Integrand size = 25, antiderivative size = 139 \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\frac {2 a^2 \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {3 a^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{d \sqrt {e \sin (c+d x)}}+\frac {a^2 \sec (c+d x) \sqrt {e \sin (c+d x)}}{d e} \]
2*a^2*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(1/2)+2*a^2*arctanh((e*sin( d*x+c))^(1/2)/e^(1/2))/d/e^(1/2)-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))*si n(d*x+c)^(1/2)/d/(e*sin(d*x+c))^(1/2)+a^2*sec(d*x+c)*(e*sin(d*x+c))^(1/2)/ d/e
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 19.81 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.18 \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\frac {a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sec ^4\left (\frac {1}{2} \arcsin (\sin (c+d x))\right ) \left (2 \arctan \left (\sqrt {\sin (c+d x)}\right ) \sqrt {\cos ^2(c+d x)}+2 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {\cos ^2(c+d x)}+\sqrt {\sin (c+d x)}+3 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\sin ^2(c+d x)\right ) \sqrt {\sin (c+d x)}\right ) \sqrt {\sin (c+d x)}}{d \sqrt {e \sin (c+d x)}} \]
(a^2*Cos[(c + d*x)/2]^4*Sec[c + d*x]*Sec[ArcSin[Sin[c + d*x]]/2]^4*(2*ArcT an[Sqrt[Sin[c + d*x]]]*Sqrt[Cos[c + d*x]^2] + 2*ArcTanh[Sqrt[Sin[c + d*x]] ]*Sqrt[Cos[c + d*x]^2] + Sqrt[Sin[c + d*x]] + 3*Sqrt[Cos[c + d*x]^2]*Hyper geometric2F1[1/4, 1/2, 5/4, Sin[c + d*x]^2]*Sqrt[Sin[c + d*x]])*Sqrt[Sin[c + d*x]])/(d*Sqrt[e*Sin[c + d*x]])
Time = 0.59 (sec) , antiderivative size = 139, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^2}{\sqrt {e \sin (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\sec ^2(c+d x) (a (-\cos (c+d x))-a)^2}{\sqrt {e \sin (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x-\frac {\pi }{2}\right )-a\right )^2}{\sin \left (c+d x-\frac {\pi }{2}\right )^2 \sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \int \left (\frac {a^2}{\sqrt {e \sin (c+d x)}}+\frac {a^2 \sec ^2(c+d x)}{\sqrt {e \sin (c+d x)}}+\frac {2 a^2 \sec (c+d x)}{\sqrt {e \sin (c+d x)}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a^2 \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {a^2 \sec (c+d x) \sqrt {e \sin (c+d x)}}{d e}+\frac {3 a^2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}\) |
(2*a^2*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*Sqrt[e]) + (2*a^2*ArcTanh[ Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*Sqrt[e]) + (3*a^2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(d*Sqrt[e*Sin[c + d*x]]) + (a^2*Sec[c + d*x ]*Sqrt[e*Sin[c + d*x]])/(d*e)
3.2.17.3.1 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 = 11.95 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.17
| method | result | size |
| default | \(\frac {a^{2} \left (-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {e}+4 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+4 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+2 \sqrt {e}\, \sin \left (d x +c \right )\right )}{2 \sqrt {e}\, \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\) | \(163\) |
| parts | \(-\frac {a^{2} \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{\cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {a^{2} \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )\right )}{2 \sqrt {-e \sin \left (d x +c \right ) \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right )}\, \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 a^{2} \left (\arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )\right )}{\sqrt {e}\, d}\) | \(251\) |
1/2/e^(1/2)/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*a^2*(-3*(-sin(d*x+c)+1)^(1/2)* (2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/ 2*2^(1/2))*e^(1/2)+4*cos(d*x+c)*(e*sin(d*x+c))^(1/2)*arctan((e*sin(d*x+c)) ^(1/2)/e^(1/2))+4*cos(d*x+c)*(e*sin(d*x+c))^(1/2)*arctanh((e*sin(d*x+c))^( 1/2)/e^(1/2))+2*e^(1/2)*sin(d*x+c))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.19 (sec) , antiderivative size = 660, normalized size of antiderivative = 4.75 \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\left [\frac {6 \, \sqrt {2} a^{2} \sqrt {-i \, e} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 6 \, \sqrt {2} a^{2} \sqrt {i \, e} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, a^{2} \sqrt {-e} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 6 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {-e}}{4 \, {\left (e \cos \left (d x + c\right )^{2} - e \sin \left (d x + c\right ) - e\right )}}\right ) \cos \left (d x + c\right ) - a^{2} \sqrt {-e} \cos \left (d x + c\right ) \log \left (\frac {e \cos \left (d x + c\right )^{4} - 72 \, e \cos \left (d x + c\right )^{2} - 8 \, {\left (7 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) - 8\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {-e} + 28 \, {\left (e \cos \left (d x + c\right )^{2} - 2 \, e\right )} \sin \left (d x + c\right ) + 72 \, e}{\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8}\right ) + 4 \, \sqrt {e \sin \left (d x + c\right )} a^{2}}{4 \, d e \cos \left (d x + c\right )}, \frac {6 \, \sqrt {2} a^{2} \sqrt {-i \, e} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 6 \, \sqrt {2} a^{2} \sqrt {i \, e} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, a^{2} \sqrt {e} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {e}}{4 \, {\left (e \cos \left (d x + c\right )^{2} + e \sin \left (d x + c\right ) - e\right )}}\right ) \cos \left (d x + c\right ) + a^{2} \sqrt {e} \cos \left (d x + c\right ) \log \left (\frac {e \cos \left (d x + c\right )^{4} - 72 \, e \cos \left (d x + c\right )^{2} - 8 \, {\left (7 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) - 8\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {e} - 28 \, {\left (e \cos \left (d x + c\right )^{2} - 2 \, e\right )} \sin \left (d x + c\right ) + 72 \, e}{\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8}\right ) + 4 \, \sqrt {e \sin \left (d x + c\right )} a^{2}}{4 \, d e \cos \left (d x + c\right )}\right ] \]
[1/4*(6*sqrt(2)*a^2*sqrt(-I*e)*cos(d*x + c)*weierstrassPInverse(4, 0, cos( d*x + c) + I*sin(d*x + c)) + 6*sqrt(2)*a^2*sqrt(I*e)*cos(d*x + c)*weierstr assPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*a^2*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))*cos(d*x + c) - a^2*sqrt(-e)*cos(d*x + c)*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)) + 4*sqrt(e*sin(d*x + c)) *a^2)/(d*e*cos(d*x + c)), 1/4*(6*sqrt(2)*a^2*sqrt(-I*e)*cos(d*x + c)*weier strassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + 6*sqrt(2)*a^2*sqrt(I *e)*cos(d*x + c)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*a^2*sqrt(e)*arctan(1/4*(cos(d*x + c)^2 + 6*sin(d*x + c) - 2)*sqrt(e*si n(d*x + c))*sqrt(e)/(e*cos(d*x + c)^2 + e*sin(d*x + c) - e))*cos(d*x + c) + a^2*sqrt(e)*cos(d*x + c)*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)) + 4 *sqrt(e*sin(d*x + c))*a^2)/(d*e*cos(d*x + c))]
\[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=a^{2} \left (\int \frac {1}{\sqrt {e \sin {\left (c + d x \right )}}}\, dx + \int \frac {2 \sec {\left (c + d x \right )}}{\sqrt {e \sin {\left (c + d x \right )}}}\, dx + \int \frac {\sec ^{2}{\left (c + d x \right )}}{\sqrt {e \sin {\left (c + d x \right )}}}\, dx\right ) \]
a**2*(Integral(1/sqrt(e*sin(c + d*x)), x) + Integral(2*sec(c + d*x)/sqrt(e *sin(c + d*x)), x) + Integral(sec(c + d*x)**2/sqrt(e*sin(c + d*x)), x))
\[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \sin \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \sin \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{\sqrt {e\,\sin \left (c+d\,x\right )}} \,d x \]