3.3.36 \(\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx\) [236]

Optimal. Leaf size=281 \[ \frac {3 C (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b d}+\frac {\sqrt {2} (5 b B-3 a C) F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b^2 d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}-\frac {\sqrt {2} \left (5 a b B-3 a^2 C-2 b^2 C\right ) F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \sin (c+d x)}{5 b^2 d \sqrt {1+\cos (c+d x)} \sqrt [3]{a+b \cos (c+d x)}} \]

[Out]

3/5*C*(a+b*cos(d*x+c))^(2/3)*sin(d*x+c)/b/d+1/5*(5*B*b-3*C*a)*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))^(2/3)*sin(d*x+c)*2^(1/2)/b^2/d/((a+b*cos(d*x+c))/(a+b))^(2/3)/(1+cos(d*x
+c))^(1/2)-1/5*(5*B*a*b-3*C*a^2-2*C*b^2)*AppellF1(1/2,1/3,1/2,3/2,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)/b^2/d/(a+b*cos(d*x+c))^(1/3)/(1+cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.22, antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3102, 2835, 2744, 144, 143} \begin {gather*} -\frac {\sqrt {2} \left (-3 a^2 C+5 a b B-2 b^2 C\right ) \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 {1}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{5 b^2 d \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}+\frac {\sqrt {2} (5 b B-3 a C) \sin (c+d x) (a+b \cos (c+d x))^{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 )}{5 b^2 d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^(1/3),x]

[Out]

(3*C*(a + b*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(5*b*d) + (Sqrt[2]*(5*b*B - 3*a*C)*AppellF1[1/2, 1/2, -2/3, 3/2,
(1 - Cos[c + d*x])/2, (b*(1 - Cos[c + d*x]))/(a + b)]*(a + b*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(5*b^2*d*Sqrt[
1 + Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(2/3)) - (Sqrt[2]*(5*a*b*B - 3*a^2*C - 2*b^2*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])/(a + b))^(1/3)*Sin
[c + d*x])/(5*b^2*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]

Rule 2835

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 3102

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

Rubi steps

\begin {align*} \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx &=\frac {3 C (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b d}+\frac {3 \int \frac {\frac {2 b C}{3}+\frac {1}{3} (5 b B-3 a C) \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx}{5 b}\\ &=\frac {3 C (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b d}+\frac {(5 b B-3 a C) \int (a+b \cos (c+d x))^{2/3} \, dx}{5 b^2}-\frac {\left (5 a b B-3 a^2 C-2 b^2 C\right ) \int \frac {1}{\sqrt [3]{a+b \cos (c+d x)}} \, dx}{5 b^2}\\ &=\frac {3 C (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b d}-\frac {((5 b B-3 a C) \sin (c+d x)) \text {Subst}\left (\int \frac {(a+b x)^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\cos (c+d x)\right )}{5 b^2 d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}}+\frac {\left (\left (5 a b B-3 a^2 C-2 b^2 C\right ) \sin (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{a+b x}} \, dx,x,\cos (c+d x)\right )}{5 b^2 d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}}\\ &=\frac {3 C (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b d}-\frac {\left ((5 b B-3 a C) (a+b \cos (c+d x))^{2/3} \sin (c+d x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\cos (c+d x)\right )}{5 b^2 d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)} \left (-\frac {a+b \cos (c+d x)}{-a-b}\right )^{2/3}}+\frac {\left (\left (5 a b B-3 a^2 C-2 b^2 C\right ) \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} \sqrt [3]{-\frac {a}{-a-b}-\frac {b x}{-a-b}}} \, dx,x,\cos (c+d x)\right )}{5 b^2 d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)} \sqrt [3]{a+b \cos (c+d x)}}\\ &=\frac {3 C (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b d}+\frac {\sqrt {2} (5 b B-3 a C) 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 ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b^2 d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}-\frac {\sqrt {2} \left (5 a b B-3 a^2 C-2 b^2 C\right ) 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]{\frac {a+b \cos (c+d x)}{a+b}} \sin (c+d x)}{5 b^2 d \sqrt {1+\cos (c+d x)} \sqrt [3]{a+b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 2.20, size = 263, normalized size = 0.94 \begin {gather*} -\frac {3 (a+b \cos (c+d x))^{2/3} \csc (c+d x) \left (5 \left (-5 a b B+3 a^2 C+2 b^2 C\right ) 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}}+2 (5 b B-3 a C) 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))-10 b^2 C \sin ^2(c+d x)\right )}{50 b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^(1/3),x]

[Out]

(-3*(a + b*Cos[c + d*x])^(2/3)*Csc[c + d*x]*(5*(-5*a*b*B + 3*a^2*C + 2*b^2*C)*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))] + 2*(5*b*B - 3*a*C)*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]) - 10*b^2*C*Sin[c + d*x]^2))/(50*b^3*d)

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Maple [F]
time = 0.15, size = 0, normalized size = 0.00 \[\int \frac {B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )}{\left (a +b \cos \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x)

[Out]

int((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),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((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/(b*cos(d*x + c) + a)^(1/3), 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((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + B*cos(d*x + c))/(b*cos(d*x + c) + a)^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**(1/3),x)

[Out]

Integral((B + C*cos(c + d*x))*cos(c + d*x)/(a + b*cos(c + d*x))**(1/3), 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((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/(b*cos(d*x + c) + a)^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(a + b*cos(c + d*x))^(1/3),x)

[Out]

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(a + b*cos(c + d*x))^(1/3), x)

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