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]

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

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Rubi [A]  time = 0.08, antiderivative size = 105, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [B]  time = 1.97, 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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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}, x\right ) \]

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

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)

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maple [F]  time = 0.80, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x +c \right )}{\left (a +b \sec \left (d x +c \right )\right )^{\frac {2}{3}}}\, dx \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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