3.3.83 \(\int \frac {1}{(e \sec (c+d x))^{2/3} \sqrt {a+a \sec (c+d x)}} \, dx\) [283]

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

[Out]

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

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Rubi [A]
time = 0.12, 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 = {3913, 3912, 129, 524} \begin {gather*} \frac {3 \tan (c+d x) F_1\left (-\frac {2}{3};\frac {1}{2},1;\frac {1}{3};\sec (c+d x),-\sec (c+d x)\right )}{2 d \sqrt {1-\sec (c+d x)} \sqrt {a \sec (c+d x)+a} (e \sec (c+d x))^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((e*Sec[c + d*x])^(2/3)*Sqrt[a + a*Sec[c + d*x]]),x]

[Out]

(3*AppellF1[-2/3, 1/2, 1, 1/3, Sec[c + d*x], -Sec[c + d*x]]*Tan[c + d*x])/(2*d*Sqrt[1 - Sec[c + d*x]]*(e*Sec[c
+ d*x])^(2/3)*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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 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 {1}{(e \sec (c+d x))^{2/3} \sqrt {a+a \sec (c+d x)}} \, dx &=\frac {\sqrt {1+\sec (c+d x)} \int \frac {1}{(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)) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} (e x)^{5/3} (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}{x^3 \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 {1}{3};\sec (c+d x),-\sec (c+d x)\right ) \tan (c+d x)}{2 d \sqrt {1-\sec (c+d x)} (e \sec (c+d x))^{2/3} \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. \(585\) vs. \(2(78)=156\).
time = 7.35, size = 585, normalized size = 7.50 \begin {gather*} \frac {\sec ^{\frac {7}{6}}(c+d x) \left (-\frac {3}{2} \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{6}}(c+d x) \left (\sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )\right )+\frac {5 \sqrt {\frac {1}{1+\cos (c+d x)}} (-1+3 \cos (c+d x)) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{5/6} \sin \left (\frac {1}{2} (c+d x)\right ) \left (-3 \cos ^{\frac {5}{6}}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt [3]{\sec ^2\left (\frac {1}{2} (c+d x)\right )}+2 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 ) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{5/6} \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{-120 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 ) \left (\frac {1}{1+\cos (c+d x)}\right )^{2/3} \left (\frac {\cos (c+d x)}{1+\cos (c+d x)}\right )^{5/6} \sin \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )+32 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 ) \left (\frac {1}{1+\cos (c+d x)}\right )^{2/3} \left (\frac {\cos (c+d x)}{1+\cos (c+d x)}\right )^{5/6} \sin \left (\frac {1}{2} (c+d x)\right ) \tan ^3\left (\frac {1}{2} (c+d x)\right )+5 \sqrt {2} \cos \left (\frac {1}{2} (c+d x)\right ) \left (3-4 \sqrt {2} 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 ) \left (\frac {1}{1+\cos (c+d x)}\right )^{2/3} \left (\frac {\cos (c+d x)}{1+\cos (c+d x)}\right )^{5/6} \tan ^4\left (\frac {1}{2} (c+d x)\right )\right )}\right )}{d (e \sec (c+d x))^{2/3} \sqrt {a (1+\sec (c+d x))}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((e*Sec[c + d*x])^(2/3)*Sqrt[a + a*Sec[c + d*x]]),x]

[Out]

(Sec[c + d*x]^(7/6)*((-3*Cos[(c + d*x)/2]*Sec[c + d*x]^(5/6)*(Sin[(c + d*x)/2] - Sin[(3*(c + d*x))/2]))/2 + (5
*Sqrt[(1 + Cos[c + d*x])^(-1)]*(-1 + 3*Cos[c + d*x])*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(5/6)*Sin[(c + d*x)/2]*
(-3*Cos[c + d*x]^(5/6)*Hypergeometric2F1[1/2, 5/6, 3/2, 2*Sin[(c + d*x)/2]^2]*(Sec[(c + d*x)/2]^2)^(1/3) + 2*A
ppellF1[3/2, 5/6, 2/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(5/6)*T
an[(c + d*x)/2]^2))/(-120*AppellF1[3/2, 5/6, 2/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*((1 + Cos[c +
d*x])^(-1))^(2/3)*(Cos[c + d*x]/(1 + Cos[c + d*x]))^(5/6)*Sin[(c + d*x)/2]*Tan[(c + d*x)/2] + 32*AppellF1[5/2,
5/6, 5/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*((1 + Cos[c + d*x])^(-1))^(2/3)*(Cos[c + d*x]/(1 + Co
s[c + d*x]))^(5/6)*Sin[(c + d*x)/2]*Tan[(c + d*x)/2]^3 + 5*Sqrt[2]*Cos[(c + d*x)/2]*(3 - 4*Sqrt[2]*AppellF1[5/
2, 11/6, 2/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*((1 + Cos[c + d*x])^(-1))^(2/3)*(Cos[c + d*x]/(1 +
Cos[c + d*x]))^(5/6)*Tan[(c + d*x)/2]^4))))/(d*(e*Sec[c + d*x])^(2/3)*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [F]
time = 0.09, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (e \sec \left (d x +c \right )\right )^{\frac {2}{3}} \sqrt {a +a \sec \left (d x +c \right )}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-2/3)*integrate(1/(sqrt(a*sec(d*x + c) + a)*sec(d*x + c)^(2/3)), 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(1/(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (e \sec {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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