3.7.76 \(\int \frac {\cos ^{\frac {2}{3}}(c+d x)}{a+b \cos (c+d x)} \, dx\) [676]

Optimal. Leaf size=176 \[ -\frac {b F_1\left (\frac {1}{2};-\frac {1}{3},1;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^{\frac {2}{3}}(c+d x) \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt [3]{\cos ^2(c+d x)}}+\frac {a F_1\left (\frac {1}{2};\frac {1}{6},1;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sqrt [6]{\cos ^2(c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt [3]{\cos (c+d x)}} \]

[Out]

-b*AppellF1(1/2,-1/3,1,3/2,sin(d*x+c)^2,-b^2*sin(d*x+c)^2/(a^2-b^2))*cos(d*x+c)^(2/3)*sin(d*x+c)/(a^2-b^2)/d/(
cos(d*x+c)^2)^(1/3)+a*AppellF1(1/2,1/6,1,3/2,sin(d*x+c)^2,-b^2*sin(d*x+c)^2/(a^2-b^2))*(cos(d*x+c)^2)^(1/6)*si
n(d*x+c)/(a^2-b^2)/d/cos(d*x+c)^(1/3)

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Rubi [A]
time = 0.14, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2902, 3268, 440} \begin {gather*} \frac {a \sin (c+d x) \sqrt [6]{\cos ^2(c+d x)} F_1\left (\frac {1}{2};\frac {1}{6},1;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right ) \sqrt [3]{\cos (c+d x)}}-\frac {b \sin (c+d x) \cos ^{\frac {2}{3}}(c+d x) F_1\left (\frac {1}{2};-\frac {1}{3},1;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right ) \sqrt [3]{\cos ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*AppellF1[1/2, -1/3, 1, 3/2, Sin[c + d*x]^2, -((b^2*Sin[c + d*x]^2)/(a^2 - b^2))]*Cos[c + d*x]^(2/3)*Sin[c
+ d*x])/((a^2 - b^2)*d*(Cos[c + d*x]^2)^(1/3))) + (a*AppellF1[1/2, 1/6, 1, 3/2, Sin[c + d*x]^2, -((b^2*Sin[c
+ d*x]^2)/(a^2 - b^2))]*(Cos[c + d*x]^2)^(1/6)*Sin[c + d*x])/((a^2 - b^2)*d*Cos[c + d*x]^(1/3))

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 2902

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[a, Int[(d*
Sin[e + f*x])^n/(a^2 - b^2*Sin[e + f*x]^2), x], x] - Dist[b/d, Int[(d*Sin[e + f*x])^(n + 1)/(a^2 - b^2*Sin[e +
f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0]

Rule 3268

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff
= FreeFactors[Cos[e + f*x], x]}, Dist[(-ff)*d^(2*IntPart[(m - 1)/2] + 1)*((d*Sin[e + f*x])^(2*FracPart[(m - 1
)/2])/(f*(Sin[e + f*x]^2)^FracPart[(m - 1)/2])), Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p,
x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {2}{3}}(c+d x)}{a+b \cos (c+d x)} \, dx &=a \int \frac {\cos ^{\frac {2}{3}}(c+d x)}{a^2-b^2 \cos ^2(c+d x)} \, dx-b \int \frac {\cos ^{\frac {5}{3}}(c+d x)}{a^2-b^2 \cos ^2(c+d x)} \, dx\\ &=-\frac {\left (b \cos ^{\frac {2}{3}}(c+d x)\right ) \text {Subst}\left (\int \frac {\sqrt [3]{1-x^2}}{a^2-b^2+b^2 x^2} \, dx,x,\sin (c+d x)\right )}{d \sqrt [3]{\cos ^2(c+d x)}}+\frac {\left (a \sqrt [6]{\cos ^2(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [6]{1-x^2} \left (a^2-b^2+b^2 x^2\right )} \, dx,x,\sin (c+d x)\right )}{d \sqrt [3]{\cos (c+d x)}}\\ &=-\frac {b F_1\left (\frac {1}{2};-\frac {1}{3},1;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^{\frac {2}{3}}(c+d x) \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt [3]{\cos ^2(c+d x)}}+\frac {a F_1\left (\frac {1}{2};\frac {1}{6},1;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sqrt [6]{\cos ^2(c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt [3]{\cos (c+d x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(4614\) vs. \(2(176)=352\).
time = 41.84, size = 4614, normalized size = 26.22 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(9*(a^2 - b^2)*Sin[c + d*x]*((a*AppellF1[1/2, 1/3, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2)
)]*Sqrt[Sec[c + d*x]^2])/(9*(a^2 - b^2)*AppellF1[1/2, 1/3, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^
2 - b^2))] - 2*(3*a^2*AppellF1[3/2, 1/3, 2, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] + (a^2
- b^2)*AppellF1[3/2, 4/3, 1, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))])*Tan[c + d*x]^2) + (b*
AppellF1[1/2, 5/6, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))])/(-9*(a^2 - b^2)*AppellF1[1/2
, 5/6, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] + (6*a^2*AppellF1[3/2, 5/6, 2, 5/2, -Tan[
c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] + 5*(a^2 - b^2)*AppellF1[3/2, 11/6, 1, 5/2, -Tan[c + d*x]^2,
-((a^2*Tan[c + d*x]^2)/(a^2 - b^2))])*Tan[c + d*x]^2)))/(d*Cos[c + d*x]^(1/3)*(a + b*Cos[c + d*x])*(Sec[c + d*
x]^2)^(5/6)*(-b^2 + a^2*Sec[c + d*x]^2)*((9*(a^2 - b^2)*(Sec[c + d*x]^2)^(1/6)*((a*AppellF1[1/2, 1/3, 1, 3/2,
-Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))]*Sqrt[Sec[c + d*x]^2])/(9*(a^2 - b^2)*AppellF1[1/2, 1/3,
1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] - 2*(3*a^2*AppellF1[3/2, 1/3, 2, 5/2, -Tan[c + d
*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] + (a^2 - b^2)*AppellF1[3/2, 4/3, 1, 5/2, -Tan[c + d*x]^2, -((a^2*T
an[c + d*x]^2)/(a^2 - b^2))])*Tan[c + d*x]^2) + (b*AppellF1[1/2, 5/6, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c +
d*x]^2)/(a^2 - b^2))])/(-9*(a^2 - b^2)*AppellF1[1/2, 5/6, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2
- b^2))] + (6*a^2*AppellF1[3/2, 5/6, 2, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] + 5*(a^2 -
b^2)*AppellF1[3/2, 11/6, 1, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))])*Tan[c + d*x]^2)))/(-b
^2 + a^2*Sec[c + d*x]^2) - (18*a^2*(a^2 - b^2)*(Sec[c + d*x]^2)^(1/6)*Tan[c + d*x]^2*((a*AppellF1[1/2, 1/3, 1,
3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))]*Sqrt[Sec[c + d*x]^2])/(9*(a^2 - b^2)*AppellF1[1/2,
1/3, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] - 2*(3*a^2*AppellF1[3/2, 1/3, 2, 5/2, -Tan
[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] + (a^2 - b^2)*AppellF1[3/2, 4/3, 1, 5/2, -Tan[c + d*x]^2, -(
(a^2*Tan[c + d*x]^2)/(a^2 - b^2))])*Tan[c + d*x]^2) + (b*AppellF1[1/2, 5/6, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Ta
n[c + d*x]^2)/(a^2 - b^2))])/(-9*(a^2 - b^2)*AppellF1[1/2, 5/6, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2
)/(a^2 - b^2))] + (6*a^2*AppellF1[3/2, 5/6, 2, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] + 5*
(a^2 - b^2)*AppellF1[3/2, 11/6, 1, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))])*Tan[c + d*x]^2)
))/(-b^2 + a^2*Sec[c + d*x]^2)^2 - (15*(a^2 - b^2)*Tan[c + d*x]^2*((a*AppellF1[1/2, 1/3, 1, 3/2, -Tan[c + d*x]
^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))]*Sqrt[Sec[c + d*x]^2])/(9*(a^2 - b^2)*AppellF1[1/2, 1/3, 1, 3/2, -Tan[
c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] - 2*(3*a^2*AppellF1[3/2, 1/3, 2, 5/2, -Tan[c + d*x]^2, -((a^2
*Tan[c + d*x]^2)/(a^2 - b^2))] + (a^2 - b^2)*AppellF1[3/2, 4/3, 1, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2
)/(a^2 - b^2))])*Tan[c + d*x]^2) + (b*AppellF1[1/2, 5/6, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2
- b^2))])/(-9*(a^2 - b^2)*AppellF1[1/2, 5/6, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] + (
6*a^2*AppellF1[3/2, 5/6, 2, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] + 5*(a^2 - b^2)*AppellF
1[3/2, 11/6, 1, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))])*Tan[c + d*x]^2)))/((Sec[c + d*x]^2
)^(5/6)*(-b^2 + a^2*Sec[c + d*x]^2)) + (9*(a^2 - b^2)*Tan[c + d*x]*((a*AppellF1[1/2, 1/3, 1, 3/2, -Tan[c + d*x
]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))]*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])/(9*(a^2 - b^2)*AppellF1[1/2, 1/3,
1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] - 2*(3*a^2*AppellF1[3/2, 1/3, 2, 5/2, -Tan[c +
d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] + (a^2 - b^2)*AppellF1[3/2, 4/3, 1, 5/2, -Tan[c + d*x]^2, -((a^2*
Tan[c + d*x]^2)/(a^2 - b^2))])*Tan[c + d*x]^2) + (a*Sqrt[Sec[c + d*x]^2]*((-2*a^2*AppellF1[3/2, 1/3, 2, 5/2, -
Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))]*Sec[c + d*x]^2*Tan[c + d*x])/(3*(a^2 - b^2)) - (2*AppellF
1[3/2, 4/3, 1, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))]*Sec[c + d*x]^2*Tan[c + d*x])/9))/(9*
(a^2 - b^2)*AppellF1[1/2, 1/3, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] - 2*(3*a^2*Appell
F1[3/2, 1/3, 2, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] + (a^2 - b^2)*AppellF1[3/2, 4/3, 1,
5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))])*Tan[c + d*x]^2) + (b*((-2*a^2*AppellF1[3/2, 5/6,
2, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))]*Sec[c + d*x]^2*Tan[c + d*x])/(3*(a^2 - b^2)) - (
5*AppellF1[3/2, 11/6, 1, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))]*Sec[c + d*x]^2*Tan[c + d*x
])/9))/(-9*(a^2 - b^2)*AppellF1[1/2, 5/6, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] + (6*a
^2*AppellF1[3/2, 5/6, 2, 5/2, -Tan[c + d*x]^2, ...

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\cos ^{\frac {2}{3}}\left (d x +c \right )}{a +b \cos \left (d x +c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(2/3)/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(2/3)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(2/3)/(a+b*cos(d*x+c)),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(2/3)/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^{2/3}}{a+b\,\cos \left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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