3.1.12 \(\int \frac {a+b \text {ArcCos}(c x)}{(d-c^2 d x^2)^2} \, dx\) [12]

Optimal. Leaf size=132 \[ \frac {b}{2 c d^2 \sqrt {1-c^2 x^2}}+\frac {x (a+b \text {ArcCos}(c x))}{2 d^2 \left (1-c^2 x^2\right )}+\frac {(a+b \text {ArcCos}(c x)) \tanh ^{-1}\left (e^{i \text {ArcCos}(c x)}\right )}{c d^2}-\frac {i b \text {PolyLog}\left (2,-e^{i \text {ArcCos}(c x)}\right )}{2 c d^2}+\frac {i b \text {PolyLog}\left (2,e^{i \text {ArcCos}(c x)}\right )}{2 c d^2} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4748, 4750, 4268, 2317, 2438, 267} \begin {gather*} \frac {x (a+b \text {ArcCos}(c x))}{2 d^2 \left (1-c^2 x^2\right )}+\frac {\tanh ^{-1}\left (e^{i \text {ArcCos}(c x)}\right ) (a+b \text {ArcCos}(c x))}{c d^2}-\frac {i b \text {Li}_2\left (-e^{i \text {ArcCos}(c x)}\right )}{2 c d^2}+\frac {i b \text {Li}_2\left (e^{i \text {ArcCos}(c x)}\right )}{2 c d^2}+\frac {b}{2 c d^2 \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 267

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

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 4268

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 4748

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

Rule 4750

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)}{\left (d-c^2 d x^2\right )^2} \, dx &=\frac {x \left (a+b \cos ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}+\frac {(b c) \int \frac {x}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{2 d^2}+\frac {\int \frac {a+b \cos ^{-1}(c x)}{d-c^2 d x^2} \, dx}{2 d}\\ &=\frac {b}{2 c d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \cos ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}-\frac {\text {Subst}\left (\int (a+b x) \csc (x) \, dx,x,\cos ^{-1}(c x)\right )}{2 c d^2}\\ &=\frac {b}{2 c d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \cos ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}+\frac {\left (a+b \cos ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )}{c d^2}+\frac {b \text {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{2 c d^2}-\frac {b \text {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{2 c d^2}\\ &=\frac {b}{2 c d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \cos ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}+\frac {\left (a+b \cos ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )}{c d^2}-\frac {(i b) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \cos ^{-1}(c x)}\right )}{2 c d^2}+\frac {(i b) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \cos ^{-1}(c x)}\right )}{2 c d^2}\\ &=\frac {b}{2 c d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \cos ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}+\frac {\left (a+b \cos ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )}{c d^2}-\frac {i b \text {Li}_2\left (-e^{i \cos ^{-1}(c x)}\right )}{2 c d^2}+\frac {i b \text {Li}_2\left (e^{i \cos ^{-1}(c x)}\right )}{2 c d^2}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 220, normalized size = 1.67 \begin {gather*} \frac {\frac {b \sqrt {1-c^2 x^2}}{c-c^2 x}+\frac {b \sqrt {1-c^2 x^2}}{c+c^2 x}-\frac {2 a x}{-1+c^2 x^2}+\frac {b \text {ArcCos}(c x)}{c-c^2 x}-\frac {b \text {ArcCos}(c x)}{c+c^2 x}-\frac {2 b \text {ArcCos}(c x) \log \left (1-e^{i \text {ArcCos}(c x)}\right )}{c}+\frac {2 b \text {ArcCos}(c x) \log \left (1+e^{i \text {ArcCos}(c x)}\right )}{c}-\frac {a \log (1-c x)}{c}+\frac {a \log (1+c x)}{c}-\frac {2 i b \text {PolyLog}\left (2,-e^{i \text {ArcCos}(c x)}\right )}{c}+\frac {2 i b \text {PolyLog}\left (2,e^{i \text {ArcCos}(c x)}\right )}{c}}{4 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.20, size = 228, normalized size = 1.73

method	result	size
derivativedivides	\(\frac {-\frac {a}{4 d^{2} \left (c x +1\right )}+\frac {a \ln \left (c x +1\right )}{4 d^{2}}-\frac {a}{4 d^{2} \left (c x -1\right )}-\frac {a \ln \left (c x -1\right )}{4 d^{2}}-\frac {b \arccos \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-c^{2} x^{2}+1}}{2 d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )}{2 d^{2}}+\frac {i b \polylog \left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2 d^{2}}+\frac {b \arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2 d^{2}}-\frac {i b \polylog \left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )}{2 d^{2}}}{c}\)	\(228\)
default	\(\frac {-\frac {a}{4 d^{2} \left (c x +1\right )}+\frac {a \ln \left (c x +1\right )}{4 d^{2}}-\frac {a}{4 d^{2} \left (c x -1\right )}-\frac {a \ln \left (c x -1\right )}{4 d^{2}}-\frac {b \arccos \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-c^{2} x^{2}+1}}{2 d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )}{2 d^{2}}+\frac {i b \polylog \left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2 d^{2}}+\frac {b \arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2 d^{2}}-\frac {i b \polylog \left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )}{2 d^{2}}}{c}\)	\(228\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(-1/4*a/d^2/(c*x+1)+1/4*a/d^2*ln(c*x+1)-1/4*a/d^2/(c*x-1)-1/4*a/d^2*ln(c*x-1)-1/2*b/d^2/(c^2*x^2-1)*arccos
(c*x)*c*x-1/2*b/d^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)-1/2*b/d^2*arccos(c*x)*ln(1-c*x-I*(-c^2*x^2+1)^(1/2))+1/2*I*
b/d^2*polylog(2,c*x+I*(-c^2*x^2+1)^(1/2))+1/2*b/d^2*arccos(c*x)*ln(1+c*x+I*(-c^2*x^2+1)^(1/2))-1/2*I*b/d^2*pol
ylog(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 \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))/(-c^2*d*x^2+d)^2,x, algorithm="maxima")

[Out]

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

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac {b \operatorname {acos}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**4*x**4 - 2*c**2*x**2 + 1), x) + Integral(b*acos(c*x)/(c**4*x**4 - 2*c**2*x**2 + 1), x))/d**2

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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