3.3.74 \(\int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{2/3}} \, dx\) [274]
Optimal. Leaf size=764 \[ \frac {3 B \tan (c+d x)}{d (a+a \sec (c+d x))^{2/3}}-\frac {3 \sqrt {2} A F_1\left (-\frac {1}{6};\frac {1}{2},1;\frac {5}{6};\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}+\frac {3 \left (1+\sqrt {3}\right ) B \sqrt [3]{1+\sec (c+d x)} \tan (c+d x)}{d (a+a \sec (c+d x))^{2/3} \left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )}-\frac {3 \sqrt [3]{2} \sqrt [4]{3} B E\left (\text {ArcCos}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{d (1-\sec (c+d x)) (a+a \sec (c+d x))^{2/3} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}}-\frac {3^{3/4} \left (1-\sqrt {3}\right ) B F\left (\text {ArcCos}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{2^{2/3} d (1-\sec (c+d x)) (a+a \sec (c+d x))^{2/3} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \]
[Out]
3*B*tan(d*x+c)/d/(a+a*sec(d*x+c))^(2/3)+3*B*(1+sec(d*x+c))^(1/3)*(1+3^(1/2))*tan(d*x+c)/d/(a+a*sec(d*x+c))^(2/
3)/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))-3*A*AppellF1(-1/6,1,1/2,5/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*2^(
1/2)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(2/3)/(1-sec(d*x+c))^(1/2)-3*2^(1/3)*3^(1/4)*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)))*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))*(1+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)/d/(1-sec(d*x+c))/(a+a*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)-1/2*3^(3/4)*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))*(1+sec(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))/(a+a*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 = 0.52, antiderivative size = 764, 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 = {4009, 3864,
3863, 141, 3913, 3912, 53, 65, 314, 231, 1895} \begin {gather*} -\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^{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 (\text {ArcCos}\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 (\text {ArcCos}\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}}+\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}} \end {gather*}
Warning: Unable to verify antiderivative.
[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 53
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 65
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 141
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)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(
b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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 231
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]/(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)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^
2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x]
Rule 314
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 1895
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]/(2*r^2*Sqrt[(r*x^2*(s + r*x^2))/
(s + (1 + Sqrt[3])*r*x^2)^2]*Sqrt[a + b*x^6]))*EllipticE[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r
*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b, c, d}, x] && EqQ[2*Rt[b/a, 3]^2*c - (1 - Sqrt[3])*d, 0]
Rule 3863
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 3864
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^IntPart[n]*((a + b*Csc[c + d*x])^FracPart
[n]/(1 + (b/a)*Csc[c + d*x])^FracPart[n]), Int[(1 + (b/a)*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && !GtQ[a, 0]
Rule 3912
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 3913
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[a^Int
Part[m]*((a + b*Csc[e + f*x])^FracPart[m]/(1 + (b/a)*Csc[e + f*x])^FracPart[m]), Int[(1 + (b/a)*Csc[e + f*x])^
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 4009
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 \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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(4066\) vs. \(2(764)=1528\).
time = 19.15, size = 4066, normalized size = 5.32 \begin {gather*} \text {Result too large to show} \end {gather*}
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)*(-1/3*(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))))/(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*(-(S
ec[(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]*(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, -Ta
n[(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,
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*(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]*S
ec[(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, Ta
n[(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*Appe
llF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan...
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Maple [F]
time = 0.25, size = 0, normalized size = 0.00 \[\int \frac {A +B \sec \left (d x +c \right )}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {2}{3}}}\, dx\]
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)
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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((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)
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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((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(2/3),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 {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 {gather*}
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)
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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((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)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{2/3}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A + B/cos(c + d*x))/(a + a/cos(c + d*x))^(2/3),x)
[Out]
int((A + B/cos(c + d*x))/(a + a/cos(c + d*x))^(2/3), x)
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