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

Optimal. Leaf size=229 \[ \frac {\sqrt {2} (A b-a B) \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\sqrt {2} B (a+b) \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {4}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}} \]

[Out]

(a+b)*B*AppellF1(1/2,-4/3,1/2,3/2,b*(1-cos(d*x+c))/(a+b),1/2-1/2*cos(d*x+c))*(a+b*cos(d*x+c))^(1/3)*sin(d*x+c)
*2^(1/2)/b/d/((a+b*cos(d*x+c))/(a+b))^(1/3)/(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))^(1/3)*sin(d*x+c)*2^(1/2)/b/d/((a+b*cos(d*x+c))/(a+b))^(1/3)
/(1+cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2756, 2665, 139, 138} \[ \frac {\sqrt {2} (A b-a B) \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\sqrt {2} B (a+b) \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {4}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*(a + b)*B*AppellF1[1/2, 1/2, -4/3, 3/2, (1 - Cos[c + d*x])/2, (b*(1 - Cos[c + d*x]))/(a + b)]*(a + b*
Cos[c + d*x])^(1/3)*Sin[c + d*x])/(b*d*Sqrt[1 + Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(1/3)) + (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])^(1/3)*Sin[c + d*x])/(b*d*Sqrt[1 + Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(1/3))

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])

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 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 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]

Rubi steps

\begin {align*} \int \sqrt [3]{a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx &=\frac {B \int (a+b \cos (c+d x))^{4/3} \, dx}{b}+\frac {(A b-a B) \int \sqrt [3]{a+b \cos (c+d x)} \, dx}{b}\\ &=-\frac {(B \sin (c+d x)) \operatorname {Subst}\left (\int \frac {(a+b x)^{4/3}}{\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 {\sqrt [3]{a+b x}}{\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 {\left ((-a-b) B \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{4/3}}{\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)} \sqrt [3]{-\frac {a+b \cos (c+d x)}{-a-b}}}-\frac {\left ((A b-a B) \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{-\frac {a}{-a-b}-\frac {b x}{-a-b}}}{\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)} \sqrt [3]{-\frac {a+b \cos (c+d x)}{-a-b}}}\\ &=\frac {\sqrt {2} (a+b) B F_1\left (\frac {1}{2};\frac {1}{2},-\frac {4}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}+\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]{a+b \cos (c+d x)} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}\\ \end {align*}

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Mathematica [A]  time = 1.92, size = 253, normalized size = 1.10 \[ -\frac {3 \csc (c+d x) \sqrt [3]{a+b \cos (c+d x)} \left (4 B \left (b^2-a^2\right ) \sqrt {-\frac {b (\cos (c+d x)-1)}{a+b}} \sqrt {-\frac {b (\cos (c+d x)+1)}{a-b}} F_1\left (\frac {1}{3};\frac {1}{2},\frac {1}{2};\frac {4}{3};\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right )+(a B+4 A b) \sqrt {-\frac {b (\cos (c+d x)-1)}{a+b}} \sqrt {\frac {b (\cos (c+d x)+1)}{b-a}} (a+b \cos (c+d x)) F_1\left (\frac {4}{3};\frac {1}{2},\frac {1}{2};\frac {7}{3};\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right )-4 b^2 B \sin ^2(c+d x)\right )}{16 b^2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3*(a + b*Cos[c + d*x])^(1/3)*Csc[c + d*x]*(4*(-a^2 + b^2)*B*AppellF1[1/3, 1/2, 1/2, 4/3, (a + b*Cos[c + d*x]
)/(a - b), (a + b*Cos[c + d*x])/(a + b)]*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[-((b*(1 + Cos[c + d*x])
)/(a - b))] + (4*A*b + a*B)*AppellF1[4/3, 1/2, 1/2, 7/3, (a + b*Cos[c + d*x])/(a - b), (a + b*Cos[c + d*x])/(a
+ b)]*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[(b*(1 + Cos[c + d*x]))/(-a + b)]*(a + b*Cos[c + d*x]) - 4
*b^2*B*Sin[c + d*x]^2))/(16*b^2*d)

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \left (a +b \cos \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +B \cos \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{1/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \cos {\left (c + d x \right )}\right ) \sqrt [3]{a + b \cos {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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