\(\int \frac {1}{(a+b \arccos (-1+d x^2))^2} \, dx\) [85]

Optimal result

Integrand size = 14, antiderivative size = 149 \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^2} \, dx=\frac {\sqrt {2 d x^2-d^2 x^4}}{2 b d x \left (a+b \arccos \left (-1+d x^2\right )\right )}-\frac {x \cos \left (\frac {a}{2 b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos \left (-1+d x^2\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}-\frac {x \sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \arccos \left (-1+d x^2\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}} \]

[Out]

-1/4*x*Ci(1/2*(a+b*arccos(d*x^2-1))/b)*cos(1/2*a/b)/b^2*2^(1/2)/(d*x^2)^(1/2)-1/4*x*Si(1/2*(a+b*arccos(d*x^2-1
))/b)*sin(1/2*a/b)/b^2*2^(1/2)/(d*x^2)^(1/2)+1/2*(-d^2*x^4+2*d*x^2)^(1/2)/b/d/x/(a+b*arccos(d*x^2-1))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {4911} \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^2} \, dx=-\frac {x \cos \left (\frac {a}{2 b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos \left (d x^2-1\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}-\frac {x \sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \arccos \left (d x^2-1\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}+\frac {\sqrt {2 d x^2-d^2 x^4}}{2 b d x \left (a+b \arccos \left (d x^2-1\right )\right )} \]

[In]

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

[Out]

Sqrt[2*d*x^2 - d^2*x^4]/(2*b*d*x*(a + b*ArcCos[-1 + d*x^2])) - (x*Cos[a/(2*b)]*CosIntegral[(a + b*ArcCos[-1 +
d*x^2])/(2*b)])/(2*Sqrt[2]*b^2*Sqrt[d*x^2]) - (x*Sin[a/(2*b)]*SinIntegral[(a + b*ArcCos[-1 + d*x^2])/(2*b)])/(
2*Sqrt[2]*b^2*Sqrt[d*x^2])

Rule 4911

Int[((a_.) + ArcCos[-1 + (d_.)*(x_)^2]*(b_.))^(-2), x_Symbol] :> Simp[Sqrt[2*d*x^2 - d^2*x^4]/(2*b*d*x*(a + b*
ArcCos[-1 + d*x^2])), x] + (-Simp[x*Cos[a/(2*b)]*(CosIntegral[(a + b*ArcCos[-1 + d*x^2])/(2*b)]/(2*Sqrt[2]*b^2
*Sqrt[d*x^2])), x] - Simp[x*Sin[a/(2*b)]*(SinIntegral[(a + b*ArcCos[-1 + d*x^2])/(2*b)]/(2*Sqrt[2]*b^2*Sqrt[d*
x^2])), x]) /; FreeQ[{a, b, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {2 d x^2-d^2 x^4}}{2 b d x \left (a+b \arccos \left (-1+d x^2\right )\right )}-\frac {x \cos \left (\frac {a}{2 b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos \left (-1+d x^2\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}-\frac {x \sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \arccos \left (-1+d x^2\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^2} \, dx=\frac {\sqrt {-d x^2 \left (-2+d x^2\right )} \left (\frac {b}{a+b \arccos \left (-1+d x^2\right )}+\frac {\sin \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \left (\cos \left (\frac {a}{2 b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos \left (-1+d x^2\right )}{2 b}\right )+\sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \arccos \left (-1+d x^2\right )}{2 b}\right )\right )}{-2+d x^2}\right )}{2 b^2 d x} \]

[In]

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

[Out]

(Sqrt[-(d*x^2*(-2 + d*x^2))]*(b/(a + b*ArcCos[-1 + d*x^2]) + (Sin[ArcCos[-1 + d*x^2]/2]*(Cos[a/(2*b)]*CosInteg
ral[(a + b*ArcCos[-1 + d*x^2])/(2*b)] + Sin[a/(2*b)]*SinIntegral[(a + b*ArcCos[-1 + d*x^2])/(2*b)]))/(-2 + d*x
^2)))/(2*b^2*d*x)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

-1/2*(2*(b^2*d*arctan2(sqrt(-d*x^2 + 2)*sqrt(d)*x, d*x^2 - 1) + a*b*d)*sqrt(d)*integrate(1/2*sqrt(-d*x^2 + 2)*
x/(a*b*d*x^2 - 2*a*b + (b^2*d*x^2 - 2*b^2)*arctan2(sqrt(-d*x^2 + 2)*sqrt(d)*x, d*x^2 - 1)), x) - sqrt(-d*x^2 +
2)*sqrt(d))/(b^2*d*arctan2(sqrt(-d*x^2 + 2)*sqrt(d)*x, d*x^2 - 1) + a*b*d)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(a + b*acos(d*x^2 - 1))^2,x)

[Out]

int(1/(a + b*acos(d*x^2 - 1))^2, x)