\(\int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx\) [797]

Optimal result

Integrand size = 25, antiderivative size = 226 \[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\frac {\sqrt {2} B \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}+\frac {\sqrt {2} (A b-a B) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} \sqrt [3]{a+b \cos (c+d x)}} \]

[Out]

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)*2^(1/
2)/b/d/((a+b*cos(d*x+c))/(a+b))^(2/3)/(1+cos(d*x+c))^(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)*2^(1/2)/b/d/(a+b*cos(d*x+c))^(1/3)/(1+cos
(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2835, 2744, 144, 143} \[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\frac {\sqrt {2} (A b-a B) \sin (c+d x) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}+\frac {\sqrt {2} B \sin (c+d x) (a+b \cos (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}} \]

[In]

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

[Out]

(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))

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(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] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 144

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((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*
(b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 2744

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[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]

Rule 2835

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(b*
c - a*d)/b, Int[(a + b*Sin[e + f*x])^m, x], x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {B \int (a+b \cos (c+d x))^{2/3} \, dx}{b}+\frac {(A b-a B) \int \frac {1}{\sqrt [3]{a+b \cos (c+d x)}} \, dx}{b} \\ & = -\frac {(B \sin (c+d x)) \text {Subst}\left (\int \frac {(a+b x)^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\cos (c+d x)\right )}{b d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}}-\frac {((A b-a B) \sin (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{a+b x}} \, dx,x,\cos (c+d x)\right )}{b d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}} \\ & = -\frac {\left (B (a+b \cos (c+d x))^{2/3} \sin (c+d x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\cos (c+d x)\right )}{b d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)} \left (-\frac {a+b \cos (c+d x)}{-a-b}\right )^{2/3}}-\frac {\left ((A b-a B) \sqrt [3]{-\frac {a+b \cos (c+d x)}{-a-b}} \sin (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{-\frac {a}{-a-b}-\frac {b x}{-a-b}}} \, dx,x,\cos (c+d x)\right )}{b d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)} \sqrt [3]{a+b \cos (c+d x)}} \\ & = \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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.84 \[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=-\frac {3 \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} (a+b \cos (c+d x))^{2/3} \left (5 (A b-a B) \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right )+2 B \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},\frac {1}{2},\frac {8}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) (a+b \cos (c+d x))\right ) \csc (c+d x)}{10 b^2 d} \]

[In]

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

[Out]

(-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*
AppellF1[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)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((B*cos(d*x + c) + A)/(b*cos(d*x + c) + a)^(1/3), x)

Sympy [F]

\[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int \frac {A + B \cos {\left (c + d x \right )}}{\sqrt [3]{a + b \cos {\left (c + d x \right )}}}\, dx \]

[In]

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))**(1/3),x)

[Out]

Integral((A + B*cos(c + d*x))/(a + b*cos(c + d*x))**(1/3), x)

Maxima [F]

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

[In]

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(1/3),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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