3.206 \(\int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx\)

Optimal. Leaf size=272 \[ \frac {\left (3 a^2 C+b^2 (4 A+C)\right ) \sin (c+d x) \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3} F_1\left (\frac {1}{2};\frac {1}{2},\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{2 \sqrt {2} b^2 d \sqrt {\cos (c+d x)+1} (a+b \cos (c+d x))^{2/3}}-\frac {3 a C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{2 \sqrt {2} b^2 d \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}+\frac {3 C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)}}{4 b d} \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 272, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3024, 2756, 2665, 139, 138} \[ \frac {\left (3 a^2 C+b^2 (4 A+C)\right ) \sin (c+d x) \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3} F_1\left (\frac {1}{2};\frac {1}{2},\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{2 \sqrt {2} b^2 d \sqrt {\cos (c+d x)+1} (a+b \cos (c+d x))^{2/3}}-\frac {3 a C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{2 \sqrt {2} b^2 d \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}+\frac {3 C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)}}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), 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 139

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 2665

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]

Rule 2756

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(b*
c - a*d)/b, Int[(a + b*Sin[e + f*x])^m, x], x] + Dist[d/b, Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{
a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 3024

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp
[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x]
)^m*Simp[A*b*(m + 2) + b*C*(m + 1) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] &&  !LtQ[
m, -1]

Rubi steps

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

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Mathematica [A]  time = 1.94, size = 256, normalized size = 0.94 \[ -\frac {3 \csc (c+d x) \sqrt [3]{a+b \cos (c+d x)} \left (4 \left (C \left (3 a^2+b^2\right )+4 A b^2\right ) \sqrt {-\frac {b (\cos (c+d x)-1)}{a+b}} \sqrt {\frac {b (\cos (c+d x)+1)}{b-a}} F_1\left (\frac {1}{3};\frac {1}{2},\frac {1}{2};\frac {4}{3};\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right )+C \left (-3 a \sqrt {-\frac {b (\cos (c+d x)-1)}{a+b}} \sqrt {\frac {b (\cos (c+d x)+1)}{b-a}} (a+b \cos (c+d x)) F_1\left (\frac {4}{3};\frac {1}{2},\frac {1}{2};\frac {7}{3};\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right )-4 b^2 \sin ^2(c+d x)\right )\right )}{16 b^3 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3*(a + b*Cos[c + d*x])^(1/3)*Csc[c + d*x]*(4*(4*A*b^2 + (3*a^2 + b^2)*C)*AppellF1[1/3, 1/2, 1/2, 4/3, (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)] + C*(-3*a*AppellF1[4/3, 1/2, 1/2, 7/3, (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]
) - 4*b^2*Sin[c + d*x]^2)))/(16*b^3*d)

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(2/3),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \frac {A +C \left (\cos ^{2}\left (d x +c \right )\right )}{\left (a +b \cos \left (d x +c \right )\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(2/3),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{2/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + C \cos ^{2}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**(2/3),x)

[Out]

Integral((A + C*cos(c + d*x)**2)/(a + b*cos(c + d*x))**(2/3), x)

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