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

Optimal. Leaf size=135 \[ \frac {(10 A+7 C) \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x))\right )}{5\ 2^{5/6} d \sqrt [6]{\cos (c+d x)+1} \sqrt [3]{a \cos (c+d x)+a}}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 a d}-\frac {9 C \sin (c+d x)}{10 d \sqrt [3]{a \cos (c+d x)+a}} \]

[Out]

-9/10*C*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/3)+3/5*C*(a+a*cos(d*x+c))^(2/3)*sin(d*x+c)/a/d+1/10*(10*A+7*C)*hyperg
eom([1/2, 5/6],[3/2],1/2-1/2*cos(d*x+c))*sin(d*x+c)*2^(1/6)/d/(1+cos(d*x+c))^(1/6)/(a+a*cos(d*x+c))^(1/3)

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Rubi [A]  time = 0.16, antiderivative size = 135, 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, 2751, 2652, 2651} \[ \frac {(10 A+7 C) \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x))\right )}{5\ 2^{5/6} d \sqrt [6]{\cos (c+d x)+1} \sqrt [3]{a \cos (c+d x)+a}}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 a d}-\frac {9 C \sin (c+d x)}{10 d \sqrt [3]{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*C*Sin[c + d*x])/(10*d*(a + a*Cos[c + d*x])^(1/3)) + (3*C*(a + a*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(5*a*d)
+ ((10*A + 7*C)*Hypergeometric2F1[1/2, 5/6, 3/2, (1 - Cos[c + d*x])/2]*Sin[c + d*x])/(5*2^(5/6)*d*(1 + Cos[c +
d*x])^(1/6)*(a + a*Cos[c + d*x])^(1/3))

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 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 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)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx &=\frac {3 C (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{5 a d}+\frac {3 \int \frac {\frac {1}{3} a (5 A+2 C)-a C \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx}{5 a}\\ &=-\frac {9 C \sin (c+d x)}{10 d \sqrt [3]{a+a \cos (c+d x)}}+\frac {3 C (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{5 a d}+\frac {1}{10} (10 A+7 C) \int \frac {1}{\sqrt [3]{a+a \cos (c+d x)}} \, dx\\ &=-\frac {9 C \sin (c+d x)}{10 d \sqrt [3]{a+a \cos (c+d x)}}+\frac {3 C (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{5 a d}+\frac {\left ((10 A+7 C) \sqrt [3]{1+\cos (c+d x)}\right ) \int \frac {1}{\sqrt [3]{1+\cos (c+d x)}} \, dx}{10 \sqrt [3]{a+a \cos (c+d x)}}\\ &=-\frac {9 C \sin (c+d x)}{10 d \sqrt [3]{a+a \cos (c+d x)}}+\frac {3 C (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{5 a d}+\frac {(10 A+7 C) \, _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)}{5\ 2^{5/6} d \sqrt [6]{1+\cos (c+d x)} \sqrt [3]{a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.47, size = 144, normalized size = 1.07 \[ -\frac {2 (10 A+7 C) \sin \left (d x-2 \tan ^{-1}\left (\cot \left (\frac {c}{2}\right )\right )\right ) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\cos ^2\left (\frac {d x}{2}-\tan ^{-1}\left (\cot \left (\frac {c}{2}\right )\right )\right )\right )+3\ 2^{5/6} C (\sin (c+d x)-\sin (2 (c+d x))) \sqrt [6]{1-\cos \left (d x-2 \tan ^{-1}\left (\cot \left (\frac {c}{2}\right )\right )\right )}}{20 d \sqrt [3]{a (\cos (c+d x)+1)} \sqrt [6]{\sin ^2\left (\frac {d x}{2}-\tan ^{-1}\left (\cot \left (\frac {c}{2}\right )\right )\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/20*(3*2^(5/6)*C*(1 - Cos[d*x - 2*ArcTan[Cot[c/2]]])^(1/6)*(Sin[c + d*x] - Sin[2*(c + d*x)]) + 2*(10*A + 7*C
)*Hypergeometric2F1[1/2, 5/6, 3/2, Cos[(d*x)/2 - ArcTan[Cot[c/2]]]^2]*Sin[d*x - 2*ArcTan[Cot[c/2]]])/(d*(a*(1
+ Cos[c + d*x]))^(1/3)*(Sin[(d*x)/2 - ArcTan[Cot[c/2]]]^2)^(1/6))

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C \cos \left (d x + c\right )^{2} + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}, x\right ) \]

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))^(1/3),x, algorithm="fricas")

[Out]

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

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

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))^(1/3),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \frac {A +C \left (\cos ^{2}\left (d x +c \right )\right )}{\left (a +a \cos \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx \]

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))^(1/3),x)

[Out]

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

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

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))^(1/3),x, algorithm="maxima")

[Out]

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

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

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))^(1/3),x)

[Out]

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

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

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))**(1/3),x)

[Out]

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

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