Integrand size = 23, antiderivative size = 191 \[ \int \sec ^9(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {315 a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2048 \sqrt {2} d}-\frac {315 a^4}{2048 d \sqrt {a+a \sin (c+d x)}}+\frac {105 a^3 \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{1024 d}+\frac {21 a^2 \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{256 d}+\frac {3 a \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{32 d}+\frac {\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d} \]
21/256*a^2*sec(d*x+c)^4*(a+a*sin(d*x+c))^(3/2)/d+3/32*a*sec(d*x+c)^6*(a+a* sin(d*x+c))^(5/2)/d+1/8*sec(d*x+c)^8*(a+a*sin(d*x+c))^(7/2)/d+315/4096*a^( 7/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d-315/204 8*a^4/d/(a+a*sin(d*x+c))^(1/2)+105/1024*a^3*sec(d*x+c)^2*(a+a*sin(d*x+c))^ (1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.23 \[ \int \sec ^9(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {a^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},5,\frac {1}{2},\frac {1}{2} (1+\sin (c+d x))\right )}{16 d \sqrt {a+a \sin (c+d x)}} \]
-1/16*(a^4*Hypergeometric2F1[-1/2, 5, 1/2, (1 + Sin[c + d*x])/2])/(d*Sqrt[ a + a*Sin[c + d*x]])
Time = 0.80 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 3154, 3042, 3154, 3042, 3154, 3042, 3154, 3042, 3146, 61, 73, 219}
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.
| \(\displaystyle \int \sec ^9(c+d x) (a \sin (c+d x)+a)^{7/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^{7/2}}{\cos (c+d x)^9}dx\) |
| \(\Big \downarrow \) 3154 |
| \(\displaystyle \frac {9}{16} a \int \sec ^7(c+d x) (\sin (c+d x) a+a)^{5/2}dx+\frac {\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {9}{16} a \int \frac {(\sin (c+d x) a+a)^{5/2}}{\cos (c+d x)^7}dx+\frac {\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}\) |
| \(\Big \downarrow \) 3154 |
| \(\displaystyle \frac {9}{16} a \left (\frac {7}{12} a \int \sec ^5(c+d x) (\sin (c+d x) a+a)^{3/2}dx+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\right )+\frac {\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {9}{16} a \left (\frac {7}{12} a \int \frac {(\sin (c+d x) a+a)^{3/2}}{\cos (c+d x)^5}dx+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\right )+\frac {\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}\) |
| \(\Big \downarrow \) 3154 |
| \(\displaystyle \frac {9}{16} a \left (\frac {7}{12} a \left (\frac {5}{8} a \int \sec ^3(c+d x) \sqrt {\sin (c+d x) a+a}dx+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\right )+\frac {\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {9}{16} a \left (\frac {7}{12} a \left (\frac {5}{8} a \int \frac {\sqrt {\sin (c+d x) a+a}}{\cos (c+d x)^3}dx+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\right )+\frac {\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}\) |
| \(\Big \downarrow \) 3154 |
| \(\displaystyle \frac {9}{16} a \left (\frac {7}{12} a \left (\frac {5}{8} a \left (\frac {3}{4} a \int \frac {\sec (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\right )+\frac {\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {9}{16} a \left (\frac {7}{12} a \left (\frac {5}{8} a \left (\frac {3}{4} a \int \frac {1}{\cos (c+d x) \sqrt {\sin (c+d x) a+a}}dx+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\right )+\frac {\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {9}{16} a \left (\frac {7}{12} a \left (\frac {5}{8} a \left (\frac {3 a^2 \int \frac {1}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}}d(a \sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\right )+\frac {\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {9}{16} a \left (\frac {7}{12} a \left (\frac {5}{8} a \left (\frac {3 a^2 \left (\frac {\int \frac {1}{(a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}}d(a \sin (c+d x))}{2 a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}\right )}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\right )+\frac {\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {9}{16} a \left (\frac {7}{12} a \left (\frac {5}{8} a \left (\frac {3 a^2 \left (\frac {\int \frac {1}{2 a-a^2 \sin ^2(c+d x)}d\sqrt {\sin (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}\right )}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\right )+\frac {\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {9}{16} a \left (\frac {7}{12} a \left (\frac {5}{8} a \left (\frac {3 a^2 \left (\frac {\text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}\right )}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\right )+\frac {\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}\) |
(Sec[c + d*x]^8*(a + a*Sin[c + d*x])^(7/2))/(8*d) + (9*a*((Sec[c + d*x]^6* (a + a*Sin[c + d*x])^(5/2))/(6*d) + (7*a*((Sec[c + d*x]^4*(a + a*Sin[c + d *x])^(3/2))/(4*d) + (5*a*((Sec[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]])/(2*d) + (3*a^2*(ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[2]]/(Sqrt[2]*a^(3/2)) - 1/(a *Sqrt[a + a*Sin[c + d*x]])))/(4*d)))/8))/12))/16
3.2.54.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p + 1))), x] + Simp[a*((m + p + 1)/(g^2*(p + 1))) Int[(g* Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ [m + 1/2, 2*p]
Time = 0.90 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.68
\[-\frac {2 a^{9} \left (\frac {1}{32 a^{5} \sqrt {a +a \sin \left (d x +c \right )}}+\frac {-\frac {\sqrt {a +a \sin \left (d x +c \right )}\, a^{3} \left (187 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-725 \left (\cos ^{2}\left (d x +c \right )\right )-1236 \sin \left (d x +c \right )+1364\right )}{128 \left (a \sin \left (d x +c \right )-a \right )^{4}}-\frac {315 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{256 \sqrt {a}}}{32 a^{5}}\right )}{d}\]
-2*a^9*(1/32/a^5/(a+a*sin(d*x+c))^(1/2)+1/32/a^5*(-1/128*(a+a*sin(d*x+c))^ (1/2)*a^3*(187*cos(d*x+c)^2*sin(d*x+c)-725*cos(d*x+c)^2-1236*sin(d*x+c)+13 64)/(a*sin(d*x+c)-a)^4-315/256*2^(1/2)/a^(1/2)*arctanh(1/2*(a+a*sin(d*x+c) )^(1/2)*2^(1/2)/a^(1/2))))/d
Time = 0.31 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.33 \[ \int \sec ^9(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {315 \, {\left (3 \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{4} - 4 \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} - {\left (\sqrt {2} a^{3} \cos \left (d x + c\right )^{4} - 4 \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) + 4 \, {\left (315 \, a^{3} \cos \left (d x + c\right )^{4} - 1722 \, a^{3} \cos \left (d x + c\right )^{2} + 896 \, a^{3} + 6 \, {\left (175 \, a^{3} \cos \left (d x + c\right )^{2} - 192 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{8192 \, {\left (3 \, d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \]
1/8192*(315*(3*sqrt(2)*a^3*cos(d*x + c)^4 - 4*sqrt(2)*a^3*cos(d*x + c)^2 - (sqrt(2)*a^3*cos(d*x + c)^4 - 4*sqrt(2)*a^3*cos(d*x + c)^2)*sin(d*x + c)) *sqrt(a)*log(-(a*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a) + 3*a)/(sin(d*x + c) - 1)) + 4*(315*a^3*cos(d*x + c)^4 - 1722*a^3*cos(d*x + c)^2 + 896*a^3 + 6*(175*a^3*cos(d*x + c)^2 - 192*a^3)*sin(d*x + c))*sqr t(a*sin(d*x + c) + a))/(3*d*cos(d*x + c)^4 - 4*d*cos(d*x + c)^2 - (d*cos(d *x + c)^4 - 4*d*cos(d*x + c)^2)*sin(d*x + c))
Timed out. \[ \int \sec ^9(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.15 \[ \int \sec ^9(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {315 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (315 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{4} a^{5} - 2310 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} a^{6} + 6132 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a^{7} - 6696 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{8} + 2048 \, a^{9}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 8 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 24 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 32 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 16 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{4}}}{8192 \, a d} \]
-1/8192*(315*sqrt(2)*a^(9/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x + c) + a))) + 4*(315*(a*sin(d*x + c ) + a)^4*a^5 - 2310*(a*sin(d*x + c) + a)^3*a^6 + 6132*(a*sin(d*x + c) + a) ^2*a^7 - 6696*(a*sin(d*x + c) + a)*a^8 + 2048*a^9)/((a*sin(d*x + c) + a)^( 9/2) - 8*(a*sin(d*x + c) + a)^(7/2)*a + 24*(a*sin(d*x + c) + a)^(5/2)*a^2 - 32*(a*sin(d*x + c) + a)^(3/2)*a^3 + 16*sqrt(a*sin(d*x + c) + a)*a^4))/(a *d)
Time = 0.32 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.84 \[ \int \sec ^9(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {\sqrt {2} a^{\frac {7}{2}} {\left (\frac {256}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {2 \, {\left (187 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 643 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 765 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 325 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}} - 315 \, \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 315 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{8192 \, d} \]
-1/8192*sqrt(2)*a^(7/2)*(256/cos(-1/4*pi + 1/2*d*x + 1/2*c) + 2*(187*cos(- 1/4*pi + 1/2*d*x + 1/2*c)^7 - 643*cos(-1/4*pi + 1/2*d*x + 1/2*c)^5 + 765*c os(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 325*cos(-1/4*pi + 1/2*d*x + 1/2*c))/(cos (-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^4 - 315*log(cos(-1/4*pi + 1/2*d*x + 1/2 *c) + 1) + 315*log(-cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d
Timed out. \[ \int \sec ^9(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\cos \left (c+d\,x\right )}^9} \,d x \]