Integrand size = 35, antiderivative size = 181 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (52 A+63 B) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {4 a^2 (52 A+63 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)} \]
2/35*a^2*(8*A+7*B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+2/ 105*a^2*(52*A+63*B)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+4 /105*a^2*(52*A+63*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+ 2/7*a*A*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(5/2)
Time = 0.32 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.51 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a^2 \left (15 A+3 (13 A+7 B) \sec (c+d x)+(52 A+63 B) \sec ^2(c+d x)+2 (52 A+63 B) \sec ^3(c+d x)\right ) \sin (c+d x)}{105 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a (1+\sec (c+d x))}} \]
(2*a^2*(15*A + 3*(13*A + 7*B)*Sec[c + d*x] + (52*A + 63*B)*Sec[c + d*x]^2 + 2*(52*A + 63*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 = 0.93 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 4505, 27, 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.
\(\displaystyle \int \frac {(a \sec (c+d x)+a)^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
\(\Big \downarrow \) 4505 |
\(\displaystyle \frac {2}{7} \int \frac {\sqrt {\sec (c+d x) a+a} (a (8 A+7 B)+a (4 A+7 B) \sec (c+d x))}{2 \sec ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \int \frac {\sqrt {\sec (c+d x) a+a} (a (8 A+7 B)+a (4 A+7 B) \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (8 A+7 B)+a (4 A+7 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4503 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} a (52 A+63 B) \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} a (52 A+63 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4292 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} a (52 A+63 B) \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} a (52 A+63 B) \left (\frac {2}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4291 |
\(\displaystyle \frac {1}{7} \left (\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {1}{5} a (52 A+63 B) \left (\frac {4 a \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )\right )+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
(2*a*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + ( (2*a^2*(8*A + 7*B)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (a*(52*A + 63*B)*((2*a*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sq rt[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
3.3.37.3.1 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[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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[n, -1]
Time = 5.14 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.56
method | result | size |
default | \(\frac {2 a \left (15 A \cos \left (d x +c \right )^{3}+39 A \cos \left (d x +c \right )^{2}+21 B \cos \left (d x +c \right )^{2}+52 A \cos \left (d x +c \right )+63 B \cos \left (d x +c \right )+104 A +126 B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{105 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(101\) |
parts | \(\frac {2 A a \left (15 \cos \left (d x +c \right )^{3}+39 \cos \left (d x +c \right )^{2}+52 \cos \left (d x +c \right )+104\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{105 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 B a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sin \left (d x +c \right )+3 \tan \left (d x +c \right )+6 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}\) | \(141\) |
2/105*a/d*(15*A*cos(d*x+c)^3+39*A*cos(d*x+c)^2+21*B*cos(d*x+c)^2+52*A*cos( d*x+c)+63*B*cos(d*x+c)+104*A+126*B)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1 )/sec(d*x+c)^(3/2)*tan(d*x+c)
Time = 0.29 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.62 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, {\left (15 \, A a \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, A + 7 \, B\right )} a \cos \left (d x + c\right )^{3} + {\left (52 \, A + 63 \, B\right )} a \cos \left (d x + c\right )^{2} + 2 \, {\left (52 \, A + 63 \, B\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]
2/105*(15*A*a*cos(d*x + c)^4 + 3*(13*A + 7*B)*a*cos(d*x + c)^3 + (52*A + 6 3*B)*a*cos(d*x + c)^2 + 2*(52*A + 63*B)*a*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. \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 514 vs. \(2 (157) = 314\).
Time = 0.54 (sec) , antiderivative size = 514, normalized size of antiderivative = 2.84 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \]
1/840*(sqrt(2)*(735*a*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) + 175*a*cos(4/7*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) + 63*a*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) - 735*a*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) )) - 175*a*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))) - 63*a*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))) + 30*a*sin(7/2*d*x + 7/2*c) + 63*a*sin(5 /7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 175*a*sin(3/7*ar ctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 735*a*sin(1/7*arctan2 (sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*A*sqrt(a) + 42*sqrt(2)*(20* a*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*a*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) - 20*a*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*a*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))) + 2*a*sin(5/2*d*x + 5 /2*c) + 5*a*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 20*a*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*B*sqrt (a))/d
\[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sec \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
Time = 16.24 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.72 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {a\,\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (910\,A\,\sin \left (c+d\,x\right )+1050\,B\,\sin \left (c+d\,x\right )+238\,A\,\sin \left (2\,c+2\,d\,x\right )+78\,A\,\sin \left (3\,c+3\,d\,x\right )+15\,A\,\sin \left (4\,c+4\,d\,x\right )+252\,B\,\sin \left (2\,c+2\,d\,x\right )+42\,B\,\sin \left (3\,c+3\,d\,x\right )\right )}{420\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \]