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

Optimal. Leaf size=110 \[ \frac {\sqrt {2} F_1\left (\frac {1}{2};\frac {1}{2},\frac {4}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \sin (c+d x)}{(a+b) d \sqrt {1+\cos (c+d x)} \sqrt [3]{a+b \cos (c+d x)}} \]

[Out]

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))^(1/3)*sin(d*x+c)*
2^(1/2)/(a+b)/d/(a+b*cos(d*x+c))^(1/3)/(1+cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2744, 144, 143} \begin {gather*} \frac {\sqrt {2} \sin (c+d x) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} F_1\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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^(-4/3),x]

[Out]

(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))

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(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] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 144

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

Rule 2744

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[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]

Rubi steps

\begin {align*} \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx &=-\frac {\sin (c+d x) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} (a+b x)^{4/3}} \, dx,x,\cos (c+d x)\right )}{d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}}\\ &=-\frac {\left (\sqrt [3]{-\frac {a+b \cos (c+d x)}{-a-b}} \sin (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{4/3}} \, dx,x,\cos (c+d x)\right )}{(a+b) d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)} \sqrt [3]{a+b \cos (c+d x)}}\\ &=\frac {\sqrt {2} F_1\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 ) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \sin (c+d x)}{(a+b) d \sqrt {1+\cos (c+d x)} \sqrt [3]{a+b \cos (c+d x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(268\) vs. \(2(110)=220\).
time = 1.37, size = 268, normalized size = 2.44 \begin {gather*} \frac {15 a F_1\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 F_1\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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a*AppellF1[2/3, 1/2, 1/2, 5/3, (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 + 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 + Cos[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))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \cos \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \cos {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{4/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cos(c + d*x))^(4/3),x)

[Out]

int(1/(a + b*cos(c + d*x))^(4/3), x)

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