Integrand size = 23, antiderivative size = 150 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {7 \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} a^{3/2} d}-\frac {7}{24 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec ^2(c+d x)}{5 d (a+a \sin (c+d x))^{3/2}}-\frac {7}{16 a d \sqrt {a+a \sin (c+d x)}}+\frac {7 \sec ^2(c+d x)}{20 a d \sqrt {a+a \sin (c+d x)}} \]
-7/24/d/(a+a*sin(d*x+c))^(3/2)-1/5*sec(d*x+c)^2/d/(a+a*sin(d*x+c))^(3/2)+7 /32*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(3/2)/d*2^(1/2)- 7/16/a/d/(a+a*sin(d*x+c))^(1/2)+7/20*sec(d*x+c)^2/a/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 = 0.28 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},2,-\frac {3}{2},\frac {1}{2} (1+\sin (c+d x))\right )}{10 d (a+a \sin (c+d x))^{5/2}} \]
-1/10*(a*Hypergeometric2F1[-5/2, 2, -3/2, (1 + Sin[c + d*x])/2])/(d*(a + a *Sin[c + d*x])^(5/2))
Time = 0.53 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {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.
| \(\displaystyle \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^3 (a \sin (c+d x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3160 |
| \(\displaystyle \frac {7 \int \frac {\sec ^3(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7 \int \frac {1}{\cos (c+d x)^3 \sqrt {\sin (c+d x) a+a}}dx}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle \frac {7 \left (\frac {5}{4} a \int \frac {\sec (c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7 \left (\frac {5}{4} a \int \frac {1}{\cos (c+d x) (\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {7 \left (\frac {5 a^2 \int \frac {1}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^{5/2}}d(a \sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {7 \left (\frac {5 a^2 \left (\frac {\int \frac {1}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}}d(a \sin (c+d x))}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {7 \left (\frac {5 a^2 \left (\frac {\frac {\int \frac {1}{(a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}}d(a \sin (c+d x))}{2 a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {7 \left (\frac {5 a^2 \left (\frac {\frac {\int \frac {1}{2 a-a^2 \sin ^2(c+d x)}d\sqrt {\sin (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {7 \left (\frac {5 a^2 \left (\frac {\frac {\text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )}{10 a}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}}\) |
-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*Sqr t[a + a*Sin[c + d*x]]))/(2*a)))/(4*d)))/(10*a)
3.2.78.3.1 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 = 0.70 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.83
| method | result | size |
| default | \(\frac {2 a^{3} \left (-\frac {3}{16 a^{4} \sqrt {a +a \sin \left (d x +c \right )}}-\frac {1}{12 a^{3} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {1}{20 a^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\frac {\sqrt {a +a \sin \left (d x +c \right )}}{2 a \sin \left (d x +c \right )-2 a}-\frac {7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4 \sqrt {a}}}{16 a^{4}}\right )}{d}\) | \(124\) |
2*a^3*(-3/16/a^4/(a+a*sin(d*x+c))^(1/2)-1/12/a^3/(a+a*sin(d*x+c))^(3/2)-1/ 20/a^2/(a+a*sin(d*x+c))^(5/2)-1/16/a^4*(1/2*(a+a*sin(d*x+c))^(1/2)/(a*sin( d*x+c)-a)-7/4*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 = 0.29 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {105 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) + 4 \, {\left (175 \, \cos \left (d x + c\right )^{2} + 21 \, {\left (5 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 36\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{960 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]
1/960*(105*sqrt(2)*(cos(d*x + c)^4 - 2*cos(d*x + c)^2*sin(d*x + c) - 2*cos (d*x + c)^2)*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*(175*cos(d*x + c)^2 + 21*(5*co s(d*x + c)^2 - 4)*sin(d*x + c) - 36)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos( d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.97 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {\frac {105 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right )}{\sqrt {a}} + \frac {4 \, {\left (105 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} - 140 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a - 56 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{2} - 48 \, a^{3}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 2 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a}}{960 \, a d} \]
-1/960*(105*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqr t(2)*sqrt(a) + sqrt(a*sin(d*x + c) + a)))/sqrt(a) + 4*(105*(a*sin(d*x + c) + a)^3 - 140*(a*sin(d*x + c) + a)^2*a - 56*(a*sin(d*x + c) + a)*a^2 - 48* a^3)/((a*sin(d*x + c) + a)^(7/2) - 2*(a*sin(d*x + c) + a)^(5/2)*a))/(a*d)
Time = 0.33 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.41 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\sqrt {a} {\left (\frac {105 \, \sqrt {2} \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {105 \, \sqrt {2} \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {30 \, \sqrt {2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {4 \, \sqrt {2} {\left (45 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3\right )}}{a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{960 \, d} \]
1/960*sqrt(a)*(105*sqrt(2)*log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^2*sg n(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 105*sqrt(2)*log(-cos(-1/4*pi + 1/2*d* x + 1/2*c) + 1)/(a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 30*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^2*sg n(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 4*sqrt(2)*(45*cos(-1/4*pi + 1/2*d*x + 1/2*c)^4 + 10*cos(-1/4*pi + 1/2*d*x + 1/2*c)^2 + 3)/(a^2*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. \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]