Integrand size = 25, antiderivative size = 101 \[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx=\frac {3 B \sin (c+d x)}{2 d \sqrt [3]{a+a \cos (c+d x)}}+\frac {(2 A-B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{2^{5/6} d \sqrt [6]{1+\cos (c+d x)} \sqrt [3]{a+a \cos (c+d x)}} \]
3/2*B*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/3)+1/2*(2*A-B)*hypergeom([1/2, 5/6] ,[3/2],1/2-1/2*cos(d*x+c))*sin(d*x+c)*2^(1/6)/d/(1+cos(d*x+c))^(1/6)/(a+a* cos(d*x+c))^(1/3)
Time = 0.38 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.32 \[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx=\frac {3\ 2^{5/6} B \sqrt [6]{1-\cos \left (d x-2 \arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )} \sin (c+d x)-2 (2 A-B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\cos ^2\left (\frac {d x}{2}-\arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )\right ) \sin \left (d x-2 \arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )}{4 d \sqrt [3]{a (1+\cos (c+d x))} \sqrt [6]{\sin ^2\left (\frac {d x}{2}-\arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )}} \]
(3*2^(5/6)*B*(1 - Cos[d*x - 2*ArcTan[Cot[c/2]]])^(1/6)*Sin[c + d*x] - 2*(2 *A - B)*Hypergeometric2F1[1/2, 5/6, 3/2, Cos[(d*x)/2 - ArcTan[Cot[c/2]]]^2 ]*Sin[d*x - 2*ArcTan[Cot[c/2]]])/(4*d*(a*(1 + Cos[c + d*x]))^(1/3)*(Sin[(d *x)/2 - ArcTan[Cot[c/2]]]^2)^(1/6))
Time = 0.41 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {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.
\(\displaystyle \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a \cos (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {1}{2} (2 A-B) \int \frac {1}{\sqrt [3]{\cos (c+d x) a+a}}dx+\frac {3 B \sin (c+d x)}{2 d \sqrt [3]{a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} (2 A-B) \int \frac {1}{\sqrt [3]{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {3 B \sin (c+d x)}{2 d \sqrt [3]{a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 3131 |
\(\displaystyle \frac {(2 A-B) \sqrt [3]{\cos (c+d x)+1} \int \frac {1}{\sqrt [3]{\cos (c+d x)+1}}dx}{2 \sqrt [3]{a \cos (c+d x)+a}}+\frac {3 B \sin (c+d x)}{2 d \sqrt [3]{a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(2 A-B) \sqrt [3]{\cos (c+d x)+1} \int \frac {1}{\sqrt [3]{\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx}{2 \sqrt [3]{a \cos (c+d x)+a}}+\frac {3 B \sin (c+d x)}{2 d \sqrt [3]{a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 3130 |
\(\displaystyle \frac {(2 A-B) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right )}{2^{5/6} d \sqrt [6]{\cos (c+d x)+1} \sqrt [3]{a \cos (c+d x)+a}}+\frac {3 B \sin (c+d x)}{2 d \sqrt [3]{a \cos (c+d x)+a}}\) |
(3*B*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(1/3)) + ((2*A - B)*Hypergeom etric2F1[1/2, 5/6, 3/2, (1 - Cos[c + d*x])/2]*Sin[c + d*x])/(2^(5/6)*d*(1 + Cos[c + d*x])^(1/6)*(a + a*Cos[c + d*x])^(1/3))
3.8.89.3.1 Defintions of rubi rules used
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)]
\[\int \frac {A +B \cos \left (d x +c \right )}{\left (a +\cos \left (d x +c \right ) a \right )^{\frac {1}{3}}}d x\]
\[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx=\int \frac {A + B \cos {\left (c + d x \right )}}{\sqrt [3]{a \left (\cos {\left (c + d x \right )} + 1\right )}}\, dx \]
\[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{1/3}} \,d x \]