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

Optimal. Leaf size=135 \[ \frac{(28 A+13 C) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a} \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x))\right )}{14 \sqrt [6]{2} d (\cos (c+d x)+1)^{5/6}}+\frac{3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}-\frac{9 C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{28 d} \]

[Out]

(-9*C*(a + a*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(28*d) + (3*C*(a + a*Cos[c + d*x])^(4/3)*Sin[c + d*x])/(7*a*d)
+ ((28*A + 13*C)*(a + a*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/6, 1/2, 3/2, (1 - Cos[c + d*x])/2]*Sin[c + d*x
])/(14*2^(1/6)*d*(1 + Cos[c + d*x])^(5/6))

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Rubi [A]  time = 0.160788, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {3024, 2751, 2652, 2651} \[ \frac{(28 A+13 C) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a} \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x))\right )}{14 \sqrt [6]{2} d (\cos (c+d x)+1)^{5/6}}+\frac{3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}-\frac{9 C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{28 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*C*(a + a*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(28*d) + (3*C*(a + a*Cos[c + d*x])^(4/3)*Sin[c + d*x])/(7*a*d)
+ ((28*A + 13*C)*(a + a*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/6, 1/2, 3/2, (1 - Cos[c + d*x])/2]*Sin[c + d*x
])/(14*2^(1/6)*d*(1 + Cos[c + d*x])^(5/6))

Rule 3024

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (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) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] &&  !LtQ[
m, -1]

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2652

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Sin[c + d*x])^FracPart
[n])/(1 + (b*Sin[c + d*x])/a)^FracPart[n], Int[(1 + (b*Sin[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 2651

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(2^(n + 1/2)*a^(n - 1/2)*b*Cos[c + d*x]*Hy
pergeometric2F1[1/2, 1/2 - n, 3/2, (1*(1 - (b*Sin[c + d*x])/a))/2])/(d*Sqrt[a + b*Sin[c + d*x]]), x] /; FreeQ[
{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rubi steps

\begin{align*} \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac{3 C (a+a \cos (c+d x))^{4/3} \sin (c+d x)}{7 a d}+\frac{3 \int \sqrt [3]{a+a \cos (c+d x)} \left (\frac{1}{3} a (7 A+4 C)-a C \cos (c+d x)\right ) \, dx}{7 a}\\ &=-\frac{9 C \sqrt [3]{a+a \cos (c+d x)} \sin (c+d x)}{28 d}+\frac{3 C (a+a \cos (c+d x))^{4/3} \sin (c+d x)}{7 a d}+\frac{1}{28} (28 A+13 C) \int \sqrt [3]{a+a \cos (c+d x)} \, dx\\ &=-\frac{9 C \sqrt [3]{a+a \cos (c+d x)} \sin (c+d x)}{28 d}+\frac{3 C (a+a \cos (c+d x))^{4/3} \sin (c+d x)}{7 a d}+\frac{\left ((28 A+13 C) \sqrt [3]{a+a \cos (c+d x)}\right ) \int \sqrt [3]{1+\cos (c+d x)} \, dx}{28 \sqrt [3]{1+\cos (c+d x)}}\\ &=-\frac{9 C \sqrt [3]{a+a \cos (c+d x)} \sin (c+d x)}{28 d}+\frac{3 C (a+a \cos (c+d x))^{4/3} \sin (c+d x)}{7 a d}+\frac{(28 A+13 C) \sqrt [3]{a+a \cos (c+d x)} \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{14 \sqrt [6]{2} d (1+\cos (c+d x))^{5/6}}\\ \end{align*}

Mathematica [C]  time = 0.83763, size = 240, normalized size = 1.78 \[ \frac{3 \sqrt [3]{a (\cos (c+d x)+1)} \left (\frac{(28 A+13 C) \csc \left (\frac{c}{4}\right ) \sec \left (\frac{c}{4}\right ) \sqrt [3]{i \sin (c) e^{i d x}+\cos (c) e^{i d x}+1} \left (2 \, _2F_1\left (-\frac{1}{3},\frac{1}{3};\frac{2}{3};-e^{i d x} (\cos (c)+i \sin (c))\right )+e^{i d x} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-e^{i d x} (\cos (c)+i \sin (c))\right )\right )}{i \sin \left (\frac{c}{2}\right ) \left (-1+e^{i d x}\right )+\cos \left (\frac{c}{2}\right ) \left (1+e^{i d x}\right )}-4 (28 A+13 C) \cot \left (\frac{c}{2}\right )+4 C \sin (c) \cos (d x)+8 C \sin (2 c) \cos (2 d x)+4 C \cos (c) \sin (d x)+8 C \cos (2 c) \sin (2 d x)\right )}{112 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a*(1 + Cos[c + d*x]))^(1/3)*(-4*(28*A + 13*C)*Cot[c/2] + 4*C*Cos[d*x]*Sin[c] + ((28*A + 13*C)*Csc[c/4]*(2*
Hypergeometric2F1[-1/3, 1/3, 2/3, -(E^(I*d*x)*(Cos[c] + I*Sin[c]))] + E^(I*d*x)*Hypergeometric2F1[1/3, 2/3, 5/
3, -(E^(I*d*x)*(Cos[c] + I*Sin[c]))])*Sec[c/4]*(1 + E^(I*d*x)*Cos[c] + I*E^(I*d*x)*Sin[c])^(1/3))/((1 + E^(I*d
*x))*Cos[c/2] + I*(-1 + E^(I*d*x))*Sin[c/2]) + 8*C*Cos[2*d*x]*Sin[2*c] + 4*C*Cos[c]*Sin[d*x] + 8*C*Cos[2*c]*Si
n[2*d*x]))/(112*d)

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Maple [F]  time = 0.302, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{a+\cos \left ( dx+c \right ) a} \left ( A+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((a+cos(d*x+c)*a)^(1/3)*(A+C*cos(d*x+c)^2),x)

[Out]

int((a+cos(d*x+c)*a)^(1/3)*(A+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} + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{1}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*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)*(a*cos(d*x + c) + a)^(1/3), 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} + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{1}{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*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)*(a*cos(d*x + c) + a)^(1/3), x)