Integrand size = 25, antiderivative size = 194 \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2} \, dx=-\frac {2 a^2 e^{5/2} \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a^2 e^{5/2} \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}-\frac {9 a^2 e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}-\frac {4 a^2 e (e \sin (c+d x))^{3/2}}{3 d}-\frac {2 a^2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {a^2 e \sec (c+d x) (e \sin (c+d x))^{3/2}}{d} \]
-2*a^2*e^(5/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d+2*a^2*e^(5/2)*arctan h((e*sin(d*x+c))^(1/2)/e^(1/2))/d-4/3*a^2*e*(e*sin(d*x+c))^(3/2)/d-2/5*a^2 *e*cos(d*x+c)*(e*sin(d*x+c))^(3/2)/d+a^2*e*sec(d*x+c)*(e*sin(d*x+c))^(3/2) /d+9/5*a^2*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)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/d/sin (d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 17.64 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.06 \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2} \, dx=\frac {2 a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sec ^4\left (\frac {1}{2} \arcsin (\sin (c+d x))\right ) (e \sin (c+d x))^{5/2} \left (-15 \arctan \left (\sqrt {\sin (c+d x)}\right ) \sqrt {\cos ^2(c+d x)}+15 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {\cos ^2(c+d x)}-9 \sin ^{\frac {3}{2}}(c+d x)-10 \sqrt {\cos ^2(c+d x)} \sin ^{\frac {3}{2}}(c+d x)+9 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},\sin ^2(c+d x)\right ) \sin ^{\frac {3}{2}}(c+d x)+3 \sin ^{\frac {7}{2}}(c+d x)\right )}{15 d \sin ^{\frac {5}{2}}(c+d x)} \]
(2*a^2*Cos[(c + d*x)/2]^4*Sec[c + d*x]*Sec[ArcSin[Sin[c + d*x]]/2]^4*(e*Si n[c + d*x])^(5/2)*(-15*ArcTan[Sqrt[Sin[c + d*x]]]*Sqrt[Cos[c + d*x]^2] + 1 5*ArcTanh[Sqrt[Sin[c + d*x]]]*Sqrt[Cos[c + d*x]^2] - 9*Sin[c + d*x]^(3/2) - 10*Sqrt[Cos[c + d*x]^2]*Sin[c + d*x]^(3/2) + 9*Sqrt[Cos[c + d*x]^2]*Hype rgeometric2F1[3/4, 3/2, 7/4, Sin[c + d*x]^2]*Sin[c + d*x]^(3/2) + 3*Sin[c + d*x]^(7/2)))/(15*d*Sin[c + d*x]^(5/2))
Time = 0.70 (sec) , antiderivative size = 194, 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.
| \(\displaystyle \int (a \sec (c+d x)+a)^2 (e \sin (c+d x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2 \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \sec ^2(c+d x) (a (-\cos (c+d x))-a)^2 (e \sin (c+d x))^{5/2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )-a\right )^2 \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \int \left (a^2 (e \sin (c+d x))^{5/2}+a^2 \sec ^2(c+d x) (e \sin (c+d x))^{5/2}+2 a^2 \sec (c+d x) (e \sin (c+d x))^{5/2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a^2 e^{5/2} \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a^2 e^{5/2} \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}-\frac {9 a^2 e^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}-\frac {4 a^2 e (e \sin (c+d x))^{3/2}}{3 d}-\frac {2 a^2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {a^2 e \sec (c+d x) (e \sin (c+d x))^{3/2}}{d}\) |
(-2*a^2*e^(5/2)*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/d + (2*a^2*e^(5/2)*A rcTanh[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/d - (9*a^2*e^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*d*Sqrt[Sin[c + d*x]]) - (4*a^2*e*(e*S in[c + d*x])^(3/2))/(3*d) - (2*a^2*e*Cos[c + d*x]*(e*Sin[c + d*x])^(3/2))/ (5*d) + (a^2*e*Sec[c + d*x]*(e*Sin[c + d*x])^(3/2))/d
3.2.14.3.1 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 = 19.13 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.37
| method | result | size |
| default | \(\frac {a^{2} \left (-60 \sqrt {e \sin \left (d x +c \right )}\, e^{\frac {5}{2}} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \cos \left (d x +c \right )+60 \sqrt {e \sin \left (d x +c \right )}\, e^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \cos \left (d x +c \right )+54 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) e^{3}-27 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) e^{3}+12 e^{3} \cos \left (d x +c \right )^{4}+40 e^{3} \cos \left (d x +c \right )^{3}-42 e^{3} \cos \left (d x +c \right )^{2}-40 \cos \left (d x +c \right ) e^{3}+30 e^{3}\right )}{30 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\) | \(265\) |
| parts | \(-\frac {a^{2} e^{3} \left (6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{4}+2 \sin \left (d x +c \right )^{2}\right )}{5 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {a^{2} e^{3} \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \cos \left (d x +c \right )^{2}+2\right )}{2 \sqrt {-e \sin \left (d x +c \right ) \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right )}\, \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 a^{2} \left (-\frac {2 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-e^{\frac {5}{2}} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+e^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )\right )}{d}\) | \(407\) |
1/30/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*a^2*(-60*(e*sin(d*x+c))^(1/2)*e^(5/2) *arctan((e*sin(d*x+c))^(1/2)/e^(1/2))*cos(d*x+c)+60*(e*sin(d*x+c))^(1/2)*e ^(5/2)*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))*cos(d*x+c)+54*(-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))*e^3-27*(-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^3+12*e^3*cos(d *x+c)^4+40*e^3*cos(d*x+c)^3-42*e^3*cos(d*x+c)^2-40*cos(d*x+c)*e^3+30*e^3)/ d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.37 (sec) , antiderivative size = 769, normalized size of antiderivative = 3.96 \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2} \, dx=\text {Too large to display} \]
[1/60*(-54*I*sqrt(2)*a^2*sqrt(-I*e)*e^2*cos(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 54*I*sqrt(2)* a^2*sqrt(I*e)*e^2*cos(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4 , 0, cos(d*x + c) - I*sin(d*x + c))) - 30*a^2*sqrt(-e)*e^2*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) + 15*a^2*sqrt(-e)*e^2*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*(6*a^2*e^2*cos(d*x + c)^2 + 20*a^2*e^2*cos(d*x + c) - 15*a^2*e^2)*sqrt(e*sin(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), 1/60*(-54*I*sqrt(2)*a^2*sqrt(-I*e)*e^2*cos(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d* x + c))) + 54*I*sqrt(2)*a^2*sqrt(I*e)*e^2*cos(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) - 30*a^2*e^(5 /2)*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) + 15*a^2*e^( 5/2)*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))*sq rt(e) - 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 72*e)/(cos(d*x + c)^...
Timed out. \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]
\[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \]
\[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2} \, dx=\int {\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2 \,d x \]