Integrand size = 25, antiderivative size = 162 \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {52 e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{21 a^2 d \sqrt {e \sin (c+d x)}}-\frac {4 e^3 \sqrt {e \sin (c+d x)}}{a^2 d}+\frac {26 e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 a^2 d}+\frac {2 e^3 \cos ^3(c+d x) \sqrt {e \sin (c+d x)}}{7 a^2 d}+\frac {4 e (e \sin (c+d x))^{5/2}}{5 a^2 d} \]
4/5*e*(e*sin(d*x+c))^(5/2)/a^2/d-52/21*e^4*(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^2/d/(e*sin(d*x+c))^(1/2)-4*e^3*(e*sin(d*x+c))^(1/2)/a ^2/d+26/21*e^3*cos(d*x+c)*(e*sin(d*x+c))^(1/2)/a^2/d+2/7*e^3*cos(d*x+c)^3* (e*sin(d*x+c))^(1/2)/a^2/d
Time = 2.54 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.58 \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+a \sec (c+d x))^2} \, dx=-\frac {e^3 \left (520 \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right )+(756-305 \cos (c+d x)+84 \cos (2 (c+d x))-15 \cos (3 (c+d x))) \sqrt {\sin (c+d x)}\right ) \sqrt {e \sin (c+d x)}}{210 a^2 d \sqrt {\sin (c+d x)}} \]
-1/210*(e^3*(520*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] + (756 - 305*Cos[c + d*x] + 84*Cos[2*(c + d*x)] - 15*Cos[3*(c + d*x)])*Sqrt[Sin[c + d*x]])*Sqrt [e*Sin[c + d*x]])/(a^2*d*Sqrt[Sin[c + d*x]])
Time = 0.85 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.04, 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.
\(\displaystyle \int \frac {(e \sin (c+d x))^{7/2}}{(a \sec (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{7/2}}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \frac {\cos ^2(c+d x) (e \sin (c+d x))^{7/2}}{(a (-\cos (c+d x))-a)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}{\left (a \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )-a\right )^2}dx\) |
\(\Big \downarrow \) 3354 |
\(\displaystyle \frac {e^4 \int \frac {\cos ^2(c+d x) (a-a \cos (c+d x))^2}{\sqrt {e \sin (c+d x)}}dx}{a^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e^4 \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x-\frac {\pi }{2}\right ) a+a\right )^2}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}dx}{a^4}\) |
\(\Big \downarrow \) 3352 |
\(\displaystyle \frac {e^4 \int \left (\frac {a^2 \cos ^4(c+d x)}{\sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos ^3(c+d x)}{\sqrt {e \sin (c+d x)}}+\frac {a^2 \cos ^2(c+d x)}{\sqrt {e \sin (c+d x)}}\right )dx}{a^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^4 \left (\frac {4 a^2 (e \sin (c+d x))^{5/2}}{5 d e^3}-\frac {4 a^2 \sqrt {e \sin (c+d x)}}{d e}+\frac {2 a^2 \cos ^3(c+d x) \sqrt {e \sin (c+d x)}}{7 d e}+\frac {26 a^2 \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 d e}+\frac {52 a^2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{21 d \sqrt {e \sin (c+d x)}}\right )}{a^4}\) |
(e^4*((52*a^2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(21*d*S qrt[e*Sin[c + d*x]]) - (4*a^2*Sqrt[e*Sin[c + d*x]])/(d*e) + (26*a^2*Cos[c + d*x]*Sqrt[e*Sin[c + d*x]])/(21*d*e) + (2*a^2*Cos[c + d*x]^3*Sqrt[e*Sin[c + d*x]])/(7*d*e) + (4*a^2*(e*Sin[c + d*x])^(5/2))/(5*d*e^3)))/a^4
3.2.27.3.1 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 = 6.87 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {2 e^{4} \left (-15 \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )+65 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+42 \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )-65 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+168 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{105 a^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\) | \(145\) |
-2/105/a^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*e^4*(-15*cos(d*x+c)^4*sin(d*x+c )+65*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Ellipti cF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+42*cos(d*x+c)^3*sin(d*x+c)-65*cos(d* x+c)^2*sin(d*x+c)+168*cos(d*x+c)*sin(d*x+c))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.77 \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {2 \, {\left (65 \, \sqrt {2} \sqrt {-i \, e} e^{3} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 65 \, \sqrt {2} \sqrt {i \, e} e^{3} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (15 \, e^{3} \cos \left (d x + c\right )^{3} - 42 \, e^{3} \cos \left (d x + c\right )^{2} + 65 \, e^{3} \cos \left (d x + c\right ) - 168 \, e^{3}\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{105 \, a^{2} d} \]
2/105*(65*sqrt(2)*sqrt(-I*e)*e^3*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + 65*sqrt(2)*sqrt(I*e)*e^3*weierstrassPInverse(4, 0, cos(d *x + c) - I*sin(d*x + c)) + (15*e^3*cos(d*x + c)^3 - 42*e^3*cos(d*x + c)^2 + 65*e^3*cos(d*x + c) - 168*e^3)*sqrt(e*sin(d*x + c)))/(a^2*d)
Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]