3.790 \(\int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx\)

Optimal. Leaf size=105 \[ \frac {3 (A-B) \sin (c+d x)}{d (a \cos (c+d x)+a)^{2/3}}-\frac {2^{5/6} (A-2 B) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x))\right )}{a d (\cos (c+d x)+1)^{5/6}} \]

[Out]

3*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(2/3)-2^(5/6)*(A-2*B)*(a+a*cos(d*x+c))^(1/3)*hypergeom([1/6, 1/2],[3/2],
1/2-1/2*cos(d*x+c))*sin(d*x+c)/a/d/(1+cos(d*x+c))^(5/6)

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Rubi [A]  time = 0.09, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2750, 2652, 2651} \[ \frac {3 (A-B) \sin (c+d x)}{d (a \cos (c+d x)+a)^{2/3}}-\frac {2^{5/6} (A-2 B) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x))\right )}{a d (\cos (c+d x)+1)^{5/6}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x])/(a + a*Cos[c + d*x])^(2/3),x]

[Out]

(3*(A - B)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^(2/3)) - (2^(5/6)*(A - 2*B)*(a + a*Cos[c + d*x])^(1/3)*Hyperg
eometric2F1[1/6, 1/2, 3/2, (1 - Cos[c + d*x])/2]*Sin[c + d*x])/(a*d*(1 + Cos[c + d*x])^(5/6))

Rule 2651

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(2^(n + 1/2)*a^(n - 1/2)*b*Cos[c + d*x]*Hy
pergeometric2F1[1/2, 1/2 - n, 3/2, (1*(1 - (b*Sin[c + d*x])/a))/2])/(d*Sqrt[a + b*Sin[c + d*x]]), x] /; FreeQ[
{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2652

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Sin[c + d*x])^FracPart
[n])/(1 + (b*Sin[c + d*x])/a)^FracPart[n], Int[(1 + (b*Sin[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 2750

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b
*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx &=\frac {3 (A-B) \sin (c+d x)}{d (a+a \cos (c+d x))^{2/3}}-\frac {(A-2 B) \int \sqrt [3]{a+a \cos (c+d x)} \, dx}{a}\\ &=\frac {3 (A-B) \sin (c+d x)}{d (a+a \cos (c+d x))^{2/3}}-\frac {\left ((A-2 B) \sqrt [3]{a+a \cos (c+d x)}\right ) \int \sqrt [3]{1+\cos (c+d x)} \, dx}{a \sqrt [3]{1+\cos (c+d x)}}\\ &=\frac {3 (A-B) \sin (c+d x)}{d (a+a \cos (c+d x))^{2/3}}-\frac {2^{5/6} (A-2 B) \sqrt [3]{a+a \cos (c+d x)} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{a d (1+\cos (c+d x))^{5/6}}\\ \end {align*}

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Mathematica [C]  time = 1.42, size = 197, normalized size = 1.88 \[ \frac {3 \cos \left (\frac {1}{2} (c+d x)\right ) \left (-4 \csc \left (\frac {c}{2}\right ) \left ((3 B-2 A) \cos \left (\frac {d x}{2}\right )+B \cos \left (c+\frac {d x}{2}\right )\right )-(A-2 B) \csc \left (\frac {c}{4}\right ) \sec \left (\frac {c}{4}\right ) e^{-\frac {1}{2} i d x} \sqrt [3]{i \sin (c) e^{i d x}+\cos (c) e^{i d x}+1} \left (2 \, _2F_1\left (-\frac {1}{3},\frac {1}{3};\frac {2}{3};-e^{i d x} (\cos (c)+i \sin (c))\right )+e^{i d x} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-e^{i d x} (\cos (c)+i \sin (c))\right )\right )\right )}{4 d (a (\cos (c+d x)+1))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Cos[(c + d*x)/2]*(-4*((-2*A + 3*B)*Cos[(d*x)/2] + B*Cos[c + (d*x)/2])*Csc[c/2] - ((A - 2*B)*Csc[c/4]*(2*Hyp
ergeometric2F1[-1/3, 1/3, 2/3, -(E^(I*d*x)*(Cos[c] + I*Sin[c]))] + E^(I*d*x)*Hypergeometric2F1[1/3, 2/3, 5/3,
-(E^(I*d*x)*(Cos[c] + I*Sin[c]))])*Sec[c/4]*(1 + E^(I*d*x)*Cos[c] + I*E^(I*d*x)*Sin[c])^(1/3))/E^((I/2)*d*x)))
/(4*d*(a*(1 + Cos[c + d*x]))^(2/3))

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fricas [F]  time = 1.05, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int \frac {A +B \cos \left (d x +c \right )}{\left (a +a \cos \left (d x +c \right )\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{2/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \cos {\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*cos(c + d*x))/(a*(cos(c + d*x) + 1))**(2/3), x)

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