3.628 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx\)

Optimal. Leaf size=466 \[ -\frac {3 \sqrt {2} A \tan (c+d x) F_1\left (-\frac {5}{6};\frac {1}{2},1;\frac {1}{6};\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{5 a^2 d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1) \sqrt [3]{a \sec (c+d x)+a}}-\frac {3 (4 A-4 B-7 C) \tan (c+d x)}{55 a^2 d (\sec (c+d x)+1) \sqrt [3]{a \sec (c+d x)+a}}+\frac {3^{3/4} (4 A-4 B-7 C) \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{55 \sqrt [3]{2} a^2 d (1-\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a}}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}} \]

[Out]

-3/11*(A-B+C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(7/3)-3/55*(4*A-4*B-7*C)*tan(d*x+c)/a^2/d/(1+sec(d*x+c))/(a+a*sec(
d*x+c))^(1/3)-3/5*A*AppellF1(-5/6,1,1/2,1/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*2^(1/2)*tan(d*x+c)/a^2/d/(1+sec(d
*x+c))/(a+a*sec(d*x+c))^(1/3)/(1-sec(d*x+c))^(1/2)+1/110*3^(3/4)*(4*A-4*B-7*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))*(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^(2/3)/a^2/d/(1-sec(d*x+c))/(a+a*sec(d*x+c))^(1/3)/(-(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)

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Rubi [A]  time = 0.52, antiderivative size = 466, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4052, 3924, 3779, 3778, 136, 3828, 3827, 51, 63, 225} \[ -\frac {3 \sqrt {2} A \tan (c+d x) F_1\left (-\frac {5}{6};\frac {1}{2},1;\frac {1}{6};\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{5 a^2 d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1) \sqrt [3]{a \sec (c+d x)+a}}-\frac {3 (4 A-4 B-7 C) \tan (c+d x)}{55 a^2 d (\sec (c+d x)+1) \sqrt [3]{a \sec (c+d x)+a}}+\frac {3^{3/4} (4 A-4 B-7 C) \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{55 \sqrt [3]{2} a^2 d (1-\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a}}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^(7/3),x]

[Out]

(-3*(A - B + C)*Tan[c + d*x])/(11*d*(a + a*Sec[c + d*x])^(7/3)) - (3*(4*A - 4*B - 7*C)*Tan[c + d*x])/(55*a^2*d
*(1 + Sec[c + d*x])*(a + a*Sec[c + d*x])^(1/3)) - (3*Sqrt[2]*A*AppellF1[-5/6, 1/2, 1, 1/6, (1 + Sec[c + d*x])/
2, 1 + Sec[c + d*x]]*Tan[c + d*x])/(5*a^2*d*Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d*x])*(a + a*Sec[c + d*x])^(1/
3)) + (3^(3/4)*(4*A - 4*B - 7*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]*(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])/(55*2^(1/3)*a^2*d*(1 - Sec[c + d*x])*(a + a*Sec[c + d*x])^(1/3)*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 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 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 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 4052

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

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx &=-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}-\frac {3 \int \frac {-\frac {11 a A}{3}+\frac {1}{3} a (4 A-4 B-7 C) \sec (c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx}{11 a^2}\\ &=-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}+\frac {A \int \frac {1}{(a+a \sec (c+d x))^{4/3}} \, dx}{a}-\frac {(4 A-4 B-7 C) \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx}{11 a}\\ &=-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}+\frac {\left (A \sqrt [3]{1+\sec (c+d x)}\right ) \int \frac {1}{(1+\sec (c+d x))^{4/3}} \, dx}{a^2 \sqrt [3]{a+a \sec (c+d x)}}-\frac {\left ((4 A-4 B-7 C) \sqrt [3]{1+\sec (c+d x)}\right ) \int \frac {\sec (c+d x)}{(1+\sec (c+d x))^{4/3}} \, dx}{11 a^2 \sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}-\frac {(A \tan (c+d x)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} x (1+x)^{11/6}} \, dx,x,\sec (c+d x)\right )}{a^2 d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}+\frac {((4 A-4 B-7 C) \tan (c+d x)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} (1+x)^{11/6}} \, dx,x,\sec (c+d x)\right )}{11 a^2 d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}-\frac {3 (4 A-4 B-7 C) \tan (c+d x)}{55 a^2 d (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 \sqrt {2} A F_1\left (-\frac {5}{6};\frac {1}{2},1;\frac {1}{6};\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{5 a^2 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}+\frac {((4 A-4 B-7 C) \tan (c+d x)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{55 a^2 d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}-\frac {3 (4 A-4 B-7 C) \tan (c+d x)}{55 a^2 d (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 \sqrt {2} A F_1\left (-\frac {5}{6};\frac {1}{2},1;\frac {1}{6};\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{5 a^2 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}+\frac {(6 (4 A-4 B-7 C) \tan (c+d x)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{55 a^2 d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}-\frac {3 (4 A-4 B-7 C) \tan (c+d x)}{55 a^2 d (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 \sqrt {2} A F_1\left (-\frac {5}{6};\frac {1}{2},1;\frac {1}{6};\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{5 a^2 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}+\frac {3^{3/4} (4 A-4 B-7 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 ) \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)}{55 \sqrt [3]{2} a^2 d (1-\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \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 = 19.72, size = 3111, normalized size = 6.68 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^(7/3),x]

[Out]

(Cos[c + d*x]^2*((1 + Cos[c + d*x])*Sec[c + d*x])^(2/3)*(1 + Sec[c + d*x])^(7/3)*(A + B*Sec[c + d*x] + C*Sec[c
+ d*x]^2)*((-6*Sec[(c + d*x)/2]*(20*A*Sin[(c + d*x)/2] - 9*B*Sin[(c + d*x)/2] - 2*C*Sin[(c + d*x)/2]))/55 - (
3*Sec[(c + d*x)/2]^5*(A*Sin[(c + d*x)/2] - B*Sin[(c + d*x)/2] + C*Sin[(c + d*x)/2]))/22 + (3*Sec[(c + d*x)/2]^
3*(25*A*Sin[(c + d*x)/2] - 14*B*Sin[(c + d*x)/2] + 3*C*Sin[(c + d*x)/2]))/55))/(d*(A + 2*C + 2*B*Cos[c + d*x]
+ A*Cos[2*c + 2*d*x])*(a*(1 + Sec[c + d*x]))^(7/3)) + (2*2^(2/3)*Cos[c + d*x]^2*(Cos[(c + d*x)/2]^2*Sec[c + d*
x])^(2/3)*(1 + Sec[c + d*x])^(7/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(A*Cos[c + d*x]*Sec[(c + d*x)/2]^2*
(1 + Sec[c + d*x])^(2/3) + Sec[(c + d*x)/2]^2*((-3*A*(1 + Sec[c + d*x])^(2/3))/11 + (4*B*(1 + Sec[c + d*x])^(2
/3))/55 + (7*C*(1 + Sec[c + d*x])^(2/3))/55))*Tan[(c + d*x)/2]*((-70*A + 4*B + 7*C)*AppellF1[3/2, 2/3, 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)^(2/3)*Tan[(c + d*x)/2]^2 + (27*(40
*A + 4*B + 7*C)*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^2)/(9*App
ellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(-3*AppellF1[3/2, 2/3, 2, 5/2, Tan[(c + d
*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*AppellF1[3/2, 5/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c
+ d*x)/2]^2)))/(165*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a*(1 + Sec[c + d*x]))^(7/3)*((2^(2/3
)*Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)*((-70*A + 4*B + 7*C)*AppellF1[3/2, 2/3, 1, 5/2, T
an[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(2/3)*Tan[(c + d*x)/2]^2 + (27*(40*A
+ 4*B + 7*C)*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^2)/(9*Appel
lF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(-3*AppellF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x
)/2]^2, -Tan[(c + d*x)/2]^2] + 2*AppellF1[3/2, 5/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c +
d*x)/2]^2)))/165 + (2*2^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)*Tan[(c + d*x)/2]*((-70*A + 4*B + 7*C)*A
ppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*(Cos[c + d*x]*Sec[(c + d
*x)/2]^2)^(2/3)*Tan[(c + d*x)/2] + (-70*A + 4*B + 7*C)*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(2/3)*Tan[(c + d*x)/2
]^2*((-3*AppellF1[5/2, 2/3, 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*AppellF1[5/2, 5/3, 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) + (2*(-70*A + 4*B + 7*C)*AppellF1[3/2, 2/3, 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)^(1/3)) - (27*(40*A + 4*B + 7*C)*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan
[(c + d*x)/2]^2]*Cos[(c + d*x)/2]*Sin[(c + d*x)/2])/(9*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c
+ d*x)/2]^2] + 2*(-3*AppellF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*AppellF1[3/2, 5/3
, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2) + (27*(40*A + 4*B + 7*C)*Cos[(c + d*x)
/2]^2*(-1/3*(AppellF1[3/2, 2/3, 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]) + (2*AppellF1[3/2, 5/3, 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))/(9*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(-3*AppellF1[3/2, 2/3,
2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*AppellF1[3/2, 5/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c +
d*x)/2]^2])*Tan[(c + d*x)/2]^2) - (27*(40*A + 4*B + 7*C)*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(
c + d*x)/2]^2]*Cos[(c + d*x)/2]^2*(2*(-3*AppellF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] +
2*AppellF1[3/2, 5/3, 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, 2/3, 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]) + (2*AppellF1[3/2, 5/3, 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*((-6*AppellF1[5/2, 2/3, 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 + (2*AppellF1[5/2, 5/3, 2, 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) + 2*((-3*AppellF1[5/2, 5/3, 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, 8/3, 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]))))/(9*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/
2]^2] + 2*(-3*AppellF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*AppellF1[3/2, 5/3, 1, 5/
2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)^2))/165 + (4*2^(2/3)*Tan[(c + d*x)/2]*((-70*A
+ 4*B + 7*C)*AppellF1[3/2, 2/3, 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)^(2/3)*Tan[(c + d*x)/2]^2 + (27*(40*A + 4*B + 7*C)*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c
+ d*x)/2]^2]*Cos[(c + d*x)/2]^2)/(9*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(
-3*AppellF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*AppellF1[3/2, 5/3, 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]) + Co
s[(c + d*x)/2]^2*Sec[c + d*x]*Tan[c + d*x]))/(495*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/3))))

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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+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(7/3),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {7}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(7/3),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/(a*sec(d*x + c) + a)^(7/3), x)

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maple [F]  time = 1.56, size = 0, normalized size = 0.00 \[ \int \frac {A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {7}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(7/3),x)

[Out]

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(7/3),x)

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maxima [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+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(7/3),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{7/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(a + a/cos(c + d*x))^(7/3),x)

[Out]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(a + a/cos(c + d*x))^(7/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {7}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(7/3),x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)/(a*(sec(c + d*x) + 1))**(7/3), x)

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