3.161 \(\int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx\)

Optimal. Leaf size=731 \[ \frac {15 \tan (c+d x)}{7 a d (a \sec (c+d x)+a)^{2/3}}+\frac {15 \left (1+\sqrt {3}\right ) \tan (c+d x) \sqrt [3]{\sec (c+d x)+1}}{7 a d \left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right ) (a \sec (c+d x)+a)^{2/3}}-\frac {3 \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}-\frac {5\ 3^{3/4} \left (1-\sqrt {3}\right ) \tan (c+d x) \sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{7\ 2^{2/3} a d (1-\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3}}-\frac {15 \sqrt [3]{2} \sqrt [4]{3} \tan (c+d x) \sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} E\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{7 a d (1-\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3}} \]

[Out]

-3/7*tan(d*x+c)/d/(a+a*sec(d*x+c))^(5/3)+15/7*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(2/3)+15/7*(1+sec(d*x+c))^(1/3)*
(1+3^(1/2))*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(2/3)/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))-15/7*2^(1/3)*3^(1
/4)*((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)/a/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)-5/14*3^(3/4)*
((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)/a/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 = 0.67, antiderivative size = 731, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3797, 3828, 3827, 51, 63, 308, 225, 1881} \[ \frac {15 \tan (c+d x)}{7 a d (a \sec (c+d x)+a)^{2/3}}+\frac {15 \left (1+\sqrt {3}\right ) \tan (c+d x) \sqrt [3]{\sec (c+d x)+1}}{7 a d \left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right ) (a \sec (c+d x)+a)^{2/3}}-\frac {3 \tan (c+d x)}{7 d (a \sec (c+d x)+a)^{5/3}}-\frac {5\ 3^{3/4} \left (1-\sqrt {3}\right ) \tan (c+d x) \sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{7\ 2^{2/3} a d (1-\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3}}-\frac {15 \sqrt [3]{2} \sqrt [4]{3} \tan (c+d x) \sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} E\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{7 a d (1-\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a + a*Sec[c + d*x])^(5/3),x]

[Out]

(-3*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^(5/3)) + (15*Tan[c + d*x])/(7*a*d*(a + a*Sec[c + d*x])^(2/3)) + (1
5*(1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3)*Tan[c + d*x])/(7*a*d*(a + a*Sec[c + d*x])^(2/3)*(2^(1/3) - (1 + Sqrt[
3])*(1 + Sec[c + d*x])^(1/3))) - (15*2^(1/3)*3^(1/4)*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 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2]*Tan[c + d*x])/(7*a*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)]) - (5*3^(3/4)*(1 - Sqrt[3])*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 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2]*Tan[c + d*x])/(7*2^(2/3)*a*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 + Sq
rt[3])*(1 + Sec[c + d*x])^(1/3))^2)])

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 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 3797

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^m)/(f*(2*m + 1)), x] + Dist[m/(b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

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]

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx &=-\frac {3 \tan (c+d x)}{7 d (a+a \sec (c+d x))^{5/3}}+\frac {5 \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{2/3}} \, dx}{7 a}\\ &=-\frac {3 \tan (c+d x)}{7 d (a+a \sec (c+d x))^{5/3}}+\frac {\left (5 (1+\sec (c+d x))^{2/3}\right ) \int \frac {\sec (c+d x)}{(1+\sec (c+d x))^{2/3}} \, dx}{7 a (a+a \sec (c+d x))^{2/3}}\\ &=-\frac {3 \tan (c+d x)}{7 d (a+a \sec (c+d x))^{5/3}}-\frac {\left (5 \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 )}{7 a d \sqrt {1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}\\ &=-\frac {3 \tan (c+d x)}{7 d (a+a \sec (c+d x))^{5/3}}+\frac {15 \tan (c+d x)}{7 a d (a+a \sec (c+d x))^{2/3}}+\frac {\left (5 \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 )}{7 a d \sqrt {1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}\\ &=-\frac {3 \tan (c+d x)}{7 d (a+a \sec (c+d x))^{5/3}}+\frac {15 \tan (c+d x)}{7 a d (a+a \sec (c+d x))^{2/3}}+\frac {\left (30 \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 )}{7 a d \sqrt {1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}\\ &=-\frac {3 \tan (c+d x)}{7 d (a+a \sec (c+d x))^{5/3}}+\frac {15 \tan (c+d x)}{7 a d (a+a \sec (c+d x))^{2/3}}-\frac {\left (15 \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 )}{7 a d \sqrt {1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}-\frac {\left (15\ 2^{2/3} \left (1-\sqrt {3}\right ) \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 )}{7 a d \sqrt {1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}\\ &=-\frac {3 \tan (c+d x)}{7 d (a+a \sec (c+d x))^{5/3}}+\frac {15 \tan (c+d x)}{7 a d (a+a \sec (c+d x))^{2/3}}+\frac {15 \left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)} \tan (c+d x)}{7 a 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 {15 \sqrt [3]{2} \sqrt [4]{3} 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)}{7 a 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 {5\ 3^{3/4} \left (1-\sqrt {3}\right ) 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)}{7\ 2^{2/3} a 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 [C]  time = 0.36, size = 90, normalized size = 0.12 \[ \frac {\tan (c+d x) \left (5\ 2^{5/6} \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sqrt [6]{\sec (c+d x)+1} \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x))\right )-3\right )}{7 d (a (\sec (c+d x)+1))^{5/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a + a*Sec[c + d*x])^(5/3),x]

[Out]

((-3 + 5*2^(5/6)*Cos[(c + d*x)/2]^2*Hypergeometric2F1[1/2, 7/6, 3/2, (1 - Sec[c + d*x])/2]*Sec[c + d*x]*(1 + S
ec[c + d*x])^(1/6))*Tan[c + d*x])/(7*d*(a*(1 + Sec[c + d*x]))^(5/3))

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fricas [F]  time = 1.14, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \sec \left (d x + c\right )^{2}}{a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+a*sec(d*x+c))^(5/3),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{2}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+a*sec(d*x+c))^(5/3),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^2/(a*sec(d*x + c) + a)^(5/3), x)

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maple [F]  time = 0.71, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}\left (d x +c \right )}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {5}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a+a*sec(d*x+c))^(5/3),x)

[Out]

int(sec(d*x+c)^2/(a+a*sec(d*x+c))^(5/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{2}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+a*sec(d*x+c))^(5/3),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^2/(a*sec(d*x + c) + a)^(5/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^2*(a + a/cos(c + d*x))^(5/3)),x)

[Out]

int(1/(cos(c + d*x)^2*(a + a/cos(c + d*x))^(5/3)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2/(a+a*sec(d*x+c))**(5/3),x)

[Out]

Integral(sec(c + d*x)**2/(a*(sec(c + d*x) + 1))**(5/3), x)

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