\(\int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx\) [555]

Optimal result

Integrand size = 27, antiderivative size = 123 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d) f}+\frac {2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d) \sqrt {c+d} f} \] Output:

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

Mathematica [C] (verified)

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

Time = 1.97 (sec) , antiderivative size = 588, normalized size of antiderivative = 4.78 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\frac {\left ((4+4 i) (-1)^{3/4} \sqrt {c+d} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right )+\sqrt {d} \text {RootSum}\left [c+4 d \text {$\#$1}+2 c \text {$\#$1}^2-4 d \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {-d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )+\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+2 \sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+3 d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^3}{-d-c \text {$\#$1}+3 d \text {$\#$1}^2-c \text {$\#$1}^3}\&\right ]-\sqrt {d} \text {RootSum}\left [c+4 d \text {$\#$1}+2 c \text {$\#$1}^2-4 d \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {-d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}-2 \sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+3 d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2+\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^3}{-d-c \text {$\#$1}+3 d \text {$\#$1}^2-c \text {$\#$1}^3}\&\right ]\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{2 (c-d) \sqrt {c+d} f \sqrt {a (1+\sin (e+f x))}} \] Input:

Integrate[1/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])),x]
 

Output:

(((4 + 4*I)*(-1)^(3/4)*Sqrt[c + d]*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Ta 
n[(e + f*x)/4])] + Sqrt[d]*RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1 
^4 & , (-(d*Log[-#1 + Tan[(e + f*x)/4]]) + Sqrt[d]*Sqrt[c + d]*Log[-#1 + T 
an[(e + f*x)/4]] - c*Log[-#1 + Tan[(e + f*x)/4]]*#1 + 2*Sqrt[d]*Sqrt[c + d 
]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + 3*d*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - 
Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - c*Log[-#1 + Tan[(e 
+ f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ] - Sqrt[d]*RootSum[c + 
 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(d*Log[-#1 + Tan[(e + f*x)/4] 
]) - Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] - c*Log[-#1 + Tan[(e 
+ f*x)/4]]*#1 - 2*Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + 3*d 
*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + 
 f*x)/4]]*#1^2 - c*Log[-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 
 - c*#1^3) & ])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/(2*(c - d)*Sqrt[c + 
 d]*f*Sqrt[a*(1 + Sin[e + f*x])])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 3259, 3042, 3128, 219, 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.

\(\displaystyle \int \frac {1}{\sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}dx\)

\(\Big \downarrow \) 3259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 3252

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

\(\Big \downarrow \) 221

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

Input:

Int[1/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])),x]
 

Output:

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

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

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 3128

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

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]
 

rule 3259

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])), x_Symbol] :> Simp[b/(b*c - a*d)   Int[1/Sqrt[a + b*Sin[e + 
 f*x]], x], x] - Simp[d/(b*c - a*d)   Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*S 
in[e + f*x]), 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 = 0.95 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.07

method	result	size
default	\(-\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \left (-2 d \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a^{\frac {3}{2}}+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \sqrt {a \left (c +d \right ) d}\right )}{\left (c -d \right ) \sqrt {a \left (c +d \right ) d}\, a^{\frac {3}{2}} \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(131\)

Input:

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

Output:

-(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(-2*d*arctanh((-a*(-1+sin(f*x+e 
)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(3/2)+2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x 
+e)))^(1/2)*2^(1/2)/a^(1/2))*a*(a*(c+d)*d)^(1/2))/(c-d)/(a*(c+d)*d)^(1/2)/ 
a^(3/2)/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. 201 vs. \(2 (100) = 200\).

Time = 0.17 (sec) , antiderivative size = 685, normalized size of antiderivative = 5.57 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.60 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\frac {\sqrt {2} {\left (\frac {2 \, \sqrt {2} d \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right )}{\sqrt {-c d - d^{2}} {\left (c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )}} + \frac {\log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )}}{2 \, \sqrt {a} f} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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