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

Optimal. Leaf size=82 \[ \frac {i \left (a+b \cos ^{-1}(c x)\right )^2}{2 b c^2 d}-\frac {\log \left (1-e^{2 i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{c^2 d}+\frac {i b \text {Li}_2\left (e^{2 i \cos ^{-1}(c x)}\right )}{2 c^2 d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4676, 3717, 2190, 2279, 2391} \[ \frac {i b \text {PolyLog}\left (2,e^{2 i \cos ^{-1}(c x)}\right )}{2 c^2 d}+\frac {i \left (a+b \cos ^{-1}(c x)\right )^2}{2 b c^2 d}-\frac {\log \left (1-e^{2 i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{c^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 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 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4676

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

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 115, normalized size = 1.40 \[ \frac {i \left (i a \log \left (1-c^2 x^2\right )+2 b \text {Li}_2\left (-e^{i \cos ^{-1}(c x)}\right )+2 b \text {Li}_2\left (e^{i \cos ^{-1}(c x)}\right )+b \cos ^{-1}(c x)^2+2 i b \cos ^{-1}(c x) \log \left (1-e^{i \cos ^{-1}(c x)}\right )+2 i b \cos ^{-1}(c x) \log \left (1+e^{i \cos ^{-1}(c x)}\right )\right )}{2 c^2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b x \arccos \left (c x\right ) + a x}{c^{2} d x^{2} - d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.12, size = 180, normalized size = 2.20 \[ -\frac {a \ln \left (c x -1\right )}{2 c^{2} d}-\frac {a \ln \left (c x +1\right )}{2 c^{2} d}+\frac {i b \arccos \left (c x \right )^{2}}{2 c^{2} d}-\frac {b \arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{c^{2} d}-\frac {b \arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )}{c^{2} d}+\frac {i b \polylog \left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )}{c^{2} d}+\frac {i b \polylog \left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )}{c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/c^2*a/d*ln(c*x-1)-1/2/c^2*a/d*ln(c*x+1)+1/2*I/c^2*b/d*arccos(c*x)^2-1/c^2*b/d*arccos(c*x)*ln(1+c*x+I*(-c^
2*x^2+1)^(1/2))-1/c^2*b/d*arccos(c*x)*ln(1-c*x-I*(-c^2*x^2+1)^(1/2))+I/c^2*b/d*polylog(2,-c*x-I*(-c^2*x^2+1)^(
1/2))+I/c^2*b/d*polylog(2,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 \[ \frac {{\left (c^{2} d \int \frac {e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )} \log \left (c x + 1\right ) + e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )} \log \left (-c x + 1\right )}{c^{5} d x^{4} - c^{3} d x^{2} - {\left (c^{3} d x^{2} - c d\right )} {\left (c x + 1\right )} {\left (c x - 1\right )}}\,{d x} - {\left (\log \left (c x + 1\right ) + \log \left (-c x + 1\right )\right )} \arctan \left (\sqrt {c x + 1} \sqrt {-c x + 1}, c x\right )\right )} b}{2 \, c^{2} d} - \frac {a \log \left (c^{2} d x^{2} - d\right )}{2 \, c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*c^2*d*integrate(1/2*(e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1))*log(c*x + 1) + e^(1/2*log(c*x + 1) + 1/2*
log(-c*x + 1))*log(-c*x + 1))/(c^5*d*x^4 - c^3*d*x^2 + (c^3*d*x^2 - c*d)*e^(log(c*x + 1) + log(-c*x + 1))), x)
- (log(c*x + 1) + log(-c*x + 1))*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x))*b/(c^2*d) - 1/2*a*log(c^2*d*x^2
- d)/(c^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a x}{c^{2} x^{2} - 1}\, dx + \int \frac {b x \operatorname {acos}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a*x/(c**2*x**2 - 1), x) + Integral(b*x*acos(c*x)/(c**2*x**2 - 1), x))/d

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