3.716 \(\int \frac {1}{\sec ^{\frac {2}{3}}(c+d x) (a+b \sec (c+d x))} \, dx\)

Optimal. Leaf size=174 \[ \frac {a \sin (c+d x) \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)} F_1\left (\frac {1}{2};-\frac {5}{6},1;\frac {3}{2};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )}-\frac {b \sin (c+d x) F_1\left (\frac {1}{2};-\frac {1}{3},1;\frac {3}{2};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right ) \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac {2}{3}}(c+d x)} \]

[Out]

-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))*sin(d*x+c)/(a^2-b^2)/d/(cos(d*x+c)^2)^(1/3
)/sec(d*x+c)^(2/3)+a*AppellF1(1/2,-5/6,1,3/2,sin(d*x+c)^2,a^2*sin(d*x+c)^2/(a^2-b^2))*(cos(d*x+c)^2)^(1/6)*sec
(d*x+c)^(1/3)*sin(d*x+c)/(a^2-b^2)/d

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Rubi [A]  time = 0.24, 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 = {3869, 2823, 3189, 429} \[ \frac {a \sin (c+d x) \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)} F_1\left (\frac {1}{2};-\frac {5}{6},1;\frac {3}{2};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )}-\frac {b \sin (c+d x) F_1\left (\frac {1}{2};-\frac {1}{3},1;\frac {3}{2};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right ) \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac {2}{3}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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)]*Sin[c + d*x])/((a^2 - b^2)*
d*(Cos[c + d*x]^2)^(1/3)*Sec[c + d*x]^(2/3))) + (a*AppellF1[1/2, -5/6, 1, 3/2, Sin[c + d*x]^2, (a^2*Sin[c + d*
x]^2)/(a^2 - b^2)]*(Cos[c + d*x]^2)^(1/6)*Sec[c + d*x]^(1/3)*Sin[c + d*x])/((a^2 - b^2)*d)

Rule 429

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 2823

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 3189

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 3869

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 {1}{\sec ^{\frac {2}{3}}(c+d x) (a+b \sec (c+d x))} \, dx &=\left (\sqrt [3]{\cos (c+d x)} \sqrt [3]{\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {5}{3}}(c+d x)}{b+a \cos (c+d x)} \, dx\\ &=-\left (\left (a \sqrt [3]{\cos (c+d x)} \sqrt [3]{\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {8}{3}}(c+d x)}{b^2-a^2 \cos ^2(c+d x)} \, dx\right )+\left (b \sqrt [3]{\cos (c+d x)} \sqrt [3]{\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {5}{3}}(c+d x)}{b^2-a^2 \cos ^2(c+d x)} \, dx\\ &=\frac {b \operatorname {Subst}\left (\int \frac {\sqrt [3]{1-x^2}}{-a^2+b^2+a^2 x^2} \, dx,x,\sin (c+d x)\right )}{d \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac {2}{3}}(c+d x)}-\frac {\left (a \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^{5/6}}{-a^2+b^2+a^2 x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\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 ) \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac {2}{3}}(c+d x)}+\frac {a F_1\left (\frac {1}{2};-\frac {5}{6},1;\frac {3}{2};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d}\\ \end {align*}

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Mathematica [B]  time = 29.26, size = 7588, normalized size = 43.61 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.95, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sec \left (d x +c \right )^{\frac {2}{3}} \left (a +b \sec \left (d x +c \right )\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/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 \[ \int \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((a + b*sec(c + d*x))*sec(c + d*x)**(2/3)), x)

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