Integrand size = 23, antiderivative size = 116 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {5 \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} \sqrt {a} d}-\frac {5 a}{12 d (a+a \sin (c+d x))^{3/2}}-\frac {5}{8 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a+a \sin (c+d x)}} \]
-5/12*a/d/(a+a*sin(d*x+c))^(3/2)+5/16*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2 ^(1/2)/a^(1/2))/d*2^(1/2)/a^(1/2)-5/8/d/(a+a*sin(d*x+c))^(1/2)+1/2*sec(d*x +c)^2/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 = 42, normalized size of antiderivative = 0.36 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2,-\frac {1}{2},\frac {1}{2} (1+\sin (c+d x))\right )}{6 d (a+a \sin (c+d x))^{3/2}} \]
-1/6*(a*Hypergeometric2F1[-3/2, 2, -1/2, (1 + Sin[c + d*x])/2])/(d*(a + a* Sin[c + d*x])^(3/2))
Time = 0.38 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {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.
\(\displaystyle \int \frac {\sec ^3(c+d x)}{\sqrt {a \sin (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (c+d x)^3 \sqrt {a \sin (c+d x)+a}}dx\) |
\(\Big \downarrow \) 3166 |
\(\displaystyle \frac {5}{4} a \int \frac {\sec (c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{4} a \int \frac {1}{\cos (c+d x) (\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {5 a^2 \int \frac {1}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^{5/2}}d(a \sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {5 a^2 \left (\frac {\int \frac {1}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}}d(a \sin (c+d x))}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {5 a^2 \left (\frac {\frac {\int \frac {1}{(a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}}d(a \sin (c+d x))}{2 a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5 a^2 \left (\frac {\frac {\int \frac {1}{2 a-a^2 \sin ^2(c+d x)}d\sqrt {\sin (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5 a^2 \left (\frac {\frac {\text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\) |
Sec[c + d*x]^2/(2*d*Sqrt[a + a*Sin[c + d*x]]) + (5*a^2*(-1/3*1/(a*(a + a*S in[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)
3.2.65.3.1 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_)/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 = 0.73 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.92
method | result | size |
default | \(\frac {2 a^{3} \left (-\frac {1}{4 a^{3} \sqrt {a +a \sin \left (d x +c \right )}}-\frac {1}{12 a^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {\frac {\sqrt {a +a \sin \left (d x +c \right )}}{4 a \sin \left (d x +c \right )-4 a}-\frac {5 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 \sqrt {a}}}{4 a^{3}}\right )}{d}\) | \(107\) |
2*a^3*(-1/4/a^3/(a+a*sin(d*x+c))^(1/2)-1/12/a^2/(a+a*sin(d*x+c))^(3/2)-1/4 /a^3*(1/4*(a+a*sin(d*x+c))^(1/2)/(a*sin(d*x+c)-a)-5/8*2^(1/2)/a^(1/2)*arct anh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))))/d
Time = 0.32 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {15 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 4 \, {\left (15 \, \cos \left (d x + c\right )^{2} - 10 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{96 \, {\left (a d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{2}\right )}} \]
1/96*(15*sqrt(2)*(cos(d*x + c)^2*sin(d*x + c) + cos(d*x + c)^2)*sqrt(a)*lo g(-(a*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a) + 3*a)/(si n(d*x + c) - 1)) - 4*(15*cos(d*x + c)^2 - 10*sin(d*x + c) - 2)*sqrt(a*sin( d*x + c) + a))/(a*d*cos(d*x + c)^2*sin(d*x + c) + a*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.14 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {15 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (15 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a - 20 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{2} - 8 \, a^{3}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 2 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a}}{96 \, a d} \]
-1/96*(15*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*(15*(a*sin(d*x + c) + a)^2*a - 20*(a*sin(d*x + c) + a)*a^2 - 8*a^3)/((a*sin(d*x + c) + a)^(5/2) - 2*(a*sin(d*x + c) + a)^(3/2)*a))/(a*d)
Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (93) = 186\).
Time = 0.37 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.68 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\sqrt {a} {\left (\frac {15 \, \sqrt {2} \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {15 \, \sqrt {2} \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {6 \, \sqrt {2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {4 \, \sqrt {2} {\left (6 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}{a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{96 \, d} \]
1/96*sqrt(a)*(15*sqrt(2)*log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a*sgn(co s(-1/4*pi + 1/2*d*x + 1/2*c))) - 15*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))) - 6*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)*a*sgn(cos(-1/4 *pi + 1/2*d*x + 1/2*c))) - 4*sqrt(2)*(6*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)^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c) )))/d
Timed out. \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]