\(\int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx\) [556]

Optimal result

Integrand size = 27, antiderivative size = 175 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d)^2 f}+\frac {\sqrt {d} (3 c+d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d)^2 (c+d)^{3/2} f}+\frac {d \cos (e+f x)}{\left (c^2-d^2\right ) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \] 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)^2/f+d^(1/2)*(3*c+d)*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)^2/(c+d)^(3/2)/f+d*cos(f*x+e)/(c 
^2-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 = 3.87 (sec) , antiderivative size = 646, normalized size of antiderivative = 3.69 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

Integrate[1/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^2),x]
 

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*((8 + 8*I)*(-1)^(3/4)*ArcTanh[(1/2 
+ I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])] + (Sqrt[d]*(3*c + 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) & ])/(c + d)^(3/2) - (Sqrt[d]*(3*c + 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) 
& ])/(c + d)^(3/2) - (4*d*(-c + d)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/ 
((c + d)*(c + d*Sin[e + f*x]))))/(4*(c - d)^2*f*Sqrt[a*(1 + Sin[e + f*x])] 
)
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3258, 3042, 3464, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3464

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 3252

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

\(\Big \downarrow \) 221

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

Input:

Int[1/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^2),x]
 

Output:

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

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 3258

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

rule 3464

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(447\) vs. \(2(150)=300\).

Time = 1.03 (sec) , antiderivative size = 448, normalized size of antiderivative = 2.56

method	result	size
default	\(-\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \left (\sin \left (f x +e \right ) d \left (-3 a^{\frac {7}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) c d -a^{\frac {7}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) d^{2}+\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, \sqrt {a \left (c +d \right ) d}\, a^{3} c +\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, \sqrt {a \left (c +d \right ) d}\, a^{3} d \right )-3 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{\frac {7}{2}} c^{2} d -\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{\frac {7}{2}} c \,d^{2}-\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a \left (c +d \right ) d}\, a^{\frac {5}{2}} c d +\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a \left (c +d \right ) d}\, a^{\frac {5}{2}} d^{2}+\sqrt {2}\, \sqrt {a \left (c +d \right ) d}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} c^{2}+\sqrt {2}\, \sqrt {a \left (c +d \right ) d}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} c d \right )}{a^{\frac {7}{2}} \left (c -d \right )^{2} \left (c +d \right ) \left (c +d \sin \left (f x +e \right )\right ) \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(448\)

Input:

int(1/(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)/a^(7/2)*(sin(f*x+e)*d*(-3*a^(7/ 
2)*arctanh((a-sin(f*x+e)*a)^(1/2)*d/(a*c*d+a*d^2)^(1/2))*c*d-a^(7/2)*arcta 
nh((a-sin(f*x+e)*a)^(1/2)*d/(a*c*d+a*d^2)^(1/2))*d^2+arctanh(1/2*(a-sin(f* 
x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*(a*(c+d)*d)^(1/2)*a^3*c+arctanh(1/2 
*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*(a*(c+d)*d)^(1/2)*a^3*d)- 
3*arctanh((a-sin(f*x+e)*a)^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(7/2)*c^2*d-arct 
anh((a-sin(f*x+e)*a)^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(7/2)*c*d^2-(a-sin(f*x 
+e)*a)^(1/2)*(a*(c+d)*d)^(1/2)*a^(5/2)*c*d+(a-sin(f*x+e)*a)^(1/2)*(a*(c+d) 
*d)^(1/2)*a^(5/2)*d^2+2^(1/2)*(a*(c+d)*d)^(1/2)*arctanh(1/2*(a-sin(f*x+e)* 
a)^(1/2)*2^(1/2)/a^(1/2))*a^3*c^2+2^(1/2)*(a*(c+d)*d)^(1/2)*arctanh(1/2*(a 
-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*a^3*c*d)/(c-d)^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. 604 vs. \(2 (150) = 300\).

Time = 0.32 (sec) , antiderivative size = 1494, normalized size of antiderivative = 8.54 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/4*((3*a*c^2 + 4*a*c*d + a*d^2 - (3*a*c*d + a*d^2)*cos(f*x + e)^2 + (3* 
a*c^2 + a*c*d)*cos(f*x + e) + (3*a*c^2 + 4*a*c*d + a*d^2 + (3*a*c*d + a*d^ 
2)*cos(f*x + e))*sin(f*x + e))*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))*si 
n(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))) + 2*sqrt(2)*(a*c^2 + 2*a*c*d + a* 
d^2 - (a*c*d + a*d^2)*cos(f*x + e)^2 + (a*c^2 + a*c*d)*cos(f*x + e) + (a*c 
^2 + 2*a*c*d + a*d^2 + (a*c*d + a*d^2)*cos(f*x + e))*sin(f*x + e))*log(-(c 
os(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)*sin(f*x + e) - cos(f*x + e) - 2))/sq 
rt(a) + 4*(c*d - d^2 + (c*d - d^2)*cos(f*x + e) - (c*d - d^2)*sin(f*x + e) 
)*sqrt(a*sin(f*x + e) + a))/((a*c^3*d - a*c^2*d^2 - a*c*d^3 + a*d^4)*f*cos 
(f*x + e)^2 - (a*c^4 - a*c^3*d - a*c^2*d^2 + a*c*d^3)*f*cos(f*x + e) - (a* 
c^4 - 2*a*c^2*d^2 + a*d^4)*f - ((a*c^3*d - a*c^2*d^2 - a*c*d^3 + a*d^4)...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 363 vs. \(2 (150) = 300\).

Time = 0.18 (sec) , antiderivative size = 363, normalized size of antiderivative = 2.07 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

1/2*sqrt(2)*(sqrt(2)*(3*c*d + d^2)*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x 
+ 1/2*e)/sqrt(-c*d - d^2))/((c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - c^2 
*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - c*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 
 1/2*e)) + d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(-c*d - d^2)) + lo 
g(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2 
*e)) - 2*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + d^2*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^2*sgn(cos( 
-1/4*pi + 1/2*f*x + 1/2*e)) - 2*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 
d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) + 2*d*sin(-1/4*pi + 1/2*f*x + 1/2 
*e)/((c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - d^2*sgn(cos(-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)))/(sqrt(a)*f 
)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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