3.2.58 \(\int \frac {\cos (c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx\) [158]

Optimal. Leaf size=75 \[ -\frac {3 \sqrt {2} F_1\left (\frac {1}{6};\frac {1}{2},2;\frac {7}{6};\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}} \]

[Out]

-3*AppellF1(1/6,2,1/2,7/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*2^(1/2)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/3)/(1-sec(
d*x+c))^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3913, 3912, 141} \begin {gather*} -\frac {3 \sqrt {2} \tan (c+d x) F_1\left (\frac {1}{6};\frac {1}{2},2;\frac {7}{6};\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{a \sec (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[2]*AppellF1[1/6, 1/2, 2, 7/6, (1 + Sec[c + d*x])/2, 1 + Sec[c + d*x]]*Tan[c + d*x])/(d*Sqrt[1 - Sec[c
+ d*x]]*(a + a*Sec[c + d*x])^(1/3))

Rule 141

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(b*e - a*f
)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(
b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 3912

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[a^2*d
*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])), Subst[Int[(d*x)^(n - 1)*((a + b*x)^(m -
1/2)/Sqrt[a - b*x]), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !In
tegerQ[m] && GtQ[a, 0]

Rule 3913

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[a^Int
Part[m]*((a + b*Csc[e + f*x])^FracPart[m]/(1 + (b/a)*Csc[e + f*x])^FracPart[m]), Int[(1 + (b/a)*Csc[e + f*x])^
m*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx &=\frac {\sqrt [3]{1+\sec (c+d x)} \int \frac {\cos (c+d x)}{\sqrt [3]{1+\sec (c+d x)}} \, dx}{\sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac {\tan (c+d x) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} x^2 (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac {3 \sqrt {2} F_1\left (\frac {1}{6};\frac {1}{2},2;\frac {7}{6};\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(240\) vs. \(2(75)=150\).
time = 2.10, size = 240, normalized size = 3.20 \begin {gather*} \frac {(a (1+\sec (c+d x)))^{2/3} \left (\frac {20 F_1\left (\frac {3}{2};\frac {2}{3},1;\frac {5}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \sin ^3\left (\frac {1}{2} (c+d x)\right )}{6 \left (3 F_1\left (\frac {5}{2};\frac {2}{3},2;\frac {7}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 F_1\left (\frac {5}{2};\frac {5}{3},1;\frac {7}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+45 F_1\left (\frac {3}{2};\frac {2}{3},1;\frac {5}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))}+\sin (c+d x)-\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a*(1 + Sec[c + d*x]))^(2/3)*((20*AppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c
+ d*x)/2]*Sin[(c + d*x)/2]^3)/(6*(3*AppellF1[5/2, 2/3, 2, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*Ap
pellF1[5/2, 5/3, 1, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 + Cos[c + d*x]) + 45*AppellF1[3/2, 2/3,
1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x])) + Sin[c + d*x] - Tan[(c + d*x)/2]))/(a*d
)

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\cos \left (d x +c \right )}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(d*x+c)/(a+a*sec(d*x+c))^(1/3),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)/(a+a*sec(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)/(a*sec(d*x + c) + a)^(1/3), 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)/(a+a*sec(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos {\left (c + d x \right )}}{\sqrt [3]{a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(a+a*sec(d*x+c))^(1/3),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)/(a*sec(d*x + c) + a)^(1/3), 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 )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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