\(\int \frac {1}{(a+b \arccos (-1+d x^2))^{3/2}} \, dx\) [98]

Optimal result

Integrand size = 16, antiderivative size = 190 \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx=\frac {\sqrt {2 d x^2-d^2 x^4}}{b d x \sqrt {a+b \arccos \left (-1+d x^2\right )}}-\frac {2 \left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {\pi }}\right )}{d x}-\frac {2 \left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )}{d x} \]

[Out]

-2*(1/b)^(3/2)*cos(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)
)*Pi^(1/2)/d/x-2*(1/b)^(3/2)*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)*Pi^(1/2)/d/x+(-d^2*x^4+2*d*x^2)^(1/2)/b/d/x/(a+b*arccos(d*x^2-1))^(1/2)

Rubi [A] (verified)

Time = 0.02 (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 = {4908} \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx=\frac {\sqrt {2 d x^2-d^2 x^4}}{b d x \sqrt {a+b \arccos \left (d x^2-1\right )}}-\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (d x^2-1\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{d x}-\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} \sin \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (d x^2-1\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{d x} \]

[In]

Int[(a + b*ArcCos[-1 + d*x^2])^(-3/2),x]

[Out]

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[A
rcCos[-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*ArcCos[-1 + d*x^2]])/Sqrt[Pi]]*Sin[a/(
2*b)])/(d*x)

Rule 4908

Int[((a_.) + ArcCos[-1 + (d_.)*(x_)^2]*(b_.))^(-3/2), x_Symbol] :> Simp[Sqrt[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)*Sqrt[Pi]*Sin[a/(2*b)]*Cos[ArcCo
s[-1 + d*x^2]/2]*(FresnelS[Sqrt[1/(Pi*b)]*Sqrt[a + b*ArcCos[-1 + d*x^2]]]/(d*x)), x]) /; FreeQ[{a, b, d}, x]

Rubi steps \begin{align*} \text {integral}& = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx=-\frac {2 \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \left (\sqrt {\pi } \sqrt {a+b \arccos \left (-1+d x^2\right )} \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right )+\sqrt {\pi } \sqrt {a+b \arccos \left (-1+d x^2\right )} \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )-\sqrt {b} \sin \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )\right )}{b^{3/2} d x \sqrt {a+b \arccos \left (-1+d x^2\right )}} \]

[In]

Integrate[(a + b*ArcCos[-1 + d*x^2])^(-3/2),x]

[Out]

(-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])] + Sqrt[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]])

Maple [F]

\[\int \frac {1}{{\left (a +b \arccos \left (d \,x^{2}-1\right )\right )}^{\frac {3}{2}}}d x\]

[In]

int(1/(a+b*arccos(d*x^2-1))^(3/2),x)

[Out]

int(1/(a+b*arccos(d*x^2-1))^(3/2),x)

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a+b*arccos(d*x^2-1))^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

\[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {acos}{\left (d x^{2} - 1 \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(a+b*acos(d*x**2-1))**(3/2),x)

[Out]

Integral((a + b*acos(d*x**2 - 1))**(-3/2), x)

Maxima [F]

\[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b \arccos \left (d x^{2} - 1\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(a+b*arccos(d*x^2-1))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*arccos(d*x^2 - 1) + a)^(-3/2), x)

Giac [F]

\[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b \arccos \left (d x^{2} - 1\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(a+b*arccos(d*x^2-1))^(3/2),x, algorithm="giac")

[Out]

integrate((b*arccos(d*x^2 - 1) + a)^(-3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {acos}\left (d\,x^2-1\right )\right )}^{3/2}} \,d x \]

[In]

int(1/(a + b*acos(d*x^2 - 1))^(3/2),x)

[Out]

int(1/(a + b*acos(d*x^2 - 1))^(3/2), x)