\(\int (a+b \cos (c+d x))^{2/3} \, dx\) [57]

Optimal result

Integrand size = 14, antiderivative size = 105 \[ \int (a+b \cos (c+d x))^{2/3} \, dx=\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2744, 144, 143} \[ \int (a+b \cos (c+d x))^{2/3} \, dx=\frac {\sqrt {2} \sin (c+d x) (a+b \cos (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}} \]

[In]

Int[(a + b*Cos[c + d*x])^(2/3),x]

[Out]

(Sqrt[2]*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])/(d*Sqrt[1 + Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(2/3))

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\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 )}{d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}} \\ & = -\frac {\left ((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 )}{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 {\sqrt {2} \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)}{d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.12 \[ \int (a+b \cos (c+d x))^{2/3} \, dx=-\frac {3 \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},\frac {1}{2},\frac {8}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} (a+b \cos (c+d x))^{5/3} \csc (c+d x)}{5 b d} \]

[In]

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

[Out]

(-3*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)]*Sqrt[-((b*(-1 + C
os[c + d*x]))/(a + b))]*Sqrt[(b*(1 + Cos[c + d*x]))/(-a + b)]*(a + b*Cos[c + d*x])^(5/3)*Csc[c + d*x])/(5*b*d)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

\[ \int (a+b \cos (c+d x))^{2/3} \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int (a+b \cos (c+d x))^{2/3} \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right )^{\frac {2}{3}}\, dx \]

[In]

integrate((a+b*cos(d*x+c))**(2/3),x)

[Out]

Integral((a + b*cos(c + d*x))**(2/3), x)

Maxima [F]

\[ \int (a+b \cos (c+d x))^{2/3} \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^(2/3),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int (a+b \cos (c+d x))^{2/3} \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^(2/3),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int((a + b*cos(c + d*x))^(2/3),x)

[Out]

int((a + b*cos(c + d*x))^(2/3), x)