Integrand size = 33, antiderivative size = 132 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {8 a^2 (35 A+19 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (35 A+19 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}-\frac {4 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 a d} \]
-4/35*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)*t an(d*x+c)/a/d+8/105*a^2*(35*A+19*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/ 105*a*(35*A+19*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 1.39 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.55 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^2 \left (175 A+104 C+(35 A+52 C) \sec (c+d x)+39 C \sec ^2(c+d x)+15 C \sec ^3(c+d x)\right ) \tan (c+d x)}{105 d \sqrt {a (1+\sec (c+d x))}} \]
(2*a^2*(175*A + 104*C + (35*A + 52*C)*Sec[c + d*x] + 39*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 = 0.72 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4571, 27, 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.
| \(\displaystyle \int \sec (c+d x) (a \sec (c+d x)+a)^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4571 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (\sec (c+d x) a+a)^{3/2} (a (7 A+5 C)-2 a C \sec (c+d x))dx}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec (c+d x) (\sec (c+d x) a+a)^{3/2} (a (7 A+5 C)-2 a C \sec (c+d x))dx}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (7 A+5 C)-2 a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {\frac {1}{5} a (35 A+19 C) \int \sec (c+d x) (\sec (c+d x) a+a)^{3/2}dx-\frac {4 a C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} a (35 A+19 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx-\frac {4 a C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 4280 |
| \(\displaystyle \frac {\frac {1}{5} a (35 A+19 C) \left (\frac {4}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} a (35 A+19 C) \left (\frac {4}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {\frac {1}{5} a (35 A+19 C) \left (\frac {8 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
(2*C*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*a*d) + ((-4*a*C*(a + a*Se c[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (a*(35*A + 19*C)*((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*a)
3.2.66.3.1 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[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)]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + 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 + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) In t[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) - a*C* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && !LtQ[m, -1]
Time = 0.64 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.76
| method | result | size |
| default | \(\frac {2 a \left (175 A \cos \left (d x +c \right )^{3}+104 C \cos \left (d x +c \right )^{3}+35 A \cos \left (d x +c \right )^{2}+52 C \cos \left (d x +c \right )^{2}+39 C \cos \left (d x +c \right )+15 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{105 d \left (\cos \left (d x +c \right )+1\right )}\) | \(100\) |
| parts | \(\frac {2 A a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (5 \sin \left (d x +c \right )+\tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C a \left (104 \cos \left (d x +c \right )^{3}+52 \cos \left (d x +c \right )^{2}+39 \cos \left (d x +c \right )+15\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{105 d \left (\cos \left (d x +c \right )+1\right )}\) | \(119\) |
2/105*a/d*(175*A*cos(d*x+c)^3+104*C*cos(d*x+c)^3+35*A*cos(d*x+c)^2+52*C*co s(d*x+c)^2+39*C*cos(d*x+c)+15*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*t an(d*x+c)*sec(d*x+c)^2
Time = 0.26 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.77 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left ({\left (175 \, A + 104 \, C\right )} a \cos \left (d x + c\right )^{3} + {\left (35 \, A + 52 \, C\right )} a \cos \left (d x + c\right )^{2} + 39 \, C a \cos \left (d x + c\right ) + 15 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]
2/105*((175*A + 104*C)*a*cos(d*x + c)^3 + (35*A + 52*C)*a*cos(d*x + c)^2 + 39*C*a*cos(d*x + c) + 15*C*a)*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)
\[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]
\[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]
-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*A*a*sin(6*d*x + 6*c) + 5*(11*A + 4*C)*a*sin(4*d*x + 4*c) + 1 3*(5*A + 4*C)*a*sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2* d*x + 2*c) + 1)) - (105*A*a*cos(6*d*x + 6*c) + 35*(11*A + 4*C)*a*cos(4*d*x + 4*c) + 91*(5*A + 4*C)*a*cos(2*d*x + 2*c) + (175*A + 104*C)*a)*sin(7/2*a rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) - 105*(3*(A*a*d*c os(2*d*x + 2*c)^4 + A*a*d*sin(2*d*x + 2*c)^4 + 4*A*a*d*cos(2*d*x + 2*c)^3 + 6*A*a*d*cos(2*d*x + 2*c)^2 + 4*A*a*d*cos(2*d*x + 2*c) + A*a*d + 2*(A*a*d *cos(2*d*x + 2*c)^2 + 2*A*a*d*cos(2*d*x + 2*c) + A*a*d)*sin(2*d*x + 2*c)^2 )*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(8*d*x + 8*c)*cos(2*d*x + 2*c) + 3*cos(6*d*x + 6*c)*cos(2 *d*x + 2*c) + 3*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + s in(8*d*x + 8*c)*sin(2*d*x + 2*c) + 3*sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 3 *sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(7/2*arctan2(s in(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 3*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 3*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 3*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 3*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(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1 )) - ((cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 3*cos(2*d*x + 2*c)*sin(6*d*x...
\[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]
Time = 19.90 (sec) , antiderivative size = 510, normalized size of antiderivative = 3.86 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (\frac {A\,a\,2{}\mathrm {i}}{d}-\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (175\,A+104\,C\right )\,2{}\mathrm {i}}{105\,d}\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}}}}{{\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 {a\,\left (7\,A-8\,C\right )\,2{}\mathrm {i}}{35\,d}+\frac {A\,a\,6{}\mathrm {i}}{5\,d}-\frac {a\,\left (A+3\,C\right )\,8{}\mathrm {i}}{5\,d}\right )-\frac {a\,\left (3\,A+8\,C\right )\,2{}\mathrm {i}}{5\,d}+\frac {a\,\left (A+C\right )\,8{}\mathrm {i}}{5\,d}-\frac {A\,a\,2{}\mathrm {i}}{5\,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\,\left (5\,A+4\,C\right )\,2{}\mathrm {i}}{7\,d}-\frac {a\,\left (7\,A+12\,C\right )\,2{}\mathrm {i}}{7\,d}+\frac {A\,a\,4{}\mathrm {i}}{7\,d}\right )+\frac {a\,\left (5\,A+4\,C\right )\,2{}\mathrm {i}}{7\,d}-\frac {a\,\left (7\,A+12\,C\right )\,2{}\mathrm {i}}{7\,d}+\frac {A\,a\,4{}\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\,\left (35\,A+52\,C\right )\,2{}\mathrm {i}}{105\,d}-\frac {A\,a\,2{}\mathrm {i}}{d}\right )-\frac {a\,\left (3\,A+4\,C\right )\,2{}\mathrm {i}}{3\,d}+\frac {A\,a\,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*2i)/d - (a*exp(c*1i + d*x*1i)*(175*A + 104*C)*2i)/(105*d))*(a + a/( exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2))/(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)*(ex p(c*1i + d*x*1i)*((a*(7*A - 8*C)*2i)/(35*d) + (A*a*6i)/(5*d) - (a*(A + 3*C )*8i)/(5*d)) - (a*(3*A + 8*C)*2i)/(5*d) + (a*(A + C)*8i)/(5*d) - (A*a*2i)/ (5*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) + ((a + a/(e xp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*( (a*(5*A + 4*C)*2i)/(7*d) - (a*(7*A + 12*C)*2i)/(7*d) + (A*a*4i)/(7*d)) + ( a*(5*A + 4*C)*2i)/(7*d) - (a*(7*A + 12*C)*2i)/(7*d) + (A*a*4i)/(7*d)))/((e xp(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*(35*A + 52*C)*2i)/(105*d) - (A*a*2i)/d) - (a*(3*A + 4*C)*2i)/(3*d) + (A*a*2i)/(3*d )))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1))