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

Optimal. Leaf size=101 \[ \frac{(2 A-B) \sin (c+d x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x))\right )}{2^{5/6} d \sqrt [6]{\cos (c+d x)+1} \sqrt [3]{a \cos (c+d x)+a}}+\frac{3 B \sin (c+d x)}{2 d \sqrt [3]{a \cos (c+d x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.0771416, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2751, 2652, 2651} \[ \frac{(2 A-B) \sin (c+d x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x))\right )}{2^{5/6} d \sqrt [6]{\cos (c+d x)+1} \sqrt [3]{a \cos (c+d x)+a}}+\frac{3 B \sin (c+d x)}{2 d \sqrt [3]{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.2581, size = 133, normalized size = 1.32 \[ \frac{3\ 2^{5/6} B \sin (c+d x) \sqrt [6]{1-\cos \left (d x-2 \tan ^{-1}\left (\cot \left (\frac{c}{2}\right )\right )\right )}-2 (2 A-B) \sin \left (d x-2 \tan ^{-1}\left (\cot \left (\frac{c}{2}\right )\right )\right ) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\cos ^2\left (\frac{d x}{2}-\tan ^{-1}\left (\cot \left (\frac{c}{2}\right )\right )\right )\right )}{4 d \sqrt [3]{a (\cos (c+d x)+1)} \sqrt [6]{\sin ^2\left (\frac{d x}{2}-\tan ^{-1}\left (\cot \left (\frac{c}{2}\right )\right )\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.328, size = 0, normalized size = 0. \begin{align*} \int{(A+B\cos \left ( dx+c \right ) ){\frac{1}{\sqrt [3]{a+\cos \left ( dx+c \right ) a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \cos \left (d x + c\right ) + A}{{\left (a \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))/(a+a*cos(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \cos{\left (c + d x \right )}}{\sqrt [3]{a \left (\cos{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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