3.274 \(\int \frac{A+B \sec (c+d x)}{(a+a \sec (c+d x))^{2/3}} \, dx\)
Optimal. Leaf size=764 \[ -\frac{3^{3/4} \left (1-\sqrt{3}\right ) B \tan (c+d x) \sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt{\frac{(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \text{EllipticF}\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{2^{2/3} d (1-\sec (c+d x)) \sqrt{-\frac{\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3}}-\frac{3 \sqrt{2} A \tan (c+d x) F_1\left (-\frac{1}{6};\frac{1}{2},1;\frac{5}{6};\frac{1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{d \sqrt{1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}+\frac{3 B \tan (c+d x)}{d (a \sec (c+d x)+a)^{2/3}}+\frac{3 \left (1+\sqrt{3}\right ) B \tan (c+d x) \sqrt [3]{\sec (c+d x)+1}}{d \left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right ) (a \sec (c+d x)+a)^{2/3}}-\frac{3 \sqrt [3]{2} \sqrt [4]{3} B \tan (c+d x) \sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt{\frac{(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} E\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{d (1-\sec (c+d x)) \sqrt{-\frac{\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3}} \]
[Out]
(3*B*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(2/3)) - (3*Sqrt[2]*A*AppellF1[-1/6, 1/2, 1, 5/6, (1 + Sec[c + d*x]
)/2, 1 + Sec[c + d*x]]*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*(a + a*Sec[c + d*x])^(2/3)) + (3*(1 + Sqrt[3])*
B*(1 + Sec[c + d*x])^(1/3)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(2/3)*(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d
*x])^(1/3))) - (3*2^(1/3)*3^(1/4)*B*EllipticE[ArcCos[(2^(1/3) - (1 - Sqrt[3])*(1 + Sec[c + d*x])^(1/3))/(2^(1/
3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))], (2 + Sqrt[3])/4]*(1 + Sec[c + d*x])^(1/3)*(2^(1/3) - (1 + Sec[c
+ d*x])^(1/3))*Sqrt[(2^(2/3) + 2^(1/3)*(1 + Sec[c + d*x])^(1/3) + (1 + Sec[c + d*x])^(2/3))/(2^(1/3) - (1 + S
qrt[3])*(1 + Sec[c + d*x])^(1/3))^2]*Tan[c + d*x])/(d*(1 - Sec[c + d*x])*(a + a*Sec[c + d*x])^(2/3)*Sqrt[-(((1
+ Sec[c + d*x])^(1/3)*(2^(1/3) - (1 + Sec[c + d*x])^(1/3)))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3)
)^2)]) - (3^(3/4)*(1 - Sqrt[3])*B*EllipticF[ArcCos[(2^(1/3) - (1 - Sqrt[3])*(1 + Sec[c + d*x])^(1/3))/(2^(1/3)
- (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))], (2 + Sqrt[3])/4]*(1 + Sec[c + d*x])^(1/3)*(2^(1/3) - (1 + Sec[c +
d*x])^(1/3))*Sqrt[(2^(2/3) + 2^(1/3)*(1 + Sec[c + d*x])^(1/3) + (1 + Sec[c + d*x])^(2/3))/(2^(1/3) - (1 + Sqr
t[3])*(1 + Sec[c + d*x])^(1/3))^2]*Tan[c + d*x])/(2^(2/3)*d*(1 - Sec[c + d*x])*(a + a*Sec[c + d*x])^(2/3)*Sqrt
[-(((1 + Sec[c + d*x])^(1/3)*(2^(1/3) - (1 + Sec[c + d*x])^(1/3)))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])
^(1/3))^2)])
________________________________________________________________________________________
Rubi [A] time = 0.733451, antiderivative size = 764, normalized size of antiderivative
= 1., number of steps used = 11, number of rules used = 11, integrand size = 25,
\(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {3924, 3779, 3778, 136, 3828, 3827, 51, 63, 308, 225, 1881}
\[ -\frac{3 \sqrt{2} A \tan (c+d x) F_1\left (-\frac{1}{6};\frac{1}{2},1;\frac{5}{6};\frac{1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{d \sqrt{1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}+\frac{3 B \tan (c+d x)}{d (a \sec (c+d x)+a)^{2/3}}+\frac{3 \left (1+\sqrt{3}\right ) B \tan (c+d x) \sqrt [3]{\sec (c+d x)+1}}{d \left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right ) (a \sec (c+d x)+a)^{2/3}}-\frac{3^{3/4} \left (1-\sqrt{3}\right ) B \tan (c+d x) \sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt{\frac{(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} F\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{2^{2/3} d (1-\sec (c+d x)) \sqrt{-\frac{\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3}}-\frac{3 \sqrt [3]{2} \sqrt [4]{3} B \tan (c+d x) \sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt{\frac{(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} E\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{d (1-\sec (c+d x)) \sqrt{-\frac{\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3}} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Sec[c + d*x])/(a + a*Sec[c + d*x])^(2/3),x]
[Out]
(3*B*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(2/3)) - (3*Sqrt[2]*A*AppellF1[-1/6, 1/2, 1, 5/6, (1 + Sec[c + d*x]
)/2, 1 + Sec[c + d*x]]*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*(a + a*Sec[c + d*x])^(2/3)) + (3*(1 + Sqrt[3])*
B*(1 + Sec[c + d*x])^(1/3)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(2/3)*(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d
*x])^(1/3))) - (3*2^(1/3)*3^(1/4)*B*EllipticE[ArcCos[(2^(1/3) - (1 - Sqrt[3])*(1 + Sec[c + d*x])^(1/3))/(2^(1/
3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))], (2 + Sqrt[3])/4]*(1 + Sec[c + d*x])^(1/3)*(2^(1/3) - (1 + Sec[c
+ d*x])^(1/3))*Sqrt[(2^(2/3) + 2^(1/3)*(1 + Sec[c + d*x])^(1/3) + (1 + Sec[c + d*x])^(2/3))/(2^(1/3) - (1 + S
qrt[3])*(1 + Sec[c + d*x])^(1/3))^2]*Tan[c + d*x])/(d*(1 - Sec[c + d*x])*(a + a*Sec[c + d*x])^(2/3)*Sqrt[-(((1
+ Sec[c + d*x])^(1/3)*(2^(1/3) - (1 + Sec[c + d*x])^(1/3)))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3)
)^2)]) - (3^(3/4)*(1 - Sqrt[3])*B*EllipticF[ArcCos[(2^(1/3) - (1 - Sqrt[3])*(1 + Sec[c + d*x])^(1/3))/(2^(1/3)
- (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))], (2 + Sqrt[3])/4]*(1 + Sec[c + d*x])^(1/3)*(2^(1/3) - (1 + Sec[c +
d*x])^(1/3))*Sqrt[(2^(2/3) + 2^(1/3)*(1 + Sec[c + d*x])^(1/3) + (1 + Sec[c + d*x])^(2/3))/(2^(1/3) - (1 + Sqr
t[3])*(1 + Sec[c + d*x])^(1/3))^2]*Tan[c + d*x])/(2^(2/3)*d*(1 - Sec[c + d*x])*(a + a*Sec[c + d*x])^(2/3)*Sqrt
[-(((1 + Sec[c + d*x])^(1/3)*(2^(1/3) - (1 + Sec[c + d*x])^(1/3)))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])
^(1/3))^2)])
Rule 3924
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Dist[c, I
nt[(a + b*Csc[e + f*x])^m, x], x] + Dist[d, Int[(a + b*Csc[e + f*x])^m*Csc[e + f*x], x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[2*m]
Rule 3779
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Csc[c + d*x])^FracPart
[n])/(1 + (b*Csc[c + d*x])/a)^FracPart[n], Int[(1 + (b*Csc[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && !GtQ[a, 0]
Rule 3778
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[(a^n*Cot[c + d*x])/(d*Sqrt[1 + Csc[c + d*x]
]*Sqrt[1 - Csc[c + d*x]]), Subst[Int[(1 + (b*x)/a)^(n - 1/2)/(x*Sqrt[1 - (b*x)/a]), x], x, Csc[c + d*x]], x] /
; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && GtQ[a, 0]
Rule 136
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(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^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && !IntegerQ[m] && !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*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 51
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
Rule 308
Int[(x_)^4/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Dist[(
(Sqrt[3] - 1)*s^2)/(2*r^2), Int[1/Sqrt[a + b*x^6], x], x] - Dist[1/(2*r^2), Int[((Sqrt[3] - 1)*s^2 - 2*r^2*x^4
)/Sqrt[a + b*x^6], x], x]] /; FreeQ[{a, b}, x]
Rule 225
Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(x*(s
+ r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[(r*x^2*(s + r*x^2))/(s + (1
+ Sqrt[3])*r*x^2)^2]), x]] /; FreeQ[{a, b}, x]
Rule 1881
Int[((c_) + (d_.)*(x_)^4)/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/
a, 3]]}, Simp[((1 + Sqrt[3])*d*s^3*x*Sqrt[a + b*x^6])/(2*a*r^2*(s + (1 + Sqrt[3])*r*x^2)), x] - Simp[(3^(1/4)*
d*s*x*(s + r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticE[ArcCos[(s + (1 - Sqrt[
3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*r^2*Sqrt[(r*x^2*(s + r*x^2))/(s + (1 + Sqrt[3])*r*
x^2)^2]*Sqrt[a + b*x^6]), x]] /; FreeQ[{a, b, c, d}, x] && EqQ[2*Rt[b/a, 3]^2*c - (1 - Sqrt[3])*d, 0]
Rubi steps
\begin{align*} \int \frac{A+B \sec (c+d x)}{(a+a \sec (c+d x))^{2/3}} \, dx &=A \int \frac{1}{(a+a \sec (c+d x))^{2/3}} \, dx+B \int \frac{\sec (c+d x)}{(a+a \sec (c+d x))^{2/3}} \, dx\\ &=\frac{\left (A (1+\sec (c+d x))^{2/3}\right ) \int \frac{1}{(1+\sec (c+d x))^{2/3}} \, dx}{(a+a \sec (c+d x))^{2/3}}+\frac{\left (B (1+\sec (c+d x))^{2/3}\right ) \int \frac{\sec (c+d x)}{(1+\sec (c+d x))^{2/3}} \, dx}{(a+a \sec (c+d x))^{2/3}}\\ &=-\frac{\left (A \sqrt [6]{1+\sec (c+d x)} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} x (1+x)^{7/6}} \, dx,x,\sec (c+d x)\right )}{d \sqrt{1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}-\frac{\left (B \sqrt [6]{1+\sec (c+d x)} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} (1+x)^{7/6}} \, dx,x,\sec (c+d x)\right )}{d \sqrt{1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}\\ &=\frac{3 B \tan (c+d x)}{d (a+a \sec (c+d x))^{2/3}}-\frac{3 \sqrt{2} A F_1\left (-\frac{1}{6};\frac{1}{2},1;\frac{5}{6};\frac{1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{d \sqrt{1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}+\frac{\left (B \sqrt [6]{1+\sec (c+d x)} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} \sqrt [6]{1+x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt{1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}\\ &=\frac{3 B \tan (c+d x)}{d (a+a \sec (c+d x))^{2/3}}-\frac{3 \sqrt{2} A F_1\left (-\frac{1}{6};\frac{1}{2},1;\frac{5}{6};\frac{1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{d \sqrt{1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}+\frac{\left (6 B \sqrt [6]{1+\sec (c+d x)} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{d \sqrt{1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}\\ &=\frac{3 B \tan (c+d x)}{d (a+a \sec (c+d x))^{2/3}}-\frac{3 \sqrt{2} A F_1\left (-\frac{1}{6};\frac{1}{2},1;\frac{5}{6};\frac{1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{d \sqrt{1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}-\frac{\left (3 B \sqrt [6]{1+\sec (c+d x)} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{2^{2/3} \left (-1+\sqrt{3}\right )-2 x^4}{\sqrt{2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{d \sqrt{1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}-\frac{\left (3\ 2^{2/3} \left (1-\sqrt{3}\right ) B \sqrt [6]{1+\sec (c+d x)} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{d \sqrt{1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}\\ &=\frac{3 B \tan (c+d x)}{d (a+a \sec (c+d x))^{2/3}}-\frac{3 \sqrt{2} A F_1\left (-\frac{1}{6};\frac{1}{2},1;\frac{5}{6};\frac{1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{d \sqrt{1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}+\frac{3 \left (1+\sqrt{3}\right ) B \sqrt [3]{1+\sec (c+d x)} \tan (c+d x)}{d (a+a \sec (c+d x))^{2/3} \left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )}-\frac{3 \sqrt [3]{2} \sqrt [4]{3} B E\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right ) \sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt{\frac{2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{d (1-\sec (c+d x)) (a+a \sec (c+d x))^{2/3} \sqrt{-\frac{\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}}-\frac{3^{3/4} \left (1-\sqrt{3}\right ) B F\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right ) \sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt{\frac{2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{2^{2/3} d (1-\sec (c+d x)) (a+a \sec (c+d x))^{2/3} \sqrt{-\frac{\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}}\\ \end{align*}
Mathematica [B] time = 19.1699, size = 4066, normalized size = 5.32 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(A + B*Sec[c + d*x])/(a + a*Sec[c + d*x])^(2/3),x]
[Out]
(Cos[c + d*x]*((1 + Cos[c + d*x])*Sec[c + d*x])^(1/3)*(1 + Sec[c + d*x])^(2/3)*(A + B*Sec[c + d*x])*(3*Sec[(c
+ d*x)/2]*(-(A*Sin[(c + d*x)/2]) + B*Sin[(c + d*x)/2]) - 3*(-A + B)*Sin[c + d*x]))/(d*(B + A*Cos[c + d*x])*(a*
(1 + Sec[c + d*x]))^(2/3)) - (2^(1/3)*Cos[c + d*x]*(1 + Sec[c + d*x])^(2/3)*(A + B*Sec[c + d*x])*(2*A*(1 + Sec
[c + d*x])^(1/3) - B*(1 + Sec[c + d*x])^(1/3) + Cos[c + d*x]*(-3*A*(1 + Sec[c + d*x])^(1/3) + 3*B*(1 + Sec[c +
d*x])^(1/3)))*Tan[(c + d*x)/2]*(((A - B)*AppellF1[3/2, 1/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*
Tan[(c + d*x)/2]^2)/(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(2/3) - (9*(3*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2
]^2, -Tan[(c + d*x)/2]^2]*(A + B + (-5*A + 7*B)*Cos[c + d*x]) + 4*(A - B)*(3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c
+ d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Cos[
c + d*x]*Tan[(c + d*x)/2]^2))/(2*(-1 + Tan[(c + d*x)/2]^2)*(-9*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2,
-Tan[(c + d*x)/2]^2] + 2*(3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[3/2
, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2))))/(3*d*(B + A*Cos[c + d*x])*(Cos
[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)*(a*(1 + Sec[c + d*x]))^(2/3)*(-(Sec[(c + d*x)/2]^2*(((A - B)*AppellF1[3/2,
1/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2)/(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(
2/3) - (9*(3*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(A + B + (-5*A + 7*B)*Cos[c +
d*x]) + 4*(A - B)*(3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[3/2, 4/3,
1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Cos[c + d*x]*Tan[(c + d*x)/2]^2))/(2*(-1 + Tan[(c + d*x)/2]
^2)*(-9*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(3*AppellF1[3/2, 1/3, 2, 5/2,
Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]
)*Tan[(c + d*x)/2]^2))))/(3*2^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)) - (2^(1/3)*Tan[(c + d*x)/2]*(((A
- B)*AppellF1[3/2, 1/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/
(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(2/3) + ((A - B)*Tan[(c + d*x)/2]^2*((-3*AppellF1[5/2, 1/3, 2, 7/2, Tan[(c +
d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/5 + (AppellF1[5/2, 4/3, 1, 7/2, Tan[(c +
d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/5))/(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(2
/3) - (2*(A - B)*AppellF1[3/2, 1/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2*(-(Sec
[(c + d*x)/2]^2*Sin[c + d*x]) + Cos[c + d*x]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*(Cos[c + d*x]*Sec[(c + d
*x)/2]^2)^(5/3)) + (9*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]*(3*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -T
an[(c + d*x)/2]^2]*(A + B + (-5*A + 7*B)*Cos[c + d*x]) + 4*(A - B)*(3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)
/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Cos[c + d*x
]*Tan[(c + d*x)/2]^2))/(2*(-1 + Tan[(c + d*x)/2]^2)^2*(-9*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[
(c + d*x)/2]^2] + 2*(3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[3/2, 4/3
, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)) + (9*(3*AppellF1[1/2, 1/3, 1, 3/2, Ta
n[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(A + B + (-5*A + 7*B)*Cos[c + d*x]) + 4*(A - B)*(3*AppellF1[3/2, 1/3, 2
, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x
)/2]^2])*Cos[c + d*x]*Tan[(c + d*x)/2]^2)*(2*(3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/
2]^2] - AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/
2] - 9*(-(AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)
/2])/3 + (AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)
/2])/9) + 2*Tan[(c + d*x)/2]^2*((3*AppellF1[5/2, 4/3, 2, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c
+ d*x)/2]^2*Tan[(c + d*x)/2])/5 - (4*AppellF1[5/2, 7/3, 1, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(
c + d*x)/2]^2*Tan[(c + d*x)/2])/5 + 3*((-6*AppellF1[5/2, 1/3, 3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]
*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/5 + (AppellF1[5/2, 4/3, 2, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]
*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/5))))/(2*(-1 + Tan[(c + d*x)/2]^2)*(-9*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c
+ d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]
- AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)^2) - (9*(-3*(-5*A +
7*B)*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sin[c + d*x] + 4*(A - B)*(3*AppellF1
[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -
Tan[(c + d*x)/2]^2])*Cos[c + d*x]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2] - 4*(A - B)*(3*AppellF1[3/2, 1/3, 2, 5/2
, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^
2])*Sin[c + d*x]*Tan[(c + d*x)/2]^2 + 3*(A + B + (-5*A + 7*B)*Cos[c + d*x])*(-(AppellF1[3/2, 1/3, 2, 5/2, Tan[
(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/3 + (AppellF1[3/2, 4/3, 1, 5/2, Tan[
(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/9) + 4*(A - B)*Cos[c + d*x]*Tan[(c +
d*x)/2]^2*((3*AppellF1[5/2, 4/3, 2, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c +
d*x)/2])/5 - (4*AppellF1[5/2, 7/3, 1, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c
+ d*x)/2])/5 + 3*((-6*AppellF1[5/2, 1/3, 3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*
Tan[(c + d*x)/2])/5 + (AppellF1[5/2, 4/3, 2, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*
Tan[(c + d*x)/2])/5))))/(2*(-1 + Tan[(c + d*x)/2]^2)*(-9*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(
c + d*x)/2]^2] + 2*(3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[3/2, 4/3,
1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2))))/(3*(Cos[(c + d*x)/2]^2*Sec[c + d*x])
^(2/3)) + (2*2^(1/3)*Tan[(c + d*x)/2]*(((A - B)*AppellF1[3/2, 1/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/
2]^2]*Tan[(c + d*x)/2]^2)/(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(2/3) - (9*(3*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c +
d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(A + B + (-5*A + 7*B)*Cos[c + d*x]) + 4*(A - B)*(3*AppellF1[3/2, 1/3, 2, 5/2,
Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]
)*Cos[c + d*x]*Tan[(c + d*x)/2]^2))/(2*(-1 + Tan[(c + d*x)/2]^2)*(-9*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/
2]^2, -Tan[(c + d*x)/2]^2] + 2*(3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - Appell
F1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)))*(-(Cos[(c + d*x)/2]*Sec[c
+ d*x]*Sin[(c + d*x)/2]) + Cos[(c + d*x)/2]^2*Sec[c + d*x]*Tan[c + d*x]))/(9*(Cos[(c + d*x)/2]^2*Sec[c + d*x]
)^(5/3))))
________________________________________________________________________________________
Maple [F] time = 0.147, size = 0, normalized size = 0. \begin{align*} \int{(A+B\sec \left ( dx+c \right ) ) \left ( a+a\sec \left ( dx+c \right ) \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(2/3),x)
[Out]
int((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(2/3),x)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \sec \left (d x + c\right ) + A}{{\left (a \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((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(2/3),x, algorithm="maxima")
[Out]
integrate((B*sec(d*x + c) + A)/(a*sec(d*x + c) + a)^(2/3), x)
________________________________________________________________________________________
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((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(2/3),x, algorithm="fricas")
[Out]
Timed out
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \sec{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{2}{3}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(2/3),x)
[Out]
Integral((A + B*sec(c + d*x))/(a*(sec(c + d*x) + 1))**(2/3), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \sec \left (d x + c\right ) + A}{{\left (a \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((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(2/3),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)/(a*sec(d*x + c) + a)^(2/3), x)