Integrand size = 35, antiderivative size = 144 \[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 (7 B-3 C) \sqrt [3]{a+a \cos (c+d x)} \sin (c+d x)}{28 d}+\frac {3 C (a+a \cos (c+d x))^{4/3} \sin (c+d x)}{7 a d}+\frac {(28 A+7 B+13 C) \sqrt [3]{a+a \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{14 \sqrt [6]{2} d (1+\cos (c+d x))^{5/6}} \]
3/28*(7*B-3*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+7*B+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)
\[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int \sqrt [3]{a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx \]
Time = 0.63 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 3502, 27, 3042, 3230, 3042, 3131, 3042, 3130}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sqrt [3]{a \cos (c+d x)+a} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt [3]{a \sin \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {3 \int \frac {1}{3} \sqrt [3]{\cos (c+d x) a+a} (a (7 A+4 C)+a (7 B-3 C) \cos (c+d x))dx}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sqrt [3]{\cos (c+d x) a+a} (a (7 A+4 C)+a (7 B-3 C) \cos (c+d x))dx}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sqrt [3]{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (7 A+4 C)+a (7 B-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {\frac {1}{4} a (28 A+7 B+13 C) \int \sqrt [3]{\cos (c+d x) a+a}dx+\frac {3 a (7 B-3 C) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 d}}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{4} a (28 A+7 B+13 C) \int \sqrt [3]{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {3 a (7 B-3 C) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 d}}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
\(\Big \downarrow \) 3131 |
\(\displaystyle \frac {\frac {a (28 A+7 B+13 C) \sqrt [3]{a \cos (c+d x)+a} \int \sqrt [3]{\cos (c+d x)+1}dx}{4 \sqrt [3]{\cos (c+d x)+1}}+\frac {3 a (7 B-3 C) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 d}}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {a (28 A+7 B+13 C) \sqrt [3]{a \cos (c+d x)+a} \int \sqrt [3]{\sin \left (c+d x+\frac {\pi }{2}\right )+1}dx}{4 \sqrt [3]{\cos (c+d x)+1}}+\frac {3 a (7 B-3 C) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 d}}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
\(\Big \downarrow \) 3130 |
\(\displaystyle \frac {\frac {a (28 A+7 B+13 C) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right )}{2 \sqrt [6]{2} d (\cos (c+d x)+1)^{5/6}}+\frac {3 a (7 B-3 C) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 d}}{7 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{4/3}}{7 a d}\) |
(3*C*(a + a*Cos[c + d*x])^(4/3)*Sin[c + d*x])/(7*a*d) + ((3*a*(7*B - 3*C)* (a + a*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(4*d) + (a*(28*A + 7*B + 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])/(2*2^(1/6)*d*(1 + Cos[c + d*x])^(5/6)))/(7*a)
3.4.85.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]]))*Hypergeome tric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && GtQ[a, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPar t[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]
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] + Simp[(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)]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
\[\int \left (a +\cos \left (d x +c \right ) a \right )^{\frac {1}{3}} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )d x\]
\[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]
\[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int \sqrt [3]{a \left (\cos {\left (c + d x \right )} + 1\right )} \left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right )\, dx \]
\[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]
\[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]
Timed out. \[ \int \sqrt [3]{a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int {\left (a+a\,\cos \left (c+d\,x\right )\right )}^{1/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]