Integrand size = 34, antiderivative size = 62 \[ \int \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a (3 B+C) \tan (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3 d} \]
Time = 0.24 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.69 \[ \int \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a (3 B+2 C+C \sec (c+d x)) \tan (c+d x)}{3 d \sqrt {a (1+\sec (c+d x))}} \]
Time = 0.33 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3042, 4542, 27, 3042, 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.
\(\displaystyle \int \sqrt {a \sec (c+d x)+a} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4542 |
\(\displaystyle \frac {2 \int \frac {1}{2} a (3 B+C) \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx}{3 a}+\frac {2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} (3 B+C) \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} (3 B+C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
\(\Big \downarrow \) 4279 |
\(\displaystyle \frac {2 a (3 B+C) \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
(2*a*(3*B + C)*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d)
3.4.60.3.1 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[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[((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 = 0.64 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.89
method | result | size |
default | \(\frac {2 \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (3 B \sin \left (d x +c \right )+2 C \sin \left (d x +c \right )+C \tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right )}\) | \(55\) |
parts | \(-\frac {2 B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{d}+\frac {2 C \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (2 \sin \left (d x +c \right )+\tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right )}\) | \(78\) |
Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.06 \[ \int \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left ({\left (3 \, B + 2 \, C\right )} \cos \left (d x + c\right ) + C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \]
2/3*((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))
\[ \int \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (B + C \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]
\[ \int \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {a \sec \left (d x + c\right ) + a} \,d x } \]
-2/3*((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*(((B + 2*C)*d*cos(2*d*x + 2*c)^ 2 + (B + 2*C)*d*sin(2*d*x + 2*c)^2 + 2*(B + 2*C)*d*cos(2*d*x + 2*c) + (B + 2*C)*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(3/2*arc tan2(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(3/2*arctan2(sin(2*d*x + 2* c), cos(2*d*x + 2*c))))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((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))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (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)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d *x + 2*c))))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))/((( 2*(2*cos(4*d*x + 4*c) + cos(2*d*x + 2*c))*cos(6*d*x + 6*c) + cos(6*d*x ...
\[ \int \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {a \sec \left (d x + c\right ) + a} \,d x } \]
Time = 17.74 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.56 \[ \int \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (6\,B\,\sin \left (c+d\,x\right )+6\,C\,\sin \left (c+d\,x\right )+6\,B\,\sin \left (2\,c+2\,d\,x\right )+6\,B\,\sin \left (3\,c+3\,d\,x\right )+3\,B\,\sin \left (4\,c+4\,d\,x\right )+8\,C\,\sin \left (2\,c+2\,d\,x\right )+6\,C\,\sin \left (3\,c+3\,d\,x\right )+2\,C\,\sin \left (4\,c+4\,d\,x\right )\right )}{3\,d\,\left (12\,\cos \left (c+d\,x\right )+8\,\cos \left (2\,c+2\,d\,x\right )+4\,\cos \left (3\,c+3\,d\,x\right )+\cos \left (4\,c+4\,d\,x\right )+7\right )} \]
(2*((a*(cos(c + d*x) + 1))/cos(c + d*x))^(1/2)*(6*B*sin(c + d*x) + 6*C*sin (c + d*x) + 6*B*sin(2*c + 2*d*x) + 6*B*sin(3*c + 3*d*x) + 3*B*sin(4*c + 4* d*x) + 8*C*sin(2*c + 2*d*x) + 6*C*sin(3*c + 3*d*x) + 2*C*sin(4*c + 4*d*x)) )/(3*d*(12*cos(c + d*x) + 8*cos(2*c + 2*d*x) + 4*cos(3*c + 3*d*x) + cos(4* c + 4*d*x) + 7))