Integrand size = 40, antiderivative size = 101 \[ \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a (5 B+7 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (5 B-2 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 a d} \]
2/5*C*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+2/15*a*(5*B+7*C)*tan(d*x+c)/d/ (a+a*sec(d*x+c))^(1/2)+2/15*(5*B-2*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 0.56 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.79 \[ \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (5 B+7 C+(5 B+4 C) \cos (c+d x)+(5 B+4 C) \cos (2 (c+d x))) \sec (c+d x) \sqrt {a (1+\sec (c+d x))} \tan (c+d x)}{15 d (1+\cos (c+d x))} \]
(2*(5*B + 7*C + (5*B + 4*C)*Cos[c + d*x] + (5*B + 4*C)*Cos[2*(c + d*x)])*S ec[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*Tan[c + d*x])/(15*d*(1 + Cos[c + d* x]))
Time = 0.69 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {3042, 4560, 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.
| \(\displaystyle \int \sec (c+d x) \sqrt {a \sec (c+d x)+a} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \sec ^2(c+d x) \sqrt {a \sec (c+d x)+a} (B+C \sec (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4498 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) \sqrt {\sec (c+d x) a+a} (3 a C+a (5 B-2 C) \sec (c+d x))dx}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec (c+d x) \sqrt {\sec (c+d x) a+a} (3 a C+a (5 B-2 C) \sec (c+d x))dx}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a C+a (5 B-2 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {\frac {1}{3} a (5 B+7 C) \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a (5 B-2 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} a (5 B+7 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a (5 B-2 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {\frac {2 a^2 (5 B+7 C) \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (5 B-2 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\) |
(2*C*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*a*d) + ((2*a^2*(5*B + 7*C )*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(5*B - 2*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/(5*a)
3.4.59.3.1 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_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.63 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.79
| method | result | size |
| default | \(\frac {2 \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (10 B \sin \left (d x +c \right )+8 C \sin \left (d x +c \right )+5 B \tan \left (d x +c \right )+4 C \tan \left (d x +c \right )+3 C \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right )}\) | \(80\) |
| parts | \(\frac {2 B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (2 \sin \left (d x +c \right )+\tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (8 \sin \left (d x +c \right )+4 \tan \left (d x +c \right )+3 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right )}\) | \(104\) |
2/15/d*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*(10*B*sin(d*x+c)+8*C*sin(d* x+c)+5*B*tan(d*x+c)+4*C*tan(d*x+c)+3*C*sec(d*x+c)*tan(d*x+c))
Time = 0.26 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.86 \[ \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (2 \, {\left (5 \, B + 4 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, B + 4 \, C\right )} \cos \left (d x + c\right ) + 3 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]
integrate(sec(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2), x, algorithm="fricas")
2/15*(2*(5*B + 4*C)*cos(d*x + c)^2 + (5*B + 4*C)*cos(d*x + c) + 3*C)*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)
\[ \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \]
\[ \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right ) \,d x } \]
integrate(sec(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2), x, algorithm="maxima")
4/15*(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)*((B*d*cos(2*d*x + 2*c)^2 + B*d*sin(2*d*x + 2*c)^2 + 2*B*d*cos(2*d* x + 2*c) + B*d)*integrate((((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 + sin(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)*co s(5/2*arctan2(sin(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)*s in(4*d*x + 4*c) - cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 3*cos(6*d*x + 6*c)*s in(2*d*x + 2*c) - 3*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(1/2*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 + 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))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)) ) - (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 + sin(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)*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x...
\[ \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right ) \,d x } \]
integrate(sec(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2), x, algorithm="giac")
Time = 21.44 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.10 \[ \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {4\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}-1\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (B\,5{}\mathrm {i}+C\,4{}\mathrm {i}+B\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,5{}\mathrm {i}+B\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,10{}\mathrm {i}+B\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,5{}\mathrm {i}+B\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,5{}\mathrm {i}+C\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,4{}\mathrm {i}+C\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,14{}\mathrm {i}+C\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,4{}\mathrm {i}+C\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,4{}\mathrm {i}\right )}{15\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \]
-(4*(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)*(B*5i + C*4i + B*exp(c*1i + d*x*1i)*5i + B*exp(c*2i + d*x *2i)*10i + B*exp(c*3i + d*x*3i)*5i + B*exp(c*4i + d*x*4i)*5i + C*exp(c*1i + d*x*1i)*4i + C*exp(c*2i + d*x*2i)*14i + C*exp(c*3i + d*x*3i)*4i + C*exp( c*4i + d*x*4i)*4i))/(15*d*(exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1 )^2)