Integrand size = 25, antiderivative size = 135 \[ \int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx=-\frac {2 e}{7 a d (e \sin (c+d x))^{7/2}}+\frac {2 e \cos (c+d x)}{7 a d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{21 a d e (e \sin (c+d x))^{3/2}}+\frac {4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{21 a d e^2 \sqrt {e \sin (c+d x)}} \]
-2/7*e/a/d/(e*sin(d*x+c))^(7/2)+2/7*e*cos(d*x+c)/a/d/(e*sin(d*x+c))^(7/2)- 4/21*cos(d*x+c)/a/d/e/(e*sin(d*x+c))^(3/2)-4/21*(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)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^ (1/2))*sin(d*x+c)^(1/2)/a/d/e^2/(e*sin(d*x+c))^(1/2)
Time = 1.74 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx=-\frac {2 \left (4+2 \cos (c+d x)+\cos (2 (c+d x))+\csc ^2\left (\frac {1}{2} (c+d x)\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right ) \sin ^{\frac {7}{2}}(c+d x)\right )}{21 a d e (1+\cos (c+d x)) (e \sin (c+d x))^{3/2}} \]
(-2*(4 + 2*Cos[c + d*x] + Cos[2*(c + d*x)] + Csc[(c + d*x)/2]^2*EllipticF[ (-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d*x]^(7/2)))/(21*a*d*e*(1 + Cos[c + d*x] )*(e*Sin[c + d*x])^(3/2))
Time = 0.86 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.07, 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, 3120}
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.
| \(\displaystyle \int \frac {1}{(a \sec (c+d x)+a) (e \sin (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right ) \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int -\frac {\cos (c+d x)}{(a (-\cos (c+d x))-a) (e \sin (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -\frac {\cos (c+d x)}{(\cos (c+d x) a+a) (e \sin (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\cos (c+d x)}{(a \cos (c+d x)+a) (e \sin (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )}{\left (a-a \sin \left (c+d x-\frac {\pi }{2}\right )\right ) \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\left (e \cos \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )^{5/2} \left (a-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )}dx\) |
| \(\Big \downarrow \) 3318 |
| \(\displaystyle -\frac {e^2 \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{9/2}}dx}{a}-\frac {e^2 \int -\frac {\cos (c+d x)}{(e \sin (c+d x))^{9/2}}dx}{a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {e^2 \int \frac {\cos (c+d x)}{(e \sin (c+d x))^{9/2}}dx}{a}-\frac {e^2 \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{9/2}}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \int \frac {\cos (c+d x)}{(e \sin (c+d x))^{9/2}}dx}{a}-\frac {e^2 \int \frac {\cos (c+d x)^2}{(e \sin (c+d x))^{9/2}}dx}{a}\) |
| \(\Big \downarrow \) 3044 |
| \(\displaystyle \frac {e \int \frac {1}{(e \sin (c+d x))^{9/2}}d(e \sin (c+d x))}{a d}-\frac {e^2 \int \frac {\cos (c+d x)^2}{(e \sin (c+d x))^{9/2}}dx}{a}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle -\frac {e^2 \int \frac {\cos (c+d x)^2}{(e \sin (c+d x))^{9/2}}dx}{a}-\frac {2 e}{7 a d (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3047 |
| \(\displaystyle -\frac {e^2 \left (-\frac {2 \int \frac {1}{(e \sin (c+d x))^{5/2}}dx}{7 e^2}-\frac {2 \cos (c+d x)}{7 d e (e \sin (c+d x))^{7/2}}\right )}{a}-\frac {2 e}{7 a d (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (-\frac {2 \int \frac {1}{(e \sin (c+d x))^{5/2}}dx}{7 e^2}-\frac {2 \cos (c+d x)}{7 d e (e \sin (c+d x))^{7/2}}\right )}{a}-\frac {2 e}{7 a d (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle -\frac {e^2 \left (-\frac {2 \left (\frac {\int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{3 e^2}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 \cos (c+d x)}{7 d e (e \sin (c+d x))^{7/2}}\right )}{a}-\frac {2 e}{7 a d (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (-\frac {2 \left (\frac {\int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{3 e^2}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 \cos (c+d x)}{7 d e (e \sin (c+d x))^{7/2}}\right )}{a}-\frac {2 e}{7 a d (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle -\frac {e^2 \left (-\frac {2 \left (\frac {\sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{3 e^2 \sqrt {e \sin (c+d x)}}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 \cos (c+d x)}{7 d e (e \sin (c+d x))^{7/2}}\right )}{a}-\frac {2 e}{7 a d (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {e^2 \left (-\frac {2 \left (\frac {\sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{3 e^2 \sqrt {e \sin (c+d x)}}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 \cos (c+d x)}{7 d e (e \sin (c+d x))^{7/2}}\right )}{a}-\frac {2 e}{7 a d (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {e^2 \left (-\frac {2 \left (\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{3 d e^2 \sqrt {e \sin (c+d x)}}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 \cos (c+d x)}{7 d e (e \sin (c+d x))^{7/2}}\right )}{a}-\frac {2 e}{7 a d (e \sin (c+d x))^{7/2}}\) |
(-2*e)/(7*a*d*(e*Sin[c + d*x])^(7/2)) - (e^2*((-2*Cos[c + d*x])/(7*d*e*(e* Sin[c + d*x])^(7/2)) - (2*((-2*Cos[c + d*x])/(3*d*e*(e*Sin[c + d*x])^(3/2) ) + (2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*e^2*Sqrt[ e*Sin[c + d*x]])))/(7*e^2)))/a
3.2.26.3.1 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[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 = 4.81 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01
| method | result | size |
| default | \(\frac {-\frac {2 e}{7 a \left (e \sin \left (d x +c \right )\right )^{\frac {7}{2}}}-\frac {2 \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {9}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{5}+5 \sin \left (d x +c \right )^{3}-3 \sin \left (d x +c \right )\right )}{21 e^{2} a \sin \left (d x +c \right )^{4} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(136\) |
(-2/7/a*e/(e*sin(d*x+c))^(7/2)-2/21/e^2*((-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+ c)+2)^(1/2)*sin(d*x+c)^(9/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))- 2*sin(d*x+c)^5+5*sin(d*x+c)^3-3*sin(d*x+c))/a/sin(d*x+c)^4/cos(d*x+c)/(e*s in(d*x+c))^(1/2))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.59 \[ \int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx=\frac {2 \, {\left ({\left (\sqrt {2} \cos \left (d x + c\right )^{3} + \sqrt {2} \cos \left (d x + c\right )^{2} - \sqrt {2} \cos \left (d x + c\right ) - \sqrt {2}\right )} \sqrt {-i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} \cos \left (d x + c\right )^{3} + \sqrt {2} \cos \left (d x + c\right )^{2} - \sqrt {2} \cos \left (d x + c\right ) - \sqrt {2}\right )} \sqrt {i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (2 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 3\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{21 \, {\left (a d e^{3} \cos \left (d x + c\right )^{3} + a d e^{3} \cos \left (d x + c\right )^{2} - a d e^{3} \cos \left (d x + c\right ) - a d e^{3}\right )}} \]
2/21*((sqrt(2)*cos(d*x + c)^3 + sqrt(2)*cos(d*x + c)^2 - sqrt(2)*cos(d*x + c) - sqrt(2))*sqrt(-I*e)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d *x + c)) + (sqrt(2)*cos(d*x + c)^3 + sqrt(2)*cos(d*x + c)^2 - sqrt(2)*cos( d*x + c) - sqrt(2))*sqrt(I*e)*weierstrassPInverse(4, 0, cos(d*x + c) - I*s in(d*x + c)) + (2*cos(d*x + c)^2 + 2*cos(d*x + c) + 3)*sqrt(e*sin(d*x + c) ))/(a*d*e^3*cos(d*x + c)^3 + a*d*e^3*cos(d*x + c)^2 - a*d*e^3*cos(d*x + c) - a*d*e^3)
Timed out. \[ \int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )}{a\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]