3.704 \(\int \frac{\sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx\)

Optimal. Leaf size=105 \[ \frac{\sqrt{2} \tan (c+d x) \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{d \sqrt{\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.0837088, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3834, 139, 138} \[ \frac{\sqrt{2} \tan (c+d x) \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{d \sqrt{\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3834

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[Cot[e + f*x]/(f*Sqr
t[1 + Csc[e + f*x]]*Sqrt[1 - Csc[e + f*x]]), Subst[Int[(a + b*x)^m/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Csc[e + f
*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*m]

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rubi steps

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

Mathematica [B]  time = 1.70522, size = 310, normalized size = 2.95 \[ \frac{24 (a-b)^2 (a+b) \cos (c+d x) \cot ^3(c+d x) (\sec (c+d x)+1) (b-b \sec (c+d x)) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac{1}{3};\frac{1}{2},\frac{1}{2};\frac{4}{3};\frac{a+b \sec (c+d x)}{a-b},\frac{a+b \sec (c+d x)}{a+b}\right )}{b^2 d (b-a) \left (3 (a-b) (a \cos (c+d x)+b) F_1\left (\frac{4}{3};\frac{1}{2},\frac{3}{2};\frac{7}{3};\frac{a+b \sec (c+d x)}{a-b},\frac{a+b \sec (c+d x)}{a+b}\right )+(a+b) \left (8 (a-b) \cos (c+d x) F_1\left (\frac{1}{3};\frac{1}{2},\frac{1}{2};\frac{4}{3};\frac{a+b \sec (c+d x)}{a-b},\frac{a+b \sec (c+d x)}{a+b}\right )+3 (a \cos (c+d x)+b) F_1\left (\frac{4}{3};\frac{3}{2},\frac{1}{2};\frac{7}{3};\frac{a+b \sec (c+d x)}{a-b},\frac{a+b \sec (c+d x)}{a+b}\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(24*(a - b)^2*(a + b)*AppellF1[1/3, 1/2, 1/2, 4/3, (a + b*Sec[c + d*x])/(a - b), (a + b*Sec[c + d*x])/(a + b)]
*Cos[c + d*x]*Cot[c + d*x]^3*(1 + Sec[c + d*x])*(b - b*Sec[c + d*x])*(a + b*Sec[c + d*x])^(1/3))/(b^2*(-a + b)
*d*(3*(a - b)*AppellF1[4/3, 1/2, 3/2, 7/3, (a + b*Sec[c + d*x])/(a - b), (a + b*Sec[c + d*x])/(a + b)]*(b + a*
Cos[c + d*x]) + (a + b)*(8*(a - b)*AppellF1[1/3, 1/2, 1/2, 4/3, (a + b*Sec[c + d*x])/(a - b), (a + b*Sec[c + d
*x])/(a + b)]*Cos[c + d*x] + 3*AppellF1[4/3, 3/2, 1/2, 7/3, (a + b*Sec[c + d*x])/(a - b), (a + b*Sec[c + d*x])
/(a + b)]*(b + a*Cos[c + d*x]))))

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Maple [F]  time = 0.126, size = 0, normalized size = 0. \begin{align*} \int{\sec \left ( dx+c \right ) \left ( a+b\sec \left ( dx+c \right ) \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sec(d*x + c)/(b*sec(d*x + c) + a)^(2/3), x)

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

[Out]

Integral(sec(c + d*x)/(a + b*sec(c + d*x))**(2/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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