\(\int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx\) [117]

Optimal result

Integrand size = 25, antiderivative size = 139 \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\frac {2 a^2 \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {3 a^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{d \sqrt {e \sin (c+d x)}}+\frac {a^2 \sec (c+d x) \sqrt {e \sin (c+d x)}}{d e} \] Output:

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*InverseJacobiAM(1/2*c-1/4*Pi+1/2*d* 
x,2^(1/2))*sin(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
 

Mathematica [C] (warning: unable to verify)

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

Time = 17.29 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.18 \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\frac {a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sec ^4\left (\frac {1}{2} \arcsin (\sin (c+d x))\right ) \left (2 \arctan \left (\sqrt {\sin (c+d x)}\right ) \sqrt {\cos ^2(c+d x)}+2 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {\cos ^2(c+d x)}+\sqrt {\sin (c+d x)}+3 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\sin ^2(c+d x)\right ) \sqrt {\sin (c+d x)}\right ) \sqrt {\sin (c+d x)}}{d \sqrt {e \sin (c+d x)}} \] Input:

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

Output:

(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]])
 

Rubi [A] (verified)

Time = 0.60 (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

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

Input:

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

Output:

(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)
 

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 = 0.28 (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	\(-\frac {a^{2} \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{\cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {a^{2} \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )\right )}{2 \sqrt {-e \sin \left (d x +c \right ) \left (\sin \left (d x +c \right )-1\right ) \left (\sin \left (d x +c \right )+1\right )}\, \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 a^{2} \left (\arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )\right )}{\sqrt {e}\, d}\)	\(251\)

Input:

int((a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-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
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 647, normalized size of antiderivative = 4.65 \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate((a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

[-1/4*(2*a^2*sqrt(-e)*arctan(1/4*(cos(d*x + c)^2 - 6*sin(d*x + c) - 2)*sqr 
t(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)) - 12*a^2*sqrt(-1/2*I*e)*cos(d*x + c)*weierstrassPInverse(4, 0, cos(d 
*x + c) + I*sin(d*x + c)) - 12*a^2*sqrt(1/2*I*e)*cos(d*x + c)*weierstrassP 
Inverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) - 4*sqrt(e*sin(d*x + c))*a^2) 
/(d*e*cos(d*x + c)), 1/4*(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)) + 12*a^2*sqrt(-1/2*I*e)*cos(d*x + c)*weierstrassPInve 
rse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + 12*a^2*sqrt(1/2*I*e)*cos(d*x + 
c)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 4*sqrt(e*sin 
(d*x + c))*a^2)/(d*e*cos(d*x + c))]
 

Sympy [F]

\[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=a^{2} \left (\int \frac {1}{\sqrt {e \sin {\left (c + d x \right )}}}\, dx + \int \frac {2 \sec {\left (c + d x \right )}}{\sqrt {e \sin {\left (c + d x \right )}}}\, dx + \int \frac {\sec ^{2}{\left (c + d x \right )}}{\sqrt {e \sin {\left (c + d x \right )}}}\, dx\right ) \] Input:

integrate((a+a*sec(d*x+c))**2/(e*sin(d*x+c))**(1/2),x)
 

Output:

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))
 

Maxima [F]

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

integrate((a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((a + a/cos(c + d*x))^2/(e*sin(c + d*x))^(1/2),x)
 

Output:

int((a + a/cos(c + d*x))^2/(e*sin(c + d*x))^(1/2), x)
 

Reduce [F]

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

int((a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(1/2),x)
                                                                                    
                                                                                    
 

Output:

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