3.272 \(\int (a+a \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx\)
Optimal. Leaf size=787 \[ \frac {3 \sqrt {2} a A \tan (c+d x) (\sec (c+d x)+1) \sqrt [3]{a \sec (c+d x)+a} F_1\left (\frac {11}{6};\frac {1}{2},1;\frac {17}{6};\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{11 d \sqrt {1-\sec (c+d x)}}+\frac {3 a B \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a}}{4 d}-\frac {15 \left (1+\sqrt {3}\right ) a B \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a}}{4 d (\sec (c+d x)+1)^{2/3} \left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )}+\frac {5\ 3^{3/4} \left (1-\sqrt {3}\right ) a B \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4\ 2^{2/3} d (1-\sec (c+d x)) (\sec (c+d x)+1)^{2/3} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}+\frac {15 \sqrt [4]{3} a B \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a} E\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2\ 2^{2/3} d (1-\sec (c+d x)) (\sec (c+d x)+1)^{2/3} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}} \]
[Out]
3/4*a*B*(a+a*sec(d*x+c))^(1/3)*tan(d*x+c)/d-15/4*a*B*(a+a*sec(d*x+c))^(1/3)*(1+3^(1/2))*tan(d*x+c)/d/(1+sec(d*
x+c))^(2/3)/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))+3/11*a*A*AppellF1(11/6,1,1/2,17/6,1+sec(d*x+c),1/2+1/2*
sec(d*x+c))*(1+sec(d*x+c))*(a+a*sec(d*x+c))^(1/3)*2^(1/2)*tan(d*x+c)/d/(1-sec(d*x+c))^(1/2)+15/4*3^(1/4)*a*B*(
(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)))^2/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)/(2^(1/3)-(1+s
ec(d*x+c))^(1/3)*(1-3^(1/2)))*(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))*EllipticE((1-(2^(1/3)-(1+sec(d*x+c))^
(1/3)*(1-3^(1/2)))^2/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(a+a*sec(d*x
+c))^(1/3)*(2^(1/3)-(1+sec(d*x+c))^(1/3))*((2^(2/3)+2^(1/3)*(1+sec(d*x+c))^(1/3)+(1+sec(d*x+c))^(2/3))/(2^(1/3
)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)*tan(d*x+c)*2^(1/3)/d/(1-sec(d*x+c))/(1+sec(d*x+c))^(2/3)/(-(1+sec
(d*x+c))^(1/3)*(2^(1/3)-(1+sec(d*x+c))^(1/3))/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)+5/8*3^(3/4)*
a*B*((2^(1/3)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)))^2/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)/(2^(1/3)
-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)))*(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))*EllipticF((1-(2^(1/3)-(1+sec(d*x
+c))^(1/3)*(1-3^(1/2)))^2/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(a+a*se
c(d*x+c))^(1/3)*(2^(1/3)-(1+sec(d*x+c))^(1/3))*(1-3^(1/2))*((2^(2/3)+2^(1/3)*(1+sec(d*x+c))^(1/3)+(1+sec(d*x+c
))^(2/3))/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)*tan(d*x+c)*2^(1/3)/d/(1-sec(d*x+c))/(1+sec(d*x+c
))^(2/3)/(-(1+sec(d*x+c))^(1/3)*(2^(1/3)-(1+sec(d*x+c))^(1/3))/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(
1/2)
________________________________________________________________________________________
Rubi [A] time = 0.84, antiderivative size = 787, normalized size of antiderivative =
1.00, number of steps used = 11, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\)
= 0.440, Rules used = {3924, 3779, 3778, 136, 3828, 3827, 50, 63, 308, 225, 1881}
\[ \frac {3 \sqrt {2} a A \tan (c+d x) (\sec (c+d x)+1) \sqrt [3]{a \sec (c+d x)+a} F_1\left (\frac {11}{6};\frac {1}{2},1;\frac {17}{6};\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{11 d \sqrt {1-\sec (c+d x)}}+\frac {3 a B \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a}}{4 d}-\frac {15 \left (1+\sqrt {3}\right ) a B \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a}}{4 d (\sec (c+d x)+1)^{2/3} \left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )}+\frac {5\ 3^{3/4} \left (1-\sqrt {3}\right ) a B \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4\ 2^{2/3} d (1-\sec (c+d x)) (\sec (c+d x)+1)^{2/3} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}+\frac {15 \sqrt [4]{3} a B \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a} E\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2\ 2^{2/3} d (1-\sec (c+d x)) (\sec (c+d x)+1)^{2/3} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x]),x]
[Out]
(3*a*B*(a + a*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(4*d) + (3*Sqrt[2]*a*A*AppellF1[11/6, 1/2, 1, 17/6, (1 + Sec[c
+ d*x])/2, 1 + Sec[c + d*x]]*(1 + Sec[c + d*x])*(a + a*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(11*d*Sqrt[1 - Sec[c
+ d*x]]) - (15*(1 + Sqrt[3])*a*B*(a + a*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(4*d*(1 + Sec[c + d*x])^(2/3)*(2^(1
/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))) + (15*3^(1/4)*a*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]*(a + a*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 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2]*Tan[c + d*x])/(2*2^(2/3)*d*(1 - Sec[c + d*
x])*(1 + 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)]) + (5*3^(3/4)*(1 - Sqrt[3])*a*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]*(a +
a*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 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2]*Tan[c + d*x])/(4*2^(2/3)*d*(1
- Sec[c + d*x])*(1 + 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 50
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && 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 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 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 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 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]
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 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 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 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 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]
Rubi steps
\begin {align*} \int (a+a \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx &=A \int (a+a \sec (c+d x))^{4/3} \, dx+B \int \sec (c+d x) (a+a \sec (c+d x))^{4/3} \, dx\\ &=\frac {\left (a A \sqrt [3]{a+a \sec (c+d x)}\right ) \int (1+\sec (c+d x))^{4/3} \, dx}{\sqrt [3]{1+\sec (c+d x)}}+\frac {\left (a B \sqrt [3]{a+a \sec (c+d x)}\right ) \int \sec (c+d x) (1+\sec (c+d x))^{4/3} \, dx}{\sqrt [3]{1+\sec (c+d x)}}\\ &=-\frac {\left (a A \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {(1+x)^{5/6}}{\sqrt {1-x} x} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{5/6}}-\frac {\left (a B \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {(1+x)^{5/6}}{\sqrt {1-x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{5/6}}\\ &=\frac {3 a B \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{4 d}+\frac {3 \sqrt {2} a A F_1\left (\frac {11}{6};\frac {1}{2},1;\frac {17}{6};\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{11 d \sqrt {1-\sec (c+d x)}}-\frac {\left (5 a B \sqrt [3]{a+a \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 )}{4 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{5/6}}\\ &=\frac {3 a B \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{4 d}+\frac {3 \sqrt {2} a A F_1\left (\frac {11}{6};\frac {1}{2},1;\frac {17}{6};\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{11 d \sqrt {1-\sec (c+d x)}}-\frac {\left (15 a B \sqrt [3]{a+a \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 )}{2 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{5/6}}\\ &=\frac {3 a B \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{4 d}+\frac {3 \sqrt {2} a A F_1\left (\frac {11}{6};\frac {1}{2},1;\frac {17}{6};\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{11 d \sqrt {1-\sec (c+d x)}}+\frac {\left (15 a B \sqrt [3]{a+a \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 )}{4 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{5/6}}+\frac {\left (15 \left (1-\sqrt {3}\right ) a B \sqrt [3]{a+a \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 )}{2 \sqrt [3]{2} d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{5/6}}\\ &=\frac {3 a B \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{4 d}+\frac {3 \sqrt {2} a A F_1\left (\frac {11}{6};\frac {1}{2},1;\frac {17}{6};\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{11 d \sqrt {1-\sec (c+d x)}}-\frac {15 \left (1+\sqrt {3}\right ) a B \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{4 d (1+\sec (c+d x))^{2/3} \left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )}+\frac {15 \sqrt [4]{3} a 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]{a+a \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^{2/3} d (1-\sec (c+d x)) (1+\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 {5\ 3^{3/4} \left (1-\sqrt {3}\right ) a 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]{a+a \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)}{4\ 2^{2/3} d (1-\sec (c+d x)) (1+\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*}
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Mathematica [B] time = 19.61, size = 4110, normalized size = 5.22 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x]),x]
[Out]
(Cos[c + d*x]*((1 + Cos[c + d*x])*Sec[c + d*x])^(1/3)*(a*(1 + Sec[c + d*x]))^(4/3)*(A + B*Sec[c + d*x])*((3*(4
*A + 5*B)*Sin[c + d*x])/4 + (3*B*Tan[c + d*x])/4))/(d*(B + A*Cos[c + d*x])*(1 + Sec[c + d*x])^(4/3)) + (Cos[c
+ d*x]*(a*(1 + Sec[c + d*x]))^(4/3)*(A + B*Sec[c + d*x])*(2*A*(1 + Sec[c + d*x])^(1/3) + (5*B*(1 + Sec[c + d*x
])^(1/3))/4 + Cos[c + d*x]*(-3*A*(1 + Sec[c + d*x])^(1/3) - (15*B*(1 + Sec[c + d*x])^(1/3))/4))*Tan[(c + d*x)/
2]*(-(((4*A + 5*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)/(Co
s[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]*(-4*A + 5*B + 5*(4*A + 7*B)*Cos[c + d*x]) - 4*(4*A + 5*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))))/(6*2^(2/3)*d*(B + A*Cos[c + d*x])*(Cos[(c
+ d*x)/2]^2*Sec[c + d*x])^(2/3)*(1 + Sec[c + d*x])^(4/3)*((Sec[(c + d*x)/2]^2*(-(((4*A + 5*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]*(-4*A + 5*B + 5*(4*A + 7*B)*Cos
[c + d*x]) - 4*(4*A + 5*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))))/(12*2^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)) + (Tan[(c + d*x)/2]*(-((
(4*A + 5*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)) - ((4*A + 5*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*(4*A + 5*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, -Tan[(c + d*x)/2]^2]*(-4*A + 5*B + 5*(4*A + 7*B)*Cos[c + d*x]) - 4*(4*A + 5*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*App
ellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(-4*A + 5*B + 5*(4*A + 7*B)*Cos[c + d*x]) - 4
*(4*A + 5*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*(-1/3*(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]) + (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, Ta
n[(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, Ta
n[(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])*T
an[(c + d*x)/2]^2)^2) - (9*(-15*(4*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*(4*A + 5*B)*(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])*Cos[c + d*x]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2
] + 4*(4*A + 5*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*(-4*A + 5*B + 5*(4*A + 7
*B)*Cos[c + d*x])*(-1/3*(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]) + (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*(4*A + 5*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, Ta
n[(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, Ta
n[(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])*T
an[(c + d*x)/2]^2))))/(6*2^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)) - (Tan[(c + d*x)/2]*(-(((4*A + 5*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]*(-4*A + 5*B + 5
*(4*A + 7*B)*Cos[c + d*x]) - 4*(4*A + 5*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*App
ellF1[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)))*(-(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*2^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(5/3))))
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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((a+a*sec(d*x+c))^(4/3)*(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 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {4}{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(4/3), x)
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maple [F] time = 1.73, size = 0, normalized size = 0.00 \[ \int \left (a +a \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)),x)
[Out]
int((a+a*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)),x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {4}{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)),x, algorithm="maxima")
[Out]
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(4/3), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{4/3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(4/3),x)
[Out]
int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(4/3), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {4}{3}} \left (A + B \sec {\left (c + d x \right )}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))**(4/3)*(A+B*sec(d*x+c)),x)
[Out]
Integral((a*(sec(c + d*x) + 1))**(4/3)*(A + B*sec(c + d*x)), x)
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