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

Optimal. Leaf size=226 \[ \frac{\sqrt{2} (A b-a B) \sin (c+d x) \sqrt [3]{\frac{a+b \cos (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt{\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}+\frac{\sqrt{2} B \sin (c+d x) (a+b \cos (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt{\cos (c+d x)+1} \left (\frac{a+b \cos (c+d x)}{a+b}\right )^{2/3}} \]

[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))

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Rubi [A]  time = 0.183003, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2756, 2665, 139, 138} \[ \frac{\sqrt{2} (A b-a B) \sin (c+d x) \sqrt [3]{\frac{a+b \cos (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt{\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}+\frac{\sqrt{2} B \sin (c+d x) (a+b \cos (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt{\cos (c+d x)+1} \left (\frac{a+b \cos (c+d x)}{a+b}\right )^{2/3}} \]

Antiderivative was successfully verified.

[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 2756

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]

Rule 2665

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 139

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 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), 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])

Rubi steps

\begin{align*} \int \frac{A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx &=\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)) \operatorname{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)) \operatorname{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 ) \operatorname{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 ) \operatorname{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 F_1\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) F_1\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]  time = 0.413311, size = 189, normalized size = 0.84 \[ -\frac{3 \csc (c+d x) \sqrt{-\frac{b (\cos (c+d x)-1)}{a+b}} \sqrt{\frac{b (\cos (c+d x)+1)}{b-a}} (a+b \cos (c+d x))^{2/3} \left (5 (A b-a B) F_1\left (\frac{2}{3};\frac{1}{2},\frac{1}{2};\frac{5}{3};\frac{a+b \cos (c+d x)}{a-b},\frac{a+b \cos (c+d x)}{a+b}\right )+2 B (a+b \cos (c+d x)) F_1\left (\frac{5}{3};\frac{1}{2},\frac{1}{2};\frac{8}{3};\frac{a+b \cos (c+d x)}{a-b},\frac{a+b \cos (c+d x)}{a+b}\right )\right )}{10 b^2 d} \]

Warning: Unable to verify antiderivative.

[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)

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Maple [F]  time = 0.404, size = 0, normalized size = 0. \begin{align*} \int{(A+B\cos \left ( dx+c \right ) ){\frac{1}{\sqrt [3]{a+b\cos \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \cos{\left (c + d x \right )}}{\sqrt [3]{a + b \cos{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)