\(\int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx\) [119]

Optimal result

Integrand size = 25, antiderivative size = 234 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\frac {2 a^2 \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}-\frac {4 a^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a^2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a^2 \sec (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac {7 a^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 d e^2 \sqrt {e \sin (c+d x)}}+\frac {5 a^2 \sec (c+d x) \sqrt {e \sin (c+d x)}}{3 d e^3} \] Output:

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*InverseJacobiAM(1/2*c-1/4*Pi+1/2*d*x,2^(1/2))*sin(d*x+c)^(1/ 
2)/d/e^2/(e*sin(d*x+c))^(1/2)+5/3*a^2*sec(d*x+c)*(e*sin(d*x+c))^(1/2)/d/e^ 
3
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 16.82 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.72 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=-\frac {a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (3+4 \sqrt {\cos ^2(c+d x)} \csc ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},\sin ^2(c+d x)\right )+4 \sqrt {\cos ^2(c+d x)} \csc ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{2},\frac {1}{4},\sin ^2(c+d x)\right )+3 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\sin ^2(c+d x)\right )\right ) \sec (c+d x) \sec ^4\left (\frac {1}{2} \arcsin (\sin (c+d x))\right ) \sqrt {e \sin (c+d x)}}{3 d e^3} \] Input:

Integrate[(a + a*Sec[c + d*x])^2/(e*Sin[c + d*x])^(5/2),x]
 

Output:

-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)
 

Rubi [A] (verified)

Time = 0.73 (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

\(\displaystyle \frac {2 a^2 \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}+\frac {5 a^2 \sec (c+d x) \sqrt {e \sin (c+d x)}}{3 d e^3}+\frac {7 a^2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{3 d e^2 \sqrt {e \sin (c+d x)}}-\frac {4 a^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a^2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a^2 \sec (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\)

Input:

Int[(a + a*Sec[c + d*x])^2/(e*Sin[c + d*x])^(5/2),x]
 

Output:

(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)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3352

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]
 

rule 4360

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]
 

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.29

method	result	size
default	\(\frac {a^{2} \left (7 \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {7}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) e^{\frac {7}{2}} \sqrt {-\sin \left (d x +c \right )+1}-14 e^{\frac {7}{2}} \cos \left (d x +c \right )^{4}-8 e^{\frac {7}{2}} \cos \left (d x +c \right )^{3}+20 e^{\frac {7}{2}} \cos \left (d x +c \right )^{2}+12 \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} \cos \left (d x +c \right )^{3} e^{2}+12 \,\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} \cos \left (d x +c \right )^{3} e^{2}+8 e^{\frac {7}{2}} \cos \left (d x +c \right )-6 e^{\frac {7}{2}}-12 \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} \cos \left (d x +c \right ) e^{2}-12 \,\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} \cos \left (d x +c \right ) e^{2}\right )}{6 e^{\frac {9}{2}} \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} \cos \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}-1\right ) d}\)	\(301\)
parts	\(-\frac {a^{2} \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {5}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {a^{2} \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (5 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {5}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+10 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )\right )}{6 e^{2} \sin \left (d x +c \right )^{2} \sqrt {-e \sin \left (d x +c \right ) \left (\sin \left (d x +c \right )-1\right ) \left (\sin \left (d x +c \right )+1\right )}\, \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 a^{2} \left (-\frac {2}{3 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {\arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {5}{2}}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {5}{2}}}\right )}{d}\)	\(330\)

Input:

int((a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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*(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)*(-sin(d*x+c)+1)^(1/2)-14*e^(7/2)*cos(d*x+c)^4-8*e^(7/2)*cos(d 
*x+c)^3+20*e^(7/2)*cos(d*x+c)^2+12*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))*(e 
*sin(d*x+c))^(3/2)*cos(d*x+c)^3*e^2+12*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2 
))*(e*sin(d*x+c))^(3/2)*cos(d*x+c)^3*e^2+8*e^(7/2)*cos(d*x+c)-6*e^(7/2)-12 
*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))*(e*sin(d*x+c))^(3/2)*cos(d*x+c)*e^2- 
12*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))*(e*sin(d*x+c))^(3/2)*cos(d*x+c)*e 
^2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 825, normalized size of antiderivative = 3.53 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

[-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)) - 28*(a^2*cos(d*x + 
 c)^2 - a^2*cos(d*x + c))*sqrt(-1/2*I*e)*weierstrassPInverse(4, 0, cos(d*x 
 + c) + I*sin(d*x + c)) - 28*(a^2*cos(d*x + c)^2 - a^2*cos(d*x + c))*sqrt( 
1/2*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)*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)) + 28*(a 
^2*cos(d*x + c)^2 - a^2*cos(d*x + c))*sqrt(-1/2*I*e)*weierstrassPInverse(4 
, 0, cos(d*x + c) + I*sin(d*x + c)) + 28*(a^2*cos(d*x + c)^2 - a^2*cos(...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sec(d*x+c))**2/(e*sin(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

integrate((a*sec(d*x + c) + a)^2/(e*sin(d*x + c))^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:

int((a + a/cos(c + d*x))^2/(e*sin(c + d*x))^(5/2),x)
 

Output:

int((a + a/cos(c + d*x))^2/(e*sin(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\frac {\sqrt {e}\, a^{2} \left (\int \frac {\sqrt {\sin \left (d x +c \right )}}{\sin \left (d x +c \right )^{3}}d x +\int \frac {\sqrt {\sin \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\sin \left (d x +c \right )^{3}}d x +2 \left (\int \frac {\sqrt {\sin \left (d x +c \right )}\, \sec \left (d x +c \right )}{\sin \left (d x +c \right )^{3}}d x \right )\right )}{e^{3}} \] Input:

int((a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(5/2),x)
                                                                                    
                                                                                    
 

Output:

(sqrt(e)*a**2*(int(sqrt(sin(c + d*x))/sin(c + d*x)**3,x) + int((sqrt(sin(c 
 + d*x))*sec(c + d*x)**2)/sin(c + d*x)**3,x) + 2*int((sqrt(sin(c + d*x))*s 
ec(c + d*x))/sin(c + d*x)**3,x)))/e**3