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

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

[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]])

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Rubi [A]  time = 0.168797, antiderivative size = 78, 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, 510} \[ \frac{3 \tan (c+d x) F_1\left (-\frac{2}{3};\frac{1}{2},1;\frac{1}{3};\sec (c+d x),-\sec (c+d x)\right )}{2 d \sqrt{1-\sec (c+d x)} \sqrt{a \sec (c+d x)+a} (e \sec (c+d x))^{2/3}} \]

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 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 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])

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)) \operatorname{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)) \operatorname{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*}

Mathematica [B]  time = 6.81286, size = 585, normalized size = 7.5 \[ \frac{\sec ^{\frac{7}{6}}(c+d x) \left (\frac{5 \sin \left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{1}{\cos (c+d x)+1}} (3 \cos (c+d x)-1) \left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^{5/6} \left (2 \tan ^2\left (\frac{1}{2} (c+d x)\right ) F_1\left (\frac{3}{2};\frac{5}{6},\frac{2}{3};\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right ) \left (\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )\right )^{5/6}-3 \cos ^{\frac{5}{6}}(c+d x) \sqrt [3]{\sec ^2\left (\frac{1}{2} (c+d x)\right )} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{6},\frac{3}{2},2 \sin ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )}{5 \sqrt{2} \cos \left (\frac{1}{2} (c+d x)\right ) \left (3-4 \sqrt{2} \left (\frac{1}{\cos (c+d x)+1}\right )^{2/3} \left (\frac{\cos (c+d x)}{\cos (c+d x)+1}\right )^{5/6} \tan ^4\left (\frac{1}{2} (c+d x)\right ) F_1\left (\frac{5}{2};\frac{11}{6},\frac{2}{3};\frac{7}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )+32 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\frac{1}{\cos (c+d x)+1}\right )^{2/3} \left (\frac{\cos (c+d x)}{\cos (c+d x)+1}\right )^{5/6} \tan ^3\left (\frac{1}{2} (c+d x)\right ) F_1\left (\frac{5}{2};\frac{5}{6},\frac{5}{3};\frac{7}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-120 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\frac{1}{\cos (c+d x)+1}\right )^{2/3} \left (\frac{\cos (c+d x)}{\cos (c+d x)+1}\right )^{5/6} \tan \left (\frac{1}{2} (c+d x)\right ) F_1\left (\frac{3}{2};\frac{5}{6},\frac{2}{3};\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}-\frac{3}{2} \left (\sin \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{3}{2} (c+d x)\right )\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{5}{6}}(c+d x)\right )}{d \sqrt{a (\sec (c+d x)+1)} (e \sec (c+d x))^{2/3}} \]

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 = 0.167, size = 0, normalized size = 0. \begin{align*} \int{ \left ( e\sec \left ( dx+c \right ) \right ) ^{-{\frac{2}{3}}}{\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(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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac{2}{3}}}\,{d x} \end{align*}

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]

integrate(1/(sqrt(a*sec(d*x + c) + a)*(e*sec(d*x + c))^(2/3)), 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(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., size = 0, normalized size = 0. \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac{2}{3}}}\,{d x} \end{align*}

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)