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

Optimal. Leaf size=144 \[ \frac {(10 A-5 B+7 C) \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x))\right )}{5\ 2^{5/6} d \sqrt [6]{\cos (c+d x)+1} \sqrt [3]{a \cos (c+d x)+a}}+\frac {3 (5 B-3 C) \sin (c+d x)}{10 d \sqrt [3]{a \cos (c+d x)+a}}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 a d} \]

[Out]

3/10*(5*B-3*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/3)+3/5*C*(a+a*cos(d*x+c))^(2/3)*sin(d*x+c)/a/d+1/10*(10*A-5*B+
7*C)*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)

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Rubi [A]  time = 0.18, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3023, 2751, 2652, 2651} \[ \frac {(10 A-5 B+7 C) \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x))\right )}{5\ 2^{5/6} d \sqrt [6]{\cos (c+d x)+1} \sqrt [3]{a \cos (c+d x)+a}}+\frac {3 (5 B-3 C) \sin (c+d x)}{10 d \sqrt [3]{a \cos (c+d x)+a}}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(5*B - 3*C)*Sin[c + d*x])/(10*d*(a + a*Cos[c + d*x])^(1/3)) + (3*C*(a + a*Cos[c + d*x])^(2/3)*Sin[c + d*x])
/(5*a*d) + ((10*A - 5*B + 7*C)*Hypergeometric2F1[1/2, 5/6, 3/2, (1 - Cos[c + d*x])/2]*Sin[c + d*x])/(5*2^(5/6)
*d*(1 + Cos[c + d*x])^(1/6)*(a + a*Cos[c + d*x])^(1/3))

Rule 2651

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(2^(n + 1/2)*a^(n - 1/2)*b*Cos[c + d*x]*Hy
pergeometric2F1[1/2, 1/2 - n, 3/2, (1*(1 - (b*Sin[c + d*x])/a))/2])/(d*Sqrt[a + b*Sin[c + d*x]]), x] /; FreeQ[
{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2652

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Sin[c + d*x])^FracPart
[n])/(1 + (b*Sin[c + d*x])/a)^FracPart[n], Int[(1 + (b*Sin[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 2751

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*Sin
[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)]

Rule 3023

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 {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx &=\frac {3 C (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{5 a d}+\frac {3 \int \frac {\frac {1}{3} a (5 A+2 C)+\frac {1}{3} a (5 B-3 C) \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx}{5 a}\\ &=\frac {3 (5 B-3 C) \sin (c+d x)}{10 d \sqrt [3]{a+a \cos (c+d x)}}+\frac {3 C (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{5 a d}+\frac {1}{10} (10 A-5 B+7 C) \int \frac {1}{\sqrt [3]{a+a \cos (c+d x)}} \, dx\\ &=\frac {3 (5 B-3 C) \sin (c+d x)}{10 d \sqrt [3]{a+a \cos (c+d x)}}+\frac {3 C (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{5 a d}+\frac {\left ((10 A-5 B+7 C) \sqrt [3]{1+\cos (c+d x)}\right ) \int \frac {1}{\sqrt [3]{1+\cos (c+d x)}} \, dx}{10 \sqrt [3]{a+a \cos (c+d x)}}\\ &=\frac {3 (5 B-3 C) \sin (c+d x)}{10 d \sqrt [3]{a+a \cos (c+d x)}}+\frac {3 C (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{5 a d}+\frac {(10 A-5 B+7 C) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{5\ 2^{5/6} d \sqrt [6]{1+\cos (c+d x)} \sqrt [3]{a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.60, size = 105, normalized size = 0.73 \[ \frac {3 \sin (c+d x) (5 B+2 C \cos (c+d x)-C)-3 i (10 A-5 B+7 C) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-e^{i (c+d x)}\right ) (i \sin (c+d x)+\cos (c+d x)+1)^{2/3}}{10 d \sqrt [3]{a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*I)*(10*A - 5*B + 7*C)*Hypergeometric2F1[1/3, 2/3, 4/3, -E^(I*(c + d*x))]*(1 + Cos[c + d*x] + I*Sin[c + d*
x])^(2/3) + 3*(5*B - C + 2*C*Cos[c + d*x])*Sin[c + d*x])/(10*d*(a*(1 + Cos[c + d*x]))^(1/3))

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fricas [F]  time = 2.16, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.47, size = 0, normalized size = 0.00 \[ \int \frac {A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )}{\left (a +a \cos \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{1/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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