Integrand size = 23, antiderivative size = 175 \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {63 \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{128 \sqrt {2} \sqrt {a} d}-\frac {21 a}{64 d (a+a \sin (c+d x))^{3/2}}-\frac {9 a \sec ^2(c+d x)}{40 d (a+a \sin (c+d x))^{3/2}}-\frac {63}{128 d \sqrt {a+a \sin (c+d x)}}+\frac {63 \sec ^2(c+d x)}{160 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x)}{4 d \sqrt {a+a \sin (c+d x)}} \]
-21/64*a/d/(a+a*sin(d*x+c))^(3/2)-9/40*a*sec(d*x+c)^2/d/(a+a*sin(d*x+c))^( 3/2)+63/256*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2)/ a^(1/2)-63/128/d/(a+a*sin(d*x+c))^(1/2)+63/160*sec(d*x+c)^2/d/(a+a*sin(d*x +c))^(1/2)+1/4*sec(d*x+c)^4/d/(a+a*sin(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.05 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.25 \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {a^2 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},3,-\frac {3}{2},\frac {1}{2} (1+\sin (c+d x))\right )}{20 d (a+a \sin (c+d x))^{5/2}} \]
-1/20*(a^2*Hypergeometric2F1[-5/2, 3, -3/2, (1 + Sin[c + d*x])/2])/(d*(a + a*Sin[c + d*x])^(5/2))
Time = 0.66 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.09, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 3166, 3042, 3160, 3042, 3166, 3042, 3146, 61, 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.
| \(\displaystyle \int \frac {\sec ^5(c+d x)}{\sqrt {a \sin (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^5 \sqrt {a \sin (c+d x)+a}}dx\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle \frac {9}{8} a \int \frac {\sec ^3(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec ^4(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {9}{8} a \int \frac {1}{\cos (c+d x)^3 (\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec ^4(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3160 |
| \(\displaystyle \frac {9}{8} a \left (\frac {7 \int \frac {\sec ^3(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^4(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {9}{8} a \left (\frac {7 \int \frac {1}{\cos (c+d x)^3 \sqrt {\sin (c+d x) a+a}}dx}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^4(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle \frac {9}{8} a \left (\frac {7 \left (\frac {5}{4} a \int \frac {\sec (c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^4(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {9}{8} a \left (\frac {7 \left (\frac {5}{4} a \int \frac {1}{\cos (c+d x) (\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^4(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {9}{8} a \left (\frac {7 \left (\frac {5 a^2 \int \frac {1}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^{5/2}}d(a \sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^4(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {9}{8} a \left (\frac {7 \left (\frac {5 a^2 \left (\frac {\int \frac {1}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}}d(a \sin (c+d x))}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^4(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {9}{8} a \left (\frac {7 \left (\frac {5 a^2 \left (\frac {\frac {\int \frac {1}{(a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}}d(a \sin (c+d x))}{2 a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^4(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {9}{8} a \left (\frac {7 \left (\frac {5 a^2 \left (\frac {\frac {\int \frac {1}{2 a-a^2 \sin ^2(c+d x)}d\sqrt {\sin (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^4(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {9}{8} a \left (\frac {7 \left (\frac {5 a^2 \left (\frac {\frac {\text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^4(c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}\) |
Sec[c + d*x]^4/(4*d*Sqrt[a + a*Sin[c + d*x]]) + (9*a*(-1/5*Sec[c + d*x]^2/ (d*(a + a*Sin[c + d*x])^(3/2)) + (7*(Sec[c + d*x]^2/(2*d*Sqrt[a + a*Sin[c + d*x]]) + (5*a^2*(-1/3*1/(a*(a + a*Sin[c + d*x])^(3/2)) + (ArcTanh[(Sqrt[ a]*Sin[c + d*x])/Sqrt[2]]/(Sqrt[2]*a^(3/2)) - 1/(a*Sqrt[a + a*Sin[c + d*x] ]))/(2*a)))/(4*d)))/(10*a)))/8
3.2.67.3.1 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*(2*m + p + 1))), x] + Simp[(m + p + 1)/(a*(2*m + p + 1)) Int[ (g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] & & IntegersQ[2*m, 2*p]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_. )*(x_)]], x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(a*f*g*(p + 1)*S qrt[a + b*Sin[e + f*x]])), x] + Simp[a*((2*p + 1)/(2*g^2*(p + 1))) Int[(g *Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e , f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*p]
Time = 0.86 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.77
| method | result | size |
| default | \(-\frac {2 a^{5} \left (\frac {3}{16 a^{5} \sqrt {a +a \sin \left (d x +c \right )}}+\frac {1}{16 a^{4} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{40 a^{3} \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {\frac {\sqrt {a +a \sin \left (d x +c \right )}\, a \left (15 \sin \left (d x +c \right )-19\right )}{16 \left (a \sin \left (d x +c \right )-a \right )^{2}}-\frac {63 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{32 \sqrt {a}}}{16 a^{5}}\right )}{d}\) | \(135\) |
-2*a^5*(3/16/a^5/(a+a*sin(d*x+c))^(1/2)+1/16/a^4/(a+a*sin(d*x+c))^(3/2)+1/ 40/a^3/(a+a*sin(d*x+c))^(5/2)+1/16/a^5*(1/16*(a+a*sin(d*x+c))^(1/2)*a*(15* sin(d*x+c)-19)/(a*sin(d*x+c)-a)^2-63/32*2^(1/2)/a^(1/2)*arctanh(1/2*(a+a*s in(d*x+c))^(1/2)*2^(1/2)/a^(1/2))))/d
Time = 0.30 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {315 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 4 \, {\left (315 \, \cos \left (d x + c\right )^{4} - 42 \, \cos \left (d x + c\right )^{2} - 6 \, {\left (35 \, \cos \left (d x + c\right )^{2} + 24\right )} \sin \left (d x + c\right ) - 16\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{2560 \, {\left (a d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{4}\right )}} \]
1/2560*(315*sqrt(2)*(cos(d*x + c)^4*sin(d*x + c) + cos(d*x + c)^4)*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*cos(d*x + c)^4 - 42*cos(d*x + c)^2 - 6*(35*co s(d*x + c)^2 + 24)*sin(d*x + c) - 16)*sqrt(a*sin(d*x + c) + a))/(a*d*cos(d *x + c)^4*sin(d*x + c) + a*d*cos(d*x + c)^4)
\[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {\sec ^{5}{\left (c + d x \right )}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.05 \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {315 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (315 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{4} a - 1050 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} a^{2} + 672 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a^{3} + 192 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{4} + 128 \, a^{5}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2}}}{2560 \, a d} \]
-1/2560*(315*sqrt(2)*sqrt(a)*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 - 1050*(a*sin(d*x + c) + a)^3*a^2 + 672*(a*sin(d*x + c) + a)^2* a^3 + 192*(a*sin(d*x + c) + a)*a^4 + 128*a^5)/((a*sin(d*x + c) + a)^(9/2) - 4*(a*sin(d*x + c) + a)^(7/2)*a + 4*(a*sin(d*x + c) + a)^(5/2)*a^2))/(a*d )
Time = 0.35 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.33 \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\sqrt {a} {\left (\frac {315 \, \sqrt {2} \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {315 \, \sqrt {2} \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {10 \, {\left (15 \, \sqrt {2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 17 \, \sqrt {2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {16 \, \sqrt {2} {\left (30 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 5 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}{a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{2560 \, d} \]
1/2560*sqrt(a)*(315*sqrt(2)*log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a*sgn (cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 315*sqrt(2)*log(-cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 10*(15*sqrt(2)*co s(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 17*sqrt(2)*cos(-1/4*pi + 1/2*d*x + 1/2*c) )/((cos(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 16*sqrt(2)*(30*cos(-1/4*pi + 1/2*d*x + 1/2*c)^4 + 5*cos(-1/4*pi + 1/2*d*x + 1/2*c)^2 + 1)/(a*cos(-1/4*pi + 1/2*d*x + 1/2*c)^5*sgn(cos(-1/ 4*pi + 1/2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^5\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]