\(\int \frac {\sqrt {e \sin (c+d x)}}{a+a \sec (c+d x)} \, dx\) [123]

Optimal result

Integrand size = 25, antiderivative size = 95 \[ \int \frac {\sqrt {e \sin (c+d x)}}{a+a \sec (c+d x)} \, dx=-\frac {2 e}{a d \sqrt {e \sin (c+d x)}}+\frac {2 e \cos (c+d x)}{a d \sqrt {e \sin (c+d x)}}+\frac {4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}} \] Output:

-2*e/a/d/(e*sin(d*x+c))^(1/2)+2*e*cos(d*x+c)/a/d/(e*sin(d*x+c))^(1/2)-4*El 
lipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/a/d/sin(d* 
x+c)^(1/2)
 

Mathematica [C] (warning: unable to verify)

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

Time = 4.20 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.21 \[ \int \frac {\sqrt {e \sin (c+d x)}}{a+a \sec (c+d x)} \, dx=\frac {2 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sqrt {e \sin (c+d x)} \left (\sec \left (\frac {c}{2}\right ) \sec (c) \left (3 \sin \left (\frac {c}{2}\right )+\sin \left (\frac {3 c}{2}\right )\right )-2 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )-2 \sqrt {\csc ^2(c)} \csc (c+d x) \csc (d x-\arctan (\cot (c))) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x-\arctan (\cot (c)))\right ) \sin (c) \sqrt {\sin ^2(d x-\arctan (\cot (c)))}-\frac {\csc (c) \csc (c+d x) \sec (c) (\sin (c+d x-\arctan (\cot (c)))+3 \sin (c-d x+\arctan (\cot (c))))}{\sqrt {\csc ^2(c)}}\right )}{a d (1+\sec (c+d x))} \] Input:

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

Output:

(2*Cos[(c + d*x)/2]^2*Sec[c + d*x]*Sqrt[e*Sin[c + d*x]]*(Sec[c/2]*Sec[c]*( 
3*Sin[c/2] + Sin[(3*c)/2]) - 2*Sec[c/2]*Sec[(c + d*x)/2]*Sin[(d*x)/2] - 2* 
Sqrt[Csc[c]^2]*Csc[c + d*x]*Csc[d*x - ArcTan[Cot[c]]]*HypergeometricPFQ[{- 
1/2, -1/4}, {3/4}, Cos[d*x - ArcTan[Cot[c]]]^2]*Sin[c]*Sqrt[Sin[d*x - ArcT 
an[Cot[c]]]^2] - (Csc[c]*Csc[c + d*x]*Sec[c]*(Sin[c + d*x - ArcTan[Cot[c]] 
] + 3*Sin[c - d*x + ArcTan[Cot[c]]]))/Sqrt[Csc[c]^2]))/(a*d*(1 + Sec[c + d 
*x]))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.08, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4360, 25, 25, 3042, 3318, 3042, 3044, 15, 3047, 3042, 3121, 3042, 3119}

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.

\(\displaystyle \int \frac {\sqrt {e \sin (c+d x)}}{a \sec (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{a-a \csc \left (c+d x-\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 4360

\(\displaystyle \int -\frac {\cos (c+d x) \sqrt {e \sin (c+d x)}}{a (-\cos (c+d x))-a}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int -\frac {\cos (c+d x) \sqrt {e \sin (c+d x)}}{\cos (c+d x) a+a}dx\)

\(\Big \downarrow \) 25

\(\displaystyle \int \frac {\cos (c+d x) \sqrt {e \sin (c+d x)}}{a \cos (c+d x)+a}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {-e \cos \left (c+d x+\frac {\pi }{2}\right )}}{a \sin \left (c+d x+\frac {\pi }{2}\right )+a}dx\)

\(\Big \downarrow \) 3318

\(\displaystyle \frac {e^2 \int \frac {\cos (c+d x)}{(e \sin (c+d x))^{3/2}}dx}{a}-\frac {e^2 \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{3/2}}dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \int \frac {\cos (c+d x)}{(e \sin (c+d x))^{3/2}}dx}{a}-\frac {e^2 \int \frac {\cos (c+d x)^2}{(e \sin (c+d x))^{3/2}}dx}{a}\)

\(\Big \downarrow \) 3044

\(\displaystyle \frac {e \int \frac {1}{(e \sin (c+d x))^{3/2}}d(e \sin (c+d x))}{a d}-\frac {e^2 \int \frac {\cos (c+d x)^2}{(e \sin (c+d x))^{3/2}}dx}{a}\)

\(\Big \downarrow \) 15

\(\displaystyle -\frac {e^2 \int \frac {\cos (c+d x)^2}{(e \sin (c+d x))^{3/2}}dx}{a}-\frac {2 e}{a d \sqrt {e \sin (c+d x)}}\)

\(\Big \downarrow \) 3047

\(\displaystyle -\frac {e^2 \left (-\frac {2 \int \sqrt {e \sin (c+d x)}dx}{e^2}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{a}-\frac {2 e}{a d \sqrt {e \sin (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {e^2 \left (-\frac {2 \int \sqrt {e \sin (c+d x)}dx}{e^2}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{a}-\frac {2 e}{a d \sqrt {e \sin (c+d x)}}\)

\(\Big \downarrow \) 3121

\(\displaystyle -\frac {e^2 \left (-\frac {2 \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{a}-\frac {2 e}{a d \sqrt {e \sin (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {e^2 \left (-\frac {2 \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{a}-\frac {2 e}{a d \sqrt {e \sin (c+d x)}}\)

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {e^2 \left (-\frac {4 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{a}-\frac {2 e}{a d \sqrt {e \sin (c+d x)}}\)

Input:

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

Output:

(-2*e)/(a*d*Sqrt[e*Sin[c + d*x]]) - (e^2*((-2*Cos[c + d*x])/(d*e*Sqrt[e*Si 
n[c + d*x]]) - (4*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/( 
d*e^2*Sqrt[Sin[c + d*x]])))/a
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

rule 3044

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ 
Symbol] :> Simp[1/(a*f)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a 
*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&  !(I 
ntegerQ[(m - 1)/2] && LtQ[0, m, n])
 

rule 3047

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[a*(a*Cos[e + f*x])^(m - 1)*((b*Sin[e + f*x])^(n + 1)/ 
(b*f*(n + 1))), x] + Simp[a^2*((m - 1)/(b^2*(n + 1)))   Int[(a*Cos[e + f*x] 
)^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ 
[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || EqQ[m + n, 0])
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3318

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( 
n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a   Int 
[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d)   Int 
[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, 
d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 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 = 0.23 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.57

method	result	size
default	\(-\frac {2 e \left (2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )\right )}{a \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(149\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.17 \[ \int \frac {\sqrt {e \sin (c+d x)}}{a+a \sec (c+d x)} \, dx=-\frac {2 \, {\left (2 \, \sqrt {-\frac {1}{2} i \, e} {\left (-i \, \cos \left (d x + c\right ) - i\right )} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 2 \, \sqrt {\frac {1}{2} i \, e} {\left (i \, \cos \left (d x + c\right ) + i\right )} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {e \sin \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{a d \cos \left (d x + c\right ) + a d} \] Input:

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

Output:

-2*(2*sqrt(-1/2*I*e)*(-I*cos(d*x + c) - I)*weierstrassZeta(4, 0, weierstra 
ssPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 2*sqrt(1/2*I*e)*(I*cos( 
d*x + c) + I)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) 
 - I*sin(d*x + c))) + sqrt(e*sin(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c) 
 + a*d)
 

Sympy [F]

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

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

Output:

Integral(sqrt(e*sin(c + d*x))/(sec(c + d*x) + 1), x)/a
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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