Integrand size = 34, antiderivative size = 138 \[ \int (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {64 a^3 (7 B+5 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (7 B+5 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d} \]
2/35*a*(7*B+5*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/7*C*(a+a*sec(d*x+c) )^(5/2)*tan(d*x+c)/d+64/105*a^3*(7*B+5*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1 /2)+16/105*a^2*(7*B+5*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 0.46 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.57 \[ \int (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^3 \left (301 B+230 C+(98 B+115 C) \sec (c+d x)+3 (7 B+20 C) \sec ^2(c+d x)+15 C \sec ^3(c+d x)\right ) \tan (c+d x)}{105 d \sqrt {a (1+\sec (c+d x))}} \]
(2*a^3*(301*B + 230*C + (98*B + 115*C)*Sec[c + d*x] + 3*(7*B + 20*C)*Sec[c + d*x]^2 + 15*C*Sec[c + d*x]^3)*Tan[c + d*x])/(105*d*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.62 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {3042, 4542, 27, 3042, 4280, 3042, 4280, 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.
| \(\displaystyle \int (a \sec (c+d x)+a)^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4542 |
| \(\displaystyle \frac {2 \int \frac {1}{2} a (7 B+5 C) \sec (c+d x) (\sec (c+d x) a+a)^{5/2}dx}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} (7 B+5 C) \int \sec (c+d x) (\sec (c+d x) a+a)^{5/2}dx+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} (7 B+5 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}dx+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4280 |
| \(\displaystyle \frac {1}{7} (7 B+5 C) \left (\frac {8}{5} a \int \sec (c+d x) (\sec (c+d x) a+a)^{3/2}dx+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} (7 B+5 C) \left (\frac {8}{5} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4280 |
| \(\displaystyle \frac {1}{7} (7 B+5 C) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} (7 B+5 C) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {1}{7} (7 B+5 C) \left (\frac {8}{5} a \left (\frac {8 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\) |
(2*C*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + ((7*B + 5*C)*((2*a*( a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (8*a*((8*a^2*Tan[c + d*x]) /(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d* x])/(3*d)))/5))/7
3.4.77.3.1 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[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ Symbol] :> Simp[(-b)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Simp[a*((2*m - 1)/m) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && Intege rQ[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)]
Time = 21.64 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.80
| method | result | size |
| default | \(\frac {2 a^{2} \left (301 B \cos \left (d x +c \right )^{3}+230 C \cos \left (d x +c \right )^{3}+98 B \cos \left (d x +c \right )^{2}+115 C \cos \left (d x +c \right )^{2}+21 B \cos \left (d x +c \right )+60 C \cos \left (d x +c \right )+15 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{105 d \left (\cos \left (d x +c \right )+1\right )}\) | \(111\) |
| parts | \(\frac {2 B \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (43 \sin \left (d x +c \right )+14 \tan \left (d x +c \right )+3 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C \,a^{2} \left (46 \cos \left (d x +c \right )^{3}+23 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{21 d \left (\cos \left (d x +c \right )+1\right )}\) | \(139\) |
2/105*a^2/d*(301*B*cos(d*x+c)^3+230*C*cos(d*x+c)^3+98*B*cos(d*x+c)^2+115*C *cos(d*x+c)^2+21*B*cos(d*x+c)+60*C*cos(d*x+c)+15*C)*(a*(1+sec(d*x+c)))^(1/ 2)/(cos(d*x+c)+1)*tan(d*x+c)*sec(d*x+c)^2
Time = 0.26 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.83 \[ \int (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left ({\left (301 \, B + 230 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (98 \, B + 115 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B + 20 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]
2/105*((301*B + 230*C)*a^2*cos(d*x + c)^3 + (98*B + 115*C)*a^2*cos(d*x + c )^2 + 3*(7*B + 20*C)*a^2*cos(d*x + c) + 15*C*a^2)*sqrt((a*cos(d*x + c) + a )/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)
\[ \int (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \left (B + C \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]
\[ \int (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \]
-2/105*((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) ^(1/4)*(7*(15*B*a^2*sin(6*d*x + 6*c) + 5*(17*B + 10*C)*a^2*sin(4*d*x + 4*c ) + (113*B + 100*C)*a^2*sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c) , cos(2*d*x + 2*c) + 1)) - (105*B*a^2*cos(6*d*x + 6*c) + 35*(17*B + 10*C)* a^2*cos(4*d*x + 4*c) + 7*(113*B + 100*C)*a^2*cos(2*d*x + 2*c) + (301*B + 2 30*C)*a^2)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt( a) - 105*(((5*B + 2*C)*a^2*d*cos(2*d*x + 2*c)^4 + (5*B + 2*C)*a^2*d*sin(2* d*x + 2*c)^4 + 4*(5*B + 2*C)*a^2*d*cos(2*d*x + 2*c)^3 + 6*(5*B + 2*C)*a^2* d*cos(2*d*x + 2*c)^2 + 4*(5*B + 2*C)*a^2*d*cos(2*d*x + 2*c) + (5*B + 2*C)* a^2*d + 2*((5*B + 2*C)*a^2*d*cos(2*d*x + 2*c)^2 + 2*(5*B + 2*C)*a^2*d*cos( 2*d*x + 2*c) + (5*B + 2*C)*a^2*d)*sin(2*d*x + 2*c)^2)*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(7/2*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(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(5 /2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*s in(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*s in(2*d*x + 2*c) - 2*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*cos(7/2*arctan2(...
\[ \int (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \]
Time = 20.40 (sec) , antiderivative size = 590, normalized size of antiderivative = 4.28 \[ \int (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {B\,a^2\,2{}\mathrm {i}}{d}-\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (301\,B+230\,C\right )\,2{}\mathrm {i}}{105\,d}\right )}{{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1}-\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {B\,a^2\,2{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (3\,B+4\,C\right )\,10{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (5\,B+2\,C\right )\,2{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (11\,B+10\,C\right )\,2{}\mathrm {i}}{7\,d}\right )+\frac {B\,a^2\,2{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (3\,B+4\,C\right )\,10{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (5\,B+2\,C\right )\,2{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (11\,B+10\,C\right )\,2{}\mathrm {i}}{7\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {a^2\,\left (5\,B+2\,C\right )\,2{}\mathrm {i}}{5\,d}-\frac {a^2\,\left (5\,B+9\,C\right )\,4{}\mathrm {i}}{5\,d}+\frac {a^2\,\left (7\,B-8\,C\right )\,2{}\mathrm {i}}{35\,d}\right )-\frac {B\,a^2\,2{}\mathrm {i}}{5\,d}-\frac {a^2\,\left (B+2\,C\right )\,2{}\mathrm {i}}{d}+\frac {a^2\,\left (B+C\right )\,4{}\mathrm {i}}{d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {a^2\,\left (5\,B+2\,C\right )\,2{}\mathrm {i}}{3\,d}-\frac {a^2\,\left (63\,B+80\,C\right )\,2{}\mathrm {i}}{105\,d}\right )-\frac {B\,a^2\,2{}\mathrm {i}}{3\,d}+\frac {a^2\,\left (9\,B+10\,C\right )\,2{}\mathrm {i}}{3\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )} \]
((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((B*a^2*2i) /d - (a^2*exp(c*1i + d*x*1i)*(301*B + 230*C)*2i)/(105*d)))/(exp(c*1i + d*x *1i) + 1) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2) *(exp(c*1i + d*x*1i)*((B*a^2*2i)/(7*d) + (a^2*(3*B + 4*C)*10i)/(7*d) - (a^ 2*(5*B + 2*C)*2i)/(7*d) - (a^2*(11*B + 10*C)*2i)/(7*d)) + (B*a^2*2i)/(7*d) + (a^2*(3*B + 4*C)*10i)/(7*d) - (a^2*(5*B + 2*C)*2i)/(7*d) - (a^2*(11*B + 10*C)*2i)/(7*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((a^2*(5*B + 2*C)*2i)/(5*d) - (a^2*(5*B + 9*C)*4i)/(5*d) + (a^2* (7*B - 8*C)*2i)/(35*d)) - (B*a^2*2i)/(5*d) - (a^2*(B + 2*C)*2i)/d + (a^2*( B + C)*4i)/d))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d* x*1i)*((a^2*(5*B + 2*C)*2i)/(3*d) - (a^2*(63*B + 80*C)*2i)/(105*d)) - (B*a ^2*2i)/(3*d) + (a^2*(9*B + 10*C)*2i)/(3*d)))/((exp(c*1i + d*x*1i) + 1)*(ex p(c*2i + d*x*2i) + 1))