Integrand size = 25, antiderivative size = 135 \[ \int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{3/2}} \, dx=-\frac {2 e}{5 a d (e \sin (c+d x))^{5/2}}+\frac {2 e \cos (c+d x)}{5 a d (e \sin (c+d x))^{5/2}}-\frac {4 \cos (c+d x)}{5 a d e \sqrt {e \sin (c+d x)}}-\frac {4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a d e^2 \sqrt {\sin (c+d x)}} \]
-2/5*e/a/d/(e*sin(d*x+c))^(5/2)+2/5*e*cos(d*x+c)/a/d/(e*sin(d*x+c))^(5/2)- 4/5*cos(d*x+c)/a/d/e/(e*sin(d*x+c))^(1/2)+4/5*(sin(1/2*c+1/4*Pi+1/2*d*x)^2 )^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1 /2))*(e*sin(d*x+c))^(1/2)/a/d/e^2/sin(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.52 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{3/2}} \, dx=\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (\cos (c+d x)+i \sin (c+d x)) \left (-6-9 \cos (c+d x)+2 \sqrt {1-e^{2 i (c+d x)}} (1+\cos (c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},e^{2 i (c+d x)}\right )+3 i \sin (c+d x)\right )}{15 a d e \sqrt {e \sin (c+d x)}} \]
(Sec[(c + d*x)/2]^2*(Cos[c + d*x] + I*Sin[c + d*x])*(-6 - 9*Cos[c + d*x] + 2*Sqrt[1 - E^((2*I)*(c + d*x))]*(1 + Cos[c + d*x])*Hypergeometric2F1[1/2, 3/4, 7/4, E^((2*I)*(c + d*x))] + (3*I)*Sin[c + d*x]))/(15*a*d*e*Sqrt[e*Si n[c + d*x]])
Time = 0.85 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.04, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {3042, 4360, 25, 25, 3042, 25, 3318, 25, 3042, 3044, 15, 3047, 3042, 3116, 3042, 3121, 3042, 3119}
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.
| \(\displaystyle \int \frac {1}{(a \sec (c+d x)+a) (e \sin (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right ) \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int -\frac {\cos (c+d x)}{(a (-\cos (c+d x))-a) (e \sin (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -\frac {\cos (c+d x)}{(\cos (c+d x) a+a) (e \sin (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\cos (c+d x)}{(a \cos (c+d x)+a) (e \sin (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )}{\left (a-a \sin \left (c+d x-\frac {\pi }{2}\right )\right ) \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\left (e \cos \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )^{3/2} \left (a-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )}dx\) |
| \(\Big \downarrow \) 3318 |
| \(\displaystyle -\frac {e^2 \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{7/2}}dx}{a}-\frac {e^2 \int -\frac {\cos (c+d x)}{(e \sin (c+d x))^{7/2}}dx}{a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {e^2 \int \frac {\cos (c+d x)}{(e \sin (c+d x))^{7/2}}dx}{a}-\frac {e^2 \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{7/2}}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \int \frac {\cos (c+d x)}{(e \sin (c+d x))^{7/2}}dx}{a}-\frac {e^2 \int \frac {\cos (c+d x)^2}{(e \sin (c+d x))^{7/2}}dx}{a}\) |
| \(\Big \downarrow \) 3044 |
| \(\displaystyle \frac {e \int \frac {1}{(e \sin (c+d x))^{7/2}}d(e \sin (c+d x))}{a d}-\frac {e^2 \int \frac {\cos (c+d x)^2}{(e \sin (c+d x))^{7/2}}dx}{a}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle -\frac {e^2 \int \frac {\cos (c+d x)^2}{(e \sin (c+d x))^{7/2}}dx}{a}-\frac {2 e}{5 a d (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3047 |
| \(\displaystyle -\frac {e^2 \left (-\frac {2 \int \frac {1}{(e \sin (c+d x))^{3/2}}dx}{5 e^2}-\frac {2 \cos (c+d x)}{5 d e (e \sin (c+d x))^{5/2}}\right )}{a}-\frac {2 e}{5 a d (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (-\frac {2 \int \frac {1}{(e \sin (c+d x))^{3/2}}dx}{5 e^2}-\frac {2 \cos (c+d x)}{5 d e (e \sin (c+d x))^{5/2}}\right )}{a}-\frac {2 e}{5 a d (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle -\frac {e^2 \left (-\frac {2 \left (-\frac {\int \sqrt {e \sin (c+d x)}dx}{e^2}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{5 e^2}-\frac {2 \cos (c+d x)}{5 d e (e \sin (c+d x))^{5/2}}\right )}{a}-\frac {2 e}{5 a d (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (-\frac {2 \left (-\frac {\int \sqrt {e \sin (c+d x)}dx}{e^2}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{5 e^2}-\frac {2 \cos (c+d x)}{5 d e (e \sin (c+d x))^{5/2}}\right )}{a}-\frac {2 e}{5 a d (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle -\frac {e^2 \left (-\frac {2 \left (-\frac {\sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{5 e^2}-\frac {2 \cos (c+d x)}{5 d e (e \sin (c+d x))^{5/2}}\right )}{a}-\frac {2 e}{5 a d (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (-\frac {2 \left (-\frac {\sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{5 e^2}-\frac {2 \cos (c+d x)}{5 d e (e \sin (c+d x))^{5/2}}\right )}{a}-\frac {2 e}{5 a d (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {e^2 \left (-\frac {2 \left (-\frac {2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{5 e^2}-\frac {2 \cos (c+d x)}{5 d e (e \sin (c+d x))^{5/2}}\right )}{a}-\frac {2 e}{5 a d (e \sin (c+d x))^{5/2}}\) |
(-2*e)/(5*a*d*(e*Sin[c + d*x])^(5/2)) - (e^2*((-2*Cos[c + d*x])/(5*d*e*(e* Sin[c + d*x])^(5/2)) - (2*((-2*Cos[c + d*x])/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(d*e^2*Sqrt[Sin[ c + d*x]])))/(5*e^2)))/a
3.2.25.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[a*(a*Cos[e + f*x])^(m - 1)*((b*Sin[e + f*x])^(n + 1)/ (b*f*(n + 1))), x] + Simp[a^2*((m - 1)/(b^2*(n + 1))) Int[(a*Cos[e + f*x] )^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ [m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || EqQ[m + n, 0])
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d) Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 5.84 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.39
| method | result | size |
| default | \(\frac {-\frac {2 e}{5 a \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {\frac {4 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {7}{2}} \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{5}-\frac {2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {7}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{5}+\frac {4 \sin \left (d x +c \right )^{5}}{5}-\frac {6 \sin \left (d x +c \right )^{3}}{5}+\frac {2 \sin \left (d x +c \right )}{5}}{e a \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(187\) |
(-2/5/a*e/(e*sin(d*x+c))^(5/2)+2/5/e*(2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c )+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-( -sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticF((-s in(d*x+c)+1)^(1/2),1/2*2^(1/2))+2*sin(d*x+c)^5-3*sin(d*x+c)^3+sin(d*x+c))/ a/sin(d*x+c)^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left ({\left (i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} \sqrt {-i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} \sqrt {i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + {\left (2 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{5 \, {\left (a d e^{2} \cos \left (d x + c\right ) + a d e^{2}\right )} \sin \left (d x + c\right )} \]
-2/5*((I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*sqrt(-I*e)*sin(d*x + c)*weierst rassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + (-I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*sqrt(I*e)*sin(d*x + c)*weierstrassZ eta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) + (2*c os(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(e*sin(d*x + c)))/((a*d*e^2*cos(d* x + c) + a*d*e^2)*sin(d*x + c))
\[ \int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{3/2}} \, dx=\frac {\int \frac {1}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec {\left (c + d x \right )} + \left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx}{a} \]
Timed out. \[ \int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{3/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )}{a\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]