3.280 \(\int \frac {(e \sec (c+d x))^{2/3}}{\sqrt {a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=78 \[ -\frac {3 \tan (c+d x) F_1\left (\frac {2}{3};\frac {1}{2},1;\frac {5}{3};\sec (c+d x),-\sec (c+d x)\right ) (e \sec (c+d x))^{2/3}}{2 d \sqrt {1-\sec (c+d x)} \sqrt {a \sec (c+d x)+a}} \]

[Out]

-3/2*AppellF1(2/3,1,1/2,5/3,-sec(d*x+c),sec(d*x+c))*(e*sec(d*x+c))^(2/3)*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 = 0.16, antiderivative size = 78, 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 = {3828, 3827, 130, 510} \[ -\frac {3 \tan (c+d x) F_1\left (\frac {2}{3};\frac {1}{2},1;\frac {5}{3};\sec (c+d x),-\sec (c+d x)\right ) (e \sec (c+d x))^{2/3}}{2 d \sqrt {1-\sec (c+d x)} \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(2/3)/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(-3*AppellF1[2/3, 1/2, 1, 5/3, Sec[c + d*x], -Sec[c + d*x]]*(e*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(2*d*Sqrt[1 -
Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]])

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 510

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

Rubi steps

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

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Mathematica [B]  time = 9.25, size = 750, normalized size = 9.62 \[ \frac {90 \cos ^2(c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} F_1\left (\frac {1}{2};\frac {1}{6},\frac {1}{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 (\tan ^2\left (\frac {1}{2} (c+d x)\right ) \left (F_1\left (\frac {3}{2};\frac {7}{6},\frac {1}{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 )-2 F_1\left (\frac {3}{2};\frac {1}{6},\frac {4}{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 )+9 F_1\left (\frac {1}{2};\frac {1}{6},\frac {1}{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 )\right ) (e \sec (c+d x))^{2/3}}{a d \left (270 \cos ^2\left (\frac {1}{2} (c+d x)\right ) (2 \cos (c+d x)+1) F_1\left (\frac {1}{2};\frac {1}{6},\frac {1}{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+10 \cos (c+d x) \tan ^4\left (\frac {1}{2} (c+d x)\right ) \left (F_1\left (\frac {3}{2};\frac {7}{6},\frac {1}{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 )-2 F_1\left (\frac {3}{2};\frac {1}{6},\frac {4}{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+3 \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) F_1\left (\frac {1}{2};\frac {1}{6},\frac {1}{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 (-10 (-9 \cos (c+d x)+\cos (2 (c+d x))+2) \cos ^2\left (\frac {1}{2} (c+d x)\right ) F_1\left (\frac {3}{2};\frac {1}{6},\frac {4}{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 (-9 \cos (c+d x)+\cos (2 (c+d x))+2) \cos ^2\left (\frac {1}{2} (c+d x)\right ) F_1\left (\frac {3}{2};\frac {7}{6},\frac {1}{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 )-6 \sin ^2\left (\frac {1}{2} (c+d x)\right ) \cos (c+d x) \left (16 F_1\left (\frac {5}{2};\frac {1}{6},\frac {7}{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 )-4 F_1\left (\frac {5}{2};\frac {7}{6},\frac {4}{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 )+7 F_1\left (\frac {5}{2};\frac {13}{6},\frac {1}{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])^(2/3)/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(90*AppellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[c + d*x]^2*(e*Sec[c + d*x])^(2/3
)*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2]*(9*AppellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*
x)/2]^2] + (-2*AppellF1[3/2, 1/6, 4/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 7/6, 1/3,
5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2))/(a*d*(270*AppellF1[1/2, 1/6, 1/3, 3/2, Ta
n[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]^2*Cos[(c + d*x)/2]^2*(1 + 2*Cos[c + d*x]) + 3*AppellF1[1/2, 1/6, 1/3, 3
/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*(-10*AppellF1[3/2, 1/6, 4/3, 5/2, Tan[(c + d*x
)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^2*(2 - 9*Cos[c + d*x] + Cos[2*(c + d*x)]) + 5*AppellF1[3/2, 7/6,
1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^2*(2 - 9*Cos[c + d*x] + Cos[2*(c + d*x)])
- 6*(16*AppellF1[5/2, 1/6, 7/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 4*AppellF1[5/2, 7/6, 4/3, 7/2
, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 7*AppellF1[5/2, 13/6, 1/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x
)/2]^2])*Cos[c + d*x]*Sin[(c + d*x)/2]^2)*Tan[(c + d*x)/2]^2 + 10*(-2*AppellF1[3/2, 1/6, 4/3, 5/2, Tan[(c + d*
x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 7/6, 1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])^2*Cos[c
+ d*x]*Tan[(c + d*x)/2]^4))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 1.16, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sec \left (d x +c \right )\right )^{\frac {2}{3}}}{\sqrt {a +a \sec \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {2}{3}}}{\sqrt {a \sec \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*sec(d*x + c))^(2/3)/sqrt(a*sec(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{2/3}}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sec {\left (c + d x \right )}\right )^{\frac {2}{3}}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((e*sec(c + d*x))**(2/3)/sqrt(a*(sec(c + d*x) + 1)), x)

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