{3}+m}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx}{(b \cos (c+d x))^{2/3}}\\ &=\frac{3 C \cos ^{1+m}(c+d x) \sin (c+d x)}{d (4+3 m) (b \cos (c+d x))^{2/3}}+\frac{\left (3 \cos ^{\frac{2}{3}}(c+d x)\right ) \int \cos ^{-\frac{2}{3}+m}(c+d x) \left (\frac{1}{3} \left (3 C \left (\frac{1}{3}+m\right )+3 A \left (\frac{4}{3}+m\right )\right )+\frac{1}{3} B (4+3 m) \cos (c+d x)\right ) \, dx}{(4+3 m) (b \cos (c+d x))^{2/3}}\\ &=\frac{3 C \cos ^{1+m}(c+d x) \sin (c+d x)}{d (4+3 m) (b \cos (c+d x))^{2/3}}+\frac{\left (B \cos ^{\frac{2}{3}}(c+d x)\right ) \int \cos ^{\frac{1}{3}+m}(c+d x) \, dx}{(b \cos (c+d x))^{2/3}}+\frac{\left ((C+3 C m+A (4+3 m)) \cos ^{\frac{2}{3}}(c+d x)\right ) \int \cos ^{-\frac{2}{3}+m}(c+d x) \, dx}{(4+3 m) (b \cos (c+d x))^{2/3}}\\ &=\frac{3 C \cos ^{1+m}(c+d x) \sin (c+d x)}{d (4+3 m) (b \cos (c+d x))^{2/3}}-\frac{3 (C+3 C m+A (4+3 m)) \cos ^{1+m}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (1+3 m);\frac{1}{6} (7+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+3 m) (4+3 m) (b \cos (c+d x))^{2/3} \sqrt{\sin ^2(c+d x)}}-\frac{3 B \cos ^{2+m}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (4+3 m);\frac{1}{6} (10+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (4+3 m) (b \cos (c+d x))^{2/3} \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.401665, size = 164, normalized size = 0.72 \[ -\frac{3 \sin (c+d x) \cos ^{m+1}(c+d x) \left ((A (3 m+4)+3 C m+C) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+1);\frac{1}{6} (3 m+7);\cos ^2(c+d x)\right )+(3 m+1) \left (B \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+4);\frac{m}{2}+\frac{5}{3};\cos ^2(c+d x)\right )-C \sqrt{\sin ^2(c+d x)}\right )\right )}{d (3 m+1) (3 m+4) \sqrt{\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Cos[c + d*x]^(1 + m)*Sin[c + d*x]*((C + 3*C*m + A*(4 + 3*m))*Hypergeometric2F1[1/2, (1 + 3*m)/6, (7 + 3*m)
/6, Cos[c + d*x]^2] + (1 + 3*m)*(B*Cos[c + d*x]*Hypergeometric2F1[1/2, (4 + 3*m)/6, 5/3 + m/2, Cos[c + d*x]^2]
- C*Sqrt[Sin[c + d*x]^2])))/(d*(1 + 3*m)*(4 + 3*m)*(b*Cos[c + d*x])^(2/3)*Sqrt[Sin[c + d*x]^2])

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Maple [F]  time = 0.328, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cos \left ( dx+c \right ) \right ) ^{m} \left ( A+B\cos \left ( dx+c \right ) +C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \left ( b\cos \left ( dx+c \right ) \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(d*x+c)^m*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(2/3),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 )} \cos \left (d x + c\right )^{m}}{\left (b \cos \left (d x + c\right )\right )^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^m/(b*cos(d*x + c))^(2/3), 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 )^{\frac{1}{3}} \cos \left (d x + c\right )^{m}}{b \cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(1/3)*cos(d*x + c)^m/(b*cos(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )} \cos \left (d x + c\right )^{m}}{\left (b \cos \left (d x + c\right )\right )^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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