Integrand size = 23, antiderivative size = 54 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {1+\cos (c+d x)}} \, dx=-\frac {\sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{1+\cos (c+d x)}\right )}{d}+\frac {2 \arcsin \left (\frac {\sin (c+d x)}{\sqrt {1+\cos (c+d x)}}\right )}{d} \] Output:
-2^(1/2)*arcsin(sin(d*x+c)/(1+cos(d*x+c)))/d+2*arcsin(sin(d*x+c)/(1+cos(d* x+c))^(1/2))/d
Time = 0.16 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.87 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {1+\cos (c+d x)}} \, dx=-\frac {\left (2 \arcsin \left (\sqrt {\cos (c+d x)}\right )-\sqrt {2} \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right )\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sin (c+d x)}{d \sqrt {-((-1+\cos (c+d x)) \cos (c+d x))}} \] Input:
Integrate[Sqrt[Cos[c + d*x]]/Sqrt[1 + Cos[c + d*x]],x]
Output:
-(((2*ArcSin[Sqrt[Cos[c + d*x]]] - Sqrt[2]*ArcTan[Sqrt[Cos[c + d*x]]/Sqrt[ Sin[(c + d*x)/2]^2]])*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sin[c + d*x])/ (d*Sqrt[-((-1 + Cos[c + d*x])*Cos[c + d*x])]))
Time = 0.40 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3256, 3042, 3253, 223, 3260, 223}
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.
\(\displaystyle \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {\cos (c+d x)+1}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx\) |
\(\Big \downarrow \) 3256 |
\(\displaystyle \int \frac {\sqrt {\cos (c+d x)+1}}{\sqrt {\cos (c+d x)}}dx-\int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx\) |
\(\Big \downarrow \) 3253 |
\(\displaystyle -\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx-\frac {2 \int \frac {1}{\sqrt {1-\frac {\sin ^2(c+d x)}{\cos (c+d x)+1}}}d\left (-\frac {\sin (c+d x)}{\sqrt {\cos (c+d x)+1}}\right )}{d}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {2 \arcsin \left (\frac {\sin (c+d x)}{\sqrt {\cos (c+d x)+1}}\right )}{d}-\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx\) |
\(\Big \downarrow \) 3260 |
\(\displaystyle \frac {\sqrt {2} \int \frac {1}{\sqrt {1-\frac {\sin ^2(c+d x)}{(\cos (c+d x)+1)^2}}}d\left (-\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}+\frac {2 \arcsin \left (\frac {\sin (c+d x)}{\sqrt {\cos (c+d x)+1}}\right )}{d}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {2 \arcsin \left (\frac {\sin (c+d x)}{\sqrt {\cos (c+d x)+1}}\right )}{d}-\frac {\sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}\) |
Input:
Int[Sqrt[Cos[c + d*x]]/Sqrt[1 + Cos[c + d*x]],x]
Output:
-((Sqrt[2]*ArcSin[Sin[c + d*x]/(1 + Cos[c + d*x])])/d) + (2*ArcSin[Sin[c + d*x]/Sqrt[1 + Cos[c + d*x]]])/d
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[-2/f Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E qQ[a^2 - b^2, 0] && EqQ[d, a/b]
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[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f _.)*(x_)]]), x_Symbol] :> Simp[-Sqrt[2]/(Sqrt[a]*f) Subst[Int[1/Sqrt[1 - x^2], x], x, b*(Cos[e + f*x]/(a + b*Sin[e + f*x]))], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && EqQ[d, a/b] && GtQ[a, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.97 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.83
method | result | size |
default | \(\frac {\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\cos \left (d x +c \right )}\, \operatorname {csgn}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right )}{d \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\) | \(99\) |
Input:
int(cos(d*x+c)^(1/2)/(cos(d*x+c)+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/d*sec(1/2*d*x+1/2*c)*cos(d*x+c)^(1/2)*csgn(cos(1/2*d*x+1/2*c))*(2^(1/2)* arctan(tan(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+arcsin(cot(d*x+c)-csc (d*x+c)))/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (50) = 100\).
Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.89 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {1+\cos (c+d x)}} \, dx=-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )}}\right ) - 2 \, \arctan \left (\frac {\sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )}\right )}{d} \] Input:
integrate(cos(d*x+c)^(1/2)/(1+cos(d*x+c))^(1/2),x, algorithm="fricas")
Output:
-(sqrt(2)*arctan(1/2*sqrt(2)*sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c))*sin (d*x + c)/(cos(d*x + c)^2 + cos(d*x + c))) - 2*arctan(sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c))*sin(d*x + c)/(cos(d*x + c)^2 + cos(d*x + c))))/d
\[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {1+\cos (c+d x)}} \, dx=\int \frac {\sqrt {\cos {\left (c + d x \right )}}}{\sqrt {\cos {\left (c + d x \right )} + 1}}\, dx \] Input:
integrate(cos(d*x+c)**(1/2)/(1+cos(d*x+c))**(1/2),x)
Output:
Integral(sqrt(cos(c + d*x))/sqrt(cos(c + d*x) + 1), x)
Result contains complex when optimal does not.
Time = 0.44 (sec) , antiderivative size = 689, normalized size of antiderivative = 12.76 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {1+\cos (c+d x)}} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^(1/2)/(1+cos(d*x+c))^(1/2),x, algorithm="maxima")
Output:
-(sqrt(2)*arctan2(((abs(2*e^(I*d*x + I*c) + 2)^4 + 16*cos(d*x + c)^4 + 16* sin(d*x + c)^4 + 8*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)* abs(2*e^(I*d*x + I*c) + 2)^2 - 64*cos(d*x + c)^3 + 32*(cos(d*x + c)^2 - 2* cos(d*x + c) + 1)*sin(d*x + c)^2 + 96*cos(d*x + c)^2 - 64*cos(d*x + c) + 1 6)^(1/4)*sin(1/2*arctan2(8*(cos(d*x + c) - 1)*sin(d*x + c)/abs(2*e^(I*d*x + I*c) + 2)^2, (abs(2*e^(I*d*x + I*c) + 2)^2 + 4*cos(d*x + c)^2 - 4*sin(d* x + c)^2 - 8*cos(d*x + c) + 4)/abs(2*e^(I*d*x + I*c) + 2)^2)) + 2*sin(d*x + c))/abs(2*e^(I*d*x + I*c) + 2), ((abs(2*e^(I*d*x + I*c) + 2)^4 + 16*cos( d*x + c)^4 + 16*sin(d*x + c)^4 + 8*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*co s(d*x + c) + 1)*abs(2*e^(I*d*x + I*c) + 2)^2 - 64*cos(d*x + c)^3 + 32*(cos (d*x + c)^2 - 2*cos(d*x + c) + 1)*sin(d*x + c)^2 + 96*cos(d*x + c)^2 - 64* cos(d*x + c) + 16)^(1/4)*cos(1/2*arctan2(8*(cos(d*x + c) - 1)*sin(d*x + c) /abs(2*e^(I*d*x + I*c) + 2)^2, (abs(2*e^(I*d*x + I*c) + 2)^2 + 4*cos(d*x + c)^2 - 4*sin(d*x + c)^2 - 8*cos(d*x + c) + 4)/abs(2*e^(I*d*x + I*c) + 2)^ 2)) + 2*cos(d*x + c) - 2)/abs(2*e^(I*d*x + I*c) + 2)) - arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arc tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arcta n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + cos(d*x + c)))/d
\[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {1+\cos (c+d x)}} \, dx=\int { \frac {\sqrt {\cos \left (d x + c\right )}}{\sqrt {\cos \left (d x + c\right ) + 1}} \,d x } \] Input:
integrate(cos(d*x+c)^(1/2)/(1+cos(d*x+c))^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(cos(d*x + c))/sqrt(cos(d*x + c) + 1), x)
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {1+\cos (c+d x)}} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}}{\sqrt {\cos \left (c+d\,x\right )+1}} \,d x \] Input:
int(cos(c + d*x)^(1/2)/(cos(c + d*x) + 1)^(1/2),x)
Output:
int(cos(c + d*x)^(1/2)/(cos(c + d*x) + 1)^(1/2), x)
\[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {1+\cos (c+d x)}} \, dx=\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )+1}d x \] Input:
int(cos(d*x+c)^(1/2)/(1+cos(d*x+c))^(1/2),x)
Output:
int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/(cos(c + d*x) + 1),x)