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

Optimal. Leaf size=138 \[ -\frac {(4 A+7 C) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x))\right )}{2 \sqrt [6]{2} a d (\cos (c+d x)+1)^{5/6}}+\frac {3 (A+C) \sin (c+d x)}{d (a \cos (c+d x)+a)^{2/3}}+\frac {3 C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 a d} \]

[Out]

3*(A+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(2/3)+3/4*C*(a+a*cos(d*x+c))^(1/3)*sin(d*x+c)/a/d-1/4*(4*A+7*C)*(a+a*cos
(d*x+c))^(1/3)*hypergeom([1/6, 1/2],[3/2],1/2-1/2*cos(d*x+c))*sin(d*x+c)*2^(5/6)/a/d/(1+cos(d*x+c))^(5/6)

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Rubi [A]  time = 0.18, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3024, 2750, 2652, 2651} \[ -\frac {(4 A+7 C) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x))\right )}{2 \sqrt [6]{2} a d (\cos (c+d x)+1)^{5/6}}+\frac {3 (A+C) \sin (c+d x)}{d (a \cos (c+d x)+a)^{2/3}}+\frac {3 C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(A + C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^(2/3)) + (3*C*(a + a*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(4*a*d
) - ((4*A + 7*C)*(a + a*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/6, 1/2, 3/2, (1 - Cos[c + d*x])/2]*Sin[c + d*x
])/(2*2^(1/6)*a*d*(1 + Cos[c + d*x])^(5/6))

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 2750

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

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+a \cos (c+d x))^{2/3}} \, dx &=\frac {3 C \sqrt [3]{a+a \cos (c+d x)} \sin (c+d x)}{4 a d}+\frac {3 \int \frac {\frac {1}{3} a (4 A+C)-a C \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx}{4 a}\\ &=\frac {3 (A+C) \sin (c+d x)}{d (a+a \cos (c+d x))^{2/3}}+\frac {3 C \sqrt [3]{a+a \cos (c+d x)} \sin (c+d x)}{4 a d}-\frac {(4 A+7 C) \int \sqrt [3]{a+a \cos (c+d x)} \, dx}{4 a}\\ &=\frac {3 (A+C) \sin (c+d x)}{d (a+a \cos (c+d x))^{2/3}}+\frac {3 C \sqrt [3]{a+a \cos (c+d x)} \sin (c+d x)}{4 a d}-\frac {\left ((4 A+7 C) \sqrt [3]{a+a \cos (c+d x)}\right ) \int \sqrt [3]{1+\cos (c+d x)} \, dx}{4 a \sqrt [3]{1+\cos (c+d x)}}\\ &=\frac {3 (A+C) \sin (c+d x)}{d (a+a \cos (c+d x))^{2/3}}+\frac {3 C \sqrt [3]{a+a \cos (c+d x)} \sin (c+d x)}{4 a d}-\frac {(4 A+7 C) \sqrt [3]{a+a \cos (c+d x)} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{2 \sqrt [6]{2} a d (1+\cos (c+d x))^{5/6}}\\ \end {align*}

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Mathematica [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C \cos \left (d x + c\right )^{2} + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int \frac {A +C \left (\cos ^{2}\left (d x +c \right )\right )}{\left (a +a \cos \left (d x +c \right )\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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