∫ 3 ∘ __________ a + a sec(c + dx) (A + B sec(c + dx) + C sec2(c + dx)) dx

3.630 \(\int \sqrt [3]{a+a \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=786 \[ \frac {3 \sqrt {2} A \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} F_1\left (\frac {5}{6};\frac {1}{2},1;\frac {11}{6};\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{5 d \sqrt {1-\sec (c+d x)}}-\frac {3 \left (1+\sqrt {3}\right ) (4 B+C) \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a}}{4 d (\sec (c+d x)+1)^{2/3} \left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )}+\frac {3^{3/4} \left (1-\sqrt {3}\right ) (4 B+C) \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4\ 2^{2/3} d (1-\sec (c+d x)) (\sec (c+d x)+1)^{2/3} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}+\frac {3 \sqrt [4]{3} (4 B+C) \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a} E\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2\ 2^{2/3} d (1-\sec (c+d x)) (\sec (c+d x)+1)^{2/3} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}+\frac {3 C \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a}}{4 d} \]

[Out]

3/4*C*(a+a*sec(d*x+c))^(1/3)*tan(d*x+c)/d-3/4*(4*B+C)*(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)))+3/5*A*AppellF1(5/6,1,1/2,11/6,1+sec(d*x+c),1/2+1/2*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)+3/4*3^(1/4)*(4*B+C)*((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))*(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)+1/8*3^(3/4)*(4*B+C)*((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*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))/(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 = 786, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {4054, 3924, 3779, 3778, 136, 3828, 3827, 63, 308, 225, 1881} \[ \frac {3 \sqrt {2} A \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} F_1\left (\frac {5}{6};\frac {1}{2},1;\frac {11}{6};\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{5 d \sqrt {1-\sec (c+d x)}}-\frac {3 \left (1+\sqrt {3}\right ) (4 B+C) \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a}}{4 d (\sec (c+d x)+1)^{2/3} \left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )}+\frac {3^{3/4} \left (1-\sqrt {3}\right ) (4 B+C) \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4\ 2^{2/3} d (1-\sec (c+d x)) (\sec (c+d x)+1)^{2/3} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}+\frac {3 \sqrt [4]{3} (4 B+C) \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a} E\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2\ 2^{2/3} d (1-\sec (c+d x)) (\sec (c+d x)+1)^{2/3} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}+\frac {3 C \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a}}{4 d} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(3*C*(a + a*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(4*d) + (3*Sqrt[2]*A*AppellF1[5/6, 1/2, 1, 11/6, (1 + Sec[c + d*

x])/2, 1 + Sec[c + d*x]]*(a + a*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(5*d*Sqrt[1 - Sec[c + d*x]]) - (3*(1 + Sqrt[

3])*(4*B + C)*(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))) + (3*3^(1/4)*(4*B + C)*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)]) + (3^(3/4)*(1 - Sqrt[3])*(4*B + C)*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 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]

Rule 4054

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +

(a_))^(m_.), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(b*(m + 1)),

Int[(a + b*Csc[e + f*x])^m*Simp[A*b*(m + 1) + (a*C*m + b*B*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,

e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \sqrt [3]{a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {3 C \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{4 d}+\frac {3 \int \sqrt [3]{a+a \sec (c+d x)} \left (\frac {4 a A}{3}+\frac {1}{3} a (4 B+C) \sec (c+d x)\right ) \, dx}{4 a}\\ &=\frac {3 C \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{4 d}+A \int \sqrt [3]{a+a \sec (c+d x)} \, dx+\frac {1}{4} (4 B+C) \int \sec (c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx\\ &=\frac {3 C \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{4 d}+\frac {\left (A \sqrt [3]{a+a \sec (c+d x)}\right ) \int \sqrt [3]{1+\sec (c+d x)} \, dx}{\sqrt [3]{1+\sec (c+d x)}}+\frac {\left ((4 B+C) \sqrt [3]{a+a \sec (c+d x)}\right ) \int \sec (c+d x) \sqrt [3]{1+\sec (c+d x)} \, dx}{4 \sqrt [3]{1+\sec (c+d x)}}\\ &=\frac {3 C \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{4 d}-\frac {\left (A \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} x \sqrt [6]{1+x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{5/6}}-\frac {\left ((4 B+C) \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 C \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{4 d}+\frac {3 \sqrt {2} A F_1\left (\frac {5}{6};\frac {1}{2},1;\frac {11}{6};\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{5 d \sqrt {1-\sec (c+d x)}}-\frac {\left (3 (4 B+C) \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 C \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{4 d}+\frac {3 \sqrt {2} A F_1\left (\frac {5}{6};\frac {1}{2},1;\frac {11}{6};\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{5 d \sqrt {1-\sec (c+d x)}}+\frac {\left (3 (4 B+C) \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 (3 \left (1-\sqrt {3}\right ) (4 B+C) \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 C \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{4 d}+\frac {3 \sqrt {2} A F_1\left (\frac {5}{6};\frac {1}{2},1;\frac {11}{6};\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{5 d \sqrt {1-\sec (c+d x)}}-\frac {3 \left (1+\sqrt {3}\right ) (4 B+C) \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 {3 \sqrt [4]{3} (4 B+C) 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 {3^{3/4} \left (1-\sqrt {3}\right ) (4 B+C) 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 = 20.64, size = 4191, normalized size = 5.33 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[c + d*x]^2*((1 + Cos[c + d*x])*Sec[c + d*x])^(1/3)*(a*(1 + Sec[c + d*x]))^(1/3)*(A + B*Sec[c + d*x] + C*S

ec[c + d*x]^2)*((3*(4*B + C)*Sin[c + d*x])/2 + (3*C*Tan[c + d*x])/2))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2

*c + 2*d*x])*(1 + Sec[c + d*x])^(1/3)) + (Cos[c + d*x]^2*(a*(1 + Sec[c + d*x]))^(1/3)*(A + B*Sec[c + d*x] + C*

Sec[c + d*x]^2)*(2*A*(1 + Sec[c + d*x])^(1/3) + 2*B*(1 + Sec[c + d*x])^(1/3) + (C*(1 + Sec[c + d*x])^(1/3))/2

+ Cos[c + d*x]*(-6*B*(1 + Sec[c + d*x])^(1/3) - (3*C*(1 + Sec[c + d*x])^(1/3))/2))*Tan[(c + d*x)/2]*(-(((4*B +

C)*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]*(8*A - 4*B

- C + (8*A - 7*(4*B + C))*Cos[c + d*x]))/2 + 2*(4*B + C)*(3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -T

an[(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))/((-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)*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos

[2*c + 2*d*x])*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)*(1 + Sec[c + d*x])^(1/3)*((Sec[(c + d*x)/2]^2*(-(((4*B

+ C)*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]*(8*A - 4*

B - C + (8*A - 7*(4*B + C))*Cos[c + d*x]))/2 + 2*(4*B + C)*(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))/((-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, Ta

n[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(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*B + C)*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*B + C)*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*B + C)*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]*(8*A - 4*B - C + (8*A - 7*(4*B + C))*Cos[c + d*x]))

/2 + 2*(4*B + C)*(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))/((-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, Ta

n[(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]*(8*A - 4*B -

C + (8*A - 7*(4*B + C))*Cos[c + d*x]))/2 + 2*(4*B + C)*(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, 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))))/((-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, Ta

n[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)^2) + (9*((-3*(8*A - 7*(4*B + C))*AppellF1[1/2, 1/3

, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sin[c + d*x])/2 + 2*(4*B + C)*(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] - 2*(4*B + C)*(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*(8*A - 4*B - C + (8*A - 7*(4*B + C))*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, Ta

n[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/9))/2 + 2*(4*B + C)*Cos[c + d*x]*T

an[(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*T

an[(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))))/((-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]*(-(((4*B + C)*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]*(8*A - 4*B - C + (8*A - 7*(4*B + C))*Cos[c + d*x]))/2 + 2*(4*

B + C)*(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))/((-1 + Tan[(c + d*x)/2]^2)*(-9*Appell

F1[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)))*(-(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))))

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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))^(1/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F] time = 1.21, size = 0, normalized size = 0.00 \[ \int \left (a +a \sec \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**(1/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((a*(sec(c + d*x) + 1))**(1/3)*(A + B*sec(c + d*x) + C*sec(c + d*x)**2), x)

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