Integrand size = 16, antiderivative size = 184 \[ \int \sqrt {a+b \arccos \left (-1+d x^2\right )} \, dx=\frac {2 \sqrt {a+b \arccos \left (-1+d x^2\right )} \cos ^2\left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )}{d x}-\frac {2 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {\pi }}\right )}{\sqrt {\frac {1}{b}} d x}-\frac {2 \sqrt {\pi } \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )}{\sqrt {\frac {1}{b}} d x} \] Output:
2*(a+b*arccos(d*x^2-1))^(1/2)*cos(1/2*arccos(d*x^2-1))^2/d/x-2*Pi^(1/2)*co s(1/2*a/b)*cos(1/2*arccos(d*x^2-1))*FresnelC((1/b)^(1/2)*(a+b*arccos(d*x^2 -1))^(1/2)/Pi^(1/2))/(1/b)^(1/2)/d/x-2*Pi^(1/2)*cos(1/2*arccos(d*x^2-1))*F resnelS((1/b)^(1/2)*(a+b*arccos(d*x^2-1))^(1/2)/Pi^(1/2))*sin(1/2*a/b)/(1/ b)^(1/2)/d/x
Time = 0.05 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.81 \[ \int \sqrt {a+b \arccos \left (-1+d x^2\right )} \, dx=-\frac {2 \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \left (-\sqrt {a+b \arccos \left (-1+d x^2\right )} \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )+\sqrt {b} \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right )+\sqrt {b} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )\right )}{d x} \] Input:
Integrate[Sqrt[a + b*ArcCos[-1 + d*x^2]],x]
Output:
(-2*Cos[ArcCos[-1 + d*x^2]/2]*(-(Sqrt[a + b*ArcCos[-1 + d*x^2]]*Cos[ArcCos [-1 + d*x^2]/2]) + Sqrt[b]*Sqrt[Pi]*Cos[a/(2*b)]*FresnelC[Sqrt[a + b*ArcCo s[-1 + d*x^2]]/(Sqrt[b]*Sqrt[Pi])] + Sqrt[b]*Sqrt[Pi]*FresnelS[Sqrt[a + b* ArcCos[-1 + d*x^2]]/(Sqrt[b]*Sqrt[Pi])]*Sin[a/(2*b)]))/(d*x)
Time = 0.25 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {5312}
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.
\(\displaystyle \int \sqrt {a+b \arccos \left (d x^2-1\right )} \, dx\) |
\(\Big \downarrow \) 5312 |
\(\displaystyle -\frac {2 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (d x^2-1\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{\sqrt {\frac {1}{b}} d x}-\frac {2 \sqrt {\pi } \sin \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (d x^2-1\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{\sqrt {\frac {1}{b}} d x}+\frac {2 \cos ^2\left (\frac {1}{2} \arccos \left (d x^2-1\right )\right ) \sqrt {a+b \arccos \left (d x^2-1\right )}}{d x}\) |
Input:
Int[Sqrt[a + b*ArcCos[-1 + d*x^2]],x]
Output:
(2*Sqrt[a + b*ArcCos[-1 + d*x^2]]*Cos[ArcCos[-1 + d*x^2]/2]^2)/(d*x) - (2* Sqrt[Pi]*Cos[a/(2*b)]*Cos[ArcCos[-1 + d*x^2]/2]*FresnelC[(Sqrt[b^(-1)]*Sqr t[a + b*ArcCos[-1 + d*x^2]])/Sqrt[Pi]])/(Sqrt[b^(-1)]*d*x) - (2*Sqrt[Pi]*C os[ArcCos[-1 + d*x^2]/2]*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[-1 + d*x ^2]])/Sqrt[Pi]]*Sin[a/(2*b)])/(Sqrt[b^(-1)]*d*x)
Int[Sqrt[(a_.) + ArcCos[-1 + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[2*Sqrt [a + b*ArcCos[-1 + d*x^2]]*(Cos[(1/2)*ArcCos[-1 + d*x^2]]^2/(d*x)), x] + (- Simp[2*Sqrt[Pi]*Cos[a/(2*b)]*Cos[ArcCos[-1 + d*x^2]/2]*(FresnelC[Sqrt[1/(Pi *b)]*Sqrt[a + b*ArcCos[-1 + d*x^2]]]/(Sqrt[1/b]*d*x)), x] - Simp[2*Sqrt[Pi] *Sin[a/(2*b)]*Cos[ArcCos[-1 + d*x^2]/2]*(FresnelS[Sqrt[1/(Pi*b)]*Sqrt[a + b *ArcCos[-1 + d*x^2]]]/(Sqrt[1/b]*d*x)), x]) /; FreeQ[{a, b, d}, x]
\[\int \sqrt {a +b \arccos \left (d \,x^{2}-1\right )}d x\]
Input:
int((a+b*arccos(d*x^2-1))^(1/2),x)
Output:
int((a+b*arccos(d*x^2-1))^(1/2),x)
Exception generated. \[ \int \sqrt {a+b \arccos \left (-1+d x^2\right )} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+b*arccos(d*x^2-1))^(1/2),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \sqrt {a+b \arccos \left (-1+d x^2\right )} \, dx=\int \sqrt {a + b \operatorname {acos}{\left (d x^{2} - 1 \right )}}\, dx \] Input:
integrate((a+b*acos(d*x**2-1))**(1/2),x)
Output:
Integral(sqrt(a + b*acos(d*x**2 - 1)), x)
\[ \int \sqrt {a+b \arccos \left (-1+d x^2\right )} \, dx=\int { \sqrt {b \arccos \left (d x^{2} - 1\right ) + a} \,d x } \] Input:
integrate((a+b*arccos(d*x^2-1))^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(b*arccos(d*x^2 - 1) + a), x)
\[ \int \sqrt {a+b \arccos \left (-1+d x^2\right )} \, dx=\int { \sqrt {b \arccos \left (d x^{2} - 1\right ) + a} \,d x } \] Input:
integrate((a+b*arccos(d*x^2-1))^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(b*arccos(d*x^2 - 1) + a), x)
Timed out. \[ \int \sqrt {a+b \arccos \left (-1+d x^2\right )} \, dx=\int \sqrt {a+b\,\mathrm {acos}\left (d\,x^2-1\right )} \,d x \] Input:
int((a + b*acos(d*x^2 - 1))^(1/2),x)
Output:
int((a + b*acos(d*x^2 - 1))^(1/2), x)
\[ \int \sqrt {a+b \arccos \left (-1+d x^2\right )} \, dx=\int \sqrt {\mathit {acos} \left (d \,x^{2}-1\right ) b +a}d x \] Input:
int((a+b*acos(d*x^2-1))^(1/2),x)
Output:
int(sqrt(acos(d*x**2 - 1)*b + a),x)