Integrand size = 31, antiderivative size = 101 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {8 a^2 (5 A+3 B) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (5 A+3 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 B (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]
2/5*B*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+8/15*a^2*(5*A+3*B)*tan(d*x+c)/d/ (a+a*sec(d*x+c))^(1/2)+2/15*a*(5*A+3*B)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/ d
Time = 0.23 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.69 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {2 a \sqrt {a (1+\sec (c+d x))} ((25 A+18 B) \sin (c+d x)+(5 A+9 B+3 B \sec (c+d x)) \tan (c+d x))}{15 d (1+\cos (c+d x))} \]
(2*a*Sqrt[a*(1 + Sec[c + d*x])]*((25*A + 18*B)*Sin[c + d*x] + (5*A + 9*B + 3*B*Sec[c + d*x])*Tan[c + d*x]))/(15*d*(1 + Cos[c + d*x]))
Time = 0.49 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3042, 4489, 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.
| \(\displaystyle \int \sec (c+d x) (a \sec (c+d x)+a)^{3/2} (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {1}{5} (5 A+3 B) \int \sec (c+d x) (\sec (c+d x) a+a)^{3/2}dx+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} (5 A+3 B) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4280 |
| \(\displaystyle \frac {1}{5} (5 A+3 B) \left (\frac {4}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} (5 A+3 B) \left (\frac {4}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {1}{5} (5 A+3 B) \left (\frac {8 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
(2*B*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + ((5*A + 3*B)*((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
3.2.29.3.1 Defintions of rubi rules used
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[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(b*(m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B , e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b *(m + 1), 0] && !LtQ[m, -2^(-1)]
Time = 3.97 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.80
| method | result | size |
| default | \(\frac {2 a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (25 A \sin \left (d x +c \right )+18 B \sin \left (d x +c \right )+5 A \tan \left (d x +c \right )+9 B \tan \left (d x +c \right )+3 B \tan \left (d x +c \right ) \sec \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right )}\) | \(81\) |
| parts | \(\frac {2 A a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (5 \sin \left (d x +c \right )+\tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (6 \sin \left (d x +c \right )+3 \tan \left (d x +c \right )+\sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right )}\) | \(105\) |
2/15*a/d*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*(25*A*sin(d*x+c)+18*B*sin (d*x+c)+5*A*tan(d*x+c)+9*B*tan(d*x+c)+3*B*tan(d*x+c)*sec(d*x+c))
Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.88 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left ({\left (25 \, A + 18 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (5 \, A + 9 \, B\right )} a \cos \left (d x + c\right ) + 3 \, B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]
2/15*((25*A + 18*B)*a*cos(d*x + c)^2 + (5*A + 9*B)*a*cos(d*x + c) + 3*B*a) *sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)
\[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (A + B \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]
\[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]
2/15*(15*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1 )^(1/4)*(((3*A + 2*B)*a*d*cos(2*d*x + 2*c)^2 + (3*A + 2*B)*a*d*sin(2*d*x + 2*c)^2 + 2*(3*A + 2*B)*a*d*cos(2*d*x + 2*c) + (3*A + 2*B)*a*d)*integrate( (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(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*x + 2*c), cos(2*d*x + 2*c) )))*cos(3/2*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(5/2*ar ctan2(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(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 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)^4 + sin(2*d*x + 2*c)^4 + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d* x + 2*c) + 1)*cos(6*d*x + 6*c)^2 + 4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + ...
\[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]
Time = 18.14 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.11 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=-\frac {2\,a\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}-1\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (A\,25{}\mathrm {i}+B\,18{}\mathrm {i}+A\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,10{}\mathrm {i}+A\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,50{}\mathrm {i}+A\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,10{}\mathrm {i}+A\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,25{}\mathrm {i}+B\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,18{}\mathrm {i}+B\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,48{}\mathrm {i}+B\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,18{}\mathrm {i}+B\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,18{}\mathrm {i}\right )}{15\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \]
-(2*a*(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)*(A*25i + B*18i + A*exp(c*1i + d*x*1i)*10i + A*exp(c*2i + d*x*2i)*50i + A*exp(c*3i + d*x*3i)*10i + A*exp(c*4i + d*x*4i)*25i + B*ex p(c*1i + d*x*1i)*18i + B*exp(c*2i + d*x*2i)*48i + B*exp(c*3i + d*x*3i)*18i + B*exp(c*4i + d*x*4i)*18i))/(15*d*(exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d *x*2i) + 1)^2)