Integrand size = 25, antiderivative size = 224 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=-\frac {2 a^2 \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}-\frac {4 a^2}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {5 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}+\frac {3 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e^3} \]
-2*a^2*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(3/2)+2*a^2*arctanh((e*sin (d*x+c))^(1/2)/e^(1/2))/d/e^(3/2)+3*a^2*sec(d*x+c)*(e*sin(d*x+c))^(3/2)/d/ e^3-4*a^2/d/e/(e*sin(d*x+c))^(1/2)-2*a^2*cos(d*x+c)/d/e/(e*sin(d*x+c))^(1/ 2)-2*a^2*sec(d*x+c)/d/e/(e*sin(d*x+c))^(1/2)+5*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)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x) ,2^(1/2))*(e*sin(d*x+c))^(1/2)/d/e^2/sin(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 19.95 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.60 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=-\frac {2 a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \cot (c+d x) \sec ^4\left (\frac {1}{2} \arcsin (\sin (c+d x))\right ) \sqrt {e \sin (c+d x)} \left (6 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},\sin ^2(c+d x)\right )+6 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {3}{4},\sin ^2(c+d x)\right )+\operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},\sin ^2(c+d x)\right ) \sin ^2(c+d x)\right )}{3 d e^2 \sqrt {\cos ^2(c+d x)}} \]
(-2*a^2*Cos[(c + d*x)/2]^4*Cot[c + d*x]*Sec[ArcSin[Sin[c + d*x]]/2]^4*Sqrt [e*Sin[c + d*x]]*(6*Hypergeometric2F1[-1/4, 1, 3/4, Sin[c + d*x]^2] + 6*Hy pergeometric2F1[-1/4, 3/2, 3/4, Sin[c + d*x]^2] + Hypergeometric2F1[3/4, 3 /2, 7/4, Sin[c + d*x]^2]*Sin[c + d*x]^2))/(3*d*e^2*Sqrt[Cos[c + d*x]^2])
Time = 0.72 (sec) , antiderivative size = 224, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^2}{(e \sin (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\sec ^2(c+d x) (a (-\cos (c+d x))-a)^2}{(e \sin (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x-\frac {\pi }{2}\right )-a\right )^2}{\sin \left (c+d x-\frac {\pi }{2}\right )^2 \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \int \left (\frac {a^2}{(e \sin (c+d x))^{3/2}}+\frac {a^2 \sec ^2(c+d x)}{(e \sin (c+d x))^{3/2}}+\frac {2 a^2 \sec (c+d x)}{(e \sin (c+d x))^{3/2}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a^2 \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}+\frac {3 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e^3}-\frac {5 a^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}-\frac {4 a^2}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}}\) |
(-2*a^2*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(3/2)) + (2*a^2*ArcTanh [Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(3/2)) - (4*a^2)/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*a^2*Cos[c + d*x])/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*a^2*Sec[c + d*x])/(d*e*Sqrt[e*Sin[c + d*x]]) - (5*a^2*EllipticE[(c - Pi/2 + d*x)/2, 2] *Sqrt[e*Sin[c + d*x]])/(d*e^2*Sqrt[Sin[c + d*x]]) + (3*a^2*Sec[c + d*x]*(e *Sin[c + d*x])^(3/2))/(d*e^3)
3.2.18.3.1 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 = 17.57 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.06
| method | result | size |
| default | \(\frac {a^{2} \left (10 e^{\frac {3}{2}} \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-5 e^{\frac {3}{2}} \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-10 e^{\frac {3}{2}} \cos \left (d x +c \right )^{2}-8 e^{\frac {3}{2}} \cos \left (d x +c \right )-4 \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \sqrt {e \sin \left (d x +c \right )}\, \cos \left (d x +c \right ) e +4 \,\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \sqrt {e \sin \left (d x +c \right )}\, \cos \left (d x +c \right ) e +2 e^{\frac {3}{2}}\right )}{2 e^{\frac {5}{2}} \sqrt {e \sin \left (d x +c \right )}\, \cos \left (d x +c \right ) d}\) | \(238\) |
| parts | \(\frac {a^{2} \left (2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \cos \left (d x +c \right )^{2}\right )}{e \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {a^{2} \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-6 \cos \left (d x +c \right )^{2}+2\right )}{2 e \sqrt {-e \sin \left (d x +c \right ) \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right )}\, \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 a^{2} \left (-\frac {\arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {3}{2}}}-\frac {2}{e \sqrt {e \sin \left (d x +c \right )}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {3}{2}}}\right )}{d}\) | \(398\) |
1/2/e^(5/2)/(e*sin(d*x+c))^(1/2)/cos(d*x+c)*a^2*(10*e^(3/2)*(-sin(d*x+c)+1 )^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE((-sin(d*x+c)+1)^ (1/2),1/2*2^(1/2))-5*e^(3/2)*(-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))-10*e^(3/2)*c os(d*x+c)^2-8*e^(3/2)*cos(d*x+c)-4*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))*(e *sin(d*x+c))^(1/2)*cos(d*x+c)*e+4*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))*(e *sin(d*x+c))^(1/2)*cos(d*x+c)*e+2*e^(3/2))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.20 (sec) , antiderivative size = 790, normalized size of antiderivative = 3.53 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
[1/4*(-10*I*sqrt(2)*a^2*sqrt(-I*e)*cos(d*x + c)*sin(d*x + c)*weierstrassZe ta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 10*I* sqrt(2)*a^2*sqrt(I*e)*cos(d*x + c)*sin(d*x + c)*weierstrassZeta(4, 0, weie rstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*a^2*sqrt(-e)*arc tan(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)*sin(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))*sin(d* x + c) - 4*(5*a^2*cos(d*x + c)^2 + 4*a^2*cos(d*x + c) - a^2)*sqrt(e*sin(d* x + c)))/(d*e^2*cos(d*x + c)*sin(d*x + c)), 1/4*(-10*I*sqrt(2)*a^2*sqrt(-I *e)*cos(d*x + c)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 10*I*sqrt(2)*a^2*sqrt(I*e)*cos(d*x + c)*sin(d*x + c)*weierstrassZeta(4, 0, 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*sin(d*x + c))*sqrt(e)/(e*cos(d*x + c)^2 + e*sin(d*x + c) - e))*cos(d*x + c)*sin(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)...
\[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=a^{2} \left (\int \frac {1}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {2 \sec {\left (c + d x \right )}}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx\right ) \]
a**2*(Integral((e*sin(c + d*x))**(-3/2), x) + Integral(2*sec(c + d*x)/(e*s in(c + d*x))**(3/2), x) + Integral(sec(c + d*x)**2/(e*sin(c + d*x))**(3/2) , x))
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]