Integrand size = 37, antiderivative size = 122 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a (7 A+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}} \]
2/15*a*(7*A+15*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/5 *A*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(3/2)+2/15*A*sin(d*x+c)* (a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)
Time = 1.94 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.56 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(19 A+30 C+8 A \cos (c+d x)+3 A \cos (2 (c+d x))) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{15 d \sqrt {\sec (c+d x)}} \]
((19*A + 30*C + 8*A*Cos[c + d*x] + 3*A*Cos[2*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(15*d*Sqrt[Sec[c + d*x]])
Time = 0.68 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {3042, 4575, 27, 3042, 4501, 3042, 4291}
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.
\(\displaystyle \int \frac {\sqrt {a \sec (c+d x)+a} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4575 |
\(\displaystyle \frac {2 \int \frac {\sqrt {\sec (c+d x) a+a} (a A+a (2 A+5 C) \sec (c+d x))}{2 \sec ^{\frac {3}{2}}(c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sqrt {\sec (c+d x) a+a} (a A+a (2 A+5 C) \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a A+a (2 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{5 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4501 |
\(\displaystyle \frac {\frac {1}{3} a (7 A+15 C) \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} a (7 A+15 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4291 |
\(\displaystyle \frac {\frac {2 a^2 (7 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
(2*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + ((2 *a^2*(7*A + 15*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]))/(5*a)
3.3.55.3.1 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[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] *(d_.)], x_Symbol] :> Simp[-2*a*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*S qrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[(a*A*m - b*B*n)/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x] , x] /; FreeQ[{a, b, d, e, f, A, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a ^2 - b^2, 0] && EqQ[m + n + 1, 0] && !LeQ[m, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co t[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/( b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b *(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Time = 0.87 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.57
method | result | size |
default | \(\frac {2 \left (3 A \cos \left (d x +c \right )^{2}+4 A \cos \left (d x +c \right )+8 A +15 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{15 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(69\) |
parts | \(\frac {2 A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (3 \sin \left (d x +c \right )+4 \tan \left (d x +c \right )+8 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 C \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{d \sqrt {\sec \left (d x +c \right )}}\) | \(110\) |
2/15/d*(3*A*cos(d*x+c)^2+4*A*cos(d*x+c)+8*A+15*C)*(a*(1+sec(d*x+c)))^(1/2) /(cos(d*x+c)+1)/sec(d*x+c)^(3/2)*tan(d*x+c)
Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \, {\left (3 \, A \cos \left (d x + c\right )^{3} + 4 \, A \cos \left (d x + c\right )^{2} + {\left (8 \, A + 15 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]
2/15*(3*A*cos(d*x + c)^3 + 4*A*cos(d*x + c)^2 + (8*A + 15*C)*cos(d*x + c)) *sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/((d*cos(d*x + c) + d )*sqrt(cos(d*x + c)))
\[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + C \sec ^{2}{\left (c + d x \right )}\right )}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (104) = 208\).
Time = 0.47 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.84 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\sqrt {2} {\left (30 \, \cos \left (\frac {4}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \cos \left (\frac {2}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 30 \, \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \sin \left (\frac {4}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) - 5 \, \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \sin \left (\frac {2}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 6 \, \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 30 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )\right )} A \sqrt {a} + 120 \, \sqrt {2} C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, d} \]
1/60*(sqrt(2)*(30*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2* c)))*sin(5/2*d*x + 5/2*c) + 5*cos(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/ 2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 30*cos(5/2*d*x + 5/2*c)*sin(4/5*ar ctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 5*cos(5/2*d*x + 5/2*c )*sin(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 6*sin(5/2 *d*x + 5/2*c) + 5*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2* c))) + 30*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*A* sqrt(a) + 120*sqrt(2)*C*sqrt(a)*sin(1/2*d*x + 1/2*c))/d
\[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {a \sec \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Time = 17.91 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (35\,A\,\sin \left (c+d\,x\right )+60\,C\,\sin \left (c+d\,x\right )+8\,A\,\sin \left (2\,c+2\,d\,x\right )+3\,A\,\sin \left (3\,c+3\,d\,x\right )\right )}{30\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \]