\(\int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx\) [789]

Optimal result

Integrand size = 25, antiderivative size = 101 \[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx=\frac {3 B \sin (c+d x)}{2 d \sqrt [3]{a+a \cos (c+d x)}}+\frac {(2 A-B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{2^{5/6} d \sqrt [6]{1+\cos (c+d x)} \sqrt [3]{a+a \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 101, 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 = {2830, 2731, 2730} \[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx=\frac {(2 A-B) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right )}{2^{5/6} d \sqrt [6]{\cos (c+d x)+1} \sqrt [3]{a \cos (c+d x)+a}}+\frac {3 B \sin (c+d x)}{2 d \sqrt [3]{a \cos (c+d x)+a}} \]

[In]

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

[Out]

(3*B*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(1/3)) + ((2*A - B)*Hypergeometric2F1[1/2, 5/6, 3/2, (1 - Cos[c +
d*x])/2]*Sin[c + d*x])/(2^(5/6)*d*(1 + Cos[c + d*x])^(1/6)*(a + a*Cos[c + d*x])^(1/3))

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 2830

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*S
in[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)]

Rubi steps \begin{align*} \text {integral}& = \frac {3 B \sin (c+d x)}{2 d \sqrt [3]{a+a \cos (c+d x)}}+\frac {1}{2} (2 A-B) \int \frac {1}{\sqrt [3]{a+a \cos (c+d x)}} \, dx \\ & = \frac {3 B \sin (c+d x)}{2 d \sqrt [3]{a+a \cos (c+d x)}}+\frac {\left ((2 A-B) \sqrt [3]{1+\cos (c+d x)}\right ) \int \frac {1}{\sqrt [3]{1+\cos (c+d x)}} \, dx}{2 \sqrt [3]{a+a \cos (c+d x)}} \\ & = \frac {3 B \sin (c+d x)}{2 d \sqrt [3]{a+a \cos (c+d x)}}+\frac {(2 A-B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{2^{5/6} d \sqrt [6]{1+\cos (c+d x)} \sqrt [3]{a+a \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.32 \[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx=\frac {3\ 2^{5/6} B \sqrt [6]{1-\cos \left (d x-2 \arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )} \sin (c+d x)-2 (2 A-B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\cos ^2\left (\frac {d x}{2}-\arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )\right ) \sin \left (d x-2 \arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )}{4 d \sqrt [3]{a (1+\cos (c+d x))} \sqrt [6]{\sin ^2\left (\frac {d x}{2}-\arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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