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

Optimal. Leaf size=277 \[ \frac{\sqrt{2} \left (3 a^2 C+b^2 (7 A+4 C)\right ) \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{7 b^2 d \sqrt{\cos (c+d x)+1} \sqrt [3]{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{3 \sqrt{2} a C (a+b) \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{4}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{7 b^2 d \sqrt{\cos (c+d x)+1} \sqrt [3]{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{3 C \sin (c+d x) (a+b \cos (c+d x))^{4/3}}{7 b d} \]

[Out]

(3*C*(a + b*Cos[c + d*x])^(4/3)*Sin[c + d*x])/(7*b*d) - (3*Sqrt[2]*a*(a + b)*C*AppellF1[1/2, 1/2, -4/3, 3/2, (
1 - Cos[c + d*x])/2, (b*(1 - Cos[c + d*x]))/(a + b)]*(a + b*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(7*b^2*d*Sqrt[1
+ Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(1/3)) + (Sqrt[2]*(3*a^2*C + b^2*(7*A + 4*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])^(1/3)*Sin[c + d*x])/(7
*b^2*d*Sqrt[1 + Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(1/3))

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Rubi [A]  time = 0.311047, antiderivative size = 277, 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} \[ \frac{\sqrt{2} \left (3 a^2 C+b^2 (7 A+4 C)\right ) \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{7 b^2 d \sqrt{\cos (c+d x)+1} \sqrt [3]{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{3 \sqrt{2} a C (a+b) \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{4}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{7 b^2 d \sqrt{\cos (c+d x)+1} \sqrt [3]{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{3 C \sin (c+d x) (a+b \cos (c+d x))^{4/3}}{7 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*(a + b*Cos[c + d*x])^(4/3)*Sin[c + d*x])/(7*b*d) - (3*Sqrt[2]*a*(a + b)*C*AppellF1[1/2, 1/2, -4/3, 3/2, (
1 - Cos[c + d*x])/2, (b*(1 - Cos[c + d*x]))/(a + b)]*(a + b*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(7*b^2*d*Sqrt[1
+ Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(1/3)) + (Sqrt[2]*(3*a^2*C + b^2*(7*A + 4*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])^(1/3)*Sin[c + d*x])/(7
*b^2*d*Sqrt[1 + Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(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 \sqrt [3]{a+b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac{3 C (a+b \cos (c+d x))^{4/3} \sin (c+d x)}{7 b d}+\frac{3 \int \sqrt [3]{a+b \cos (c+d x)} \left (\frac{1}{3} b (7 A+4 C)-a C \cos (c+d x)\right ) \, dx}{7 b}\\ &=\frac{3 C (a+b \cos (c+d x))^{4/3} \sin (c+d x)}{7 b d}-\frac{(3 a C) \int (a+b \cos (c+d x))^{4/3} \, dx}{7 b^2}+\frac{1}{7} \left (7 A+\left (4+\frac{3 a^2}{b^2}\right ) C\right ) \int \sqrt [3]{a+b \cos (c+d x)} \, dx\\ &=\frac{3 C (a+b \cos (c+d x))^{4/3} \sin (c+d x)}{7 b d}+\frac{(3 a C \sin (c+d x)) \operatorname{Subst}\left (\int \frac{(a+b x)^{4/3}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\cos (c+d x)\right )}{7 b^2 d \sqrt{1-\cos (c+d x)} \sqrt{1+\cos (c+d x)}}+\frac{\left (\left (-7 A-\left (4+\frac{3 a^2}{b^2}\right ) C\right ) \sin (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt [3]{a+b x}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\cos (c+d x)\right )}{7 d \sqrt{1-\cos (c+d x)} \sqrt{1+\cos (c+d x)}}\\ &=\frac{3 C (a+b \cos (c+d x))^{4/3} \sin (c+d x)}{7 b d}-\frac{\left (3 a (-a-b) C \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^{4/3}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\cos (c+d x)\right )}{7 b^2 d \sqrt{1-\cos (c+d x)} \sqrt{1+\cos (c+d x)} \sqrt [3]{-\frac{a+b \cos (c+d x)}{-a-b}}}+\frac{\left (\left (-7 A-\left (4+\frac{3 a^2}{b^2}\right ) C\right ) \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt [3]{-\frac{a}{-a-b}-\frac{b x}{-a-b}}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\cos (c+d x)\right )}{7 d \sqrt{1-\cos (c+d x)} \sqrt{1+\cos (c+d x)} \sqrt [3]{-\frac{a+b \cos (c+d x)}{-a-b}}}\\ &=\frac{3 C (a+b \cos (c+d x))^{4/3} \sin (c+d x)}{7 b d}-\frac{3 \sqrt{2} a (a+b) C F_1\left (\frac{1}{2};\frac{1}{2},-\frac{4}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{7 b^2 d \sqrt{1+\cos (c+d x)} \sqrt [3]{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\sqrt{2} \left (7 A+\left (4+\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]{a+b \cos (c+d x)} \sin (c+d x)}{7 d \sqrt{1+\cos (c+d x)} \sqrt [3]{\frac{a+b \cos (c+d x)}{a+b}}}\\ \end{align*}

Mathematica [A]  time = 2.51439, size = 276, normalized size = 1. \[ -\frac{3 \csc (c+d x) \sqrt [3]{a+b \cos (c+d x)} \left (\left (-3 a^2 C+28 A b^2+16 b^2 C\right ) \sqrt{-\frac{b (\cos (c+d x)-1)}{a+b}} \sqrt{\frac{b (\cos (c+d x)+1)}{b-a}} (a+b \cos (c+d x)) F_1\left (\frac{4}{3};\frac{1}{2},\frac{1}{2};\frac{7}{3};\frac{a+b \cos (c+d x)}{a-b},\frac{a+b \cos (c+d x)}{a+b}\right )+12 a C \left (a^2-b^2\right ) \sqrt{-\frac{b (\cos (c+d x)-1)}{a+b}} \sqrt{-\frac{b (\cos (c+d x)+1)}{a-b}} F_1\left (\frac{1}{3};\frac{1}{2},\frac{1}{2};\frac{4}{3};\frac{a+b \cos (c+d x)}{a-b},\frac{a+b \cos (c+d x)}{a+b}\right )-4 b^2 C \sin ^2(c+d x) (a+4 b \cos (c+d x))\right )}{112 b^3 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3*(a + b*Cos[c + d*x])^(1/3)*Csc[c + d*x]*(12*a*(a^2 - b^2)*C*AppellF1[1/3, 1/2, 1/2, 4/3, (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))] + (28*A*b^2 - 3*a^2*C + 16*b^2*C)*AppellF1[4/3, 1/2, 1/2, 7/3, (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]) - 4*b^2*C*(a + 4*b*Cos[c + d*x])*Sin[c + d*x]^2))/(112*b^3*d)

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Maple [F]  time = 0.27, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{a+b\cos \left ( dx+c \right ) } \left ( A+C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + A\right )}{\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+b*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),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 ({\left (C \cos \left (d x + c\right )^{2} + A\right )}{\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+b*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),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+b*cos(d*x+c))**(1/3)*(A+C*cos(d*x+c)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + A\right )}{\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+b*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

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