3.1.28 \(\int \frac {\text {ArcCos}(a+b x)}{x} \, dx\) [28]

Optimal. Leaf size=177 \[ -\frac {1}{2} i \text {ArcCos}(a+b x)^2+\text {ArcCos}(a+b x) \log \left (1-\frac {e^{i \text {ArcCos}(a+b x)}}{a-i \sqrt {1-a^2}}\right )+\text {ArcCos}(a+b x) \log \left (1-\frac {e^{i \text {ArcCos}(a+b x)}}{a+i \sqrt {1-a^2}}\right )-i \text {PolyLog}\left (2,\frac {e^{i \text {ArcCos}(a+b x)}}{a-i \sqrt {1-a^2}}\right )-i \text {PolyLog}\left (2,\frac {e^{i \text {ArcCos}(a+b x)}}{a+i \sqrt {1-a^2}}\right ) \]

[Out]

-1/2*I*arccos(b*x+a)^2+arccos(b*x+a)*ln(1-(b*x+a+I*(1-(b*x+a)^2)^(1/2))/(a-I*(-a^2+1)^(1/2)))+arccos(b*x+a)*ln
(1-(b*x+a+I*(1-(b*x+a)^2)^(1/2))/(a+I*(-a^2+1)^(1/2)))-I*polylog(2,(b*x+a+I*(1-(b*x+a)^2)^(1/2))/(a-I*(-a^2+1)
^(1/2)))-I*polylog(2,(b*x+a+I*(1-(b*x+a)^2)^(1/2))/(a+I*(-a^2+1)^(1/2)))

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Rubi [A]
time = 0.20, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4890, 4826, 4618, 2221, 2317, 2438} \begin {gather*} -i \text {Li}_2\left (\frac {e^{i \text {ArcCos}(a+b x)}}{a-i \sqrt {1-a^2}}\right )-i \text {Li}_2\left (\frac {e^{i \text {ArcCos}(a+b x)}}{a+i \sqrt {1-a^2}}\right )+\text {ArcCos}(a+b x) \log \left (1-\frac {e^{i \text {ArcCos}(a+b x)}}{a-i \sqrt {1-a^2}}\right )+\text {ArcCos}(a+b x) \log \left (1-\frac {e^{i \text {ArcCos}(a+b x)}}{a+i \sqrt {1-a^2}}\right )-\frac {1}{2} i \text {ArcCos}(a+b x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]/x,x]

[Out]

(-1/2*I)*ArcCos[a + b*x]^2 + ArcCos[a + b*x]*Log[1 - E^(I*ArcCos[a + b*x])/(a - I*Sqrt[1 - a^2])] + ArcCos[a +
b*x]*Log[1 - E^(I*ArcCos[a + b*x])/(a + I*Sqrt[1 - a^2])] - I*PolyLog[2, E^(I*ArcCos[a + b*x])/(a - I*Sqrt[1
- a^2])] - I*PolyLog[2, E^(I*ArcCos[a + b*x])/(a + I*Sqrt[1 - a^2])]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4618

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)])/(Cos[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :>
Simp[I*((e + f*x)^(m + 1)/(b*f*(m + 1))), x] + (Int[(e + f*x)^m*(E^(I*(c + d*x))/(I*a - Rt[-a^2 + b^2, 2] + I*
b*E^(I*(c + d*x)))), x] + Int[(e + f*x)^m*(E^(I*(c + d*x))/(I*a + Rt[-a^2 + b^2, 2] + I*b*E^(I*(c + d*x)))), x
]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && NegQ[a^2 - b^2]

Rule 4826

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Subst[Int[(a + b*x)^n*(Sin[x]/
(c*d + e*Cos[x])), x], x, ArcCos[c*x]] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[n, 0]

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} \, dx &=\frac {\text {Subst}\left (\int \frac {\cos ^{-1}(x)}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\text {Subst}\left (\int \frac {x \sin (x)}{-\frac {a}{b}+\frac {\cos (x)}{b}} \, dx,x,\cos ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1}{2} i \cos ^{-1}(a+b x)^2-\frac {\text {Subst}\left (\int \frac {e^{i x} x}{-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}+\frac {i e^{i x}}{b}} \, dx,x,\cos ^{-1}(a+b x)\right )}{b}-\frac {\text {Subst}\left (\int \frac {e^{i x} x}{-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}+\frac {i e^{i x}}{b}} \, dx,x,\cos ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1}{2} i \cos ^{-1}(a+b x)^2+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt {1-a^2}}\right )+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt {1-a^2}}\right )-\text {Subst}\left (\int \log \left (1+\frac {i e^{i x}}{\left (-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}\right ) b}\right ) \, dx,x,\cos ^{-1}(a+b x)\right )-\text {Subst}\left (\int \log \left (1+\frac {i e^{i x}}{\left (-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}\right ) b}\right ) \, dx,x,\cos ^{-1}(a+b x)\right )\\ &=-\frac {1}{2} i \cos ^{-1}(a+b x)^2+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt {1-a^2}}\right )+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt {1-a^2}}\right )+i \text {Subst}\left (\int \frac {\log \left (1+\frac {i x}{\left (-\frac {i a}{b}-\frac {\sqrt {1-a^2}}{b}\right ) b}\right )}{x} \, dx,x,e^{i \cos ^{-1}(a+b x)}\right )+i \text {Subst}\left (\int \frac {\log \left (1+\frac {i x}{\left (-\frac {i a}{b}+\frac {\sqrt {1-a^2}}{b}\right ) b}\right )}{x} \, dx,x,e^{i \cos ^{-1}(a+b x)}\right )\\ &=-\frac {1}{2} i \cos ^{-1}(a+b x)^2+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt {1-a^2}}\right )+\cos ^{-1}(a+b x) \log \left (1-\frac {e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt {1-a^2}}\right )-i \text {Li}_2\left (\frac {e^{i \cos ^{-1}(a+b x)}}{a-i \sqrt {1-a^2}}\right )-i \text {Li}_2\left (\frac {e^{i \cos ^{-1}(a+b x)}}{a+i \sqrt {1-a^2}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 228, normalized size = 1.29 \begin {gather*} -\frac {1}{2} i \text {ArcCos}(a+b x)^2-4 i \text {ArcSin}\left (\frac {\sqrt {1-a}}{\sqrt {2}}\right ) \text {ArcTan}\left (\frac {(1+a) \tan \left (\frac {1}{2} \text {ArcCos}(a+b x)\right )}{\sqrt {-1+a^2}}\right )+\left (\text {ArcCos}(a+b x)-2 \text {ArcSin}\left (\frac {\sqrt {1-a}}{\sqrt {2}}\right )\right ) \log \left (1+\left (-a+\sqrt {-1+a^2}\right ) e^{i \text {ArcCos}(a+b x)}\right )+\left (\text {ArcCos}(a+b x)+2 \text {ArcSin}\left (\frac {\sqrt {1-a}}{\sqrt {2}}\right )\right ) \log \left (1-\left (a+\sqrt {-1+a^2}\right ) e^{i \text {ArcCos}(a+b x)}\right )-i \left (\text {PolyLog}\left (2,\left (a-\sqrt {-1+a^2}\right ) e^{i \text {ArcCos}(a+b x)}\right )+\text {PolyLog}\left (2,\left (a+\sqrt {-1+a^2}\right ) e^{i \text {ArcCos}(a+b x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a + b*x]/x,x]

[Out]

(-1/2*I)*ArcCos[a + b*x]^2 - (4*I)*ArcSin[Sqrt[1 - a]/Sqrt[2]]*ArcTan[((1 + a)*Tan[ArcCos[a + b*x]/2])/Sqrt[-1
+ a^2]] + (ArcCos[a + b*x] - 2*ArcSin[Sqrt[1 - a]/Sqrt[2]])*Log[1 + (-a + Sqrt[-1 + a^2])*E^(I*ArcCos[a + b*x
])] + (ArcCos[a + b*x] + 2*ArcSin[Sqrt[1 - a]/Sqrt[2]])*Log[1 - (a + Sqrt[-1 + a^2])*E^(I*ArcCos[a + b*x])] -
I*(PolyLog[2, (a - Sqrt[-1 + a^2])*E^(I*ArcCos[a + b*x])] + PolyLog[2, (a + Sqrt[-1 + a^2])*E^(I*ArcCos[a + b*
x])])

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Maple [A]
time = 0.83, size = 199, normalized size = 1.12

method	result	size
derivativedivides	\(-\frac {i \arccos \left (b x +a \right )^{2}}{2}+\arccos \left (b x +a \right ) \ln \left (\frac {\sqrt {a^{2}-1}-b x -i \sqrt {1-\left (b x +a \right )^{2}}}{a +\sqrt {a^{2}-1}}\right )+\arccos \left (b x +a \right ) \ln \left (\frac {\sqrt {a^{2}-1}+b x +i \sqrt {1-\left (b x +a \right )^{2}}}{-a +\sqrt {a^{2}-1}}\right )-i \dilog \left (\frac {\sqrt {a^{2}-1}+b x +i \sqrt {1-\left (b x +a \right )^{2}}}{-a +\sqrt {a^{2}-1}}\right )-i \dilog \left (\frac {\sqrt {a^{2}-1}-b x -i \sqrt {1-\left (b x +a \right )^{2}}}{a +\sqrt {a^{2}-1}}\right )\)	\(199\)
default	\(-\frac {i \arccos \left (b x +a \right )^{2}}{2}+\arccos \left (b x +a \right ) \ln \left (\frac {\sqrt {a^{2}-1}-b x -i \sqrt {1-\left (b x +a \right )^{2}}}{a +\sqrt {a^{2}-1}}\right )+\arccos \left (b x +a \right ) \ln \left (\frac {\sqrt {a^{2}-1}+b x +i \sqrt {1-\left (b x +a \right )^{2}}}{-a +\sqrt {a^{2}-1}}\right )-i \dilog \left (\frac {\sqrt {a^{2}-1}+b x +i \sqrt {1-\left (b x +a \right )^{2}}}{-a +\sqrt {a^{2}-1}}\right )-i \dilog \left (\frac {\sqrt {a^{2}-1}-b x -i \sqrt {1-\left (b x +a \right )^{2}}}{a +\sqrt {a^{2}-1}}\right )\)	\(199\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccos(b*x + a)/x, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccos(b*x + a)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acos(a + b*x)/x, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccos(b*x + a)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(a + b*x)/x,x)

[Out]

int(acos(a + b*x)/x, x)

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