Integrand size = 25, antiderivative size = 234 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\frac {2 a^2 \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}-\frac {4 a^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a^2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a^2 \sec (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac {7 a^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 d e^2 \sqrt {e \sin (c+d x)}}+\frac {5 a^2 \sec (c+d x) \sqrt {e \sin (c+d x)}}{3 d e^3} \]
2*a^2*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(5/2)+2*a^2*arctanh((e*sin( d*x+c))^(1/2)/e^(1/2))/d/e^(5/2)-4/3*a^2/d/e/(e*sin(d*x+c))^(3/2)-2/3*a^2* cos(d*x+c)/d/e/(e*sin(d*x+c))^(3/2)-2/3*a^2*sec(d*x+c)/d/e/(e*sin(d*x+c))^ (3/2)-7/3*a^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*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/d/e^2/(e*s in(d*x+c))^(1/2)+5/3*a^2*sec(d*x+c)*(e*sin(d*x+c))^(1/2)/d/e^3
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 18.13 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.72 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=-\frac {a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (3+4 \sqrt {\cos ^2(c+d x)} \csc ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},\sin ^2(c+d x)\right )+4 \sqrt {\cos ^2(c+d x)} \csc ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{2},\frac {1}{4},\sin ^2(c+d x)\right )+3 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\sin ^2(c+d x)\right )\right ) \sec (c+d x) \sec ^4\left (\frac {1}{2} \arcsin (\sin (c+d x))\right ) \sqrt {e \sin (c+d x)}}{3 d e^3} \]
-1/3*(a^2*Cos[(c + d*x)/2]^4*(3 + 4*Sqrt[Cos[c + d*x]^2]*Csc[c + d*x]^2*Hy pergeometric2F1[-3/4, 1, 1/4, Sin[c + d*x]^2] + 4*Sqrt[Cos[c + d*x]^2]*Csc [c + d*x]^2*Hypergeometric2F1[-3/4, 3/2, 1/4, Sin[c + d*x]^2] + 3*Sqrt[Cos [c + d*x]^2]*Hypergeometric2F1[1/4, 1/2, 5/4, Sin[c + d*x]^2])*Sec[c + d*x ]*Sec[ArcSin[Sin[c + d*x]]/2]^4*Sqrt[e*Sin[c + d*x]])/(d*e^3)
Time = 0.70 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4360, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^2}{(e \sin (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\sec ^2(c+d x) (a (-\cos (c+d x))-a)^2}{(e \sin (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x-\frac {\pi }{2}\right )-a\right )^2}{\sin \left (c+d x-\frac {\pi }{2}\right )^2 \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \int \left (\frac {a^2}{(e \sin (c+d x))^{5/2}}+\frac {a^2 \sec ^2(c+d x)}{(e \sin (c+d x))^{5/2}}+\frac {2 a^2 \sec (c+d x)}{(e \sin (c+d x))^{5/2}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a^2 \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}+\frac {5 a^2 \sec (c+d x) \sqrt {e \sin (c+d x)}}{3 d e^3}+\frac {7 a^2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{3 d e^2 \sqrt {e \sin (c+d x)}}-\frac {4 a^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a^2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a^2 \sec (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\) |
(2*a^2*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(5/2)) + (2*a^2*ArcTanh[ Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(5/2)) - (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*S ec[c + d*x])/(3*d*e*(e*Sin[c + d*x])^(3/2)) + (7*a^2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*e^2*Sqrt[e*Sin[c + d*x]]) + (5*a^2*Se c[c + d*x]*Sqrt[e*Sin[c + d*x]])/(3*d*e^3)
3.2.19.3.1 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_.)*(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 = 17.50 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.29
| method | result | size |
| default | \(\frac {a^{2} \left (7 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {7}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) e^{\frac {7}{2}}-14 e^{\frac {7}{2}} \cos \left (d x +c \right )^{4}-8 e^{\frac {7}{2}} \cos \left (d x +c \right )^{3}+12 \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \cos \left (d x +c \right )^{3} e^{2}+12 \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \cos \left (d x +c \right )^{3} e^{2}+20 e^{\frac {7}{2}} \cos \left (d x +c \right )^{2}+8 e^{\frac {7}{2}} \cos \left (d x +c \right )-12 \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \cos \left (d x +c \right ) e^{2}-12 \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \cos \left (d x +c \right ) e^{2}-6 e^{\frac {7}{2}}\right )}{6 e^{\frac {9}{2}} \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} \cos \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}-1\right ) d}\) | \(301\) |
| parts | \(-\frac {a^{2} \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {5}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {a^{2} \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (5 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {5}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+10 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )\right )}{6 e^{2} \sin \left (d x +c \right )^{2} \sqrt {-e \sin \left (d x +c \right ) \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right )}\, \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 a^{2} \left (-\frac {2}{3 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {\arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {5}{2}}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {5}{2}}}\right )}{d}\) | \(330\) |
1/6/e^(9/2)/(e*sin(d*x+c))^(3/2)/cos(d*x+c)/(cos(d*x+c)^2-1)*a^2*(7*(-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))*e^(7/2)-14*e^(7/2)*cos(d*x+c)^4-8*e^(7/2)*cos(d *x+c)^3+12*(e*sin(d*x+c))^(3/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))*cos(d *x+c)^3*e^2+12*(e*sin(d*x+c))^(3/2)*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))* cos(d*x+c)^3*e^2+20*e^(7/2)*cos(d*x+c)^2+8*e^(7/2)*cos(d*x+c)-12*(e*sin(d* x+c))^(3/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))*cos(d*x+c)*e^2-12*(e*sin( d*x+c))^(3/2)*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))*cos(d*x+c)*e^2-6*e^(7/ 2))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.21 (sec) , antiderivative size = 849, normalized size of antiderivative = 3.63 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
[-1/12*(6*(a^2*cos(d*x + c)^2 - a^2*cos(d*x + c))*sqrt(-e)*arctan(1/4*(cos (d*x + c)^2 - 6*sin(d*x + c) - 2)*sqrt(e*sin(d*x + c))*sqrt(-e)/(e*cos(d*x + c)^2 - e*sin(d*x + c) - e)) + 3*(a^2*cos(d*x + c)^2 - a^2*cos(d*x + c)) *sqrt(-e)*log((e*cos(d*x + c)^4 - 72*e*cos(d*x + c)^2 - 8*(7*cos(d*x + c)^ 2 - (cos(d*x + c)^2 - 8)*sin(d*x + c) - 8)*sqrt(e*sin(d*x + c))*sqrt(-e) + 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 72*e)/(cos(d*x + c)^4 - 8*cos( d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) - 14*(sqrt(2)*a^2*c os(d*x + c)^2 - sqrt(2)*a^2*cos(d*x + c))*sqrt(-I*e)*weierstrassPInverse(4 , 0, cos(d*x + c) + I*sin(d*x + c)) - 14*(sqrt(2)*a^2*cos(d*x + c)^2 - sqr t(2)*a^2*cos(d*x + c))*sqrt(I*e)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) - 4*(7*a^2*cos(d*x + c) - 3*a^2)*sqrt(e*sin(d*x + c)))/(d* e^3*cos(d*x + c)^2 - d*e^3*cos(d*x + c)), 1/12*(6*(a^2*cos(d*x + c)^2 - a^ 2*cos(d*x + c))*sqrt(e)*arctan(1/4*(cos(d*x + c)^2 + 6*sin(d*x + c) - 2)*s qrt(e*sin(d*x + c))*sqrt(e)/(e*cos(d*x + c)^2 + e*sin(d*x + c) - e)) + 3*( a^2*cos(d*x + c)^2 - a^2*cos(d*x + c))*sqrt(e)*log((e*cos(d*x + c)^4 - 72* e*cos(d*x + c)^2 - 8*(7*cos(d*x + c)^2 + (cos(d*x + c)^2 - 8)*sin(d*x + c) - 8)*sqrt(e*sin(d*x + c))*sqrt(e) - 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 72*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*si n(d*x + c) + 8)) + 14*(sqrt(2)*a^2*cos(d*x + c)^2 - sqrt(2)*a^2*cos(d*x + c))*sqrt(-I*e)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))...
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]