Integrand size = 35, antiderivative size = 175 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (6 A+7 B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a (6 A+7 B) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a (6 A+7 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}} \] Output:
2/7*a*A*sin(d*x+c)/d/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2)+2/35*a*(6*A+7 *B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+8/105*a*(6*A+7*B) *sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+16/105*a*(6*A+7*B)*s ec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)
Time = 0.24 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.52 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a \left (15 A+3 (6 A+7 B) \sec (c+d x)+4 (6 A+7 B) \sec ^2(c+d x)+8 (6 A+7 B) \sec ^3(c+d x)\right ) \sin (c+d x)}{105 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(7/ 2),x]
Output:
(2*a*(15*A + 3*(6*A + 7*B)*Sec[c + d*x] + 4*(6*A + 7*B)*Sec[c + d*x]^2 + 8 *(6*A + 7*B)*Sec[c + d*x]^3)*Sin[c + d*x])/(105*d*Sec[c + d*x]^(5/2)*Sqrt[ a*(1 + Sec[c + d*x])])
Time = 0.77 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3042, 4503, 3042, 4292, 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.
\(\displaystyle \int \frac {\sqrt {a \sec (c+d x)+a} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
\(\Big \downarrow \) 4503 |
\(\displaystyle \frac {1}{7} (6 A+7 B) \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} (6 A+7 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 4292 |
\(\displaystyle \frac {1}{7} (6 A+7 B) \left (\frac {4}{5} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} (6 A+7 B) \left (\frac {4}{5} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 4292 |
\(\displaystyle \frac {1}{7} (6 A+7 B) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} (6 A+7 B) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 4291 |
\(\displaystyle \frac {1}{7} (6 A+7 B) \left (\frac {2 a \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {4}{5} \left (\frac {4 a \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )\right )+\frac {2 a A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\) |
Input:
Int[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(7/2),x]
Output:
(2*a*A*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + ( (6*A + 7*B)*((2*a*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (4*((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))/7
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 = 1.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.55
method | result | size |
default | \(\frac {2 \left (\left (15 \cos \left (d x +c \right )^{3}+18 \cos \left (d x +c \right )^{2}+24 \cos \left (d x +c \right )+48\right ) A +\left (21 \cos \left (d x +c \right )^{2}+28 \cos \left (d x +c \right )+56\right ) B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{105 d \left (1+\cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(97\) |
parts | \(\frac {2 A \left (5 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+8 \cos \left (d x +c \right )+16\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{35 d \left (1+\cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {B \left (6 \sin \left (d x +c \right )+8 \tan \left (d x +c \right )+16 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (15 \cos \left (d x +c \right )+15\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}\) | \(142\) |
Input:
int((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x,method=_RET URNVERBOSE)
Output:
2/105/d*((15*cos(d*x+c)^3+18*cos(d*x+c)^2+24*cos(d*x+c)+48)*A+(21*cos(d*x+ c)^2+28*cos(d*x+c)+56)*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 = 110, normalized size of antiderivative = 0.63 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, {\left (15 \, A \cos \left (d x + c\right )^{4} + 3 \, {\left (6 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (6 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (6 \, A + 7 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x, algo rithm="fricas")
Output:
2/105*(15*A*cos(d*x + c)^4 + 3*(6*A + 7*B)*cos(d*x + c)^3 + 4*(6*A + 7*B)* cos(d*x + c)^2 + 8*(6*A + 7*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)))
Timed out. \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))**(1/2)*(A+B*sec(d*x+c))/sec(d*x+c)**(7/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 498 vs. \(2 (151) = 302\).
Time = 0.27 (sec) , antiderivative size = 498, normalized size of antiderivative = 2.85 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x, algo rithm="maxima")
Output:
1/840*(3*sqrt(2)*(105*cos(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 35*cos(4/7*arctan2(sin(7/2*d*x + 7/2*c), c os(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 7*cos(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 105*cos(7/2*d*x + 7/2*c)*sin(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 35* cos(7/2*d*x + 7/2*c)*sin(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7 /2*c))) - 7*cos(7/2*d*x + 7/2*c)*sin(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos (7/2*d*x + 7/2*c))) + 10*sin(7/2*d*x + 7/2*c) + 7*sin(5/7*arctan2(sin(7/2* d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 35*sin(3/7*arctan2(sin(7/2*d*x + 7/ 2*c), cos(7/2*d*x + 7/2*c))) + 105*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), c os(7/2*d*x + 7/2*c))))*A*sqrt(a) + 14*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*arct an2(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*arctan2(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), co s(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))))*B*sqrt(a))/d
Time = 0.87 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, {\left (105 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (105 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 175 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (147 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 119 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (27 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 49 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {7}{2}} d} \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x, algo rithm="giac")
Output:
2/105*(105*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 105*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + (105*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 175*sqrt(2)*B*a^4*sgn(cos( d*x + c)) + (147*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 119*sqrt(2)*B*a^4*sgn(c os(d*x + c)) + (27*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 49*sqrt(2)*B*a^4*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)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 + a)^(7/2)*d)
Time = 13.17 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (420\,A\,\sin \left (c+d\,x\right )+490\,B\,\sin \left (c+d\,x\right )+126\,A\,\sin \left (2\,c+2\,d\,x\right )+36\,A\,\sin \left (3\,c+3\,d\,x\right )+15\,A\,\sin \left (4\,c+4\,d\,x\right )+112\,B\,\sin \left (2\,c+2\,d\,x\right )+42\,B\,\sin \left (3\,c+3\,d\,x\right )\right )}{420\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \] Input:
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(1/2))/(1/cos(c + d*x))^(7/ 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)*(420*A*sin(c + d*x) + 490*B*sin(c + d*x) + 126*A*sin(2*c + 2*d*x) + 36*A*sin(3*c + 3*d*x) + 15*A*sin(4*c + 4*d*x) + 112*B*sin(2*c + 2*d*x) + 42*B*sin(3*c + 3*d*x)))/(420*d*(cos(c + d*x) + 1))
\[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}}d x \right ) b \right ) \] Input:
int((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x)
Output:
sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**4,x )*a + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**3,x)*b )