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

Optimal. Leaf size=274 \[ \frac{\sqrt{2} \left (3 a^2 C+b^2 (5 A+2 C)\right ) \sin (c+d x) \sqrt [3]{\frac{a+b \cos (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{5 b^2 d \sqrt{\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}-\frac{3 \sqrt{2} a C \sin (c+d x) (a+b \cos (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{5 b^2 d \sqrt{\cos (c+d x)+1} \left (\frac{a+b \cos (c+d x)}{a+b}\right )^{2/3}}+\frac{3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d} \]

[Out]

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

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Rubi [A]  time = 0.309737, antiderivative size = 274, normalized size of antiderivative = 1., 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} \[ \frac{\sqrt{2} \left (3 a^2 C+b^2 (5 A+2 C)\right ) \sin (c+d x) \sqrt [3]{\frac{a+b \cos (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{5 b^2 d \sqrt{\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}-\frac{3 \sqrt{2} a C \sin (c+d x) (a+b \cos (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{5 b^2 d \sqrt{\cos (c+d x)+1} \left (\frac{a+b \cos (c+d x)}{a+b}\right )^{2/3}}+\frac{3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d} \]

Antiderivative was successfully verified.

[In]

Int[(A + 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) - (3*Sqrt[2]*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]*(3*a^2*C + b^2*(5*A + 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 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 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 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 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 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])

Rubi steps

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

Mathematica [A]  time = 1.69111, size = 256, normalized size = 0.93 \[ -\frac{3 \csc (c+d x) (a+b \cos (c+d x))^{2/3} \left (5 \left (3 a^2 C+5 A b^2+2 b^2 C\right ) \sqrt{-\frac{b (\cos (c+d x)-1)}{a+b}} \sqrt{\frac{b (\cos (c+d x)+1)}{b-a}} F_1\left (\frac{2}{3};\frac{1}{2},\frac{1}{2};\frac{5}{3};\frac{a+b \cos (c+d x)}{a-b},\frac{a+b \cos (c+d x)}{a+b}\right )-6 a C \sqrt{-\frac{b (\cos (c+d x)-1)}{a+b}} \sqrt{\frac{b (\cos (c+d x)+1)}{b-a}} (a+b \cos (c+d x)) F_1\left (\frac{5}{3};\frac{1}{2},\frac{1}{2};\frac{8}{3};\frac{a+b \cos (c+d x)}{a-b},\frac{a+b \cos (c+d x)}{a+b}\right )-10 b^2 C \sin ^2(c+d x)\right )}{50 b^3 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + 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^2 + 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)] - 6*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 = 0.286, size = 0, normalized size = 0. \begin{align*} \int{(A+C \left ( \cos \left ( dx+c \right ) \right ) ^{2}){\frac{1}{\sqrt [3]{a+b\cos \left ( dx+c \right ) }}}}\, dx \end{align*}

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))^(1/3),x)

[Out]

int((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),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 (b \cos \left (d x + c\right ) + a\right )}^{\frac{1}{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+b*cos(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)/(b*cos(d*x + c) + a)^(1/3), 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 (b \cos \left (d x + c\right ) + a\right )}^{\frac{1}{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+b*cos(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + A)/(b*cos(d*x + c) + a)^(1/3), 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+b*cos(d*x+c))**(1/3),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 (b \cos \left (d x + c\right ) + a\right )}^{\frac{1}{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+b*cos(d*x+c))^(1/3),x, algorithm="giac")

[Out]

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