\(\int \frac {1}{a+b \arccos (-1+d x^2)} \, dx\) [84]

Optimal result

Integrand size = 14, antiderivative size = 98 \[ \int \frac {1}{a+b \arccos \left (-1+d x^2\right )} \, dx=\frac {x \operatorname {CosIntegral}\left (\frac {a+b \arccos \left (-1+d x^2\right )}{2 b}\right ) \sin \left (\frac {a}{2 b}\right )}{\sqrt {2} b \sqrt {d x^2}}-\frac {x \cos \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \arccos \left (-1+d x^2\right )}{2 b}\right )}{\sqrt {2} b \sqrt {d x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 98, 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 = {4902} \[ \int \frac {1}{a+b \arccos \left (-1+d x^2\right )} \, dx=\frac {x \sin \left (\frac {a}{2 b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos \left (d x^2-1\right )}{2 b}\right )}{\sqrt {2} b \sqrt {d x^2}}-\frac {x \cos \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \arccos \left (d x^2-1\right )}{2 b}\right )}{\sqrt {2} b \sqrt {d x^2}} \]

[In]

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

[Out]

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

Rule 4902

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

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.87 \[ \int \frac {1}{a+b \arccos \left (-1+d x^2\right )} \, dx=\frac {\cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \left (\operatorname {CosIntegral}\left (\frac {a+b \arccos \left (-1+d x^2\right )}{2 b}\right ) \sin \left (\frac {a}{2 b}\right )-\cos \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \arccos \left (-1+d x^2\right )}{2 b}\right )\right )}{b d x} \]

[In]

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

[Out]

(Cos[ArcCos[-1 + d*x^2]/2]*(CosIntegral[(a + b*ArcCos[-1 + d*x^2])/(2*b)]*Sin[a/(2*b)] - Cos[a/(2*b)]*SinInteg
ral[(a + b*ArcCos[-1 + d*x^2])/(2*b)]))/(b*d*x)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(a + b*acos(d*x^2 - 1)),x)

[Out]

int(1/(a + b*acos(d*x^2 - 1)), x)