\(\int \frac {1}{(a+b \arccos (1+d x^2))^3} \, dx\) [79]

Optimal result

Integrand size = 14, antiderivative size = 173 \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^3} \, dx=\frac {\sqrt {-2 d x^2-d^2 x^4}}{4 b d x \left (a+b \arccos \left (1+d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a+b \arccos \left (1+d x^2\right )\right )}-\frac {x \cos \left (\frac {a}{2 b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos \left (1+d x^2\right )}{2 b}\right )}{8 \sqrt {2} b^3 \sqrt {-d x^2}}-\frac {x \sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \arccos \left (1+d x^2\right )}{2 b}\right )}{8 \sqrt {2} b^3 \sqrt {-d x^2}} \]

[Out]

1/8*x/b^2/(a+b*arccos(d*x^2+1))-1/16*x*Ci(1/2*(a+b*arccos(d*x^2+1))/b)*cos(1/2*a/b)/b^3*2^(1/2)/(-d*x^2)^(1/2)
-1/16*x*Si(1/2*(a+b*arccos(d*x^2+1))/b)*sin(1/2*a/b)/b^3*2^(1/2)/(-d*x^2)^(1/2)+1/4*(-d^2*x^4-2*d*x^2)^(1/2)/b
/d/x/(a+b*arccos(d*x^2+1))^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4913, 4901} \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^3} \, dx=-\frac {x \cos \left (\frac {a}{2 b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos \left (d x^2+1\right )}{2 b}\right )}{8 \sqrt {2} b^3 \sqrt {-d x^2}}-\frac {x \sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \arccos \left (d x^2+1\right )}{2 b}\right )}{8 \sqrt {2} b^3 \sqrt {-d x^2}}+\frac {x}{8 b^2 \left (a+b \arccos \left (d x^2+1\right )\right )}+\frac {\sqrt {-d^2 x^4-2 d x^2}}{4 b d x \left (a+b \arccos \left (d x^2+1\right )\right )^2} \]

[In]

Int[(a + b*ArcCos[1 + d*x^2])^(-3),x]

[Out]

Sqrt[-2*d*x^2 - d^2*x^4]/(4*b*d*x*(a + b*ArcCos[1 + d*x^2])^2) + x/(8*b^2*(a + b*ArcCos[1 + d*x^2])) - (x*Cos[
a/(2*b)]*CosIntegral[(a + b*ArcCos[1 + d*x^2])/(2*b)])/(8*Sqrt[2]*b^3*Sqrt[-(d*x^2)]) - (x*Sin[a/(2*b)]*SinInt
egral[(a + b*ArcCos[1 + d*x^2])/(2*b)])/(8*Sqrt[2]*b^3*Sqrt[-(d*x^2)])

Rule 4901

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

Rule 4913

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*((a + b*ArcCos[c + d*x^2])^(n + 2)/(
4*b^2*(n + 1)*(n + 2))), x] + (-Dist[1/(4*b^2*(n + 1)*(n + 2)), Int[(a + b*ArcCos[c + d*x^2])^(n + 2), x], x]
- Simp[Sqrt[-2*c*d*x^2 - d^2*x^4]*((a + b*ArcCos[c + d*x^2])^(n + 1)/(2*b*d*(n + 1)*x)), x]) /; FreeQ[{a, b, c
, d}, x] && EqQ[c^2, 1] && LtQ[n, -1] && NeQ[n, -2]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {-2 d x^2-d^2 x^4}}{4 b d x \left (a+b \arccos \left (1+d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a+b \arccos \left (1+d x^2\right )\right )}-\frac {\int \frac {1}{a+b \arccos \left (1+d x^2\right )} \, dx}{8 b^2} \\ & = \frac {\sqrt {-2 d x^2-d^2 x^4}}{4 b d x \left (a+b \arccos \left (1+d x^2\right )\right )^2}+\frac {x}{8 b^2 \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 )}{8 \sqrt {2} b^3 \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 )}{8 \sqrt {2} b^3 \sqrt {-d x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^3} \, dx=\frac {\frac {2 b^2 \sqrt {-d x^2 \left (2+d x^2\right )}}{d \left (a+b \arccos \left (1+d x^2\right )\right )^2}+\frac {b x^2}{a+b \arccos \left (1+d x^2\right )}+\frac {\sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right ) \left (\cos \left (\frac {a}{2 b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos \left (1+d x^2\right )}{2 b}\right )+\sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \arccos \left (1+d x^2\right )}{2 b}\right )\right )}{d}}{8 b^3 x} \]

[In]

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

[Out]

((2*b^2*Sqrt[-(d*x^2*(2 + d*x^2))])/(d*(a + b*ArcCos[1 + d*x^2])^2) + (b*x^2)/(a + b*ArcCos[1 + d*x^2]) + (Sin
[ArcCos[1 + d*x^2]/2]*(Cos[a/(2*b)]*CosIntegral[(a + b*ArcCos[1 + d*x^2])/(2*b)] + Sin[a/(2*b)]*SinIntegral[(a
+ b*ArcCos[1 + d*x^2])/(2*b)]))/d)/(8*b^3*x)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: sign: argument cannot be imaginary; found sqrt((-_SAGE_VAR_d*_SAGE
_VAR_x^2)-2)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(a + b*acos(d*x^2 + 1))^3,x)

[Out]

int(1/(a + b*acos(d*x^2 + 1))^3, x)