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

Optimal. Leaf size=95 \[ \frac {3 C \sin (c+d x) (b \cos (c+d x))^{7/3}}{10 b^2 d}-\frac {3 (10 A+7 C) \sin (c+d x) (b \cos (c+d x))^{7/3} \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right )}{70 b^2 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

3/10*C*(b*cos(d*x+c))^(7/3)*sin(d*x+c)/b^2/d-3/70*(10*A+7*C)*(b*cos(d*x+c))^(7/3)*hypergeom([1/2, 7/6],[13/6],
cos(d*x+c)^2)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {16, 3014, 2643} \[ \frac {3 C \sin (c+d x) (b \cos (c+d x))^{7/3}}{10 b^2 d}-\frac {3 (10 A+7 C) \sin (c+d x) (b \cos (c+d x))^{7/3} \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right )}{70 b^2 d \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*(b*Cos[c + d*x])^(7/3)*Sin[c + d*x])/(10*b^2*d) - (3*(10*A + 7*C)*(b*Cos[c + d*x])^(7/3)*Hypergeometric2F
1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x])/(70*b^2*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 3014

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

Rubi steps

\begin {align*} \int \cos (c+d x) \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac {\int (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \, dx}{b}\\ &=\frac {3 C (b \cos (c+d x))^{7/3} \sin (c+d x)}{10 b^2 d}+\frac {(10 A+7 C) \int (b \cos (c+d x))^{4/3} \, dx}{10 b}\\ &=\frac {3 C (b \cos (c+d x))^{7/3} \sin (c+d x)}{10 b^2 d}-\frac {3 (10 A+7 C) (b \cos (c+d x))^{7/3} \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{70 b^2 d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 91, normalized size = 0.96 \[ -\frac {3 \sqrt {\sin ^2(c+d x)} \cot (c+d x) (b \cos (c+d x))^{4/3} \left (13 A \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right )+7 C \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {13}{6};\frac {19}{6};\cos ^2(c+d x)\right )\right )}{91 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b*Cos[c + d*x])^(4/3)*Cot[c + d*x]*(13*A*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2] + 7*C*Cos[c +
d*x]^2*Hypergeometric2F1[1/2, 13/6, 19/6, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(91*b*d)

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{3} + A \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int \cos \left (d x +c \right ) \left (b \cos \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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