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[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))

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Rubi [A]  time = 0.0894841, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, 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 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)]

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 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]

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*}

Mathematica [C]  time = 1.30371, 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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Maple [F]  time = 0.219, size = 0, normalized size = 0. \begin{align*} \int{(A+B\cos \left ( dx+c \right ) ) \left ( a+\cos \left ( dx+c \right ) a \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^(2/3),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 (a \cos \left (d x + c\right ) + a\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

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

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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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 (a \cos \left (d x + c\right ) + a\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

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)