Integrand size = 25, antiderivative size = 406 \[ \int (a+a \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\frac {3 \sqrt {2} A \operatorname {AppellF1}\left (\frac {7}{6},\frac {1}{2},1,\frac {13}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{7 d \sqrt {1-\sec (c+d x)}}+\frac {3 B (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{2 d (1+\sec (c+d x))}-\frac {3^{3/4} B \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) (a+a \sec (c+d x))^{2/3} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{2 \sqrt [3]{2} d (1-\sec (c+d x)) (1+\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \]
3/2*B*(a+a*sec(d*x+c))^(2/3)*tan(d*x+c)/d/(1+sec(d*x+c))+3/7*A*AppellF1(7/ 6,1,1/2,13/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*(a+a*sec(d*x+c))^(2/3)*2^(1/ 2)*tan(d*x+c)/d/(1-sec(d*x+c))^(1/2)-1/4*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))* (a+a*sec(d*x+c))^(2/3)*(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^(2/3)/d/(1-sec(d*x+c))/(1+sec(d*x+c))/(-(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)
Leaf count is larger than twice the leaf count of optimal. \(4445\) vs. \(2(406)=812\).
Time = 17.41 (sec) , antiderivative size = 4445, normalized size of antiderivative = 10.95 \[ \int (a+a \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\text {Result too large to show} \]
(3*B*Cos[c + d*x]*((1 + Cos[c + d*x])*Sec[c + d*x])^(2/3)*(a*(1 + Sec[c + d*x]))^(2/3)*(A + B*Sec[c + d*x])*Tan[(c + d*x)/2])/(2*d*(B + A*Cos[c + d* x])*(1 + Sec[c + d*x])^(2/3)) + (Cos[c + d*x]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)*(a*(1 + Sec[c + d*x]))^(2/3)*(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*( (A*(1 + Sec[c + d*x])^(2/3))/2 + (B*(1 + Sec[c + d*x])^(2/3))/4))*Tan[(c + d*x)/2]*(2*B*AppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x )/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])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(2/3)*Tan[(c + d*x)/2]^4 + 9*Appell F1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(3*(4*A + B) *Cos[(c + d*x)/2]^2 + B*AppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Ta n[(c + d*x)/2]^2]*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(2/3)*Tan[(c + d*x)/2] ^2)))/(3*2^(1/3)*d*(B + A*Cos[c + d*x])*(1 + Sec[c + d*x])^(2/3)*(9*Appell F1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(-3*Appe llF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*Appell F1[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)*((Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)*(2*B *AppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/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*...
Time = 0.91 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.06, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4412, 3042, 4266, 3042, 4265, 149, 25, 1012, 4315, 3042, 4314, 60, 73, 766}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int (a \sec (c+d x)+a)^{2/3} (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{2/3} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4412 |
| \(\displaystyle A \int (\sec (c+d x) a+a)^{2/3}dx+B \int \sec (c+d x) (\sec (c+d x) a+a)^{2/3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle A \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx\) |
| \(\Big \downarrow \) 4266 |
| \(\displaystyle \frac {A (a \sec (c+d x)+a)^{2/3} \int (\sec (c+d x)+1)^{2/3}dx}{(\sec (c+d x)+1)^{2/3}}+B \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A (a \sec (c+d x)+a)^{2/3} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{2/3}dx}{(\sec (c+d x)+1)^{2/3}}+B \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx\) |
| \(\Big \downarrow \) 4265 |
| \(\displaystyle B \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx-\frac {A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \int \frac {\cos (c+d x) \sqrt [6]{\sec (c+d x)+1}}{\sqrt {1-\sec (c+d x)}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\) |
| \(\Big \downarrow \) 149 |
| \(\displaystyle B \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx-\frac {6 A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \int \frac {\cos (c+d x) (\sec (c+d x)+1)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {6 A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \int -\frac {\cos (c+d x) (\sec (c+d x)+1)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}+B \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx\) |
| \(\Big \downarrow \) 1012 |
| \(\displaystyle B \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx+\frac {3 \sqrt {2} A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},1,\frac {1}{2},\frac {13}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 d \sqrt {1-\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4315 |
| \(\displaystyle \frac {B (a \sec (c+d x)+a)^{2/3} \int \sec (c+d x) (\sec (c+d x)+1)^{2/3}dx}{(\sec (c+d x)+1)^{2/3}}+\frac {3 \sqrt {2} A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},1,\frac {1}{2},\frac {13}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 d \sqrt {1-\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B (a \sec (c+d x)+a)^{2/3} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{2/3}dx}{(\sec (c+d x)+1)^{2/3}}+\frac {3 \sqrt {2} A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},1,\frac {1}{2},\frac {13}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 d \sqrt {1-\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4314 |
| \(\displaystyle \frac {3 \sqrt {2} A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},1,\frac {1}{2},\frac {13}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 d \sqrt {1-\sec (c+d x)}}-\frac {B \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \int \frac {\sqrt [6]{\sec (c+d x)+1}}{\sqrt {1-\sec (c+d x)}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3 \sqrt {2} A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},1,\frac {1}{2},\frac {13}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 d \sqrt {1-\sec (c+d x)}}-\frac {B \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}d\sec (c+d x)-\frac {3}{2} \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1}\right )}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {3 \sqrt {2} A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},1,\frac {1}{2},\frac {13}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 d \sqrt {1-\sec (c+d x)}}-\frac {B \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \left (3 \int \frac {1}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}-\frac {3}{2} \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1}\right )}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\) |
| \(\Big \downarrow \) 766 |
| \(\displaystyle \frac {3 \sqrt {2} A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},1,\frac {1}{2},\frac {13}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 d \sqrt {1-\sec (c+d x)}}-\frac {B \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \left (\frac {3^{3/4} \sqrt [6]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [3]{2} \sqrt {1-\sec (c+d x)} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}-\frac {3}{2} \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1}\right )}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\) |
(3*Sqrt[2]*A*AppellF1[7/6, 1, 1/2, 13/6, 1 + Sec[c + d*x], (1 + Sec[c + d* x])/2]*(a + a*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(7*d*Sqrt[1 - Sec[c + d*x] ]) - (B*(a + a*Sec[c + d*x])^(2/3)*((-3*Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d*x])^(1/6))/2 + (3^(3/4)*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])^(1/6)*(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])/(2*2^(1/3) *Sqrt[1 - Sec[c + d*x]]*Sqrt[-(((1 + Sec[c + d*x])^(1/3)*(2^(1/3) - (1 + S ec[c + d*x])^(1/3)))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2) ]))*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d*x])^(7/6))
3.3.69.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) )^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b Subst[Int[x^(k*(m + 1 ) - 1)*(c - a*(d/b) + d*(x^k/b))^n*(e - a*(f/b) + f*(x^k/b))^p, x], x, (a + b*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[2*n] && IntegerQ[p]
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 ]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[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] /; Fr eeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && GtQ[a, 0 ]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[a^IntPar t[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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[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] && !IntegerQ[m] && GtQ[a, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a^IntPart[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]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d _.) + (c_)), x_Symbol] :> Simp[c Int[(a + b*Csc[e + f*x])^m, x], x] + Sim p[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]
\[\int \left (a +a \sec \left (d x +c \right )\right )^{\frac {2}{3}} \left (A +B \sec \left (d x +c \right )\right )d x\]
Timed out. \[ \int (a+a \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
\[ \int (a+a \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {2}{3}} \left (A + B \sec {\left (c + d x \right )}\right )\, dx \]
\[ \int (a+a \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \]
\[ \int (a+a \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \]
Timed out. \[ \int (a+a \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{2/3} \,d x \]