\(\int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx\) [790]

Optimal result

Integrand size = 25, antiderivative size = 105 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\frac {3 (A-B) \sin (c+d x)}{d (a+a \cos (c+d x))^{2/3}}-\frac {2^{5/6} (A-2 B) \sqrt [3]{a+a \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{a d (1+\cos (c+d x))^{5/6}} \]

[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)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2829, 2731, 2730} \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\frac {3 (A-B) \sin (c+d x)}{d (a \cos (c+d x)+a)^{2/3}}-\frac {2^{5/6} (A-2 B) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right )}{a d (\cos (c+d x)+1)^{5/6}} \]

[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 2730

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

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

Rule 2829

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*} \text {integral}& = \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)} \operatorname {Hypergeometric2F1}\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 [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(254\) vs. \(2(105)=210\).

Time = 1.99 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.42 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (4 (A-2 B) \, _2F_1\left (-\frac {1}{2},-\frac {1}{6};\frac {5}{6};\cos ^2\left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )-\csc \left (\frac {c}{2}\right ) \left (5 (A-2 B) \cos \left (\frac {1}{2} \left (c-d x-2 \arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \sec \left (\frac {c}{2}\right )+(A-2 B) \cos \left (\frac {1}{2} \left (c+d x+2 \arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \sec \left (\frac {c}{2}\right )+3 \left ((-2 A+3 B) \cos \left (\frac {d x}{2}\right )+B \cos \left (c+\frac {d x}{2}\right )\right ) \sqrt {\sec ^2\left (\frac {c}{2}\right )}\right ) \sqrt {\sin ^2\left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )}\right )}{d (a (1+\cos (c+d x)))^{2/3} \sqrt {\sec ^2\left (\frac {c}{2}\right )} \sqrt {\sin ^2\left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )}} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]*(4*(A - 2*B)*HypergeometricPFQ[{-1/2, -1/6}, {5/6}, Cos[(d*x)/2 + ArcTan[Tan[c/2]]]^2]*Sec[c
/2]*Sin[(d*x)/2 + ArcTan[Tan[c/2]]] - Csc[c/2]*(5*(A - 2*B)*Cos[(c - d*x - 2*ArcTan[Tan[c/2]])/2]*Sec[c/2] + (
A - 2*B)*Cos[(c + d*x + 2*ArcTan[Tan[c/2]])/2]*Sec[c/2] + 3*((-2*A + 3*B)*Cos[(d*x)/2] + B*Cos[c + (d*x)/2])*S
qrt[Sec[c/2]^2])*Sqrt[Sin[(d*x)/2 + ArcTan[Tan[c/2]]]^2]))/(d*(a*(1 + Cos[c + d*x]))^(2/3)*Sqrt[Sec[c/2]^2]*Sq
rt[Sin[(d*x)/2 + ArcTan[Tan[c/2]]]^2])

Maple [F]

\[\int \frac {A +B \cos \left (d x +c \right )}{\left (a +\cos \left (d x +c \right ) a \right )^{\frac {2}{3}}}d x\]

[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)

Fricas [F]

\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\int \frac {A + B \cos {\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {2}{3}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{2/3}} \,d x \]

[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)