Integrand size = 35, antiderivative size = 446 \[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 C (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {3 \sqrt {2} A \operatorname {AppellF1}\left (\frac {7}{6},\frac {1}{2},1,\frac {13}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{7 d \sqrt {1-\sec (c+d x)}}+\frac {3 (5 B+2 C) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{10 d (1+\sec (c+d x))}-\frac {3^{3/4} (5 B+2 C) \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) (a+a \sec (c+d x))^{2/3} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{10 \sqrt [3]{2} d (1-\sec (c+d x)) (1+\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \] Output:
3/5*C*(a+a*sec(d*x+c))^(2/3)*tan(d*x+c)/d+3/7*2^(1/2)*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)*tan(d*x+c)/d /(1-sec(d*x+c))^(1/2)+3/10*(5*B+2*C)*(a+a*sec(d*x+c))^(2/3)*tan(d*x+c)/d/( 1+sec(d*x+c))-1/20*3^(3/4)*(5*B+2*C)*InverseJacobiAM(arccos((2^(1/3)-(1-3^ (1/2))*(1+sec(d*x+c))^(1/3))/(2^(1/3)-(1+3^(1/2))*(1+sec(d*x+c))^(1/3))),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+3^(1/2))*(1+sec(d*x+c))^(1/3))^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+3^(1/2))*(1+sec(d*x+c))^(1/3))^2)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(5449\) vs. \(2(446)=892\).
Time = 20.61 (sec) , antiderivative size = 5449, normalized size of antiderivative = 12.22 \[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \] Input:
Integrate[(a + a*Sec[c + d*x])^(2/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^ 2),x]
Output:
Result too large to show
Time = 1.19 (sec) , antiderivative size = 475, normalized size of antiderivative = 1.07, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 4542, 27, 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.
| \(\displaystyle \int (a \sec (c+d x)+a)^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{2/3} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4542 |
| \(\displaystyle \frac {3 \int \frac {1}{3} (\sec (c+d x) a+a)^{2/3} (5 a A+a (5 B+2 C) \sec (c+d x))dx}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\sec (c+d x) a+a)^{2/3} (5 a A+a (5 B+2 C) \sec (c+d x))dx}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3} \left (5 a A+a (5 B+2 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 4412 |
| \(\displaystyle \frac {5 a A \int (\sec (c+d x) a+a)^{2/3}dx+a (5 B+2 C) \int \sec (c+d x) (\sec (c+d x) a+a)^{2/3}dx}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 a A \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx+a (5 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 4266 |
| \(\displaystyle \frac {\frac {5 a A (a \sec (c+d x)+a)^{2/3} \int (\sec (c+d x)+1)^{2/3}dx}{(\sec (c+d x)+1)^{2/3}}+a (5 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {5 a A (a \sec (c+d x)+a)^{2/3} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{2/3}dx}{(\sec (c+d x)+1)^{2/3}}+a (5 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 4265 |
| \(\displaystyle \frac {a (5 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx-\frac {5 a A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \int \frac {\cos (c+d x) \sqrt [6]{\sec (c+d x)+1}}{\sqrt {1-\sec (c+d x)}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 149 |
| \(\displaystyle \frac {a (5 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx-\frac {30 a A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \int \frac {\cos (c+d x) (\sec (c+d x)+1)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {30 a A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \int -\frac {\cos (c+d x) (\sec (c+d x)+1)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}+a (5 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 1012 |
| \(\displaystyle \frac {a (5 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx+\frac {15 \sqrt {2} a A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},1,\frac {1}{2},\frac {13}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 d \sqrt {1-\sec (c+d x)}}}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 4315 |
| \(\displaystyle \frac {\frac {a (5 B+2 C) (a \sec (c+d x)+a)^{2/3} \int \sec (c+d x) (\sec (c+d x)+1)^{2/3}dx}{(\sec (c+d x)+1)^{2/3}}+\frac {15 \sqrt {2} a A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},1,\frac {1}{2},\frac {13}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 d \sqrt {1-\sec (c+d x)}}}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a (5 B+2 C) (a \sec (c+d x)+a)^{2/3} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{2/3}dx}{(\sec (c+d x)+1)^{2/3}}+\frac {15 \sqrt {2} a A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},1,\frac {1}{2},\frac {13}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 d \sqrt {1-\sec (c+d x)}}}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 4314 |
| \(\displaystyle \frac {\frac {15 \sqrt {2} a A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},1,\frac {1}{2},\frac {13}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 d \sqrt {1-\sec (c+d x)}}-\frac {a (5 B+2 C) \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \int \frac {\sqrt [6]{\sec (c+d x)+1}}{\sqrt {1-\sec (c+d x)}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {\frac {15 \sqrt {2} a A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},1,\frac {1}{2},\frac {13}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 d \sqrt {1-\sec (c+d x)}}-\frac {a (5 B+2 C) \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}d\sec (c+d x)-\frac {3}{2} \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1}\right )}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {15 \sqrt {2} a A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},1,\frac {1}{2},\frac {13}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 d \sqrt {1-\sec (c+d x)}}-\frac {a (5 B+2 C) \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \left (3 \int \frac {1}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}-\frac {3}{2} \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1}\right )}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
| \(\Big \downarrow \) 766 |
| \(\displaystyle \frac {\frac {15 \sqrt {2} a A \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},1,\frac {1}{2},\frac {13}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 d \sqrt {1-\sec (c+d x)}}-\frac {a (5 B+2 C) \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \left (\frac {3^{3/4} \sqrt [6]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [3]{2} \sqrt {1-\sec (c+d x)} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}-\frac {3}{2} \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1}\right )}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 d}\) |
Input:
Int[(a + a*Sec[c + d*x])^(2/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(3*C*(a + a*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(5*d) + ((15*Sqrt[2]*a*A*App ellF1[7/6, 1, 1/2, 13/6, 1 + Sec[c + d*x], (1 + Sec[c + d*x])/2]*(a + a*Se c[c + d*x])^(2/3)*Tan[c + d*x])/(7*d*Sqrt[1 - Sec[c + d*x]]) - (a*(5*B + 2 *C)*(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)*Sqr t[1 - Sec[c + d*x]]*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)]))* Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d*x])^(7/6)))/(5*a)
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] :> 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[((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)]
\[\int \left (a +a \sec \left (d x +c \right )\right )^{\frac {2}{3}} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )d x\]
Input:
int((a+a*sec(d*x+c))^(2/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Output:
int((a+a*sec(d*x+c))^(2/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Timed out. \[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))^(2/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algori thm="fricas")
Output:
Timed out
\[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {2}{3}} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \] Input:
integrate((a+a*sec(d*x+c))**(2/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
Output:
Integral((a*(sec(c + d*x) + 1))**(2/3)*(A + B*sec(c + d*x) + C*sec(c + d*x )**2), x)
\[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^(2/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algori thm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(2/ 3), x)
\[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^(2/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algori thm="giac")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(2/ 3), x)
Timed out. \[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{2/3}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \] Input:
int((a + a/cos(c + d*x))^(2/3)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
Output:
int((a + a/cos(c + d*x))^(2/3)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)
\[ \int (a+a \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^{\frac {2}{3}} \left (\left (\int \left (\sec \left (d x +c \right )+1\right )^{\frac {2}{3}}d x \right ) a +\left (\int \left (\sec \left (d x +c \right )+1\right )^{\frac {2}{3}} \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \left (\sec \left (d x +c \right )+1\right )^{\frac {2}{3}} \sec \left (d x +c \right )d x \right ) b \right ) \] Input:
int((a+a*sec(d*x+c))^(2/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Output:
a**(2/3)*(int((sec(c + d*x) + 1)**(2/3),x)*a + int((sec(c + d*x) + 1)**(2/ 3)*sec(c + d*x)**2,x)*c + int((sec(c + d*x) + 1)**(2/3)*sec(c + d*x),x)*b)