Integrand size = 16, antiderivative size = 190 \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx=\frac {\sqrt {2 d x^2-d^2 x^4}}{b d x \sqrt {a+b \arccos \left (-1+d x^2\right )}}-\frac {2 \left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {\pi }}\right )}{d x}-\frac {2 \left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )}{d x} \] Output:
(-d^2*x^4+2*d*x^2)^(1/2)/b/d/x/(a+b*arccos(d*x^2-1))^(1/2)-2*(1/b)^(3/2)*P i^(1/2)*cos(1/2*a/b)*cos(1/2*arccos(d*x^2-1))*FresnelC((1/b)^(1/2)*(a+b*ar ccos(d*x^2-1))^(1/2)/Pi^(1/2))/d/x-2*(1/b)^(3/2)*Pi^(1/2)*cos(1/2*arccos(d *x^2-1))*FresnelS((1/b)^(1/2)*(a+b*arccos(d*x^2-1))^(1/2)/Pi^(1/2))*sin(1/ 2*a/b)/d/x
Time = 0.44 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx=-\frac {2 \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \left (\sqrt {\pi } \sqrt {a+b \arccos \left (-1+d x^2\right )} \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right )+\sqrt {\pi } \sqrt {a+b \arccos \left (-1+d x^2\right )} \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )-\sqrt {b} \sin \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )\right )}{b^{3/2} d x \sqrt {a+b \arccos \left (-1+d x^2\right )}} \] Input:
Integrate[(a + b*ArcCos[-1 + d*x^2])^(-3/2),x]
Output:
(-2*Cos[ArcCos[-1 + d*x^2]/2]*(Sqrt[Pi]*Sqrt[a + b*ArcCos[-1 + d*x^2]]*Cos [a/(2*b)]*FresnelC[Sqrt[a + b*ArcCos[-1 + d*x^2]]/(Sqrt[b]*Sqrt[Pi])] + Sq rt[Pi]*Sqrt[a + b*ArcCos[-1 + d*x^2]]*FresnelS[Sqrt[a + b*ArcCos[-1 + d*x^ 2]]/(Sqrt[b]*Sqrt[Pi])]*Sin[a/(2*b)] - Sqrt[b]*Sin[ArcCos[-1 + d*x^2]/2])) /(b^(3/2)*d*x*Sqrt[a + b*ArcCos[-1 + d*x^2]])
Time = 0.27 (sec) , antiderivative size = 190, 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 = {5323}
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.
\(\displaystyle \int \frac {1}{\left (a+b \arccos \left (d x^2-1\right )\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 5323 |
\(\displaystyle \frac {\sqrt {2 d x^2-d^2 x^4}}{b d x \sqrt {a+b \arccos \left (d x^2-1\right )}}-\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (d x^2-1\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{d x}-\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} \sin \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (d x^2-1\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{d x}\) |
Input:
Int[(a + b*ArcCos[-1 + d*x^2])^(-3/2),x]
Output:
Sqrt[2*d*x^2 - d^2*x^4]/(b*d*x*Sqrt[a + b*ArcCos[-1 + d*x^2]]) - (2*(b^(-1 ))^(3/2)*Sqrt[Pi]*Cos[a/(2*b)]*Cos[ArcCos[-1 + d*x^2]/2]*FresnelC[(Sqrt[b^ (-1)]*Sqrt[a + b*ArcCos[-1 + d*x^2]])/Sqrt[Pi]])/(d*x) - (2*(b^(-1))^(3/2) *Sqrt[Pi]*Cos[ArcCos[-1 + d*x^2]/2]*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcC os[-1 + d*x^2]])/Sqrt[Pi]]*Sin[a/(2*b)])/(d*x)
Int[((a_.) + ArcCos[-1 + (d_.)*(x_)^2]*(b_.))^(-3/2), x_Symbol] :> Simp[Sqr t[2*d*x^2 - d^2*x^4]/(b*d*x*Sqrt[a + b*ArcCos[-1 + d*x^2]]), x] + (-Simp[2* (1/b)^(3/2)*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]]]/(d*x)), x] - Simp[2*(1/b)^(3/2)*S qrt[Pi]*Sin[a/(2*b)]*Cos[ArcCos[-1 + d*x^2]/2]*(FresnelS[Sqrt[1/(Pi*b)]*Sqr t[a + b*ArcCos[-1 + d*x^2]]]/(d*x)), x]) /; FreeQ[{a, b, d}, x]
\[\int \frac {1}{{\left (a +b \arccos \left (d \,x^{2}-1\right )\right )}^{\frac {3}{2}}}d x\]
Input:
int(1/(a+b*arccos(d*x^2-1))^(3/2),x)
Output:
int(1/(a+b*arccos(d*x^2-1))^(3/2),x)
Exception generated. \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a+b*arccos(d*x^2-1))^(3/2),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {acos}{\left (d x^{2} - 1 \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(a+b*acos(d*x**2-1))**(3/2),x)
Output:
Integral((a + b*acos(d*x**2 - 1))**(-3/2), x)
\[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b \arccos \left (d x^{2} - 1\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(a+b*arccos(d*x^2-1))^(3/2),x, algorithm="maxima")
Output:
integrate((b*arccos(d*x^2 - 1) + a)^(-3/2), x)
\[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b \arccos \left (d x^{2} - 1\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(a+b*arccos(d*x^2-1))^(3/2),x, algorithm="giac")
Output:
integrate((b*arccos(d*x^2 - 1) + a)^(-3/2), x)
Timed out. \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {acos}\left (d\,x^2-1\right )\right )}^{3/2}} \,d x \] Input:
int(1/(a + b*acos(d*x^2 - 1))^(3/2),x)
Output:
int(1/(a + b*acos(d*x^2 - 1))^(3/2), x)
\[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx =\text {Too large to display} \] Input:
int(1/(a+b*acos(d*x^2-1))^(3/2),x)
Output:
( - sqrt(d)*sqrt(acos(d*x**2 - 1)*b + a)*sqrt( - d*x**2 + 2)*acos(d*x**2 - 1) + sqrt(d)*acos(d*x**2 - 1)*int((sqrt(acos(d*x**2 - 1)*b + a)*sqrt( - d *x**2 + 2)*acos(d*x**2 - 1)*x)/(acos(d*x**2 - 1)**2*b**2*d*x**2 - 2*acos(d *x**2 - 1)**2*b**2 + 2*acos(d*x**2 - 1)*a*b*d*x**2 - 4*acos(d*x**2 - 1)*a* b + a**2*d*x**2 - 2*a**2),x)*a*b*d + sqrt(d)*acos(d*x**2 - 1)*int((sqrt(ac os(d*x**2 - 1)*b + a)*sqrt( - d*x**2 + 2)*acos(d*x**2 - 1)**2*x)/(acos(d*x **2 - 1)**2*b**2*d*x**2 - 2*acos(d*x**2 - 1)**2*b**2 + 2*acos(d*x**2 - 1)* a*b*d*x**2 - 4*acos(d*x**2 - 1)*a*b + a**2*d*x**2 - 2*a**2),x)*b**2*d - ac os(d*x**2 - 1)*int((sqrt(acos(d*x**2 - 1)*b + a)*acos(d*x**2 - 1)*x**2)/(a cos(d*x**2 - 1)**2*b**2*d*x**2 - 2*acos(d*x**2 - 1)**2*b**2 + 2*acos(d*x** 2 - 1)*a*b*d*x**2 - 4*acos(d*x**2 - 1)*a*b + a**2*d*x**2 - 2*a**2),x)*b**2 *d**2 + 2*acos(d*x**2 - 1)*int((sqrt(acos(d*x**2 - 1)*b + a)*acos(d*x**2 - 1))/(acos(d*x**2 - 1)**2*b**2*d*x**2 - 2*acos(d*x**2 - 1)**2*b**2 + 2*aco s(d*x**2 - 1)*a*b*d*x**2 - 4*acos(d*x**2 - 1)*a*b + a**2*d*x**2 - 2*a**2), x)*b**2*d + sqrt(d)*int((sqrt(acos(d*x**2 - 1)*b + a)*sqrt( - d*x**2 + 2)* acos(d*x**2 - 1)*x)/(acos(d*x**2 - 1)**2*b**2*d*x**2 - 2*acos(d*x**2 - 1)* *2*b**2 + 2*acos(d*x**2 - 1)*a*b*d*x**2 - 4*acos(d*x**2 - 1)*a*b + a**2*d* x**2 - 2*a**2),x)*a**2*d + sqrt(d)*int((sqrt(acos(d*x**2 - 1)*b + a)*sqrt( - d*x**2 + 2)*acos(d*x**2 - 1)**2*x)/(acos(d*x**2 - 1)**2*b**2*d*x**2 - 2 *acos(d*x**2 - 1)**2*b**2 + 2*acos(d*x**2 - 1)*a*b*d*x**2 - 4*acos(d*x*...