\(\int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx\) [61]

Optimal result

Integrand size = 14, antiderivative size = 105 \[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {4}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{4/3} \sin (c+d x)}{d \sqrt {1+\cos (c+d x)} (a+b \cos (c+d x))^{4/3}} \] Output:

2^(1/2)*AppellF1(1/2,4/3,1/2,3/2,b*(1-cos(d*x+c))/(a+b),1/2-1/2*cos(d*x+c) 
)*((a+b*cos(d*x+c))/(a+b))^(4/3)*sin(d*x+c)/d/(1+cos(d*x+c))^(1/2)/(a+b*co 
s(d*x+c))^(4/3)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(268\) vs. \(2(105)=210\).

Time = 2.38 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.55 \[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\frac {15 a \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {-\frac {b (1+\cos (c+d x))}{a-b}} (a+b \cos (c+d x)) \csc (c+d x)-6 \left (5 b^2+2 \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},\frac {1}{2},\frac {8}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} (a+b \cos (c+d x))^2 \csc ^2(c+d x)\right ) \sin (c+d x)}{10 b \left (a^2-b^2\right ) d \sqrt [3]{a+b \cos (c+d x)}} \] Input:

Integrate[(a + b*Cos[c + d*x])^(-4/3),x]
 

Output:

(15*a*AppellF1[2/3, 1/2, 1/2, 5/3, (a + b*Cos[c + d*x])/(a - b), (a + b*Co 
s[c + d*x])/(a + b)]*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[-((b*(1 
 + Cos[c + d*x]))/(a - b))]*(a + b*Cos[c + d*x])*Csc[c + d*x] - 6*(5*b^2 + 
 2*AppellF1[5/3, 1/2, 1/2, 8/3, (a + b*Cos[c + d*x])/(a - b), (a + b*Cos[c 
 + d*x])/(a + b)]*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[(b*(1 + Co 
s[c + d*x]))/(-a + b)]*(a + b*Cos[c + d*x])^2*Csc[c + d*x]^2)*Sin[c + d*x] 
)/(10*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(1/3))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3144, 156, 155}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}}dx\)

\(\Big \downarrow \) 3144

\(\displaystyle -\frac {\sin (c+d x) \int \frac {1}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} (a+b \cos (c+d x))^{4/3}}d\cos (c+d x)}{d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 156

\(\displaystyle -\frac {\sin (c+d x) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \left (\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}\right )^{4/3}}d\cos (c+d x)}{d (a+b) \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 155

\(\displaystyle \frac {\sqrt {2} \sin (c+d x) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {4}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{d (a+b) \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}\)

Input:

Int[(a + b*Cos[c + d*x])^(-4/3),x]
 

Output:

(Sqrt[2]*AppellF1[1/2, 1/2, 4/3, 3/2, (1 - Cos[c + d*x])/2, (b*(1 - Cos[c 
+ d*x]))/(a + b)]*((a + b*Cos[c + d*x])/(a + b))^(1/3)*Sin[c + d*x])/((a + 
 b)*d*Sqrt[1 + Cos[c + d*x]]*(a + b*Cos[c + d*x])^(1/3))
 

Defintions of rubi rules used

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p 
]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p])   Int[(a + b*x)^m*(c + d*x)^n*Si 
mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
& GtQ[Simplify[b/(b*c - a*d)], 0] &&  !GtQ[Simplify[b/(b*e - a*f)], 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3144

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

Maple [F]

\[\int \frac {1}{\left (a +\cos \left (d x +c \right ) b \right )^{\frac {4}{3}}}d x\]

Input:

int(1/(a+cos(d*x+c)*b)^(4/3),x)
 

Output:

int(1/(a+cos(d*x+c)*b)^(4/3),x)
 

Fricas [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \] Input:

integrate(1/(a+b*cos(d*x+c))^(4/3),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\int \frac {1}{\left (a + b \cos {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \] Input:

integrate(1/(a+b*cos(d*x+c))**(4/3),x)
 

Output:

Integral((a + b*cos(c + d*x))**(-4/3), x)
 

Maxima [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \] Input:

integrate(1/(a+b*cos(d*x+c))^(4/3),x, algorithm="maxima")
 

Output:

integrate((b*cos(d*x + c) + a)^(-4/3), x)
 

Giac [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \] Input:

integrate(1/(a+b*cos(d*x+c))^(4/3),x, algorithm="giac")
 

Output:

integrate((b*cos(d*x + c) + a)^(-4/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\int \frac {1}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{4/3}} \,d x \] Input:

int(1/(a + b*cos(c + d*x))^(4/3),x)
 

Output:

int(1/(a + b*cos(c + d*x))^(4/3), x)
 

Reduce [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\int \frac {1}{\left (\cos \left (d x +c \right ) b +a \right )^{\frac {1}{3}} \cos \left (d x +c \right ) b +\left (\cos \left (d x +c \right ) b +a \right )^{\frac {1}{3}} a}d x \] Input:

int(1/(a+b*cos(d*x+c))^(4/3),x)
 

Output:

int(1/((cos(c + d*x)*b + a)**(1/3)*cos(c + d*x)*b + (cos(c + d*x)*b + a)** 
(1/3)*a),x)