3.929 \(\int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=167 \[ -\frac{3 A \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+4);\frac{1}{6} (3 m+10);\cos ^2(c+d x)\right )}{d (3 m+4) \sqrt{\sin ^2(c+d x)}}-\frac{3 B \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+7);\frac{1}{6} (3 m+13);\cos ^2(c+d x)\right )}{d (3 m+7) \sqrt{\sin ^2(c+d x)}} \]

[Out]

(-3*A*Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/2, (4 + 3*m)/6, (10 + 3*m)/6, Cos[c + d*
x]^2]*Sin[c + d*x])/(d*(4 + 3*m)*Sqrt[Sin[c + d*x]^2]) - (3*B*Cos[c + d*x]^(2 + m)*(b*Cos[c + d*x])^(1/3)*Hype
rgeometric2F1[1/2, (7 + 3*m)/6, (13 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(7 + 3*m)*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.0863866, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {20, 2748, 2643} \[ -\frac{3 A \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+4);\frac{1}{6} (3 m+10);\cos ^2(c+d x)\right )}{d (3 m+4) \sqrt{\sin ^2(c+d x)}}-\frac{3 B \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+7);\frac{1}{6} (3 m+13);\cos ^2(c+d x)\right )}{d (3 m+7) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^m*(b*Cos[c + d*x])^(1/3)*(A + B*Cos[c + d*x]),x]

[Out]

(-3*A*Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/2, (4 + 3*m)/6, (10 + 3*m)/6, Cos[c + d*
x]^2]*Sin[c + d*x])/(d*(4 + 3*m)*Sqrt[Sin[c + d*x]^2]) - (3*B*Cos[c + d*x]^(2 + m)*(b*Cos[c + d*x])^(1/3)*Hype
rgeometric2F1[1/2, (7 + 3*m)/6, (13 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(7 + 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 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) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx &=\frac{\sqrt [3]{b \cos (c+d x)} \int \cos ^{\frac{1}{3}+m}(c+d x) (A+B \cos (c+d x)) \, dx}{\sqrt [3]{\cos (c+d x)}}\\ &=\frac{\left (A \sqrt [3]{b \cos (c+d x)}\right ) \int \cos ^{\frac{1}{3}+m}(c+d x) \, dx}{\sqrt [3]{\cos (c+d x)}}+\frac{\left (B \sqrt [3]{b \cos (c+d x)}\right ) \int \cos ^{\frac{4}{3}+m}(c+d x) \, dx}{\sqrt [3]{\cos (c+d x)}}\\ &=-\frac{3 A \cos ^{1+m}(c+d x) \sqrt [3]{b \cos (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) \sqrt{\sin ^2(c+d x)}}-\frac{3 B \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \, _2F_1\left (\frac{1}{2},\frac{1}{6} (7+3 m);\frac{1}{6} (13+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (7+3 m) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.28472, size = 140, normalized size = 0.84 \[ -\frac{3 \sqrt{\sin ^2(c+d x)} \csc (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+1}(c+d x) \left (A (3 m+7) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+4);\frac{m}{2}+\frac{5}{3};\cos ^2(c+d x)\right )+B (3 m+4) \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+7);\frac{1}{6} (3 m+13);\cos ^2(c+d x)\right )\right )}{d (3 m+4) (3 m+7)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^m*(b*Cos[c + d*x])^(1/3)*(A + B*Cos[c + d*x]),x]

[Out]

(-3*Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(1/3)*Csc[c + d*x]*(A*(7 + 3*m)*Hypergeometric2F1[1/2, (4 + 3*m)/6,
5/3 + m/2, Cos[c + d*x]^2] + B*(4 + 3*m)*Cos[c + d*x]*Hypergeometric2F1[1/2, (7 + 3*m)/6, (13 + 3*m)/6, Cos[c
+ d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(4 + 3*m)*(7 + 3*m))

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Maple [F]  time = 0.217, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{m}\sqrt [3]{b\cos \left ( dx+c \right ) } \left ( A+B\cos \left ( dx+c \right ) \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))^(1/3)*(A+B*cos(d*x+c)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (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}\,{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))^(1/3)*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(1/3)*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 (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}, 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))^(1/3)*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (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}\,{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))^(1/3)*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

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