Integrand size = 23, antiderivative size = 95 \[ \int \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {3 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} d}-\frac {3 a}{4 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{2 d} \]
3/8*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^(1/2)/d- 3/4*a/d/(a+a*sin(d*x+c))^(1/2)+1/2*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 = 0.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.44 \[ \int \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\frac {1}{2} (1+\sin (c+d x))\right )}{2 d \sqrt {a+a \sin (c+d x)}} \]
Time = 0.40 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {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.
\(\displaystyle \int \sec ^3(c+d x) \sqrt {a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a \sin (c+d x)+a}}{\cos (c+d x)^3}dx\) |
\(\Big \downarrow \) 3154 |
\(\displaystyle \frac {3}{4} a \int \frac {\sec (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{4} a \int \frac {1}{\cos (c+d x) \sqrt {\sin (c+d x) a+a}}dx+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {3 a^2 \int \frac {1}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}}d(a \sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {3 a^2 \left (\frac {\int \frac {1}{(a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}}d(a \sin (c+d x))}{2 a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}\right )}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {3 a^2 \left (\frac {\int \frac {1}{2 a-a^2 \sin ^2(c+d x)}d\sqrt {\sin (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}\right )}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {3 a^2 \left (\frac {\text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}\right )}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\) |
(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)
3.2.10.3.1 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 = 0.40 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.95
method | result | size |
default | \(\frac {2 a^{3} \left (-\frac {1}{4 a^{2} \sqrt {a +a \sin \left (d x +c \right )}}-\frac {\frac {\sqrt {a +a \sin \left (d x +c \right )}}{2 a \sin \left (d x +c \right )-2 a}-\frac {3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4 \sqrt {a}}}{4 a^{2}}\right )}{d}\) | \(90\) |
2*a^3*(-1/4/a^2/(a+a*sin(d*x+c))^(1/2)-1/4/a^2*(1/2*(a+a*sin(d*x+c))^(1/2) /(a*sin(d*x+c)-a)-3/4*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 = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04 \[ \int \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {3 \, \sqrt {2} \sqrt {a} \cos \left (d x + c\right )^{2} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) + 4 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (3 \, \sin \left (d x + c\right ) - 1\right )}}{16 \, d \cos \left (d x + c\right )^{2}} \]
1/16*(3*sqrt(2)*sqrt(a)*cos(d*x + c)^2*log(-(a*sin(d*x + c) + 2*sqrt(2)*sq rt(a*sin(d*x + c) + a)*sqrt(a) + 3*a)/(sin(d*x + c) - 1)) + 4*sqrt(a*sin(d *x + c) + a)*(3*sin(d*x + c) - 1))/(d*cos(d*x + c)^2)
\[ \int \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \sec ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.23 \[ \int \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {3 \, \sqrt {2} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (3 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{2} - 4 \, a^{3}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} - 2 \, \sqrt {a \sin \left (d x + c\right ) + a} a}}{16 \, a d} \]
-1/16*(3*sqrt(2)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a)) /(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x + c) + a))) + 4*(3*(a*sin(d*x + c) + a) *a^2 - 4*a^3)/((a*sin(d*x + c) + a)^(3/2) - 2*sqrt(a*sin(d*x + c) + a)*a)) /(a*d)
Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.18 \[ \int \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\sqrt {2} \sqrt {a} {\left (\frac {2 \, {\left (3 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - 3 \, \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 3 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{16 \, d} \]
-1/16*sqrt(2)*sqrt(a)*(2*(3*cos(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 2)/(cos(-1/ 4*pi + 1/2*d*x + 1/2*c)^3 - cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 3*log(cos(-1 /4*pi + 1/2*d*x + 1/2*c) + 1) + 3*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. \[ \int \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \frac {\sqrt {a+a\,\sin \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^3} \,d x \]