Integrand size = 35, antiderivative size = 390 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 \sqrt {2} A \operatorname {AppellF1}\left (\frac {1}{6},\frac {1}{2},1,\frac {7}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}-\frac {3^{3/4} (2 B-C) \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{2 \sqrt [3]{2} d (1-\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \]
3/2*C*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/3)+3*A*AppellF1(1/6,1,1/2,7/6,1+sec (d*x+c),1/2+1/2*sec(d*x+c))*2^(1/2)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/3)/(1 -sec(d*x+c))^(1/2)-1/4*3^(3/4)*(2*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))*(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))/(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)
Leaf count is larger than twice the leaf count of optimal. \(2931\) vs. \(2(390)=780\).
Time = 20.09 (sec) , antiderivative size = 2931, normalized size of antiderivative = 7.52 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\text {Result too large to show} \]
(3*C*Cos[c + d*x]^2*((1 + Cos[c + d*x])*Sec[c + d*x])^(2/3)*(1 + Sec[c + d *x])^(1/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Tan[(c + d*x)/2])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a*(1 + Sec[c + d*x]))^(1/ 3)) + (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])^(1/3)*(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*(B*(1 + Sec[c + d*x])^(2/3) - (C*(1 + Sec[c + d*x])^(2/3))/2))*Tan[(c + d*x)/2]* (-((2*A - 2*B + 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*(2*A + 2*B - 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)))/(3*d*(A + 2*C + 2*B *Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a*(1 + Sec[c + d*x]))^(1/3)*((Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)*(-((2*A - 2*B + C)*App ellF1[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*(2*A + 2*B - C)*A ppellF1[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 ...
Time = 1.03 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.02, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4542, 27, 3042, 4412, 3042, 4266, 3042, 4265, 149, 25, 936, 4315, 3042, 4314, 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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt [3]{a \sec (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt [3]{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
| \(\Big \downarrow \) 4542 |
| \(\displaystyle \frac {3 \int \frac {2 a A+a (2 B-C) \sec (c+d x)}{3 \sqrt [3]{\sec (c+d x) a+a}}dx}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {2 a A+a (2 B-C) \sec (c+d x)}{\sqrt [3]{\sec (c+d x) a+a}}dx}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 a A+a (2 B-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4412 |
| \(\displaystyle \frac {2 a A \int \frac {1}{\sqrt [3]{\sec (c+d x) a+a}}dx+a (2 B-C) \int \frac {\sec (c+d x)}{\sqrt [3]{\sec (c+d x) a+a}}dx}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a A \int \frac {1}{\sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+a (2 B-C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4266 |
| \(\displaystyle \frac {\frac {2 a A \sqrt [3]{\sec (c+d x)+1} \int \frac {1}{\sqrt [3]{\sec (c+d x)+1}}dx}{\sqrt [3]{a \sec (c+d x)+a}}+a (2 B-C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 a A \sqrt [3]{\sec (c+d x)+1} \int \frac {1}{\sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right )+1}}dx}{\sqrt [3]{a \sec (c+d x)+a}}+a (2 B-C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4265 |
| \(\displaystyle \frac {a (2 B-C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a A \tan (c+d x) \int \frac {\cos (c+d x)}{\sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 149 |
| \(\displaystyle \frac {a (2 B-C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {12 a A \tan (c+d x) \int \frac {\cos (c+d x)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {12 a A \tan (c+d x) \int -\frac {\cos (c+d x)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}+a (2 B-C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 936 |
| \(\displaystyle \frac {a (2 B-C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {6 \sqrt {2} a A \tan (c+d x) \operatorname {AppellF1}\left (\frac {1}{6},1,\frac {1}{2},\frac {7}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{a \sec (c+d x)+a}}}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4315 |
| \(\displaystyle \frac {\frac {a (2 B-C) \sqrt [3]{\sec (c+d x)+1} \int \frac {\sec (c+d x)}{\sqrt [3]{\sec (c+d x)+1}}dx}{\sqrt [3]{a \sec (c+d x)+a}}+\frac {6 \sqrt {2} a A \tan (c+d x) \operatorname {AppellF1}\left (\frac {1}{6},1,\frac {1}{2},\frac {7}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{a \sec (c+d x)+a}}}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a (2 B-C) \sqrt [3]{\sec (c+d x)+1} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right )+1}}dx}{\sqrt [3]{a \sec (c+d x)+a}}+\frac {6 \sqrt {2} a A \tan (c+d x) \operatorname {AppellF1}\left (\frac {1}{6},1,\frac {1}{2},\frac {7}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{a \sec (c+d x)+a}}}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4314 |
| \(\displaystyle \frac {\frac {6 \sqrt {2} a A \tan (c+d x) \operatorname {AppellF1}\left (\frac {1}{6},1,\frac {1}{2},\frac {7}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{a \sec (c+d x)+a}}-\frac {a (2 B-C) \tan (c+d x) \int \frac {1}{\sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {6 \sqrt {2} a A \tan (c+d x) \operatorname {AppellF1}\left (\frac {1}{6},1,\frac {1}{2},\frac {7}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{a \sec (c+d x)+a}}-\frac {6 a (2 B-C) \tan (c+d x) \int \frac {1}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 766 |
| \(\displaystyle \frac {\frac {6 \sqrt {2} a A \tan (c+d x) \operatorname {AppellF1}\left (\frac {1}{6},1,\frac {1}{2},\frac {7}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{a \sec (c+d x)+a}}-\frac {3^{3/4} a (2 B-C) \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [3]{2} d (1-\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a}}}{2 a}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{a \sec (c+d x)+a}}\) |
(3*C*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(1/3)) + ((6*Sqrt[2]*a*A*Appe llF1[1/6, 1, 1/2, 7/6, 1 + Sec[c + d*x], (1 + Sec[c + d*x])/2]*Tan[c + d*x ])/(d*Sqrt[1 - Sec[c + d*x]]*(a + a*Sec[c + d*x])^(1/3)) - (3^(3/4)*a*(2*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]*(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^(1/3)*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)]))/(2*a)
3.7.26.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) ], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[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[((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] + Simp[1/(b*(m + 1)) I nt[(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)]
\[\int \frac {A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int \frac {A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\sqrt [3]{a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \]