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

Optimal. Leaf size=76 \[ -\frac{3 \tan (c+d x) F_1\left (\frac{1}{3};\frac{1}{2},1;\frac{4}{3};\sec (c+d x),-\sec (c+d x)\right ) \sqrt [3]{e \sec (c+d x)}}{d \sqrt{1-\sec (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

[Out]

(-3*AppellF1[1/3, 1/2, 1, 4/3, Sec[c + d*x], -Sec[c + d*x]]*(e*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(d*Sqrt[1 - S
ec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.142169, antiderivative size = 76, normalized size of antiderivative = 1., 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 = {3828, 3827, 130, 429} \[ -\frac{3 \tan (c+d x) F_1\left (\frac{1}{3};\frac{1}{2},1;\frac{4}{3};\sec (c+d x),-\sec (c+d x)\right ) \sqrt [3]{e \sec (c+d x)}}{d \sqrt{1-\sec (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*AppellF1[1/3, 1/2, 1, 4/3, Sec[c + d*x], -Sec[c + d*x]]*(e*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(d*Sqrt[1 - S
ec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]])

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 130

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + (b*x^k)/e)^m*(c + (d*x^k)/e)^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 429

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])

Rubi steps

\begin{align*} \int \frac{\sqrt [3]{e \sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx &=\frac{\sqrt{1+\sec (c+d x)} \int \frac{\sqrt [3]{e \sec (c+d x)}}{\sqrt{1+\sec (c+d x)}} \, dx}{\sqrt{a+a \sec (c+d x)}}\\ &=-\frac{(e \tan (c+d x)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} (e x)^{2/3} (1+x)} \, dx,x,\sec (c+d x)\right )}{d \sqrt{1-\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}\\ &=-\frac{(3 \tan (c+d x)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^3}{e}} \left (1+\frac{x^3}{e}\right )} \, dx,x,\sqrt [3]{e \sec (c+d x)}\right )}{d \sqrt{1-\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}\\ &=-\frac{3 F_1\left (\frac{1}{3};\frac{1}{2},1;\frac{4}{3};\sec (c+d x),-\sec (c+d x)\right ) \sqrt [3]{e \sec (c+d x)} \tan (c+d x)}{d \sqrt{1-\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 7.51445, size = 749, normalized size = 9.86 \[ \frac{720 e \sin \left (\frac{1}{2} (c+d x)\right ) \cos \left (\frac{1}{2} (c+d x)\right ) (\cos (c+d x)+1)^2 F_1\left (\frac{1}{2};-\frac{1}{6},\frac{2}{3};\frac{3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right ) \left (9 F_1\left (\frac{1}{2};-\frac{1}{6},\frac{2}{3};\frac{3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-\tan ^2\left (\frac{1}{2} (c+d x)\right ) \left (4 F_1\left (\frac{3}{2};-\frac{1}{6},\frac{5}{3};\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )+F_1\left (\frac{3}{2};\frac{5}{6},\frac{2}{3};\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{d \sqrt{a (\sec (c+d x)+1)} (e \sec (c+d x))^{2/3} \left (4320 (4 \cos (c+d x)-1) \cos ^6\left (\frac{1}{2} (c+d x)\right ) F_1\left (\frac{1}{2};-\frac{1}{6},\frac{2}{3};\frac{3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right ){}^2+160 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \cos (c+d x) \left (4 F_1\left (\frac{3}{2};-\frac{1}{6},\frac{5}{3};\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )+F_1\left (\frac{3}{2};\frac{5}{6},\frac{2}{3};\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right ){}^2+12 \sin ^2\left (\frac{1}{2} (c+d x)\right ) F_1\left (\frac{1}{2};-\frac{1}{6},\frac{2}{3};\frac{3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right ) \left (20 (14 \cos (c+d x)+5 \cos (2 (c+d x))-2 \cos (3 (c+d x))+7) F_1\left (\frac{3}{2};-\frac{1}{6},\frac{5}{3};\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )+5 (14 \cos (c+d x)+5 \cos (2 (c+d x))-2 \cos (3 (c+d x))+7) F_1\left (\frac{3}{2};\frac{5}{6},\frac{2}{3};\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-24 \sin ^2\left (\frac{1}{2} (c+d x)\right ) \cos (c+d x) \left (40 F_1\left (\frac{5}{2};-\frac{1}{6},\frac{8}{3};\frac{7}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )+8 F_1\left (\frac{5}{2};\frac{5}{6},\frac{5}{3};\frac{7}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-5 F_1\left (\frac{5}{2};\frac{11}{6},\frac{2}{3};\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 )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(720*e*AppellF1[1/2, -1/6, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]*(1 + Cos[c + d*
x])^2*Sin[(c + d*x)/2]*(9*AppellF1[1/2, -1/6, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - (4*AppellF1
[3/2, -1/6, 5/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 5/6, 2/3, 5/2, Tan[(c + d*x)/2]
^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2))/(d*(e*Sec[c + d*x])^(2/3)*Sqrt[a*(1 + Sec[c + d*x])]*(4320*Appe
llF1[1/2, -1/6, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]^2*Cos[(c + d*x)/2]^6*(-1 + 4*Cos[c + d*x])
+ 160*(4*AppellF1[3/2, -1/6, 5/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 5/6, 2/3, 5/2,
Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])^2*Cos[c + d*x]*Sin[(c + d*x)/2]^4 + 12*AppellF1[1/2, -1/6, 2/3, 3/2
, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sin[(c + d*x)/2]^2*(20*AppellF1[3/2, -1/6, 5/3, 5/2, Tan[(c + d*x)/
2]^2, -Tan[(c + d*x)/2]^2]*(7 + 14*Cos[c + d*x] + 5*Cos[2*(c + d*x)] - 2*Cos[3*(c + d*x)]) + 5*AppellF1[3/2, 5
/6, 2/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(7 + 14*Cos[c + d*x] + 5*Cos[2*(c + d*x)] - 2*Cos[3*(c
+ d*x)]) - 24*(40*AppellF1[5/2, -1/6, 8/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 8*AppellF1[5/2, 5/6
, 5/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 5*AppellF1[5/2, 11/6, 2/3, 7/2, Tan[(c + d*x)/2]^2, -Ta
n[(c + d*x)/2]^2])*Cos[c + d*x]*Sin[(c + d*x)/2]^2)))

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Maple [F]  time = 0.165, size = 0, normalized size = 0. \begin{align*} \int{\sqrt [3]{e\sec \left ( dx+c \right ) }{\frac{1}{\sqrt{a+a\sec \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(1/3)/(a+a*sec(d*x+c))^(1/2),x)

[Out]

int((e*sec(d*x+c))^(1/3)/(a+a*sec(d*x+c))^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/3)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((e*sec(d*x + c))^(1/3)/sqrt(a*sec(d*x + c) + a), 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((e*sec(d*x+c))^(1/3)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt [3]{e \sec{\left (c + d x \right )}}}{\sqrt{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((e*sec(d*x+c))**(1/3)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/3)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(1/3)/sqrt(a*sec(d*x + c) + a), x)