\(\int \sqrt [3]{a+a \cos (c+d x)} (A+C \cos ^2(c+d x)) \, dx\) [200]

Optimal result

Integrand size = 27, antiderivative size = 135 \[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {9 C \sqrt [3]{a+a \cos (c+d x)} \sin (c+d x)}{28 d}+\frac {3 C (a+a \cos (c+d x))^{4/3} \sin (c+d x)}{7 a d}+\frac {(28 A+13 C) \sqrt [3]{a+a \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{14 \sqrt [6]{2} d (1+\cos (c+d x))^{5/6}} \]

[Out]

-9/28*C*(a+a*cos(d*x+c))^(1/3)*sin(d*x+c)/d+3/7*C*(a+a*cos(d*x+c))^(4/3)*sin(d*x+c)/a/d+1/28*(28*A+13*C)*(a+a*
cos(d*x+c))^(1/3)*hypergeom([1/6, 1/2],[3/2],1/2-1/2*cos(d*x+c))*sin(d*x+c)*2^(5/6)/d/(1+cos(d*x+c))^(5/6)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3103, 2830, 2731, 2730} \[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {(28 A+13 C) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right )}{14 \sqrt [6]{2} d (\cos (c+d x)+1)^{5/6}}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}-\frac {9 C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{28 d} \]

[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 2730

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

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

Rule 2830

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*S
in[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 3103

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] &&  !Lt
Q[m, -1]

Rubi steps \begin{align*} \text {integral}& = \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)} \operatorname {Hypergeometric2F1}\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 [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(289\) vs. \(2(135)=270\).

Time = 2.82 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.14 \[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt [3]{a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (-2 (28 A+13 C) \, _2F_1\left (-\frac {1}{2},-\frac {1}{6};\frac {5}{6};\cos ^2\left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )+\frac {1}{2} \left (5 (28 A+13 C) \cos \left (\frac {1}{2} \left (c-d x-2 \arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )+(28 A+13 C) \cos \left (\frac {1}{2} \left (c+d x+2 \arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )+6 \cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec ^2\left (\frac {c}{2}\right )} \left (-\left ((28 A+13 C) \cot \left (\frac {c}{2}\right )\right )+C (\sin (c+d x)+2 \sin (2 (c+d x)))\right )\right ) \sqrt {\sin ^2\left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )}\right )}{28 d \sqrt {\sec ^2\left (\frac {c}{2}\right )} \sqrt {\sin ^2\left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )}} \]

[In]

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

[Out]

((a*(1 + Cos[c + d*x]))^(1/3)*Sec[(c + d*x)/2]*(-2*(28*A + 13*C)*HypergeometricPFQ[{-1/2, -1/6}, {5/6}, Cos[(d
*x)/2 + ArcTan[Tan[c/2]]]^2]*Sec[c/2]*Sin[(d*x)/2 + ArcTan[Tan[c/2]]] + ((5*(28*A + 13*C)*Cos[(c - d*x - 2*Arc
Tan[Tan[c/2]])/2]*Csc[c/2]*Sec[c/2] + (28*A + 13*C)*Cos[(c + d*x + 2*ArcTan[Tan[c/2]])/2]*Csc[c/2]*Sec[c/2] +
6*Cos[(c + d*x)/2]*Sqrt[Sec[c/2]^2]*(-((28*A + 13*C)*Cot[c/2]) + C*(Sin[c + d*x] + 2*Sin[2*(c + d*x)])))*Sqrt[
Sin[(d*x)/2 + ArcTan[Tan[c/2]]]^2])/2))/(28*d*Sqrt[Sec[c/2]^2]*Sqrt[Sin[(d*x)/2 + ArcTan[Tan[c/2]]]^2])

Maple [F]

\[\int \left (a +\cos \left (d x +c \right ) a \right )^{\frac {1}{3}} \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )d x\]

[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)

Fricas [F]

\[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]

[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)

Sympy [F]

\[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int \sqrt [3]{a \left (\cos {\left (c + d x \right )} + 1\right )} \left (A + C \cos ^{2}{\left (c + d x \right )}\right )\, dx \]

[In]

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

[Out]

Integral((a*(cos(c + d*x) + 1))**(1/3)*(A + C*cos(c + d*x)**2), x)

Maxima [F]

\[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]

[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)

Giac [F]

\[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{1/3} \,d x \]

[In]

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

[Out]

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