3.887 \(\int \cos ^2(c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=119 \[ -\frac {3 A \sin (c+d x) (b \cos (c+d x))^{10/3} \, _2F_1\left (\frac {1}{2},\frac {5}{3};\frac {8}{3};\cos ^2(c+d x)\right )}{10 b^3 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{13/3} \, _2F_1\left (\frac {1}{2},\frac {13}{6};\frac {19}{6};\cos ^2(c+d x)\right )}{13 b^4 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

-3/10*A*(b*cos(d*x+c))^(10/3)*hypergeom([1/2, 5/3],[8/3],cos(d*x+c)^2)*sin(d*x+c)/b^3/d/(sin(d*x+c)^2)^(1/2)-3
/13*B*(b*cos(d*x+c))^(13/3)*hypergeom([1/2, 13/6],[19/6],cos(d*x+c)^2)*sin(d*x+c)/b^4/d/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 119, normalized size of antiderivative = 1.00, 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 = {16, 2748, 2643} \[ -\frac {3 A \sin (c+d x) (b \cos (c+d x))^{10/3} \, _2F_1\left (\frac {1}{2},\frac {5}{3};\frac {8}{3};\cos ^2(c+d x)\right )}{10 b^3 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{13/3} \, _2F_1\left (\frac {1}{2},\frac {13}{6};\frac {19}{6};\cos ^2(c+d x)\right )}{13 b^4 d \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(b*Cos[c + d*x])^(1/3)*(A + B*Cos[c + d*x]),x]

[Out]

(-3*A*(b*Cos[c + d*x])^(10/3)*Hypergeometric2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2]*Sin[c + d*x])/(10*b^3*d*Sqrt[Si
n[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(13/3)*Hypergeometric2F1[1/2, 13/6, 19/6, Cos[c + d*x]^2]*Sin[c + d*x])
/(13*b^4*d*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 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]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx &=\frac {\int (b \cos (c+d x))^{7/3} (A+B \cos (c+d x)) \, dx}{b^2}\\ &=\frac {A \int (b \cos (c+d x))^{7/3} \, dx}{b^2}+\frac {B \int (b \cos (c+d x))^{10/3} \, dx}{b^3}\\ &=-\frac {3 A (b \cos (c+d x))^{10/3} \, _2F_1\left (\frac {1}{2},\frac {5}{3};\frac {8}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{10 b^3 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{13/3} \, _2F_1\left (\frac {1}{2},\frac {13}{6};\frac {19}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{13 b^4 d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 94, normalized size = 0.79 \[ -\frac {3 \sqrt {\sin ^2(c+d x)} \cos ^2(c+d x) \cot (c+d x) \sqrt [3]{b \cos (c+d x)} \left (13 A \, _2F_1\left (\frac {1}{2},\frac {5}{3};\frac {8}{3};\cos ^2(c+d x)\right )+10 B \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {13}{6};\frac {19}{6};\cos ^2(c+d x)\right )\right )}{130 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(b*Cos[c + d*x])^(1/3)*(A + B*Cos[c + d*x]),x]

[Out]

(-3*Cos[c + d*x]^2*(b*Cos[c + d*x])^(1/3)*Cot[c + d*x]*(13*A*Hypergeometric2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2]
+ 10*B*Cos[c + d*x]*Hypergeometric2F1[1/2, 13/6, 19/6, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(130*d)

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \cos \left (d x + c\right )^{3} + A \cos \left (d x + c\right )^{2}\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(b*cos(d*x+c))^(1/3)*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(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)^2, x)

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maple [F]  time = 0.40, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (d x +c \right )\right ) \left (b \cos \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +B \cos \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(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)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^2\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{1/3}\,\left (A+B\,\cos \left (c+d\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(b*cos(d*x+c))**(1/3)*(A+B*cos(d*x+c)),x)

[Out]

Timed out

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