Integrand size = 35, antiderivative size = 100 \[ \int \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \sqrt {a} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a (3 B+C) \tan (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3 d} \]
2*A*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*a^(1/2)/d+2/3*a*(3*B +C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/3*C*(a+a*sec(d*x+c))^(1/2)*tan(d *x+c)/d
Time = 1.74 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.01 \[ \int \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sqrt {a (1+\sec (c+d x))} \left (3 \sqrt {2} A \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {3}{2}}(c+d x)+2 (C+(3 B+2 C) \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{3 d} \]
(Sec[(c + d*x)/2]*Sec[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(3*Sqrt[2]*A*Arc Sin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(3/2) + 2*(C + (3*B + 2*C)*Cos[ c + d*x])*Sin[(c + d*x)/2]))/(3*d)
Time = 0.59 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 4542, 27, 3042, 4403, 3042, 4261, 216, 4279}
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.
| \(\displaystyle \int \sqrt {a \sec (c+d x)+a} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4542 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sqrt {\sec (c+d x) a+a} (3 a A+a (3 B+C) \sec (c+d x))dx}{3 a}+\frac {2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sqrt {\sec (c+d x) a+a} (3 a A+a (3 B+C) \sec (c+d x))dx}{3 a}+\frac {2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a A+a (3 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{3 a}+\frac {2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 4403 |
| \(\displaystyle \frac {3 a A \int \sqrt {\sec (c+d x) a+a}dx+a (3 B+C) \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx}{3 a}+\frac {2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 a A \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+a (3 B+C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {a (3 B+C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {6 a^2 A \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{3 a}+\frac {2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {a (3 B+C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {6 a^{3/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}+\frac {2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {\frac {6 a^{3/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^2 (3 B+C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}+\frac {2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
(2*C*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d) + ((6*a^(3/2)*A*ArcTan[( Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*(3*B + C)*Tan[ c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(3*a)
3.5.86.3.1 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_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_ .) + (c_)), x_Symbol] :> Simp[c Int[Sqrt[a + b*Csc[e + f*x]], x], x] + Si mp[d Int[Sqrt[a + b*Csc[e + f*x]]*Csc[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
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)]
Time = 1.01 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.54
| method | result | size |
| parts | \(\frac {2 A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )}{d}-\frac {2 B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{d}+\frac {2 C \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (2 \sin \left (d x +c \right )+\tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right )}\) | \(154\) |
| default | \(\frac {2 \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (3 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+3 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+3 B \sin \left (d x +c \right )+2 C \sin \left (d x +c \right )+C \tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right )}\) | \(183\) |
2*A/d*(a*(1+sec(d*x+c)))^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh( sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-2*B/d*(a*(1+ sec(d*x+c)))^(1/2)*(cot(d*x+c)-csc(d*x+c))+2/3*C/d*(a*(1+sec(d*x+c)))^(1/2 )/(cos(d*x+c)+1)*(2*sin(d*x+c)+tan(d*x+c))
Time = 0.32 (sec) , antiderivative size = 304, normalized size of antiderivative = 3.04 \[ \int \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {3 \, {\left (A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left ({\left (3 \, B + 2 \, C\right )} \cos \left (d x + c\right ) + C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, -\frac {2 \, {\left (3 \, {\left (A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left ({\left (3 \, B + 2 \, C\right )} \cos \left (d x + c\right ) + C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \]
[1/3*(3*(A*cos(d*x + c)^2 + A*cos(d*x + c))*sqrt(-a)*log((2*a*cos(d*x + c) ^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d *x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*((3*B + 2*C)*cos(d*x + c) + C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d* x + c)^2 + d*cos(d*x + c)), -2/3*(3*(A*cos(d*x + c)^2 + A*cos(d*x + c))*sq rt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a) *sin(d*x + c))) - ((3*B + 2*C)*cos(d*x + c) + C)*sqrt((a*cos(d*x + c) + a) /cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c))]
\[ \int \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \]
\[ \int \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \,d x } \]
-1/6*(4*(3*B*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin( 2*d*x + 2*c) - (3*B*cos(2*d*x + 2*c) + 3*B + 2*C)*sin(3/2*arctan2(sin(2*d* x + 2*c), cos(2*d*x + 2*c) + 1)))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + 3*((A*cos(2*d*x + 2*c)^2 + A*si n(2*d*x + 2*c)^2 + 2*A*cos(2*d*x + 2*c) + A)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2* d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c) ^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2 *d*x + 2*c) + 1)) + 1) - (A*cos(2*d*x + 2*c)^2 + A*sin(2*d*x + 2*c)^2 + 2* A*cos(2*d*x + 2*c) + A)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2* c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1 ) - 2*(A*d*cos(2*d*x + 2*c)^2 + A*d*sin(2*d*x + 2*c)^2 + 2*A*d*cos(2*d*x + 2*c) + A*d)*integrate((((cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 2*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(5/2 *arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)*sin(6*d* x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2*d* x + 2*c) - 2*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*sin(5/2*arctan2(sin(2*d...
\[ \int \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]