Integrand size = 25, antiderivative size = 226 \[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\frac {\sqrt {2} B \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}+\frac {\sqrt {2} (A b-a B) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} \sqrt [3]{a+b \cos (c+d x)}} \] Output:
2^(1/2)*B*AppellF1(1/2,-2/3,1/2,3/2,b*(1-cos(d*x+c))/(a+b),1/2-1/2*cos(d*x +c))*(a+b*cos(d*x+c))^(2/3)*sin(d*x+c)/b/d/(1+cos(d*x+c))^(1/2)/((a+b*cos( d*x+c))/(a+b))^(2/3)+2^(1/2)*(A*b-B*a)*AppellF1(1/2,1/3,1/2,3/2,b*(1-cos(d *x+c))/(a+b),1/2-1/2*cos(d*x+c))*((a+b*cos(d*x+c))/(a+b))^(1/3)*sin(d*x+c) /b/d/(1+cos(d*x+c))^(1/2)/(a+b*cos(d*x+c))^(1/3)
Time = 0.47 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.84 \[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=-\frac {3 \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} (a+b \cos (c+d x))^{2/3} \left (5 (A b-a B) \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right )+2 B \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},\frac {1}{2},\frac {8}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) (a+b \cos (c+d x))\right ) \csc (c+d x)}{10 b^2 d} \] Input:
Integrate[(A + B*Cos[c + d*x])/(a + b*Cos[c + d*x])^(1/3),x]
Output:
(-3*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[(b*(1 + Cos[c + d*x]))/( -a + b)]*(a + b*Cos[c + d*x])^(2/3)*(5*(A*b - a*B)*AppellF1[2/3, 1/2, 1/2, 5/3, (a + b*Cos[c + d*x])/(a - b), (a + b*Cos[c + d*x])/(a + b)] + 2*B*Ap pellF1[5/3, 1/2, 1/2, 8/3, (a + b*Cos[c + d*x])/(a - b), (a + b*Cos[c + d* x])/(a + b)]*(a + b*Cos[c + d*x]))*Csc[c + d*x])/(10*b^2*d)
Time = 0.45 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3235, 3042, 3144, 156, 155}
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.
\(\displaystyle \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3235 |
\(\displaystyle \frac {(A b-a B) \int \frac {1}{\sqrt [3]{a+b \cos (c+d x)}}dx}{b}+\frac {B \int (a+b \cos (c+d x))^{2/3}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A b-a B) \int \frac {1}{\sqrt [3]{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {B \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}dx}{b}\) |
\(\Big \downarrow \) 3144 |
\(\displaystyle -\frac {(A b-a B) \sin (c+d x) \int \frac {1}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}-\frac {B \sin (c+d x) \int \frac {(a+b \cos (c+d x))^{2/3}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle -\frac {(A b-a B) \sin (c+d x) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}-\frac {B \sin (c+d x) (a+b \cos (c+d x))^{2/3} \int \frac {\left (\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}\right )^{2/3}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {\sqrt {2} (A b-a B) \sin (c+d x) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}+\frac {\sqrt {2} B \sin (c+d x) (a+b \cos (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}\) |
Input:
Int[(A + B*Cos[c + d*x])/(a + b*Cos[c + d*x])^(1/3),x]
Output:
(Sqrt[2]*B*AppellF1[1/2, 1/2, -2/3, 3/2, (1 - Cos[c + d*x])/2, (b*(1 - Cos [c + d*x]))/(a + b)]*(a + b*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(b*d*Sqrt[1 + Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(2/3)) + (Sqrt[2]*(A*b - a* B)*AppellF1[1/2, 1/2, 1/3, 3/2, (1 - Cos[c + d*x])/2, (b*(1 - Cos[c + d*x] ))/(a + b)]*((a + b*Cos[c + d*x])/(a + b))^(1/3)*Sin[c + d*x])/(b*d*Sqrt[1 + Cos[c + d*x]]*(a + b*Cos[c + d*x])^(1/3))
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ (b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] && GtQ[Sim plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] && !(GtQ[Simpl ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d *x, a + b*x]) && !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c - e*d)], 0] && SimplerQ[e + f*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p ]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]) Int[(a + b*x)^m*(c + d*x)^n*Si mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & GtQ[Simplify[b/(b*c - a*d)], 0] && !GtQ[Simplify[b/(b*e - a*f)], 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt[1 - Sin[c + d*x]]) Subst[Int[(a + b*x )^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0] && !IntegerQ[2*n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)/b Int[(a + b*Sin[e + f*x])^m, x], x] + Simp[d/b Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
\[\int \frac {A +B \cos \left (d x +c \right )}{\left (a +\cos \left (d x +c \right ) b \right )^{\frac {1}{3}}}d x\]
Input:
int((A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^(1/3),x)
Output:
int((A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^(1/3),x)
\[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(1/3),x, algorithm="fricas")
Output:
integral((B*cos(d*x + c) + A)/(b*cos(d*x + c) + a)^(1/3), x)
\[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int \frac {A + B \cos {\left (c + d x \right )}}{\sqrt [3]{a + b \cos {\left (c + d x \right )}}}\, dx \] Input:
integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))**(1/3),x)
Output:
Integral((A + B*cos(c + d*x))/(a + b*cos(c + d*x))**(1/3), x)
\[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(1/3),x, algorithm="maxima")
Output:
integrate((B*cos(d*x + c) + A)/(b*cos(d*x + c) + a)^(1/3), x)
\[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(1/3),x, algorithm="giac")
Output:
integrate((B*cos(d*x + c) + A)/(b*cos(d*x + c) + a)^(1/3), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{1/3}} \,d x \] Input:
int((A + B*cos(c + d*x))/(a + b*cos(c + d*x))^(1/3),x)
Output:
int((A + B*cos(c + d*x))/(a + b*cos(c + d*x))^(1/3), x)
\[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\left (\int \frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right ) b +a \right )^{\frac {1}{3}}}d x \right ) b +\left (\int \frac {1}{\left (\cos \left (d x +c \right ) b +a \right )^{\frac {1}{3}}}d x \right ) a \] Input:
int((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(1/3),x)
Output:
int(cos(c + d*x)/(cos(c + d*x)*b + a)**(1/3),x)*b + int(1/(cos(c + d*x)*b + a)**(1/3),x)*a