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

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

[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))

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Rubi [A]  time = 0.0943261, antiderivative size = 75, normalized size of antiderivative = 1., 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 = {3828, 3827, 136} \[ -\frac{3 \sqrt{2} \tan (c+d x) F_1\left (\frac{1}{6};\frac{1}{2},2;\frac{7}{6};\frac{1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{d \sqrt{1-\sec (c+d x)} \sqrt [3]{a \sec (c+d x)+a}} \]

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 3828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(1 + (b*Csc[e + f*x])/a)^FracPart[m], Int[(1 + (b*Csc[e + f*x])/a)^
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]

Rule 3827

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 136

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)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), 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])

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) \operatorname{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*}

Mathematica [B]  time = 1.97296, size = 240, normalized size = 3.2 \[ \frac{(a (\sec (c+d x)+1))^{2/3} \left (\frac{20 \sin ^3\left (\frac{1}{2} (c+d x)\right ) \cos \left (\frac{1}{2} (c+d x)\right ) F_1\left (\frac{3}{2};\frac{2}{3},1;\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{6 (\cos (c+d x)-1) \left (3 F_1\left (\frac{5}{2};\frac{2}{3},2;\frac{7}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-2 F_1\left (\frac{5}{2};\frac{5}{3},1;\frac{7}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )+45 (\cos (c+d x)+1) F_1\left (\frac{3}{2};\frac{2}{3},1;\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}+\sin (c+d x)-\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a d} \]

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 = 0.109, size = 0, normalized size = 0. \begin{align*} \int{\cos \left ( dx+c \right ){\frac{1}{\sqrt [3]{a+a\sec \left ( dx+c \right ) }}}}\, dx \end{align*}

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

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )}{{\left (a \sec \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(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)