\(\int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx\) [536]

Optimal result

Integrand size = 27, antiderivative size = 154 \[ \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx=-\frac {3 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{4 \sqrt {d} (c+d)^{5/2} f}-\frac {a \cos (e+f x)}{2 (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {3 a \cos (e+f x)}{4 (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \] Output:

-3/4*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)^(5/2)/f-1/2*a*cos(f*x+e)/(c+d)/f/(a+a*sin(f*x+e))^ 
(1/2)/(c+d*sin(f*x+e))^2-3/4*a*cos(f*x+e)/(c+d)^2/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 = 7.27 (sec) , antiderivative size = 920, normalized size of antiderivative = 5.97 \[ \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

Integrate[Sqrt[a + a*Sin[e + f*x]]/(c + d*Sin[e + f*x])^3,x]
 

Output:

((1/16 + I/16)*Sqrt[a*(1 + Sin[e + f*x])]*((3*(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)^(5/2)*(Cos[e] + 
I*(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) + (3*(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)*Sq 
rt[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*Sqrt[d]*Sqrt[c + d]*E^(I*e)*f*x*#1^3 + 2*Sqrt[d]*Sqrt[c + d]*E^(I*e)*Lo 
g[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)^(5/2)*(Cos[e] + I* 
(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) - ((4 - 4*I)*(Cos[(e + f*x)/2] - S 
in[(e + f*x)/2]))/((c + d)*f*(c + d*Sin[e + f*x])^2) - ((6 - 6*I)*(Cos[...
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 3251, 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.

\(\displaystyle \int \frac {\sqrt {a \sin (e+f x)+a}}{(c+d \sin (e+f x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a \sin (e+f x)+a}}{(c+d \sin (e+f x))^3}dx\)

\(\Big \downarrow \) 3251

\(\displaystyle \frac {3 \int \frac {\sqrt {\sin (e+f x) a+a}}{(c+d \sin (e+f x))^2}dx}{4 (c+d)}-\frac {a \cos (e+f x)}{2 f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \int \frac {\sqrt {\sin (e+f x) a+a}}{(c+d \sin (e+f x))^2}dx}{4 (c+d)}-\frac {a \cos (e+f x)}{2 f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 3251

\(\displaystyle \frac {3 \left (\frac {\int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{2 (c+d)}-\frac {a \cos (e+f x)}{f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}\right )}{4 (c+d)}-\frac {a \cos (e+f x)}{2 f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {\int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{2 (c+d)}-\frac {a \cos (e+f x)}{f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}\right )}{4 (c+d)}-\frac {a \cos (e+f x)}{2 f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {3 \left (-\frac {a \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f (c+d)}-\frac {a \cos (e+f x)}{f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}\right )}{4 (c+d)}-\frac {a \cos (e+f x)}{2 f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {3 \left (-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {d} f (c+d)^{3/2}}-\frac {a \cos (e+f x)}{f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}\right )}{4 (c+d)}-\frac {a \cos (e+f x)}{2 f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}\)

Input:

Int[Sqrt[a + a*Sin[e + f*x]]/(c + d*Sin[e + f*x])^3,x]
 

Output:

-1/2*(a*Cos[e + f*x])/((c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f 
*x])^2) + (3*(-((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]))))/(4*(c + d))
 

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 = 0.68 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.65

method	result	size
default	\(-\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \left (3 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sin \left (f x +e \right )^{2} a \,d^{2}+6 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sin \left (f x +e \right ) a c d +3 \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {a \left (c +d \right ) d}\, \sin \left (f x +e \right ) d +3 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a \,c^{2}+5 \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {a \left (c +d \right ) d}\, c +2 \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {a \left (c +d \right ) d}\, d \right )}{4 \left (c +d \right )^{2} \left (c +d \sin \left (f x +e \right )\right )^{2} \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(254\)

Input:

int((a+sin(f*x+e)*a)^(1/2)/(c+d*sin(f*x+e))^3,x,method=_RETURNVERBOSE)
 

Output:

-1/4*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(3*arctanh((-a*(-1+sin(f*x+ 
e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)^2*a*d^2+6*arctanh((-a*(-1+sin(f 
*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)*a*c*d+3*(-a*(-1+sin(f*x+e))) 
^(1/2)*(a*(c+d)*d)^(1/2)*sin(f*x+e)*d+3*arctanh((-a*(-1+sin(f*x+e)))^(1/2) 
*d/(a*(c+d)*d)^(1/2))*a*c^2+5*(-a*(-1+sin(f*x+e)))^(1/2)*(a*(c+d)*d)^(1/2) 
*c+2*(-a*(-1+sin(f*x+e)))^(1/2)*(a*(c+d)*d)^(1/2)*d)/(c+d)^2/(c+d*sin(f*x+ 
e))^2/(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. 467 vs. \(2 (130) = 260\).

Time = 0.20 (sec) , antiderivative size = 1250, normalized size of antiderivative = 8.12 \[ \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^3,x, algorithm="fricas")
 

Output:

[1/16*(3*(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))*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*(3*d*cos(f*x + e)^2 + (5*c + 2*d)*cos(f*x + e) + (3 
*d*cos(f*x + e) - 5*c + d)*sin(f*x + e) + 5*c - d)*sqrt(a*sin(f*x + e) + a 
))/((c^2*d^2 + 2*c*d^3 + d^4)*f*cos(f*x + e)^3 + (2*c^3*d + 5*c^2*d^2 + 4* 
c*d^3 + d^4)*f*cos(f*x + e)^2 - (c^4 + 2*c^3*d + 2*c^2*d^2 + 2*c*d^3 + d^4 
)*f*cos(f*x + e) - (c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4)*f + ((c^2*d 
^2 + 2*c*d^3 + d^4)*f*cos(f*x + e)^2 - 2*(c^3*d + 2*c^2*d^2 + c*d^3)*f*cos 
(f*x + e) - (c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4)*f)*sin(f*x + e)), 
-1/8*(3*(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 +...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**(1/2)/(c+d*sin(f*x+e))**3,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^3,x, algorithm="maxima")
 

Output:

integrate(sqrt(a*sin(f*x + e) + a)/(d*sin(f*x + e) + c)^3, x)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx=-\frac {\sqrt {2} \sqrt {a} {\left (\frac {3 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (c^{2} + 2 \, c d + d^{2}\right )} \sqrt {-c d - d^{2}}} + \frac {2 \, {\left (6 \, d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 5 \, c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 5 \, d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (2 \, d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )}^{2} {\left (c^{2} + 2 \, c d + d^{2}\right )}}\right )}}{8 \, f} \] Input:

integrate((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^3,x, algorithm="giac")
 

Output:

-1/8*sqrt(2)*sqrt(a)*(3*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))/((c^2 + 2*c*d 
+ d^2)*sqrt(-c*d - d^2)) + 2*(6*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin( 
-1/4*pi + 1/2*f*x + 1/2*e)^3 - 5*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin 
(-1/4*pi + 1/2*f*x + 1/2*e) - 5*d*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 
)^2*(c^2 + 2*c*d + d^2)))/f
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx=\int \frac {\sqrt {a+a\,\sin \left (e+f\,x\right )}}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^3} \,d x \] Input:

int((a + a*sin(e + f*x))^(1/2)/(c + d*sin(e + f*x))^3,x)
 

Output:

int((a + a*sin(e + f*x))^(1/2)/(c + d*sin(e + f*x))^3, x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^3} \, dx=\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{3} d^{3}+3 \sin \left (f x +e \right )^{2} c \,d^{2}+3 \sin \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) \] Input:

int((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^3,x)
 

Output:

sqrt(a)*int(sqrt(sin(e + f*x) + 1)/(sin(e + f*x)**3*d**3 + 3*sin(e + f*x)* 
*2*c*d**2 + 3*sin(e + f*x)*c**2*d + c**3),x)