3.8.13 \(\int \frac {\sec ^{\frac {2}{3}}(c+d x)}{a+b \sec (c+d x)} \, dx\) [713]

Optimal. Leaf size=174 \[ \frac {a F_1\left (\frac {1}{2};-\frac {1}{6},1;\frac {3}{2};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)}}-\frac {b F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac {2}{3}}(c+d x) \sin (c+d x)}{\left (a^2-b^2\right ) d} \]

[Out]

a*AppellF1(1/2,-1/6,1,3/2,sin(d*x+c)^2,a^2*sin(d*x+c)^2/(a^2-b^2))*sin(d*x+c)/(a^2-b^2)/d/(cos(d*x+c)^2)^(1/6)
/sec(d*x+c)^(1/3)-b*AppellF1(1/2,1/3,1,3/2,sin(d*x+c)^2,a^2*sin(d*x+c)^2/(a^2-b^2))*(cos(d*x+c)^2)^(1/3)*sec(d
*x+c)^(2/3)*sin(d*x+c)/(a^2-b^2)/d

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Rubi [A]
time = 0.18, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3954, 2902, 3268, 440} \begin {gather*} \frac {a \sin (c+d x) F_1\left (\frac {1}{2};-\frac {1}{6},1;\frac {3}{2};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right ) \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)}}-\frac {b \sin (c+d x) \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac {2}{3}}(c+d x) F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*AppellF1[1/2, -1/6, 1, 3/2, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)]*Sin[c + d*x])/((a^2 - b^2)*d*
(Cos[c + d*x]^2)^(1/6)*Sec[c + d*x]^(1/3)) - (b*AppellF1[1/2, 1/3, 1, 3/2, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2
)/(a^2 - b^2)]*(Cos[c + d*x]^2)^(1/3)*Sec[c + d*x]^(2/3)*Sin[c + d*x])/((a^2 - b^2)*d)

Rule 440

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

Rule 2902

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[a, Int[(d*
Sin[e + f*x])^n/(a^2 - b^2*Sin[e + f*x]^2), x], x] - Dist[b/d, Int[(d*Sin[e + f*x])^(n + 1)/(a^2 - b^2*Sin[e +
f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0]

Rule 3268

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff
= FreeFactors[Cos[e + f*x], x]}, Dist[(-ff)*d^(2*IntPart[(m - 1)/2] + 1)*((d*Sin[e + f*x])^(2*FracPart[(m - 1
)/2])/(f*(Sin[e + f*x]^2)^FracPart[(m - 1)/2])), Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p,
x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] &&  !IntegerQ[m]

Rule 3954

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Dist[Sin[
e + f*x]^n*(d*Csc[e + f*x])^n, Int[(b + a*Sin[e + f*x])^m/Sin[e + f*x]^(m + n), x], x] /; FreeQ[{a, b, d, e, f
, n}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sec ^{\frac {2}{3}}(c+d x)}{a+b \sec (c+d x)} \, dx &=\left (\cos ^{\frac {2}{3}}(c+d x) \sec ^{\frac {2}{3}}(c+d x)\right ) \int \frac {\sqrt [3]{\cos (c+d x)}}{b+a \cos (c+d x)} \, dx\\ &=-\left (\left (a \cos ^{\frac {2}{3}}(c+d x) \sec ^{\frac {2}{3}}(c+d x)\right ) \int \frac {\cos ^{\frac {4}{3}}(c+d x)}{b^2-a^2 \cos ^2(c+d x)} \, dx\right )+\left (b \cos ^{\frac {2}{3}}(c+d x) \sec ^{\frac {2}{3}}(c+d x)\right ) \int \frac {\sqrt [3]{\cos (c+d x)}}{b^2-a^2 \cos ^2(c+d x)} \, dx\\ &=-\frac {a \text {Subst}\left (\int \frac {\sqrt [6]{1-x^2}}{-a^2+b^2+a^2 x^2} \, dx,x,\sin (c+d x)\right )}{d \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)}}+\frac {\left (b \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac {2}{3}}(c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{1-x^2} \left (-a^2+b^2+a^2 x^2\right )} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {a F_1\left (\frac {1}{2};-\frac {1}{6},1;\frac {3}{2};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)}}-\frac {b F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac {2}{3}}(c+d x) \sin (c+d x)}{\left (a^2-b^2\right ) d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(4543\) vs. \(2(174)=348\).
time = 39.54, size = 4543, normalized size = 26.11 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(9*(a^2 - b^2)*Sec[c + d*x]^(5/3)*Sin[c + d*x]*((b*AppellF1[1/2, 1/6, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*
x]^2)/(a^2 - b^2)]*Sqrt[Sec[c + d*x]^2])/(9*(a^2 - b^2)*AppellF1[1/2, 1/6, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c
+ d*x]^2)/(a^2 - b^2)] + (6*b^2*AppellF1[3/2, 1/6, 2, 5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]
+ (-a^2 + b^2)*AppellF1[3/2, 7/6, 1, 5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])*Tan[c + d*x]^2)
+ (a*AppellF1[1/2, 2/3, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])/(-9*(a^2 - b^2)*AppellF1[
1/2, 2/3, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] - 2*(3*b^2*AppellF1[3/2, 2/3, 2, 5/2, -Ta
n[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + 2*(-a^2 + b^2)*AppellF1[3/2, 5/3, 1, 5/2, -Tan[c + d*x]^2, (
b^2*Tan[c + d*x]^2)/(a^2 - b^2)])*Tan[c + d*x]^2)))/(d*(Sec[c + d*x]^2)^(2/3)*(a + b*Sec[c + d*x])*(-a^2 + b^2
*Sec[c + d*x]^2)*((9*(a^2 - b^2)*(Sec[c + d*x]^2)^(1/3)*((b*AppellF1[1/2, 1/6, 1, 3/2, -Tan[c + d*x]^2, (b^2*T
an[c + d*x]^2)/(a^2 - b^2)]*Sqrt[Sec[c + d*x]^2])/(9*(a^2 - b^2)*AppellF1[1/2, 1/6, 1, 3/2, -Tan[c + d*x]^2, (
b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + (6*b^2*AppellF1[3/2, 1/6, 2, 5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^
2 - b^2)] + (-a^2 + b^2)*AppellF1[3/2, 7/6, 1, 5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])*Tan[c
+ d*x]^2) + (a*AppellF1[1/2, 2/3, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])/(-9*(a^2 - b^2)*
AppellF1[1/2, 2/3, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] - 2*(3*b^2*AppellF1[3/2, 2/3, 2,
5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + 2*(-a^2 + b^2)*AppellF1[3/2, 5/3, 1, 5/2, -Tan[c +
d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])*Tan[c + d*x]^2)))/(-a^2 + b^2*Sec[c + d*x]^2) - (18*b^2*(a^2 - b^2)
*(Sec[c + d*x]^2)^(1/3)*Tan[c + d*x]^2*((b*AppellF1[1/2, 1/6, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a
^2 - b^2)]*Sqrt[Sec[c + d*x]^2])/(9*(a^2 - b^2)*AppellF1[1/2, 1/6, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^
2)/(a^2 - b^2)] + (6*b^2*AppellF1[3/2, 1/6, 2, 5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + (-a^2
+ b^2)*AppellF1[3/2, 7/6, 1, 5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])*Tan[c + d*x]^2) + (a*Ap
pellF1[1/2, 2/3, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])/(-9*(a^2 - b^2)*AppellF1[1/2, 2/3
, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] - 2*(3*b^2*AppellF1[3/2, 2/3, 2, 5/2, -Tan[c + d*
x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + 2*(-a^2 + b^2)*AppellF1[3/2, 5/3, 1, 5/2, -Tan[c + d*x]^2, (b^2*Tan[
c + d*x]^2)/(a^2 - b^2)])*Tan[c + d*x]^2)))/(-a^2 + b^2*Sec[c + d*x]^2)^2 - (12*(a^2 - b^2)*Tan[c + d*x]^2*((b
*AppellF1[1/2, 1/6, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Sqrt[Sec[c + d*x]^2])/(9*(a^2 -
b^2)*AppellF1[1/2, 1/6, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + (6*b^2*AppellF1[3/2, 1/6
, 2, 5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + (-a^2 + b^2)*AppellF1[3/2, 7/6, 1, 5/2, -Tan[c
+ d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])*Tan[c + d*x]^2) + (a*AppellF1[1/2, 2/3, 1, 3/2, -Tan[c + d*x]^2,
(b^2*Tan[c + d*x]^2)/(a^2 - b^2)])/(-9*(a^2 - b^2)*AppellF1[1/2, 2/3, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*
x]^2)/(a^2 - b^2)] - 2*(3*b^2*AppellF1[3/2, 2/3, 2, 5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] +
2*(-a^2 + b^2)*AppellF1[3/2, 5/3, 1, 5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])*Tan[c + d*x]^2))
)/((Sec[c + d*x]^2)^(2/3)*(-a^2 + b^2*Sec[c + d*x]^2)) + (9*(a^2 - b^2)*Tan[c + d*x]*((b*AppellF1[1/2, 1/6, 1,
3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])/(9*(a^2 - b^2)*App
ellF1[1/2, 1/6, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + (6*b^2*AppellF1[3/2, 1/6, 2, 5/2,
-Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + (-a^2 + b^2)*AppellF1[3/2, 7/6, 1, 5/2, -Tan[c + d*x]^2,
(b^2*Tan[c + d*x]^2)/(a^2 - b^2)])*Tan[c + d*x]^2) + (b*Sqrt[Sec[c + d*x]^2]*((2*b^2*AppellF1[3/2, 1/6, 2, 5/
2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Sec[c + d*x]^2*Tan[c + d*x])/(3*(a^2 - b^2)) - (AppellF1
[3/2, 7/6, 1, 5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Sec[c + d*x]^2*Tan[c + d*x])/9))/(9*(a^2
- b^2)*AppellF1[1/2, 1/6, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + (6*b^2*AppellF1[3/2, 1
/6, 2, 5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + (-a^2 + b^2)*AppellF1[3/2, 7/6, 1, 5/2, -Tan[
c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])*Tan[c + d*x]^2) + (a*((2*b^2*AppellF1[3/2, 2/3, 2, 5/2, -Tan[c
+ d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Sec[c + d*x]^2*Tan[c + d*x])/(3*(a^2 - b^2)) - (4*AppellF1[3/2, 5/
3, 1, 5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Sec[c + d*x]^2*Tan[c + d*x])/9))/(-9*(a^2 - b^2)
*AppellF1[1/2, 2/3, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] - 2*(3*b^2*AppellF1[3/2, 2/3, 2
, 5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + 2*(-a^2 + b^2)*AppellF1[3/2, 5/3, 1, 5/2, -Tan[c +
d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])*Tan...

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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {\sec ^{\frac {2}{3}}\left (d x +c \right )}{a +b \sec \left (d x +c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(sec(d*x + c)^(2/3)/(b*sec(d*x + c) + a), 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(sec(d*x+c)^(2/3)/(a+b*sec(d*x+c)),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 {\sec ^{\frac {2}{3}}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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