Integrand size = 29, antiderivative size = 135 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{\sqrt {3+3 \sin (e+f x)}} \, dx=-\frac {2 \sqrt {d} \arctan \left (\frac {\sqrt {3} \sqrt {d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {3} f}-\frac {\sqrt {\frac {2}{3}} \sqrt {c-d} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} \sqrt {c-d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{f} \]
-arctanh(1/2*cos(f*x+e)*a^(1/2)*(c-d)^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2) /(c+d*sin(f*x+e))^(1/2))*2^(1/2)*(c-d)^(1/2)/f/a^(1/2)-2*arctan(cos(f*x+e) *a^(1/2)*d^(1/2)/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2))*d^(1/2)/f/ a^(1/2)
Result contains complex when optimal does not.
Time = 14.20 (sec) , antiderivative size = 1251, normalized size of antiderivative = 9.27 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{\sqrt {3+3 \sin (e+f x)}} \, dx =\text {Too large to display} \]
((Sqrt[2]*Sqrt[c - d]*Log[1 + Tan[(e + f*x)/2]] - Sqrt[2]*Sqrt[c - d]*Log[ c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x ]] + (-c + d)*Tan[(e + f*x)/2]] - I*Sqrt[d]*(Log[(2*(c - I*d + (1 - I)*Sqr t[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + ((-I )*c + d)*Tan[(e + f*x)/2]))/(d^(3/2)*(I + Tan[(e + f*x)/2]))] - Log[(2*(c + I*d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*S in[e + f*x]] + (I*c + d)*Tan[(e + f*x)/2]))/(d^(3/2)*(-I + Tan[(e + f*x)/2 ]))]))*Sqrt[c + d*Sin[e + f*x]])/(f*Sqrt[3 + 3*Sin[e + f*x]]*((Sqrt[c - d] *Sec[(e + f*x)/2]^2)/(Sqrt[2]*(1 + Tan[(e + f*x)/2])) - (Sqrt[2]*Sqrt[c - d]*(((-c + d)*Sec[(e + f*x)/2]^2)/2 + (Sqrt[c - d]*d*Cos[e + f*x]*Sqrt[(1 + Cos[e + f*x])^(-1)])/Sqrt[c + d*Sin[e + f*x]] + Sqrt[c - d]*((1 + Cos[e + f*x])^(-1))^(3/2)*Sin[e + f*x]*Sqrt[c + d*Sin[e + f*x]]))/(c - d + 2*Sqr t[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (-c + d) *Tan[(e + f*x)/2]) - I*Sqrt[d]*((d^(3/2)*(I + Tan[(e + f*x)/2])*((2*((((-I )*c + d)*Sec[(e + f*x)/2]^2)/2 + ((1 - I)*d^(3/2)*Cos[e + f*x]*Sqrt[(1 + C os[e + f*x])^(-1)])/(Sqrt[2]*Sqrt[c + d*Sin[e + f*x]]) + ((1 - I)*Sqrt[d]* ((1 + Cos[e + f*x])^(-1))^(3/2)*Sin[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/Sqr t[2]))/(d^(3/2)*(I + Tan[(e + f*x)/2])) - (Sec[(e + f*x)/2]^2*(c - I*d + ( 1 - I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f* x]] + ((-I)*c + d)*Tan[(e + f*x)/2]))/(d^(3/2)*(I + Tan[(e + f*x)/2])^2...
Time = 0.59 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 3256, 3042, 3254, 218, 3261, 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.
\(\displaystyle \int \frac {\sqrt {c+d \sin (e+f x)}}{\sqrt {a \sin (e+f x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {c+d \sin (e+f x)}}{\sqrt {a \sin (e+f x)+a}}dx\) |
\(\Big \downarrow \) 3256 |
\(\displaystyle (c-d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx+\frac {d \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (c-d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx+\frac {d \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{a}\) |
\(\Big \downarrow \) 3254 |
\(\displaystyle (c-d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {2 d \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle (c-d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {2 \sqrt {d} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f}\) |
\(\Big \downarrow \) 3261 |
\(\displaystyle -\frac {2 a (c-d) \int \frac {1}{2 a^2-\frac {a^3 (c-d) \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}-\frac {2 \sqrt {d} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 \sqrt {d} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f}-\frac {\sqrt {2} \sqrt {c-d} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f}\) |
(-2*Sqrt[d]*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x] ]*Sqrt[c + d*Sin[e + f*x]])])/(Sqrt[a]*f) - (Sqrt[2]*Sqrt[c - d]*ArcTanh[( Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/(Sqrt[a]*f)
3.6.90.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b + d*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*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]
Int[Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[d/b Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[ c + d*Sin[e + f*x]], x], x] + Simp[(b*c - a*d)/b Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*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]
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S in[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]
Leaf count of result is larger than twice the leaf count of optimal. \(3360\) vs. \(2(114)=228\).
Time = 1.48 (sec) , antiderivative size = 3361, normalized size of antiderivative = 24.90
\[\text {output too large to display}\]
-1/2/f*(c^2*(2*c-2*d)^(1/2)*ln(2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e) )/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)- d*cos(f*x+e)-c+d)/(-cos(f*x+e)+1+sin(f*x+e)))*(-(d^2/c^2)^(1/2)*c)^(1/2)*2 ^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*d*sin(f*x+e)-2*c*(2*c-2*d)^ (1/2)*ln(2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2 )*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(-co s(f*x+e)+1+sin(f*x+e)))*(-(d^2/c^2)^(1/2)*c)^(1/2)*2^(1/2)*((c+d*sin(f*x+e ))/(cos(f*x+e)+1))^(1/2)*d^2*sin(f*x+e)+(2*c-2*d)^(1/2)*ln(2*((2*c-2*d)^(1 /2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e )-d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(-cos(f*x+e)+1+sin(f*x+e)))* (-(d^2/c^2)^(1/2)*c)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2) *d^3*sin(f*x+e)+((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*(2*c-2*d)^(1/2)*(- (d^2/c^2)^(1/2)*c)^(1/2)*2^(1/2)*ln(2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f *x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f* x+e)-d*cos(f*x+e)-c+d)/(-cos(f*x+e)+1+sin(f*x+e)))*c^2*cos(f*x+e)*d-2*((c+ d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*(2*c-2*d)^(1/2)*(-(d^2/c^2)^(1/2)*c)^( 1/2)*2^(1/2)*ln(2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1 ))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+ d)/(-cos(f*x+e)+1+sin(f*x+e)))*c*cos(f*x+e)*d^2+((c+d*sin(f*x+e))/(cos(f*x +e)+1))^(1/2)*(2*c-2*d)^(1/2)*(-(d^2/c^2)^(1/2)*c)^(1/2)*2^(1/2)*ln(2*(...
Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (114) = 228\).
Time = 0.62 (sec) , antiderivative size = 1944, normalized size of antiderivative = 14.40 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\text {Too large to display} \]
[1/4*(2*sqrt(2)*sqrt((c - d)/a)*log((2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sq rt(d*sin(f*x + e) + c)*sqrt((c - d)/a)*(cos(f*x + e) - sin(f*x + e) + 1) - (c - 3*d)*cos(f*x + e)^2 - (3*c - d)*cos(f*x + e) + ((c - 3*d)*cos(f*x + e) - 2*c - 2*d)*sin(f*x + e) - 2*c - 2*d)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + sqrt(-d/a)*log((128*d^4*cos(f*x + e)^5 + 128*(2*c*d^3 - d^4)*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c^2*d^2 + 4 *c*d^3 + d^4 - 32*(5*c^2*d^2 - 14*c*d^3 + 13*d^4)*cos(f*x + e)^3 - 32*(c^3 *d - 2*c^2*d^2 + 9*c*d^3 - 4*d^4)*cos(f*x + e)^2 - 8*(16*d^3*cos(f*x + e)^ 4 + 24*(c*d^2 - d^3)*cos(f*x + e)^3 - c^3 + 17*c^2*d - 59*c*d^2 + 51*d^3 - 2*(5*c^2*d - 26*c*d^2 + 33*d^3)*cos(f*x + e)^2 - (c^3 - 7*c^2*d + 31*c*d^ 2 - 25*d^3)*cos(f*x + e) + (16*d^3*cos(f*x + e)^3 + c^3 - 17*c^2*d + 59*c* d^2 - 51*d^3 - 8*(3*c*d^2 - 5*d^3)*cos(f*x + e)^2 - 2*(5*c^2*d - 14*c*d^2 + 13*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin( f*x + e) + c)*sqrt(-d/a) + (c^4 - 28*c^3*d + 230*c^2*d^2 - 476*c*d^3 + 289 *d^4)*cos(f*x + e) + (128*d^4*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4 - 256*(c*d^3 - d^4)*cos(f*x + e)^3 - 32*(5*c^2*d^2 - 6*c*d^ 3 + 5*d^4)*cos(f*x + e)^2 + 32*(c^3*d - 7*c^2*d^2 + 15*c*d^3 - 9*d^4)*cos( f*x + e))*sin(f*x + e))/(cos(f*x + e) + sin(f*x + e) + 1)))/f, 1/2*(sqrt(2 )*sqrt((c - d)/a)*log((2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt((c - d)/a)*(cos(f*x + e) - sin(f*x + e) + 1) - (c - 3*d)*...
\[ \int \frac {\sqrt {c+d \sin (e+f x)}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\int \frac {\sqrt {c + d \sin {\left (e + f x \right )}}}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \]
\[ \int \frac {\sqrt {c+d \sin (e+f x)}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\int { \frac {\sqrt {d \sin \left (f x + e\right ) + c}}{\sqrt {a \sin \left (f x + e\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\int \frac {\sqrt {c+d\,\sin \left (e+f\,x\right )}}{\sqrt {a+a\,\sin \left (e+f\,x\right )}} \,d x \]