Integrand size = 35, antiderivative size = 137 \[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a (35 A+27 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (35 A+18 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 C \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{7 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d} \]
2/35*C*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+2/105*a*(35*A+27*C)*tan(d*x+c )/d/(a+a*sec(d*x+c))^(1/2)+2/105*(35*A+18*C)*(a+a*sec(d*x+c))^(1/2)*tan(d* x+c)/d+2/7*C*sec(d*x+c)^2*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 0.67 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.52 \[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a \left (70 A+48 C+(35 A+24 C) \sec (c+d x)+18 C \sec ^2(c+d x)+15 C \sec ^3(c+d x)\right ) \tan (c+d x)}{105 d \sqrt {a (1+\sec (c+d x))}} \]
(2*a*(70*A + 48*C + (35*A + 24*C)*Sec[c + d*x] + 18*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 = 0.86 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4577, 27, 3042, 4498, 27, 3042, 4489, 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.
| \(\displaystyle \int \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a} \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4577 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec ^2(c+d x) \sqrt {\sec (c+d x) a+a} (a (7 A+4 C)+a C \sec (c+d x))dx}{7 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec ^2(c+d x) \sqrt {\sec (c+d x) a+a} (a (7 A+4 C)+a C \sec (c+d x))dx}{7 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (7 A+4 C)+a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 4498 |
| \(\displaystyle \frac {\frac {2 \int \frac {1}{2} \sec (c+d x) \sqrt {\sec (c+d x) a+a} \left (3 C a^2+(35 A+18 C) \sec (c+d x) a^2\right )dx}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \sec (c+d x) \sqrt {\sec (c+d x) a+a} \left (3 C a^2+(35 A+18 C) \sec (c+d x) a^2\right )dx}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 C a^2+(35 A+18 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {\frac {\frac {1}{3} a^2 (35 A+27 C) \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^2 (35 A+18 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {1}{3} a^2 (35 A+27 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^2 (35 A+18 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {\frac {\frac {2 a^3 (35 A+27 C) \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (35 A+18 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{7 d}\) |
(2*C*Sec[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(7*d) + ((2*C*( a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + ((2*a^3*(35*A + 27*C)*Tan[ c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(35*A + 18*C)*Sqrt[a + a *Sec[c + d*x]]*Tan[c + d*x])/(3*d))/(5*a))/(7*a)
3.2.57.3.1 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_)*(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_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]* ((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int [Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B) *Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a *B, 0] && !LtQ[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[(-C) *Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n *Simp[A*b*(m + n + 1) + b*C*n + a*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m, n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Time = 0.57 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.72
| method | result | size |
| default | \(\frac {2 \left (70 A \cos \left (d x +c \right )^{3}+48 C \cos \left (d x +c \right )^{3}+35 A \cos \left (d x +c \right )^{2}+24 C \cos \left (d x +c \right )^{2}+18 C \cos \left (d x +c \right )+15 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{105 d \left (\cos \left (d x +c \right )+1\right )}\) | \(99\) |
| parts | \(\frac {2 A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (2 \sin \left (d x +c \right )+\tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C \left (16 \cos \left (d x +c \right )^{3}+8 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+5\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{35 d \left (\cos \left (d x +c \right )+1\right )}\) | \(117\) |
2/105/d*(70*A*cos(d*x+c)^3+48*C*cos(d*x+c)^3+35*A*cos(d*x+c)^2+24*C*cos(d* x+c)^2+18*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 = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.72 \[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (2 \, {\left (35 \, A + 24 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (35 \, A + 24 \, C\right )} \cos \left (d x + c\right )^{2} + 18 \, C \cos \left (d x + c\right ) + 15 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]
2/105*(2*(35*A + 24*C)*cos(d*x + c)^3 + (35*A + 24*C)*cos(d*x + c)^2 + 18* C*cos(d*x + c) + 15*C)*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)
\[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{2} \,d x } \]
-4/105*((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*(5*A*sin(4*d*x + 4*c) + 2*(5*A + 6*C)*sin(2*d*x + 2*c))*cos(7/2* arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (35*A*cos(4*d*x + 4*c) + 14*(5*A + 6*C)*cos(2*d*x + 2*c) + 35*A + 24*C)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) - 105*((A*d*cos(2*d*x + 2*c)^4 + A*d*sin(2*d*x + 2*c)^4 + 4*A*d*cos(2*d*x + 2*c)^3 + 6*A*d*cos(2*d*x + 2*c) ^2 + 4*A*d*cos(2*d*x + 2*c) + 2*(A*d*cos(2*d*x + 2*c)^2 + 2*A*d*cos(2*d*x + 2*c) + A*d)*sin(2*d*x + 2*c)^2 + A*d)*integrate((((cos(10*d*x + 10*c)*co s(2*d*x + 2*c) + 4*cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 6*cos(6*d*x + 6*c)* cos(2*d*x + 2*c) + 4*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^ 2 + sin(10*d*x + 10*c)*sin(2*d*x + 2*c) + 4*sin(8*d*x + 8*c)*sin(2*d*x + 2 *c) + 6*sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 4*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(10*d*x + 10*c) + 4*cos(2*d*x + 2*c)*sin(8*d *x + 8*c) + 6*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 4*cos(2*d*x + 2*c)*sin(4 *d*x + 4*c) - cos(10*d*x + 10*c)*sin(2*d*x + 2*c) - 4*cos(8*d*x + 8*c)*sin (2*d*x + 2*c) - 6*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 4*cos(4*d*x + 4*c)*s in(2*d*x + 2*c))*sin(7/2*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(10*d*x + 10*c) + 4*cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 6*cos(2*d*x...
\[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{2} \,d x } \]
Time = 19.93 (sec) , antiderivative size = 423, normalized size of antiderivative = 3.09 \[ \int \sec ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,4{}\mathrm {i}}{3\,d}-\frac {C\,16{}\mathrm {i}}{35\,d}\right )+\frac {A\,4{}\mathrm {i}}{3\,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}}}}{\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 )}+\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 {A\,8{}\mathrm {i}}{7\,d}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,8{}\mathrm {i}}{7\,d}-\frac {\left (8\,A+16\,C\right )\,1{}\mathrm {i}}{7\,d}\right )-\frac {\left (8\,A+16\,C\right )\,1{}\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\,4{}\mathrm {i}}{5\,d}+\frac {C\,16{}\mathrm {i}}{35\,d}+\frac {\left (28\,A+112\,C\right )\,1{}\mathrm {i}}{35\,d}\right )-\frac {A\,4{}\mathrm {i}}{5\,d}+\frac {\left (28\,A+112\,C\right )\,1{}\mathrm {i}}{35\,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 {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\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 (140\,A+96\,C\right )\,1{}\mathrm {i}}{105\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )} \]
((exp(c*1i + d*x*1i)*((A*4i)/(3*d) - (C*16i)/(35*d)) + (A*4i)/(3*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)*(exp(c*2i + d*x*2i) + 1)) + ((a + a/(exp(- c*1i - d*x*1i)/2 + ex p(c*1i + d*x*1i)/2))^(1/2)*((A*8i)/(7*d) + exp(c*1i + d*x*1i)*((A*8i)/(7*d ) - ((8*A + 16*C)*1i)/(7*d)) - ((8*A + 16*C)*1i)/(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 + e xp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((C*16i)/(35*d) - (A*4i)/( 5*d) + ((28*A + 112*C)*1i)/(35*d)) - (A*4i)/(5*d) + ((28*A + 112*C)*1i)/(3 5*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - (exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(14 0*A + 96*C)*1i)/(105*d*(exp(c*1i + d*x*1i) + 1))