Integrand size = 25, antiderivative size = 189 \[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {4 e^3}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {2 e^3 \cos (c+d x)}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {2 e^3 \cos ^3(c+d x)}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {4 e^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{a^2 d \sqrt {e \sin (c+d x)}}+\frac {4 e \sqrt {e \sin (c+d x)}}{a^2 d}-\frac {4 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 a^2 d} \]
4/3*e^3/a^2/d/(e*sin(d*x+c))^(3/2)-2/3*e^3*cos(d*x+c)/a^2/d/(e*sin(d*x+c)) ^(3/2)-2/3*e^3*cos(d*x+c)^3/a^2/d/(e*sin(d*x+c))^(3/2)+4*e^2*(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*P i+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/a^2/d/(e*sin(d*x+c))^(1/2)+4*e*(e*sin (d*x+c))^(1/2)/a^2/d-4/3*e*cos(d*x+c)*(e*sin(d*x+c))^(1/2)/a^2/d
Time = 2.84 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.63 \[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left ((15+10 \cos (c+d x)-\cos (2 (c+d x))) \csc (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )+\frac {24 \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right )}{\sin ^{\frac {3}{2}}(c+d x)}\right ) (e \sin (c+d x))^{3/2}}{3 a^2 d (1+\sec (c+d x))^2} \]
(2*Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*((15 + 10*Cos[c + d*x] - Cos[2*(c + d *x)])*Csc[c + d*x]*Sec[(c + d*x)/2]^2 + (24*EllipticF[(-2*c + Pi - 2*d*x)/ 4, 2])/Sin[c + d*x]^(3/2))*(e*Sin[c + d*x])^(3/2))/(3*a^2*d*(1 + Sec[c + d *x])^2)
Time = 0.86 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4360, 3042, 3354, 3042, 3352, 2009}
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.
\(\displaystyle \int \frac {(e \sin (c+d x))^{3/2}}{(a \sec (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \frac {\cos ^2(c+d x) (e \sin (c+d x))^{3/2}}{(a (-\cos (c+d x))-a)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}{\left (a \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )-a\right )^2}dx\) |
\(\Big \downarrow \) 3354 |
\(\displaystyle \frac {e^4 \int \frac {\cos ^2(c+d x) (a-a \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}}dx}{a^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e^4 \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x-\frac {\pi }{2}\right ) a+a\right )^2}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}dx}{a^4}\) |
\(\Big \downarrow \) 3352 |
\(\displaystyle \frac {e^4 \int \left (\frac {a^2 \cos ^4(c+d x)}{(e \sin (c+d x))^{5/2}}-\frac {2 a^2 \cos ^3(c+d x)}{(e \sin (c+d x))^{5/2}}+\frac {a^2 \cos ^2(c+d x)}{(e \sin (c+d x))^{5/2}}\right )dx}{a^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^4 \left (\frac {4 a^2 \sqrt {e \sin (c+d x)}}{d e^3}-\frac {4 a^2 \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d e^3}-\frac {4 a^2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d e^2 \sqrt {e \sin (c+d x)}}+\frac {4 a^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a^2 \cos ^3(c+d x)}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a^2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )}{a^4}\) |
(e^4*((4*a^2)/(3*d*e*(e*Sin[c + d*x])^(3/2)) - (2*a^2*Cos[c + d*x])/(3*d*e *(e*Sin[c + d*x])^(3/2)) - (2*a^2*Cos[c + d*x]^3)/(3*d*e*(e*Sin[c + d*x])^ (3/2)) - (4*a^2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(d*e^ 2*Sqrt[e*Sin[c + d*x]]) + (4*a^2*Sqrt[e*Sin[c + d*x]])/(d*e^3) - (4*a^2*Co s[c + d*x]*Sqrt[e*Sin[c + d*x]])/(3*d*e^3)))/a^4
3.2.29.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(a/g)^(2* m) Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e + f*x] )^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 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 = 5.38 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.81
method | result | size |
default | \(-\frac {2 e^{3} \left (3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {7}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\cos \left (d x +c \right )^{6}+6 \cos \left (d x +c \right )^{5}+4 \cos \left (d x +c \right )^{4}-14 \cos \left (d x +c \right )^{3}-3 \cos \left (d x +c \right )^{2}+8 \cos \left (d x +c \right )\right )}{3 a^{2} \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} \cos \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}-1\right ) d}\) | \(153\) |
-2/3/a^2/(e*sin(d*x+c))^(3/2)/cos(d*x+c)/(cos(d*x+c)^2-1)*e^3*(3*(-sin(d*x +c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticF((-sin(d*x+c )+1)^(1/2),1/2*2^(1/2))-cos(d*x+c)^6+6*cos(d*x+c)^5+4*cos(d*x+c)^4-14*cos( d*x+c)^3-3*cos(d*x+c)^2+8*cos(d*x+c))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.74 \[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+a \sec (c+d x))^2} \, dx=-\frac {2 \, {\left (3 \, {\left (\sqrt {2} e \cos \left (d x + c\right ) + \sqrt {2} e\right )} \sqrt {-i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} e \cos \left (d x + c\right ) + \sqrt {2} e\right )} \sqrt {i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (e \cos \left (d x + c\right )^{2} - 5 \, e \cos \left (d x + c\right ) - 8 \, e\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{3 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
-2/3*(3*(sqrt(2)*e*cos(d*x + c) + sqrt(2)*e)*sqrt(-I*e)*weierstrassPInvers e(4, 0, cos(d*x + c) + I*sin(d*x + c)) + 3*(sqrt(2)*e*cos(d*x + c) + sqrt( 2)*e)*sqrt(I*e)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + (e*cos(d*x + c)^2 - 5*e*cos(d*x + c) - 8*e)*sqrt(e*sin(d*x + c)))/(a^2*d* cos(d*x + c) + a^2*d)
Timed out. \[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]