3.6 \(\int \frac {a+b \cos ^{-1}(c x)}{x^2 (d-c^2 d x^2)} \, dx\)

Optimal. Leaf size=107 \[ -\frac {a+b \cos ^{-1}(c x)}{d x}+\frac {2 c \tanh ^{-1}\left (e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{d}+\frac {b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d}-\frac {i b c \text {Li}_2\left (-e^{i \cos ^{-1}(c x)}\right )}{d}+\frac {i b c \text {Li}_2\left (e^{i \cos ^{-1}(c x)}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4702, 4658, 4183, 2279, 2391, 266, 63, 208} \[ -\frac {i b c \text {PolyLog}\left (2,-e^{i \cos ^{-1}(c x)}\right )}{d}+\frac {i b c \text {PolyLog}\left (2,e^{i \cos ^{-1}(c x)}\right )}{d}-\frac {a+b \cos ^{-1}(c x)}{d x}+\frac {2 c \tanh ^{-1}\left (e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{d}+\frac {b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[c*x])/(x^2*(d - c^2*d*x^2)),x]

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2279

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 2391

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 4183

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

Rule 4658

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

Rule 4702

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcCos[c*x])^n)/(d*f*(m + 1)), x] + (Dist[(c^2*(m + 2*p + 3))/(f^2*(m
+ 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcCos[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(f*(m + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCos[c*x
])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1] && Inte
gerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 158, normalized size = 1.48 \[ -\frac {a c x \log (1-c x)-a c x \log (c x+1)+2 a-2 b c x \log \left (\sqrt {1-c^2 x^2}+1\right )+2 i b c x \text {Li}_2\left (-e^{i \cos ^{-1}(c x)}\right )-2 i b c x \text {Li}_2\left (e^{i \cos ^{-1}(c x)}\right )+2 b c x \log (x)+2 b \cos ^{-1}(c x)+2 b c x \cos ^{-1}(c x) \log \left (1-e^{i \cos ^{-1}(c x)}\right )-2 b c x \cos ^{-1}(c x) \log \left (1+e^{i \cos ^{-1}(c x)}\right )}{2 d x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \arccos \left (c x\right ) + a}{c^{2} d x^{4} - d x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b*arccos(c*x) + a)/(c^2*d*x^4 - d*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {b \arccos \left (c x\right ) + a}{{\left (c^{2} d x^{2} - d\right )} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.38, size = 166, normalized size = 1.55 \[ \frac {c a \ln \left (c x +1\right )}{2 d}-\frac {a}{d x}-\frac {c a \ln \left (c x -1\right )}{2 d}-\frac {b \arccos \left (c x \right )}{d x}-\frac {2 i c b \arctan \left (c x +i \sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {i c b \dilog \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {c b \arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {i c b \dilog \left (c x +i \sqrt {-c^{2} x^{2}+1}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccos(c*x))/x^2/(-c^2*d*x^2+d),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a {\left (\frac {c \log \left (c x + 1\right )}{d} - \frac {c \log \left (c x - 1\right )}{d} - \frac {2}{d x}\right )} - \frac {{\left (d x \int \frac {{\left (c^{2} x \log \left (c x + 1\right ) - c^{2} x \log \left (-c x + 1\right ) - 2 \, c\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{2} d x^{3} - d x}\,{d x} - {\left (c x \log \left (c x + 1\right ) - c x \log \left (-c x + 1\right ) - 2\right )} \arctan \left (\sqrt {c x + 1} \sqrt {-c x + 1}, c x\right )\right )} b}{2 \, d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(c*log(c*x + 1)/d - c*log(c*x - 1)/d - 2/(d*x)) - 1/2*(2*d*x*integrate(1/2*(c^2*x*log(c*x + 1) - c^2*x*l
og(-c*x + 1) - 2*c)*sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^2*d*x^3 - d*x), x) - (c*x*log(c*x + 1) - c*x*log(-c*x + 1)
- 2)*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x))*b/(d*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(c*x))/(x^2*(d - c^2*d*x^2)),x)

[Out]

int((a + b*acos(c*x))/(x^2*(d - c^2*d*x^2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a}{c^{2} x^{4} - x^{2}}\, dx + \int \frac {b \operatorname {acos}{\left (c x \right )}}{c^{2} x^{4} - x^{2}}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a/(c**2*x**4 - x**2), x) + Integral(b*acos(c*x)/(c**2*x**4 - x**2), x))/d

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