Integrand size = 35, antiderivative size = 130 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (4 A+5 B) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {4 a (4 A+5 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}} \] Output:
2/5*a*A*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+2/15*a*(4*A+5 *B)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+4/15*a*(4*A+5*B)* sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)
Time = 0.23 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.55 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a (19 A+20 B+2 (4 A+5 B) \cos (c+d x)+3 A \cos (2 (c+d x))) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(5/ 2),x]
Output:
(a*(19*A + 20*B + 2*(4*A + 5*B)*Cos[c + d*x] + 3*A*Cos[2*(c + d*x)])*Sqrt[ Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.62 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3042, 4503, 3042, 4292, 3042, 4291}
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.
| \(\displaystyle \int \frac {\sqrt {a \sec (c+d x)+a} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {1}{5} (4 A+5 B) \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} (4 A+5 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {1}{5} (4 A+5 B) \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} (4 A+5 B) \left (\frac {2}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4291 |
| \(\displaystyle \frac {1}{5} (4 A+5 B) \left (\frac {4 a \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\) |
Input:
Int[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(5/2),x]
Output:
(2*a*A*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + ( (4*A + 5*B)*((2*a*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (4*a*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d* x]])))/5
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] *(d_.)], x_Symbol] :> Simp[-2*a*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*S qrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[a*((2*n + 1)/(2*b*d*n)) Int[Sqrt[a + b*Csc [e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1)] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp [(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a *B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && LtQ[n, 0]
Time = 1.21 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.60
| method | result | size |
| default | \(\frac {2 \left (3 A \cos \left (d x +c \right )^{2}+4 A \cos \left (d x +c \right )+5 B \cos \left (d x +c \right )+8 A +10 B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{15 d \left (1+\cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(78\) |
| parts | \(\frac {A \left (6 \sin \left (d x +c \right )+8 \tan \left (d x +c \right )+16 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (15 \cos \left (d x +c \right )+15\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}+\frac {B \left (2 \sin \left (d x +c \right )+4 \tan \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (3 \cos \left (d x +c \right )+3\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(124\) |
Input:
int((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x,method=_RET URNVERBOSE)
Output:
2/15/d*(3*A*cos(d*x+c)^2+4*A*cos(d*x+c)+5*B*cos(d*x+c)+8*A+10*B)*(a*(1+sec (d*x+c)))^(1/2)/(1+cos(d*x+c))/sec(d*x+c)^(3/2)*tan(d*x+c)
Time = 0.08 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \, {\left (3 \, A \cos \left (d x + c\right )^{3} + {\left (4 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (4 \, A + 5 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x, algo rithm="fricas")
Output:
2/15*(3*A*cos(d*x + c)^3 + (4*A + 5*B)*cos(d*x + c)^2 + 2*(4*A + 5*B)*cos( d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/((d*cos(d*x + c) + d)*sqrt(cos(d*x + c)))
\[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )}\right )}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \] Input:
integrate((a+a*sec(d*x+c))**(1/2)*(A+B*sec(d*x+c))/sec(d*x+c)**(5/2),x)
Output:
Integral(sqrt(a*(sec(c + d*x) + 1))*(A + B*sec(c + d*x))/sec(c + d*x)**(5/ 2), x)
Leaf count of result is larger than twice the leaf count of optimal. 317 vs. \(2 (112) = 224\).
Time = 0.28 (sec) , antiderivative size = 317, normalized size of antiderivative = 2.44 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x, algo rithm="maxima")
Output:
1/60*(sqrt(2)*(30*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2* c)))*sin(5/2*d*x + 5/2*c) + 5*cos(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/ 2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 30*cos(5/2*d*x + 5/2*c)*sin(4/5*ar ctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 5*cos(5/2*d*x + 5/2*c )*sin(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 6*sin(5/2 *d*x + 5/2*c) + 5*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2* c))) + 30*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*A* sqrt(a) + 10*sqrt(2)*(3*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(3/2*d*x + 3/2*c) - 3*cos(3/2*d*x + 3/2*c)*sin(2/3*arctan2(s in(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sin(3/2*d*x + 3/2*c) + 3*s in(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*B*sqrt(a))/d
Time = 0.71 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.18 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \, {\left (15 \, \sqrt {2} A a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 15 \, \sqrt {2} B a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (10 \, \sqrt {2} A a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 20 \, \sqrt {2} B a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (7 \, \sqrt {2} A a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, \sqrt {2} B a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{15 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {5}{2}} d} \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x, algo rithm="giac")
Output:
2/15*(15*sqrt(2)*A*a^3*sgn(cos(d*x + c)) + 15*sqrt(2)*B*a^3*sgn(cos(d*x + c)) + (10*sqrt(2)*A*a^3*sgn(cos(d*x + c)) + 20*sqrt(2)*B*a^3*sgn(cos(d*x + c)) + (7*sqrt(2)*A*a^3*sgn(cos(d*x + c)) + 5*sqrt(2)*B*a^3*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*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)^(5/2)*d)
Time = 12.14 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (35\,A\,\sin \left (c+d\,x\right )+40\,B\,\sin \left (c+d\,x\right )+8\,A\,\sin \left (2\,c+2\,d\,x\right )+3\,A\,\sin \left (3\,c+3\,d\,x\right )+10\,B\,\sin \left (2\,c+2\,d\,x\right )\right )}{30\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \] Input:
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(1/2))/(1/cos(c + d*x))^(5/ 2),x)
Output:
(cos(c + d*x)*(1/cos(c + d*x))^(1/2)*((a*(cos(c + d*x) + 1))/cos(c + d*x)) ^(1/2)*(35*A*sin(c + d*x) + 40*B*sin(c + d*x) + 8*A*sin(2*c + 2*d*x) + 3*A *sin(3*c + 3*d*x) + 10*B*sin(2*c + 2*d*x)))/(30*d*(cos(c + d*x) + 1))
\[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{2}}d x \right ) b \right ) \] Input:
int((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x)
Output:
sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**3,x )*a + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**2,x)*b )