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[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))

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Rubi [A]  time = 0.176594, antiderivative size = 138, normalized size of antiderivative = 1., 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 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]

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 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 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]

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*}

Mathematica [F]  time = 0.133988, size = 0, normalized size = 0. \[ \int \frac{A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx \]

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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Maple [F]  time = 0.286, size = 0, normalized size = 0. \begin{align*} \int{(A+C \left ( \cos \left ( dx+c \right ) \right ) ^{2}) \left ( a+\cos \left ( dx+c \right ) a \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)