3.3.81 \(\int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx\) [281]

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

[Out]

-3*AppellF1(1/3,1,1/2,4/3,-sec(d*x+c),sec(d*x+c))*(e*sec(d*x+c))^(1/3)*tan(d*x+c)/d/(1-sec(d*x+c))^(1/2)/(a+a*
sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 76, 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 = {3913, 3912, 129, 440} \begin {gather*} -\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}} \end {gather*}

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 129

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 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 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 {\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)) \text {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)) \text {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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(749\) vs. \(2(76)=152\).
time = 7.79, size = 749, normalized size = 9.86 \begin {gather*} \frac {720 e 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 ) \cos \left (\frac {1}{2} (c+d x)\right ) (1+\cos (c+d x))^2 \sin \left (\frac {1}{2} (c+d x)\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 )-\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 ) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{d (e \sec (c+d x))^{2/3} \sqrt {a (1+\sec (c+d x))} \left (4320 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 \cos ^6\left (\frac {1}{2} (c+d x)\right ) (-1+4 \cos (c+d x))+160 \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 \cos (c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )+12 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 ) \sin ^2\left (\frac {1}{2} (c+d x)\right ) \left (20 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 ) (7+14 \cos (c+d x)+5 \cos (2 (c+d x))-2 \cos (3 (c+d x)))+5 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 ) (7+14 \cos (c+d x)+5 \cos (2 (c+d x))-2 \cos (3 (c+d x)))-24 \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 ) \cos (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )} \end {gather*}

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 = 0.09, size = 0, normalized size = 0.00 \[\int \frac {\left (e \sec \left (d x +c \right )\right )^{\frac {1}{3}}}{\sqrt {a +a \sec \left (d x +c \right )}}\, dx\]

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 = 0.00, 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((e*sec(d*x+c))^(1/3)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

e^(1/3)*integrate(sec(d*x + c)^(1/3)/sqrt(a*sec(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((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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [3]{e \sec {\left (c + d x \right )}}}{\sqrt {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((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 = 0.00, 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((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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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