\(\int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^2} \, dx\) [535]

Optimal result

Integrand size = 27, antiderivative size = 105 \[ \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^2} \, dx=-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {d} (c+d)^{3/2} f}-\frac {a \cos (e+f x)}{(c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \] Output:

-a^(1/2)*arctanh(a^(1/2)*d^(1/2)*cos(f*x+e)/(c+d)^(1/2)/(a+a*sin(f*x+e))^( 
1/2))/d^(1/2)/(c+d)^(3/2)/f-a*cos(f*x+e)/(c+d)/f/(a+a*sin(f*x+e))^(1/2)/(c 
+d*sin(f*x+e))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 5.51 (sec) , antiderivative size = 871, normalized size of antiderivative = 8.30 \[ \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((1/4 + I/4)*Sqrt[a*(1 + Sin[e + f*x])]*(((Cos[e/2] + I*Sin[e/2])*((-1 + I 
)*x*Cos[e] + (RootSum[-d + (2*I)*c*E^(I*e)*#1^2 + d*E^((2*I)*e)*#1^4 & , ( 
(1 + I)*d*Sqrt[E^((-I)*e)]*f*x - (2 - 2*I)*d*Sqrt[E^((-I)*e)]*Log[E^((I/2) 
*f*x) - #1] - I*Sqrt[d]*Sqrt[c + d]*f*x*#1 + 2*Sqrt[d]*Sqrt[c + d]*Log[E^( 
(I/2)*f*x) - #1]*#1 + ((1 - I)*c*f*x*#1^2)/Sqrt[E^((-I)*e)] + ((2 + 2*I)*c 
*Log[E^((I/2)*f*x) - #1]*#1^2)/Sqrt[E^((-I)*e)] - Sqrt[d]*Sqrt[c + d]*E^(I 
*e)*f*x*#1^3 - (2*I)*Sqrt[d]*Sqrt[c + d]*E^(I*e)*Log[E^((I/2)*f*x) - #1]*# 
1^3)/(d - I*c*E^(I*e)*#1^2) & ]*(Cos[e] + I*(-1 + Sin[e]))*Sqrt[Cos[e] - I 
*Sin[e]])/(4*f) + (1 + I)*x*Sin[e]))/(Sqrt[d]*(c + d)^(3/2)*(Cos[e] + I*(- 
1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) + ((Cos[e/2] + I*Sin[e/2])*((1 - I)* 
x*Cos[e] - (1 + I)*x*Sin[e] + (RootSum[-d + (2*I)*c*E^(I*e)*#1^2 + d*E^((2 
*I)*e)*#1^4 & , ((1 - I)*d*Sqrt[E^((-I)*e)]*f*x + (2 + 2*I)*d*Sqrt[E^((-I) 
*e)]*Log[E^((I/2)*f*x) - #1] + Sqrt[d]*Sqrt[c + d]*f*x*#1 + (2*I)*Sqrt[d]* 
Sqrt[c + d]*Log[E^((I/2)*f*x) - #1]*#1 - ((1 + I)*c*f*x*#1^2)/Sqrt[E^((-I) 
*e)] + ((2 - 2*I)*c*Log[E^((I/2)*f*x) - #1]*#1^2)/Sqrt[E^((-I)*e)] - I*Sqr 
t[d]*Sqrt[c + d]*E^(I*e)*f*x*#1^3 + 2*Sqrt[d]*Sqrt[c + d]*E^(I*e)*Log[E^(( 
I/2)*f*x) - #1]*#1^3)/(d - I*c*E^(I*e)*#1^2) & ]*Sqrt[Cos[e] - I*Sin[e]]*( 
-1 - I*Cos[e] + Sin[e]))/(4*f)))/(Sqrt[d]*(c + d)^(3/2)*(Cos[e] + I*(-1 + 
Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) - ((2 - 2*I)*(Cos[(e + f*x)/2] - Sin[(e 
+ f*x)/2]))/((c + d)*f*(c + d*Sin[e + f*x]))))/(Cos[(e + f*x)/2] + Sin[...
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3251, 3042, 3252, 221}

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.

\(\displaystyle \int \frac {\sqrt {a \sin (e+f x)+a}}{(c+d \sin (e+f x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a \sin (e+f x)+a}}{(c+d \sin (e+f x))^2}dx\)

\(\Big \downarrow \) 3251

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3252

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

\(\Big \downarrow \) 221

\(\displaystyle -\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {d} f (c+d)^{3/2}}-\frac {a \cos (e+f x)}{f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}\)

Input:

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

Output:

-((Sqrt[a]*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a* 
Sin[e + f*x]])])/(Sqrt[d]*(c + d)^(3/2)*f)) - (a*Cos[e + f*x])/((c + d)*f* 
Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x]))
 

Defintions of rubi rules used

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042

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

rule 3251

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e 
+ f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim 
p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2)))   Int[Sqrt[a + b*Sin[e 
+ f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x 
] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 
-1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
 

rule 3252

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b*c + a*d - d*x^2), 
x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, 
 e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.48

method	result	size
default	\(-\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sin \left (f x +e \right ) a d +\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a c +\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {a \left (c +d \right ) d}\right )}{\left (c +d \right ) \left (c +d \sin \left (f x +e \right )\right ) \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(155\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (89) = 178\).

Time = 0.16 (sec) , antiderivative size = 786, normalized size of antiderivative = 7.49 \[ \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/4*((d*cos(f*x + e)^2 - c*cos(f*x + e) - (d*cos(f*x + e) + c + d)*sin(f* 
x + e) - c - d)*sqrt(a/(c*d + d^2))*log((a*d^2*cos(f*x + e)^3 - a*c^2 - 2* 
a*c*d - a*d^2 - (6*a*c*d + 7*a*d^2)*cos(f*x + e)^2 + 4*(c^2*d + 4*c*d^2 + 
3*d^3 - (c*d^2 + d^3)*cos(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x + 
 e) - (c^2*d + 4*c*d^2 + 3*d^3 + (c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e)) 
*sqrt(a*sin(f*x + e) + a)*sqrt(a/(c*d + d^2)) - (a*c^2 + 8*a*c*d + 9*a*d^2 
)*cos(f*x + e) + (a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a* 
c*d + 4*a*d^2)*cos(f*x + e))*sin(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + 
d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f*x + e) + (d^2* 
cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 
4*sqrt(a*sin(f*x + e) + a)*(cos(f*x + e) - sin(f*x + e) + 1))/((c*d + d^2) 
*f*cos(f*x + e)^2 - (c^2 + c*d)*f*cos(f*x + e) - (c^2 + 2*c*d + d^2)*f - ( 
(c*d + d^2)*f*cos(f*x + e) + (c^2 + 2*c*d + d^2)*f)*sin(f*x + e)), -1/2*(( 
d*cos(f*x + e)^2 - c*cos(f*x + e) - (d*cos(f*x + e) + c + d)*sin(f*x + e) 
- c - d)*sqrt(-a/(c*d + d^2))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*sin(f 
*x + e) - c - 2*d)*sqrt(-a/(c*d + d^2))/(a*cos(f*x + e))) - 2*sqrt(a*sin(f 
*x + e) + a)*(cos(f*x + e) - sin(f*x + e) + 1))/((c*d + d^2)*f*cos(f*x + e 
)^2 - (c^2 + c*d)*f*cos(f*x + e) - (c^2 + 2*c*d + d^2)*f - ((c*d + d^2)*f* 
cos(f*x + e) + (c^2 + 2*c*d + d^2)*f)*sin(f*x + e))]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.28 \[ \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^2} \, dx=-\frac {\sqrt {2} \sqrt {a} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\sqrt {-c d - d^{2}} {\left (c + d\right )}} + \frac {2 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (2 \, d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )} {\left (c + d\right )}}\right )}}{2 \, f} \] Input:

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

Output:

-1/2*sqrt(2)*sqrt(a)*(sqrt(2)*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 1/2 
*e)/sqrt(-c*d - d^2))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))/(sqrt(-c*d - d^2 
)*(c + d)) + 2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 
 1/2*e)/((2*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - c - d)*(c + d)))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*int(sqrt(sin(e + f*x) + 1)/(sin(e + f*x)**2*d**2 + 2*sin(e + f*x)* 
c*d + c**2),x)