\(\int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx\) [271]

Optimal result

Integrand size = 25, antiderivative size = 415 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx=\frac {3 B \tan (c+d x)}{5 a d (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 \sqrt {2} A \operatorname {AppellF1}\left (-\frac {5}{6},\frac {1}{2},1,\frac {1}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{5 a d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {3^{3/4} B \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{5 \sqrt [3]{2} a d (1-\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \]

[Out]

3/5*B*tan(d*x+c)/a/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/d/(1+sec(d*x+c))/(a+a*sec(d*x+c))^(1/3)/(1-sec(d*x+c))^(1/2)-1/10*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+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)))*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/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)

Rubi [A] (warning: unable to verify)

Time = 0.51 (sec) , antiderivative size = 415, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4009, 3864, 3863, 141, 3913, 3912, 53, 65, 231} \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx=-\frac {3 \sqrt {2} A \tan (c+d x) \operatorname {AppellF1}\left (-\frac {5}{6},\frac {1}{2},1,\frac {1}{6},\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{5 a d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1) \sqrt [3]{a \sec (c+d x)+a}}-\frac {3^{3/4} B \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{5 \sqrt [3]{2} a d (1-\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a}}+\frac {3 B \tan (c+d x)}{5 a d (\sec (c+d x)+1) \sqrt [3]{a \sec (c+d x)+a}} \]

[In]

Int[(A + B*Sec[c + d*x])/(a + a*Sec[c + d*x])^(4/3),x]

[Out]

(3*B*Tan[c + d*x])/(5*a*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*d*Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d*x])*(
a + a*Sec[c + d*x])^(1/3)) - (3^(3/4)*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]*(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])/(5*2^(1/3)*a*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 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 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*} \text {integral}& = A \int \frac {1}{(a+a \sec (c+d x))^{4/3}} \, dx+B \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx \\ & = \frac {\left (A \sqrt [3]{1+\sec (c+d x)}\right ) \int \frac {1}{(1+\sec (c+d x))^{4/3}} \, dx}{a \sqrt [3]{a+a \sec (c+d x)}}+\frac {\left (B \sqrt [3]{1+\sec (c+d x)}\right ) \int \frac {\sec (c+d x)}{(1+\sec (c+d x))^{4/3}} \, dx}{a \sqrt [3]{a+a \sec (c+d x)}} \\ & = -\frac {(A \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} x (1+x)^{11/6}} \, dx,x,\sec (c+d x)\right )}{a d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}-\frac {(B \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} (1+x)^{11/6}} \, dx,x,\sec (c+d x)\right )}{a d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}} \\ & = \frac {3 B \tan (c+d x)}{5 a d (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 \sqrt {2} A \operatorname {AppellF1}\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 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {(B \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{5 a d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}} \\ & = \frac {3 B \tan (c+d x)}{5 a d (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 \sqrt {2} A \operatorname {AppellF1}\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 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {(6 B \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{5 a d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}} \\ & = \frac {3 B \tan (c+d x)}{5 a d (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 \sqrt {2} A \operatorname {AppellF1}\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 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {3^{3/4} B \operatorname {EllipticF}\left (\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 ) \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)}{5 \sqrt [3]{2} a 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*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2901\) vs. \(2(415)=830\).

Time = 16.69 (sec) , antiderivative size = 2901, normalized size of antiderivative = 6.99 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx=\text {Result too large to show} \]

[In]

Integrate[(A + B*Sec[c + d*x])/(a + a*Sec[c + d*x])^(4/3),x]

[Out]

(Cos[c + d*x]*((1 + Cos[c + d*x])*Sec[c + d*x])^(2/3)*(1 + Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x])*((3*Sec[(c
+ d*x)/2]*(-(A*Sin[(c + d*x)/2]) + B*Sin[(c + d*x)/2]))/5 - (3*Sec[(c + d*x)/2]^3*(-(A*Sin[(c + d*x)/2]) + B*
Sin[(c + d*x)/2]))/10))/(d*(B + A*Cos[c + d*x])*(a*(1 + Sec[c + d*x]))^(4/3)) + (2^(2/3)*Cos[c + d*x]*(Cos[(c
+ d*x)/2]^2*Sec[c + d*x])^(2/3)*(1 + Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x])*((A*Cos[c + d*x]*Sec[(c + d*x)/2
]^2*(1 + Sec[c + d*x])^(2/3))/2 + Sec[(c + d*x)/2]^2*(-1/10*(A*(1 + Sec[c + d*x])^(2/3)) + (B*(1 + Sec[c + d*x
])^(2/3))/10))*Tan[(c + d*x)/2]*((-6*A + B)*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*(4*A + B)*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)))/(15*d*(B + A*Cos[c + d*x])*(a*(1
+ Sec[c + d*x]))^(4/3)*((Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)*((-6*A + B)*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*(4*A + B)*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)))/(15*2^(1/3)) + (2^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)*Tan[(c + d*x)/2]*((-6*A + B
)*AppellF1[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] + (-6*A + B)*(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*(-6*A + B)*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*(-(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)^(1/3)) - (27*(4*A + B)*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*App
ellF1[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*(4*A + B)*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*(4*A + B)*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*A
ppellF1[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*App
ellF1[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*A
ppellF1[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) + 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, T
an[(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))/15 + (2*2^(2/3)*Tan[(c + d*x)/2]*((-6*A + B)*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*(4*A + B)*Appel
lF1[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]) + Cos[(c + d*x)/2]^2*Sec[c + d*x]*Tan[c + d*x]))/(45*(Cos[(c + d*
x)/2]^2*Sec[c + d*x])^(1/3))))

Maple [F]

\[\int \frac {A +B \sec \left (d x +c \right )}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {4}{3}}}d x\]

[In]

int((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(4/3),x)

[Out]

int((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(4/3),x)

Fricas [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx=\text {Timed out} \]

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx=\int \frac {A + B \sec {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {4}{3}}}\, dx \]

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(4/3),x)

[Out]

Integral((A + B*sec(c + d*x))/(a*(sec(c + d*x) + 1))**(4/3), x)

Maxima [F]

\[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \]

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)/(a*sec(d*x + c) + a)^(4/3), x)

Giac [F]

\[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \]

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)/(a*sec(d*x + c) + a)^(4/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{4/3}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{4/3}} \,d x \]

[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)