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

Optimal. Leaf size=76 \[ \frac {2 \tanh ^{-1}\left (e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{c d}-\frac {i b \text {Li}_2\left (-e^{i \cos ^{-1}(c x)}\right )}{c d}+\frac {i b \text {Li}_2\left (e^{i \cos ^{-1}(c x)}\right )}{c d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4658, 4183, 2279, 2391} \[ -\frac {i b \text {PolyLog}\left (2,-e^{i \cos ^{-1}(c x)}\right )}{c d}+\frac {i b \text {PolyLog}\left (2,e^{i \cos ^{-1}(c x)}\right )}{c d}+\frac {2 \tanh ^{-1}\left (e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 107, normalized size = 1.41 \[ \frac {-a \log (1-c x)+a \log (c x+1)-2 i b \text {Li}_2\left (-e^{i \cos ^{-1}(c x)}\right )+2 i b \text {Li}_2\left (e^{i \cos ^{-1}(c x)}\right )-2 b \cos ^{-1}(c x) \log \left (1-e^{i \cos ^{-1}(c x)}\right )+2 b \cos ^{-1}(c x) \log \left (1+e^{i \cos ^{-1}(c x)}\right )}{2 c d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \arccos \left (c x\right ) + a}{c^{2} d x^{2} - d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.21, size = 170, normalized size = 2.24 \[ \frac {a \arctanh \left (c x \right )}{c d}+\frac {i b \arctanh \left (c x \right ) \ln \left (1-\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c d}-\frac {i b \arctanh \left (c x \right ) \ln \left (1+\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c d}+\frac {b \arctanh \left (c x \right ) \arccos \left (c x \right )}{c d}+\frac {i b \dilog \left (1-\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c d}-\frac {i b \dilog \left (1+\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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 \[ - \frac {\int \frac {a}{c^{2} x^{2} - 1}\, dx + \int \frac {b \operatorname {acos}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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