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

Optimal. Leaf size=104 \[ \frac{2 \sqrt [6]{2} (5 A+2 B) \sin (c+d x) (a \cos (c+d x)+a)^{2/3} \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x))\right )}{5 d (\cos (c+d x)+1)^{7/6}}+\frac{3 B \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 d} \]

[Out]

(3*B*(a + a*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(5*d) + (2*2^(1/6)*(5*A + 2*B)*(a + a*Cos[c + d*x])^(2/3)*Hyperg
eometric2F1[-1/6, 1/2, 3/2, (1 - Cos[c + d*x])/2]*Sin[c + d*x])/(5*d*(1 + Cos[c + d*x])^(7/6))

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Rubi [A]  time = 0.0819791, antiderivative size = 104, 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 = {2751, 2652, 2651} \[ \frac{2 \sqrt [6]{2} (5 A+2 B) \sin (c+d x) (a \cos (c+d x)+a)^{2/3} \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x))\right )}{5 d (\cos (c+d x)+1)^{7/6}}+\frac{3 B \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*B*(a + a*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(5*d) + (2*2^(1/6)*(5*A + 2*B)*(a + a*Cos[c + d*x])^(2/3)*Hyperg
eometric2F1[-1/6, 1/2, 3/2, (1 - Cos[c + d*x])/2]*Sin[c + d*x])/(5*d*(1 + Cos[c + d*x])^(7/6))

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, 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 (a+a \cos (c+d x))^{2/3} (A+B \cos (c+d x)) \, dx &=\frac{3 B (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{5 d}+\frac{1}{5} (5 A+2 B) \int (a+a \cos (c+d x))^{2/3} \, dx\\ &=\frac{3 B (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{5 d}+\frac{\left ((5 A+2 B) (a+a \cos (c+d x))^{2/3}\right ) \int (1+\cos (c+d x))^{2/3} \, dx}{5 (1+\cos (c+d x))^{2/3}}\\ &=\frac{3 B (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{5 d}+\frac{2 \sqrt [6]{2} (5 A+2 B) (a+a \cos (c+d x))^{2/3} \, _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)}{5 d (1+\cos (c+d x))^{7/6}}\\ \end{align*}

Mathematica [A]  time = 0.591884, size = 164, normalized size = 1.58 \[ \frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a (\cos (c+d x)+1))^{2/3} \left (3\ 2^{5/6} \sin (c+d x) (5 A+2 B \cos (c+d x)+4 B) \sqrt [6]{1-\cos \left (d x-2 \tan ^{-1}\left (\cot \left (\frac{c}{2}\right )\right )\right )}-2 (5 A+2 B) \sin \left (d x-2 \tan ^{-1}\left (\cot \left (\frac{c}{2}\right )\right )\right ) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\cos ^2\left (\frac{d x}{2}-\tan ^{-1}\left (\cot \left (\frac{c}{2}\right )\right )\right )\right )\right )}{20\ 2^{5/6} d \sqrt [6]{1-\cos \left (d x-2 \tan ^{-1}\left (\cot \left (\frac{c}{2}\right )\right )\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Cos[c + d*x]))^(2/3)*Sec[(c + d*x)/2]^2*(3*2^(5/6)*(5*A + 4*B + 2*B*Cos[c + d*x])*(1 - Cos[d*x - 2*Ar
cTan[Cot[c/2]]])^(1/6)*Sin[c + d*x] - 2*(5*A + 2*B)*Hypergeometric2F1[1/2, 5/6, 3/2, Cos[(d*x)/2 - ArcTan[Cot[
c/2]]]^2]*Sin[d*x - 2*ArcTan[Cot[c/2]]]))/(20*2^(5/6)*d*(1 - Cos[d*x - 2*ArcTan[Cot[c/2]]])^(1/6))

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Maple [F]  time = 0.246, size = 0, normalized size = 0. \begin{align*} \int \left ( a+\cos \left ( dx+c \right ) a \right ) ^{{\frac{2}{3}}} \left ( A+B\cos \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \cos \left (d x + c\right ) + A\right )}{\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+a*cos(d*x+c))^(2/3)*(A+B*cos(d*x+c)),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 ({\left (B \cos \left (d x + c\right ) + A\right )}{\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+a*cos(d*x+c))^(2/3)*(A+B*cos(d*x+c)),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+a*cos(d*x+c))**(2/3)*(A+B*cos(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \cos \left (d x + c\right ) + A\right )}{\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+a*cos(d*x+c))^(2/3)*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

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