3.30 \(\int \frac {\cos ^{-1}(a+b x)}{x^3} \, dx\)

Optimal. Leaf size=103 \[ \frac {a b^2 \tanh ^{-1}\left (\frac {1-a (a+b x)}{\sqrt {1-a^2} \sqrt {1-(a+b x)^2}}\right )}{2 \left (1-a^2\right )^{3/2}}+\frac {b \sqrt {1-(a+b x)^2}}{2 \left (1-a^2\right ) x}-\frac {\cos ^{-1}(a+b x)}{2 x^2} \]

[Out]

-1/2*arccos(b*x+a)/x^2+1/2*a*b^2*arctanh((1-a*(b*x+a))/(-a^2+1)^(1/2)/(1-(b*x+a)^2)^(1/2))/(-a^2+1)^(3/2)+1/2*
b*(1-(b*x+a)^2)^(1/2)/(-a^2+1)/x

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Rubi [A]  time = 0.11, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4806, 4744, 731, 725, 206} \[ \frac {a b^2 \tanh ^{-1}\left (\frac {1-a (a+b x)}{\sqrt {1-a^2} \sqrt {1-(a+b x)^2}}\right )}{2 \left (1-a^2\right )^{3/2}}+\frac {b \sqrt {1-(a+b x)^2}}{2 \left (1-a^2\right ) x}-\frac {\cos ^{-1}(a+b x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]/x^3,x]

[Out]

(b*Sqrt[1 - (a + b*x)^2])/(2*(1 - a^2)*x) - ArcCos[a + b*x]/(2*x^2) + (a*b^2*ArcTanh[(1 - a*(a + b*x))/(Sqrt[1
- a^2]*Sqrt[1 - (a + b*x)^2])])/(2*(1 - a^2)^(3/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 731

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d)/(c*d^2 + a*e^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0]

Rule 4744

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*ArcCos[c*x])^n)/(e*(m + 1)), x] + Dist[(b*c*n)/(e*(m + 1)), Int[((d + e*x)^(m + 1)*(a + b*ArcCos[c*x])^(
n - 1))/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4806

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcCos[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

\begin {align*} \int \frac {\cos ^{-1}(a+b x)}{x^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\cos ^{-1}(x)}{\left (-\frac {a}{b}+\frac {x}{b}\right )^3} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\cos ^{-1}(a+b x)}{2 x^2}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\left (-\frac {a}{b}+\frac {x}{b}\right )^2 \sqrt {1-x^2}} \, dx,x,a+b x\right )\\ &=\frac {b \sqrt {1-(a+b x)^2}}{2 \left (1-a^2\right ) x}-\frac {\cos ^{-1}(a+b x)}{2 x^2}-\frac {(a b) \operatorname {Subst}\left (\int \frac {1}{\left (-\frac {a}{b}+\frac {x}{b}\right ) \sqrt {1-x^2}} \, dx,x,a+b x\right )}{2 \left (1-a^2\right )}\\ &=\frac {b \sqrt {1-(a+b x)^2}}{2 \left (1-a^2\right ) x}-\frac {\cos ^{-1}(a+b x)}{2 x^2}+\frac {(a b) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{b^2}-\frac {a^2}{b^2}-x^2} \, dx,x,\frac {\frac {1}{b}-\frac {a (a+b x)}{b}}{\sqrt {1-(a+b x)^2}}\right )}{2 \left (1-a^2\right )}\\ &=\frac {b \sqrt {1-(a+b x)^2}}{2 \left (1-a^2\right ) x}-\frac {\cos ^{-1}(a+b x)}{2 x^2}+\frac {a b^2 \tanh ^{-1}\left (\frac {1-a (a+b x)}{\sqrt {1-a^2} \sqrt {1-(a+b x)^2}}\right )}{2 \left (1-a^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 126, normalized size = 1.22 \[ -\frac {\cos ^{-1}(a+b x)-\frac {b x \left (\sqrt {1-a^2} \sqrt {-a^2-2 a b x-b^2 x^2+1}+a b x \log \left (\sqrt {1-a^2} \sqrt {-a^2-2 a b x-b^2 x^2+1}-a^2-a b x+1\right )-a b x \log (x)\right )}{\left (1-a^2\right )^{3/2}}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a + b*x]/x^3,x]

[Out]

-1/2*(ArcCos[a + b*x] - (b*x*(Sqrt[1 - a^2]*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2] - a*b*x*Log[x] + a*b*x*Log[1 - a
^2 - a*b*x + Sqrt[1 - a^2]*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]]))/(1 - a^2)^(3/2))/x^2

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fricas [B]  time = 0.54, size = 482, normalized size = 4.68 \[ \left [-\frac {\sqrt {-a^{2} + 1} a b^{2} x^{2} \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x + 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {-a^{2} + 1} - 4 \, a^{2} + 2}{x^{2}}\right ) + 2 \, {\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )} b x + 2 \, {\left (a^{4} - {\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2} - 2 \, a^{2} + 1\right )} \arccos \left (b x + a\right )}{4 \, {\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2}}, \frac {\sqrt {a^{2} - 1} a b^{2} x^{2} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \, {\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) - {\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )} b x - {\left (a^{4} - {\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2} - 2 \, a^{2} + 1\right )} \arccos \left (b x + a\right )}{2 \, {\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(sqrt(-a^2 + 1)*a*b^2*x^2*log(((2*a^2 - 1)*b^2*x^2 + 2*a^4 + 4*(a^3 - a)*b*x + 2*sqrt(-b^2*x^2 - 2*a*b*x
- a^2 + 1)*(a*b*x + a^2 - 1)*sqrt(-a^2 + 1) - 4*a^2 + 2)/x^2) + 2*(a^4 - 2*a^2 + 1)*x^2*arctan(sqrt(-b^2*x^2
- 2*a*b*x - a^2 + 1)*(b*x + a)/(b^2*x^2 + 2*a*b*x + a^2 - 1)) + 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a^2 - 1)
*b*x + 2*(a^4 - (a^4 - 2*a^2 + 1)*x^2 - 2*a^2 + 1)*arccos(b*x + a))/((a^4 - 2*a^2 + 1)*x^2), 1/2*(sqrt(a^2 - 1
)*a*b^2*x^2*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a*b*x + a^2 - 1)*sqrt(a^2 - 1)/((a^2 - 1)*b^2*x^2 + a^4
+ 2*(a^3 - a)*b*x - 2*a^2 + 1)) - (a^4 - 2*a^2 + 1)*x^2*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)/(
b^2*x^2 + 2*a*b*x + a^2 - 1)) - sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a^2 - 1)*b*x - (a^4 - (a^4 - 2*a^2 + 1)*x^
2 - 2*a^2 + 1)*arccos(b*x + a))/((a^4 - 2*a^2 + 1)*x^2)]

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giac [B]  time = 0.30, size = 242, normalized size = 2.35 \[ {\left (\frac {a b^{2} \arctan \left (\frac {\frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a}{b^{2} x + a b} - 1}{\sqrt {a^{2} - 1}}\right )}{{\left (a^{2} {\left | b \right |} - {\left | b \right |}\right )} \sqrt {a^{2} - 1}} - \frac {a b^{2} - \frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} b^{2}}{b^{2} x + a b}}{{\left (a^{3} {\left | b \right |} - a {\left | b \right |}\right )} {\left (\frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{2} a}{{\left (b^{2} x + a b\right )}^{2}} + a - \frac {2 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}}{b^{2} x + a b}\right )}}\right )} b - \frac {\arccos \left (b x + a\right )}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*b^2*arctan(((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a/(b^2*x + a*b) - 1)/sqrt(a^2 - 1))/((a^2*abs(b
) - abs(b))*sqrt(a^2 - 1)) - (a*b^2 - (sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*b^2/(b^2*x + a*b))/((a^3
*abs(b) - a*abs(b))*((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a/(b^2*x + a*b)^2 + a - 2*(sqrt(-b^2*x^
2 - 2*a*b*x - a^2 + 1)*abs(b) + b)/(b^2*x + a*b))))*b - 1/2*arccos(b*x + a)/x^2

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maple [A]  time = 0.01, size = 118, normalized size = 1.15 \[ -\frac {\arccos \left (b x +a \right )}{2 x^{2}}+\frac {b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{2 \left (-a^{2}+1\right ) x}+\frac {b^{2} a \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{b x}\right )}{2 \left (-a^{2}+1\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*arccos(b*x+a)/x^2+1/2*b/(-a^2+1)/x*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+1/2*b^2*a/(-a^2+1)^(3/2)*ln((-2*a^2+2-2
*a*b*x+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/b/x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a-1>0)', see `assume?` for mor
e details)Is a-1 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acos}\left (a+b\,x\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(a + b*x)/x^3,x)

[Out]

int(acos(a + b*x)/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acos}{\left (a + b x \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acos(a + b*x)/x**3, x)

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