3.2.52 \(\int \frac {(a+b \text {ArcCos}(c x))^2}{x^2} \, dx\) [152]

Optimal. Leaf size=89 \[ -\frac {(a+b \text {ArcCos}(c x))^2}{x}-4 i b c (a+b \text {ArcCos}(c x)) \text {ArcTan}\left (e^{i \text {ArcCos}(c x)}\right )+2 i b^2 c \text {PolyLog}\left (2,-i e^{i \text {ArcCos}(c x)}\right )-2 i b^2 c \text {PolyLog}\left (2,i e^{i \text {ArcCos}(c x)}\right ) \]

[Out]

-(a+b*arccos(c*x))^2/x-4*I*b*c*(a+b*arccos(c*x))*arctan(c*x+I*(-c^2*x^2+1)^(1/2))+2*I*b^2*c*polylog(2,-I*(c*x+
I*(-c^2*x^2+1)^(1/2)))-2*I*b^2*c*polylog(2,I*(c*x+I*(-c^2*x^2+1)^(1/2)))

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Rubi [A]
time = 0.09, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4724, 4804, 4266, 2317, 2438} \begin {gather*} -4 i b c \text {ArcTan}\left (e^{i \text {ArcCos}(c x)}\right ) (a+b \text {ArcCos}(c x))-\frac {(a+b \text {ArcCos}(c x))^2}{x}+2 i b^2 c \text {Li}_2\left (-i e^{i \text {ArcCos}(c x)}\right )-2 i b^2 c \text {Li}_2\left (i e^{i \text {ArcCos}(c x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[c*x])^2/x^2,x]

[Out]

-((a + b*ArcCos[c*x])^2/x) - (4*I)*b*c*(a + b*ArcCos[c*x])*ArcTan[E^(I*ArcCos[c*x])] + (2*I)*b^2*c*PolyLog[2,
(-I)*E^(I*ArcCos[c*x])] - (2*I)*b^2*c*PolyLog[2, I*E^(I*ArcCos[c*x])]

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 4266

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4724

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

Rule 4804

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

Rubi steps

\begin {align*} \int \frac {\left (a+b \cos ^{-1}(c x)\right )^2}{x^2} \, dx &=-\frac {\left (a+b \cos ^{-1}(c x)\right )^2}{x}-(2 b c) \int \frac {a+b \cos ^{-1}(c x)}{x \sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {\left (a+b \cos ^{-1}(c x)\right )^2}{x}+(2 b c) \text {Subst}\left (\int (a+b x) \sec (x) \, dx,x,\cos ^{-1}(c x)\right )\\ &=-\frac {\left (a+b \cos ^{-1}(c x)\right )^2}{x}-4 i b c \left (a+b \cos ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )-\left (2 b^2 c\right ) \text {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )+\left (2 b^2 c\right ) \text {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )\\ &=-\frac {\left (a+b \cos ^{-1}(c x)\right )^2}{x}-4 i b c \left (a+b \cos ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )+\left (2 i b^2 c\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \cos ^{-1}(c x)}\right )-\left (2 i b^2 c\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \cos ^{-1}(c x)}\right )\\ &=-\frac {\left (a+b \cos ^{-1}(c x)\right )^2}{x}-4 i b c \left (a+b \cos ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )+2 i b^2 c \text {Li}_2\left (-i e^{i \cos ^{-1}(c x)}\right )-2 i b^2 c \text {Li}_2\left (i e^{i \cos ^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 134, normalized size = 1.51 \begin {gather*} -\frac {a^2+2 a b \left (\text {ArcCos}(c x)-c x \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )\right )+b^2 \left (\text {ArcCos}(c x)^2-2 c x \left (\text {ArcCos}(c x) \left (\log \left (1-i e^{i \text {ArcCos}(c x)}\right )-\log \left (1+i e^{i \text {ArcCos}(c x)}\right )\right )+i \left (\text {PolyLog}\left (2,-i e^{i \text {ArcCos}(c x)}\right )-\text {PolyLog}\left (2,i e^{i \text {ArcCos}(c x)}\right )\right )\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCos[c*x])^2/x^2,x]

[Out]

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

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Maple [A]
time = 0.12, size = 192, normalized size = 2.16

method	result	size
derivativedivides	\(c \left (-\frac {a^{2}}{c x}-\frac {b^{2} \arccos \left (c x \right )^{2}}{c x}-2 b^{2} \arccos \left (c x \right ) \ln \left (1+i \left (c x +i \sqrt {-c^{2} x^{2}+1}\right )\right )+2 b^{2} \arccos \left (c x \right ) \ln \left (1-i \left (c x +i \sqrt {-c^{2} x^{2}+1}\right )\right )+2 i b^{2} \dilog \left (1+i \left (c x +i \sqrt {-c^{2} x^{2}+1}\right )\right )-2 i b^{2} \dilog \left (1-i \left (c x +i \sqrt {-c^{2} x^{2}+1}\right )\right )+2 a b \left (-\frac {\arccos \left (c x \right )}{c x}+\arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )\right )\)	\(192\)
default	\(c \left (-\frac {a^{2}}{c x}-\frac {b^{2} \arccos \left (c x \right )^{2}}{c x}-2 b^{2} \arccos \left (c x \right ) \ln \left (1+i \left (c x +i \sqrt {-c^{2} x^{2}+1}\right )\right )+2 b^{2} \arccos \left (c x \right ) \ln \left (1-i \left (c x +i \sqrt {-c^{2} x^{2}+1}\right )\right )+2 i b^{2} \dilog \left (1+i \left (c x +i \sqrt {-c^{2} x^{2}+1}\right )\right )-2 i b^{2} \dilog \left (1-i \left (c x +i \sqrt {-c^{2} x^{2}+1}\right )\right )+2 a b \left (-\frac {\arccos \left (c x \right )}{c x}+\arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )\right )\)	\(192\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccos(c*x))^2/x^2,x,method=_RETURNVERBOSE)

[Out]

c*(-a^2/c/x-b^2/c/x*arccos(c*x)^2-2*b^2*arccos(c*x)*ln(1+I*(c*x+I*(-c^2*x^2+1)^(1/2)))+2*b^2*arccos(c*x)*ln(1-
I*(c*x+I*(-c^2*x^2+1)^(1/2)))+2*I*b^2*dilog(1+I*(c*x+I*(-c^2*x^2+1)^(1/2)))-2*I*b^2*dilog(1-I*(c*x+I*(-c^2*x^2
+1)^(1/2)))+2*a*b*(-1/c/x*arccos(c*x)+arctanh(1/(-c^2*x^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((a+b*arccos(c*x))^2/x^2,x, algorithm="maxima")

[Out]

2*(c*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) - arccos(c*x)/x)*a*b + (2*c*x*integrate(sqrt(c*x + 1)*sqrt(-c
*x + 1)*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x)/(c^2*x^3 - x), x) - arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c
*x)^2)*b^2/x - a^2/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((a+b*arccos(c*x))^2/x^2,x, algorithm="fricas")

[Out]

integral((b^2*arccos(c*x)^2 + 2*a*b*arccos(c*x) + a^2)/x^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {acos}{\left (c x \right )}\right )^{2}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acos(c*x))**2/x**2,x)

[Out]

Integral((a + b*acos(c*x))**2/x**2, 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((a+b*arccos(c*x))^2/x^2,x, algorithm="giac")

[Out]

integrate((b*arccos(c*x) + a)^2/x^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^2}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(c*x))^2/x^2,x)

[Out]

int((a + b*acos(c*x))^2/x^2, x)

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