Integrand size = 31, antiderivative size = 62 \[ \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 a (3 A+B) \tan (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {2 B \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3 d} \] Output:
2/3*a*(3*A+B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/3*B*(a+a*sec(d*x+c))^( 1/2)*tan(d*x+c)/d
Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.85 \[ \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 (B+(3 A+2 B) \cos (c+d x)) \sqrt {a (1+\sec (c+d x))} \tan (c+d x)}{3 d (1+\cos (c+d x))} \] Input:
Integrate[Sec[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]),x]
Output:
(2*(B + (3*A + 2*B)*Cos[c + d*x])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[c + d*x]) /(3*d*(1 + Cos[c + d*x]))
Time = 0.37 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {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.
\(\displaystyle \int \sec (c+d x) \sqrt {a \sec (c+d x)+a} (A+B \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 4489 |
\(\displaystyle \frac {1}{3} (3 A+B) \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 B \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} (3 A+B) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 B \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
\(\Big \downarrow \) 4279 |
\(\displaystyle \frac {2 a (3 A+B) \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 B \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
Input:
Int[Sec[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]),x]
Output:
(2*a*(3*A + B)*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*B*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d)
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)]
Time = 0.79 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92
method | result | size |
default | \(\frac {\left (6 A \sin \left (d x +c \right )+4 B \sin \left (d x +c \right )+2 B \tan \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (3 \cos \left (d x +c \right )+3\right )}\) | \(57\) |
parts | \(-\frac {2 A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{d}+\frac {B \left (4 \sin \left (d x +c \right )+2 \tan \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (3 \cos \left (d x +c \right )+3\right )}\) | \(81\) |
Input:
int(sec(d*x+c)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x,method=_RETURNVER BOSE)
Output:
1/d*(6*A*sin(d*x+c)+4*B*sin(d*x+c)+2*B*tan(d*x+c))/(3*cos(d*x+c)+3)*(a*(1+ sec(d*x+c)))^(1/2)
Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.06 \[ \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left ({\left (3 \, A + 2 \, B\right )} \cos \left (d x + c\right ) + B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \] Input:
integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorithm= "fricas")
Output:
2/3*((3*A + 2*B)*cos(d*x + c) + B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)) *sin(d*x + c)/(d*cos(d*x + c)^2 + d*cos(d*x + c))
\[ \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)*(a+a*sec(d*x+c))**(1/2)*(A+B*sec(d*x+c)),x)
Output:
Integral(sqrt(a*(sec(c + d*x) + 1))*(A + B*sec(c + d*x))*sec(c + d*x), x)
\[ \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right ) \,d x } \] Input:
integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorithm= "maxima")
Output:
-2/3*((3*A*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(2* d*x + 2*c) - (3*A*cos(2*d*x + 2*c) + 3*A + 2*B)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) - 3*(((A + 2*B)*d*cos(2*d*x + 2*c)^ 2 + (A + 2*B)*d*sin(2*d*x + 2*c)^2 + 2*(A + 2*B)*d*cos(2*d*x + 2*c) + (A + 2*B)*d)*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(3/2*arc tan2(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(3/2*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(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(3/2*arctan2(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(3/2*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 + 2*c) + 1)))/((( 2*(2*cos(4*d*x + 4*c) + cos(2*d*x + 2*c))*cos(6*d*x + 6*c) + cos(6*d*x ...
Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (54) = 108\).
Time = 0.37 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.08 \[ \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=-\frac {2 \, {\left (3 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (3 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{3 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \] Input:
integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorithm= "giac")
Output:
-2/3*(3*sqrt(2)*A*a^2*sgn(cos(d*x + c)) + 3*sqrt(2)*B*a^2*sgn(cos(d*x + c) ) - (3*sqrt(2)*A*a^2*sgn(cos(d*x + c)) + sqrt(2)*B*a^2*sgn(cos(d*x + c)))* tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)
Time = 1.94 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.56 \[ \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (6\,A\,\sin \left (c+d\,x\right )+6\,B\,\sin \left (c+d\,x\right )+6\,A\,\sin \left (2\,c+2\,d\,x\right )+6\,A\,\sin \left (3\,c+3\,d\,x\right )+3\,A\,\sin \left (4\,c+4\,d\,x\right )+8\,B\,\sin \left (2\,c+2\,d\,x\right )+6\,B\,\sin \left (3\,c+3\,d\,x\right )+2\,B\,\sin \left (4\,c+4\,d\,x\right )\right )}{3\,d\,\left (12\,\cos \left (c+d\,x\right )+8\,\cos \left (2\,c+2\,d\,x\right )+4\,\cos \left (3\,c+3\,d\,x\right )+\cos \left (4\,c+4\,d\,x\right )+7\right )} \] Input:
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(1/2))/cos(c + d*x),x)
Output:
(2*((a*(cos(c + d*x) + 1))/cos(c + d*x))^(1/2)*(6*A*sin(c + d*x) + 6*B*sin (c + d*x) + 6*A*sin(2*c + 2*d*x) + 6*A*sin(3*c + 3*d*x) + 3*A*sin(4*c + 4* d*x) + 8*B*sin(2*c + 2*d*x) + 6*B*sin(3*c + 3*d*x) + 2*B*sin(4*c + 4*d*x)) )/(3*d*(12*cos(c + d*x) + 8*cos(2*c + 2*d*x) + 4*cos(3*c + 3*d*x) + cos(4* c + 4*d*x) + 7))
\[ \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\sqrt {a}\, \left (\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )d x \right ) a \right ) \] Input:
int(sec(d*x+c)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x)
Output:
sqrt(a)*(int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2,x)*b + int(sqrt(sec(c + d*x) + 1)*sec(c + d*x),x)*a)