3.1.31 \(\int \frac {\text {ArcCos}(a+b x)}{x^4} \, dx\) [31]

Optimal. Leaf size=144 \[ \frac {b \sqrt {1-(a+b x)^2}}{6 \left (1-a^2\right ) x^2}+\frac {a b^2 \sqrt {1-(a+b x)^2}}{2 \left (1-a^2\right )^2 x}-\frac {\text {ArcCos}(a+b x)}{3 x^3}+\frac {\left (1+2 a^2\right ) b^3 \tanh ^{-1}\left (\frac {1-a (a+b x)}{\sqrt {1-a^2} \sqrt {1-(a+b x)^2}}\right )}{6 \left (1-a^2\right )^{5/2}} \]

[Out]

-1/3*arccos(b*x+a)/x^3+1/6*(2*a^2+1)*b^3*arctanh((1-a*(b*x+a))/(-a^2+1)^(1/2)/(1-(b*x+a)^2)^(1/2))/(-a^2+1)^(5
/2)+1/6*b*(1-(b*x+a)^2)^(1/2)/(-a^2+1)/x^2+1/2*a*b^2*(1-(b*x+a)^2)^(1/2)/(-a^2+1)^2/x

________________________________________________________________________________________

Rubi [A]
time = 0.12, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4890, 4828, 759, 821, 739, 212} \begin {gather*} \frac {\left (2 a^2+1\right ) b^3 \tanh ^{-1}\left (\frac {1-a (a+b x)}{\sqrt {1-a^2} \sqrt {1-(a+b x)^2}}\right )}{6 \left (1-a^2\right )^{5/2}}+\frac {a b^2 \sqrt {1-(a+b x)^2}}{2 \left (1-a^2\right )^2 x}+\frac {b \sqrt {1-(a+b x)^2}}{6 \left (1-a^2\right ) x^2}-\frac {\text {ArcCos}(a+b x)}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]/x^4,x]

[Out]

(b*Sqrt[1 - (a + b*x)^2])/(6*(1 - a^2)*x^2) + (a*b^2*Sqrt[1 - (a + b*x)^2])/(2*(1 - a^2)^2*x) - ArcCos[a + b*x
]/(3*x^3) + ((1 + 2*a^2)*b^3*ArcTanh[(1 - a*(a + b*x))/(Sqrt[1 - a^2]*Sqrt[1 - (a + b*x)^2])])/(6*(1 - a^2)^(5
/2))

Rule 212

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

Rule 739

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 759

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/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*Simp[d*(m + 1)
- e*(m + 2*p + 3)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[
m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ[p]) || ILtQ
[Simplify[m + 2*p + 3], 0])

Rule 821

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

Rule 4828

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 4890

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^4} \, dx &=\frac {\text {Subst}\left (\int \frac {\cos ^{-1}(x)}{\left (-\frac {a}{b}+\frac {x}{b}\right )^4} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\cos ^{-1}(a+b x)}{3 x^3}-\frac {1}{3} \text {Subst}\left (\int \frac {1}{\left (-\frac {a}{b}+\frac {x}{b}\right )^3 \sqrt {1-x^2}} \, dx,x,a+b x\right )\\ &=\frac {b \sqrt {1-(a+b x)^2}}{6 \left (1-a^2\right ) x^2}-\frac {\cos ^{-1}(a+b x)}{3 x^3}-\frac {b^2 \text {Subst}\left (\int \frac {\frac {2 a}{b}+\frac {x}{b}}{\left (-\frac {a}{b}+\frac {x}{b}\right )^2 \sqrt {1-x^2}} \, dx,x,a+b x\right )}{6 \left (1-a^2\right )}\\ &=\frac {b \sqrt {1-(a+b x)^2}}{6 \left (1-a^2\right ) x^2}+\frac {a b^2 \sqrt {1-(a+b x)^2}}{2 \left (1-a^2\right )^2 x}-\frac {\cos ^{-1}(a+b x)}{3 x^3}-\frac {\left (\left (1+2 a^2\right ) b^2\right ) \text {Subst}\left (\int \frac {1}{\left (-\frac {a}{b}+\frac {x}{b}\right ) \sqrt {1-x^2}} \, dx,x,a+b x\right )}{6 \left (1-a^2\right )^2}\\ &=\frac {b \sqrt {1-(a+b x)^2}}{6 \left (1-a^2\right ) x^2}+\frac {a b^2 \sqrt {1-(a+b x)^2}}{2 \left (1-a^2\right )^2 x}-\frac {\cos ^{-1}(a+b x)}{3 x^3}+\frac {\left (\left (1+2 a^2\right ) b^2\right ) \text {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 )}{6 \left (1-a^2\right )^2}\\ &=\frac {b \sqrt {1-(a+b x)^2}}{6 \left (1-a^2\right ) x^2}+\frac {a b^2 \sqrt {1-(a+b x)^2}}{2 \left (1-a^2\right )^2 x}-\frac {\cos ^{-1}(a+b x)}{3 x^3}+\frac {\left (1+2 a^2\right ) b^3 \tanh ^{-1}\left (\frac {1-a (a+b x)}{\sqrt {1-a^2} \sqrt {1-(a+b x)^2}}\right )}{6 \left (1-a^2\right )^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.13, size = 168, normalized size = 1.17 \begin {gather*} \frac {\sqrt {1-a^2} b x \left (1-a^2+3 a b x\right ) \sqrt {1-a^2-2 a b x-b^2 x^2}-2 \left (1-a^2\right )^{5/2} \text {ArcCos}(a+b x)-\left (1+2 a^2\right ) b^3 x^3 \log (x)+\left (1+2 a^2\right ) b^3 x^3 \log \left (1-a^2-a b x+\sqrt {1-a^2} \sqrt {1-a^2-2 a b x-b^2 x^2}\right )}{6 \left (1-a^2\right )^{5/2} x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a + b*x]/x^4,x]

[Out]

(Sqrt[1 - a^2]*b*x*(1 - a^2 + 3*a*b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2] - 2*(1 - a^2)^(5/2)*ArcCos[a + b*x] -
(1 + 2*a^2)*b^3*x^3*Log[x] + (1 + 2*a^2)*b^3*x^3*Log[1 - a^2 - a*b*x + Sqrt[1 - a^2]*Sqrt[1 - a^2 - 2*a*b*x -
b^2*x^2]])/(6*(1 - a^2)^(5/2)*x^3)

________________________________________________________________________________________

Maple [A]
time = 0.00, size = 240, normalized size = 1.67

method	result	size
derivativedivides	\(b^{3} \left (-\frac {\arccos \left (b x +a \right )}{3 b^{3} x^{3}}+\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{6 \left (-a^{2}+1\right ) b^{2} x^{2}}-\frac {a \left (-\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{\left (-a^{2}+1\right ) b x}-\frac {a \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{b x}\right )}{\left (-a^{2}+1\right )^{\frac {3}{2}}}\right )}{2 \left (-a^{2}+1\right )}+\frac {\ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{b x}\right )}{6 \left (-a^{2}+1\right )^{\frac {3}{2}}}\right )\)	\(240\)
default	\(b^{3} \left (-\frac {\arccos \left (b x +a \right )}{3 b^{3} x^{3}}+\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{6 \left (-a^{2}+1\right ) b^{2} x^{2}}-\frac {a \left (-\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{\left (-a^{2}+1\right ) b x}-\frac {a \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{b x}\right )}{\left (-a^{2}+1\right )^{\frac {3}{2}}}\right )}{2 \left (-a^{2}+1\right )}+\frac {\ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{b x}\right )}{6 \left (-a^{2}+1\right )^{\frac {3}{2}}}\right )\)	\(240\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(b*x+a)/x^4,x,method=_RETURNVERBOSE)

[Out]

b^3*(-1/3*arccos(b*x+a)/b^3/x^3+1/6/(-a^2+1)/b^2/x^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/2*a/(-a^2+1)*(-1/(-a^2+1
)/b/x*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-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))+1/6/(-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))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(b*x+a)/x^4,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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (122) = 244\).
time = 1.70, size = 580, normalized size = 4.03 \begin {gather*} \left [-\frac {{\left (2 \, a^{2} + 1\right )} \sqrt {-a^{2} + 1} b^{3} x^{3} \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 ) + 4 \, {\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} x^{3} \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 ) + 4 \, {\left (a^{6} - 3 \, a^{4} - {\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} x^{3} + 3 \, a^{2} - 1\right )} \arccos \left (b x + a\right ) - 2 \, {\left (3 \, {\left (a^{3} - a\right )} b^{2} x^{2} - {\left (a^{4} - 2 \, a^{2} + 1\right )} b x\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{12 \, {\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} x^{3}}, -\frac {{\left (2 \, a^{2} + 1\right )} \sqrt {a^{2} - 1} b^{3} x^{3} \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 ) + 2 \, {\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} x^{3} \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 \, {\left (a^{6} - 3 \, a^{4} - {\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} x^{3} + 3 \, a^{2} - 1\right )} \arccos \left (b x + a\right ) - {\left (3 \, {\left (a^{3} - a\right )} b^{2} x^{2} - {\left (a^{4} - 2 \, a^{2} + 1\right )} b x\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{6 \, {\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(b*x+a)/x^4,x, algorithm="fricas")

[Out]

[-1/12*((2*a^2 + 1)*sqrt(-a^2 + 1)*b^3*x^3*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) + 4*(a^6 - 3*a^4 + 3*a^2 - 1)*x^3*ar
ctan(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)) + 4*(a^6 - 3*a^4 - (a^6 - 3*a
^4 + 3*a^2 - 1)*x^3 + 3*a^2 - 1)*arccos(b*x + a) - 2*(3*(a^3 - a)*b^2*x^2 - (a^4 - 2*a^2 + 1)*b*x)*sqrt(-b^2*x
^2 - 2*a*b*x - a^2 + 1))/((a^6 - 3*a^4 + 3*a^2 - 1)*x^3), -1/6*((2*a^2 + 1)*sqrt(a^2 - 1)*b^3*x^3*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)) + 2*(a^6 - 3*a^4 + 3*a^2 - 1)*x^3*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*(a^6 - 3*a^4 - (a^6 - 3*a^4 + 3*a^2 - 1)*x^3 + 3*a^2 - 1)*arccos(b*x + a) - (3*(a^3 - a)*b
^2*x^2 - (a^4 - 2*a^2 + 1)*b*x)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/((a^6 - 3*a^4 + 3*a^2 - 1)*x^3)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {acos}{\left (a + b x \right )}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(b*x+a)/x**4,x)

[Out]

Integral(acos(a + b*x)/x**4, x)

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 557 vs. \(2 (122) = 244\).
time = 0.44, size = 557, normalized size = 3.87 \begin {gather*} -\frac {1}{3} \, b {\left (\frac {{\left (2 \, a^{2} b^{3} + b^{3}\right )} \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^{4} {\left | b \right |} - 2 \, a^{2} {\left | b \right |} + {\left | b \right |}\right )} \sqrt {a^{2} - 1}} - \frac {\frac {4 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{2} a^{4} b^{3}}{{\left (b^{2} x + a b\right )}^{2}} + 4 \, a^{4} b^{3} - \frac {11 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a^{3} b^{3}}{b^{2} x + a b} - \frac {5 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{3} a^{3} b^{3}}{{\left (b^{2} x + a b\right )}^{3}} + \frac {7 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{2} a^{2} b^{3}}{{\left (b^{2} x + a b\right )}^{2}} - a^{2} b^{3} + \frac {2 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a b^{3}}{b^{2} x + a b} + \frac {2 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{3} a b^{3}}{{\left (b^{2} x + a b\right )}^{3}} - \frac {2 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{2} b^{3}}{{\left (b^{2} x + a b\right )}^{2}}}{{\left (a^{6} {\left | b \right |} - 2 \, a^{4} {\left | b \right |} + a^{2} {\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 )}^{2}}\right )} - \frac {\arccos \left (b x + a\right )}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(b*x+a)/x^4,x, algorithm="giac")

[Out]

-1/3*b*((2*a^2*b^3 + b^3)*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^4*abs(b) - 2*a^2*abs(b) + abs(b))*sqrt(a^2 - 1)) - (4*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) +
b)^2*a^4*b^3/(b^2*x + a*b)^2 + 4*a^4*b^3 - 11*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a^3*b^3/(b^2*x
+ a*b) - 5*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*a^3*b^3/(b^2*x + a*b)^3 + 7*(sqrt(-b^2*x^2 - 2*a*
b*x - a^2 + 1)*abs(b) + b)^2*a^2*b^3/(b^2*x + a*b)^2 - a^2*b^3 + 2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b)
+ b)*a*b^3/(b^2*x + a*b) + 2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*a*b^3/(b^2*x + a*b)^3 - 2*(sqrt
(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*b^3/(b^2*x + a*b)^2)/((a^6*abs(b) - 2*a^4*abs(b) + a^2*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))^2)) - 1/3*arccos(b*x + a)/x^3

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {acos}\left (a+b\,x\right )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(a + b*x)/x^4,x)

[Out]

int(acos(a + b*x)/x^4, x)

________________________________________________________________________________________