3.6.48 \(\int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx\) [548]

3.6.48.1 Optimal result
3.6.48.2 Mathematica [C] (verified)
3.6.48.3 Rubi [A] (verified)
3.6.48.4 Maple [B] (verified)
3.6.48.5 Fricas [B] (verification not implemented)
3.6.48.6 Sympy [F(-1)]
3.6.48.7 Maxima [F]
3.6.48.8 Giac [B] (verification not implemented)
3.6.48.9 Mupad [F(-1)]

3.6.48.1 Optimal result

Integrand size = 27, antiderivative size = 169 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx=-\frac {\sqrt {\frac {2}{3}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)}}\right )}{(c-d)^2 f}+\frac {\sqrt {d} (3 c+d) \text {arctanh}\left (\frac {\sqrt {3} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {3+3 \sin (e+f x)}}\right )}{\sqrt {3} (c-d)^2 (c+d)^{3/2} f}+\frac {d \cos (e+f x)}{\left (c^2-d^2\right ) f \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))} \]

output
-arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))*2^(1/2)/(c 
-d)^2/f/a^(1/2)+(3*c+d)*arctanh(cos(f*x+e)*a^(1/2)*d^(1/2)/(c+d)^(1/2)/(a+ 
a*sin(f*x+e))^(1/2))*d^(1/2)/(c-d)^2/(c+d)^(3/2)/f/a^(1/2)+d*cos(f*x+e)/(c 
^2-d^2)/f/(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2)
 
3.6.48.2 Mathematica [C] (verified)

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

Time = 3.59 (sec) , antiderivative size = 649, normalized size of antiderivative = 3.84 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left ((8+8 i) (-1)^{3/4} \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 )+\frac {\sqrt {d} (3 c+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 ]}{(c+d)^{3/2}}-\frac {\sqrt {d} (3 c+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 ]}{(c+d)^{3/2}}-\frac {4 d (-c+d) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d) (c+d \sin (e+f x))}\right )}{4 \sqrt {3} (c-d)^2 f \sqrt {1+\sin (e+f x)}} \]

input
Integrate[1/(Sqrt[3 + 3*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*Sqrt[3]*(c - d)^2*f*Sqrt[1 + Sin[e + f 
*x]])
 
3.6.48.3 Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.17, 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]))
 

3.6.48.3.1 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]
 
3.6.48.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(448\) vs. \(2(150)=300\).

Time = 0.98 (sec) , antiderivative size = 449, normalized size of antiderivative = 2.66

method result size
default \(\frac {\left (\sin \left (f x +e \right )+1\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (\sin \left (f x +e \right ) d \left (3 \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{\frac {7}{2}} c d +\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{\frac {7}{2}} d^{2}-\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, \sqrt {a \left (c +d \right ) d}\, a^{3} c -\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, \sqrt {a \left (c +d \right ) d}\, a^{3} d \right )+3 a^{\frac {7}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) c^{2} d +a^{\frac {7}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) c \,d^{2}+a^{\frac {5}{2}} \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, c d -a^{\frac {5}{2}} \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, d^{2}-\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {a \left (c +d \right ) d}\, a^{3} c^{2}-\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {a \left (c +d \right ) d}\, a^{3} c d \right )}{a^{\frac {7}{2}} \left (c -d \right )^{2} \left (c +d \right ) \left (c +d \sin \left (f x +e \right )\right ) \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) \(449\)

input
int(1/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^2,x,method=_RETURNVERBOSE)
 
output
(sin(f*x+e)+1)*(-a*(sin(f*x+e)-1))^(1/2)/a^(7/2)*(sin(f*x+e)*d*(3*arctanh( 
(a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(7/2)*c*d+arctanh((a-a*sin 
(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(7/2)*d^2-arctanh(1/2*(a-a*sin(f*x 
+e))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*(a*(c+d)*d)^(1/2)*a^3*c-arctanh(1/2*(a 
-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*(a*(c+d)*d)^(1/2)*a^3*d)+3*a 
^(7/2)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*c^2*d+a^(7/2) 
*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*c*d^2+a^(5/2)*(a-a* 
sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*c*d-a^(5/2)*(a-a*sin(f*x+e))^(1/2)*(a* 
(c+d)*d)^(1/2)*d^2-2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1 
/2))*(a*(c+d)*d)^(1/2)*a^3*c^2-2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)* 
2^(1/2)/a^(1/2))*(a*(c+d)*d)^(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+a*sin(f*x+e))^(1/2)/f
 
3.6.48.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 604 vs. \(2 (150) = 300\).

Time = 0.50 (sec) , antiderivative size = 1494, normalized size of antiderivative = 8.84 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \]

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)...
 
3.6.48.6 Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx=\text {Timed out} \]

input
integrate(1/(a+a*sin(f*x+e))**(1/2)/(c+d*sin(f*x+e))**2,x)
 
output
Timed out
 
3.6.48.7 Maxima [F]

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

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)
 
3.6.48.8 Giac [B] (verification not implemented)

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

Time = 0.52 (sec) , antiderivative size = 363, normalized size of antiderivative = 2.15 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx=\frac {\sqrt {2} {\left (\frac {\sqrt {2} {\left (3 \, c d + d^{2}\right )} \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 )}{{\left (c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - c^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - c d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + d^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {-c d - d^{2}}} + \frac {\log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + d^{2} \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^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {2 \, d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\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 )}}\right )}}{2 \, \sqrt {a} f} \]

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 
)
 
3.6.48.9 Mupad [F(-1)]

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

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)