3.1.14 \(\int \frac {a+b \text {ArcCos}(c x)}{x^2 (d-c^2 d x^2)^2} \, dx\) [14]

Optimal. Leaf size=177 \[ \frac {b c}{2 d^2 \sqrt {1-c^2 x^2}}-\frac {a+b \text {ArcCos}(c x)}{d^2 x \left (1-c^2 x^2\right )}+\frac {3 c^2 x (a+b \text {ArcCos}(c x))}{2 d^2 \left (1-c^2 x^2\right )}+\frac {3 c (a+b \text {ArcCos}(c x)) \tanh ^{-1}\left (e^{i \text {ArcCos}(c x)}\right )}{d^2}+\frac {b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d^2}-\frac {3 i b c \text {PolyLog}\left (2,-e^{i \text {ArcCos}(c x)}\right )}{2 d^2}+\frac {3 i b c \text {PolyLog}\left (2,e^{i \text {ArcCos}(c x)}\right )}{2 d^2} \]

[Out]

(-a-b*arccos(c*x))/d^2/x/(-c^2*x^2+1)+3/2*c^2*x*(a+b*arccos(c*x))/d^2/(-c^2*x^2+1)+3*c*(a+b*arccos(c*x))*arcta
nh(c*x+I*(-c^2*x^2+1)^(1/2))/d^2+b*c*arctanh((-c^2*x^2+1)^(1/2))/d^2-3/2*I*b*c*polylog(2,-c*x-I*(-c^2*x^2+1)^(
1/2))/d^2+3/2*I*b*c*polylog(2,c*x+I*(-c^2*x^2+1)^(1/2))/d^2+1/2*b*c/d^2/(-c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {4790, 4748, 4750, 4268, 2317, 2438, 267, 272, 53, 65, 214} \begin {gather*} \frac {3 c^2 x (a+b \text {ArcCos}(c x))}{2 d^2 \left (1-c^2 x^2\right )}-\frac {a+b \text {ArcCos}(c x)}{d^2 x \left (1-c^2 x^2\right )}+\frac {3 c \tanh ^{-1}\left (e^{i \text {ArcCos}(c x)}\right ) (a+b \text {ArcCos}(c x))}{d^2}-\frac {3 i b c \text {Li}_2\left (-e^{i \text {ArcCos}(c x)}\right )}{2 d^2}+\frac {3 i b c \text {Li}_2\left (e^{i \text {ArcCos}(c x)}\right )}{2 d^2}+\frac {b c}{2 d^2 \sqrt {1-c^2 x^2}}+\frac {b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c)/(2*d^2*Sqrt[1 - c^2*x^2]) - (a + b*ArcCos[c*x])/(d^2*x*(1 - c^2*x^2)) + (3*c^2*x*(a + b*ArcCos[c*x]))/(2
*d^2*(1 - c^2*x^2)) + (3*c*(a + b*ArcCos[c*x])*ArcTanh[E^(I*ArcCos[c*x])])/d^2 + (b*c*ArcTanh[Sqrt[1 - c^2*x^2
]])/d^2 - (((3*I)/2)*b*c*PolyLog[2, -E^(I*ArcCos[c*x])])/d^2 + (((3*I)/2)*b*c*PolyLog[2, E^(I*ArcCos[c*x])])/d
^2

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

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 214

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

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 272

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 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]

Rule 4790

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/(f*(m + 1)))*Simp[(d + e*x
^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCos[c*x])^(n - 1), x], x]) /; Free
Q[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && ILtQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.25, size = 251, normalized size = 1.42 \begin {gather*} \frac {-\frac {4 a}{x}+\frac {b c \sqrt {1-c^2 x^2}}{1-c x}+\frac {b c \sqrt {1-c^2 x^2}}{1+c x}-\frac {2 a c^2 x}{-1+c^2 x^2}-\frac {4 b \text {ArcCos}(c x)}{x}+\frac {b c \text {ArcCos}(c x)}{1-c x}-\frac {b c \text {ArcCos}(c x)}{1+c x}-6 b c \text {ArcCos}(c x) \log \left (1-e^{i \text {ArcCos}(c x)}\right )+6 b c \text {ArcCos}(c x) \log \left (1+e^{i \text {ArcCos}(c x)}\right )-4 b c \log (x)-3 a c \log (1-c x)+3 a c \log (1+c x)+4 b c \log \left (1+\sqrt {1-c^2 x^2}\right )-6 i b c \text {PolyLog}\left (2,-e^{i \text {ArcCos}(c x)}\right )+6 i b c \text {PolyLog}\left (2,e^{i \text {ArcCos}(c x)}\right )}{4 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*a)/x + (b*c*Sqrt[1 - c^2*x^2])/(1 - c*x) + (b*c*Sqrt[1 - c^2*x^2])/(1 + c*x) - (2*a*c^2*x)/(-1 + c^2*x^2)
- (4*b*ArcCos[c*x])/x + (b*c*ArcCos[c*x])/(1 - c*x) - (b*c*ArcCos[c*x])/(1 + c*x) - 6*b*c*ArcCos[c*x]*Log[1 -
E^(I*ArcCos[c*x])] + 6*b*c*ArcCos[c*x]*Log[1 + E^(I*ArcCos[c*x])] - 4*b*c*Log[x] - 3*a*c*Log[1 - c*x] + 3*a*c
*Log[1 + c*x] + 4*b*c*Log[1 + Sqrt[1 - c^2*x^2]] - (6*I)*b*c*PolyLog[2, -E^(I*ArcCos[c*x])] + (6*I)*b*c*PolyLo
g[2, E^(I*ArcCos[c*x])])/(4*d^2)

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Maple [A]
time = 0.63, size = 257, normalized size = 1.45

method	result	size
derivativedivides	\(c \left (-\frac {a}{4 d^{2} \left (c x +1\right )}+\frac {3 a \ln \left (c x +1\right )}{4 d^{2}}-\frac {a}{4 d^{2} \left (c x -1\right )}-\frac {3 a \ln \left (c x -1\right )}{4 d^{2}}-\frac {a}{d^{2} c x}-\frac {3 b \arccos \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-c^{2} x^{2}+1}}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \arccos \left (c x \right )}{d^{2} \left (c^{2} x^{2}-1\right ) c x}-\frac {2 i b \arctan \left (c x +i \sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}-\frac {3 i b \dilog \left (c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2 d^{2}}-\frac {3 i b \dilog \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2 d^{2}}+\frac {3 b \arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2 d^{2}}\right )\)	\(257\)
default	\(c \left (-\frac {a}{4 d^{2} \left (c x +1\right )}+\frac {3 a \ln \left (c x +1\right )}{4 d^{2}}-\frac {a}{4 d^{2} \left (c x -1\right )}-\frac {3 a \ln \left (c x -1\right )}{4 d^{2}}-\frac {a}{d^{2} c x}-\frac {3 b \arccos \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-c^{2} x^{2}+1}}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \arccos \left (c x \right )}{d^{2} \left (c^{2} x^{2}-1\right ) c x}-\frac {2 i b \arctan \left (c x +i \sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}-\frac {3 i b \dilog \left (c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2 d^{2}}-\frac {3 i b \dilog \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2 d^{2}}+\frac {3 b \arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2 d^{2}}\right )\)	\(257\)

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)^2,x,method=_RETURNVERBOSE)

[Out]

c*(-1/4*a/d^2/(c*x+1)+3/4*a/d^2*ln(c*x+1)-1/4*a/d^2/(c*x-1)-3/4*a/d^2*ln(c*x-1)-a/d^2/c/x-3/2*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)+b/d^2/(c^2*x^2-1)/c/x*arccos(c*x)-2*I*b/d^2*arctan(
c*x+I*(-c^2*x^2+1)^(1/2))-3/2*I*b/d^2*dilog(c*x+I*(-c^2*x^2+1)^(1/2))-3/2*I*b/d^2*dilog(1+c*x+I*(-c^2*x^2+1)^(
1/2))+3/2*b/d^2*arccos(c*x)*ln(1+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))/x^2/(-c^2*d*x^2+d)^2,x, algorithm="maxima")

[Out]

-1/4*a*(2*(3*c^2*x^2 - 2)/(c^2*d^2*x^3 - d^2*x) - 3*c*log(c*x + 1)/d^2 + 3*c*log(c*x - 1)/d^2) - 1/4*((6*c^2*x
^2 - 3*(c^3*x^3 - c*x)*log(c*x + 1) + 3*(c^3*x^3 - c*x)*log(-c*x + 1) - 4)*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1
), c*x) + 4*(c^2*d^2*x^3 - d^2*x)*integrate(-1/4*(6*c^3*x^2 - 3*(c^4*x^3 - c^2*x)*log(c*x + 1) + 3*(c^4*x^3 -
c^2*x)*log(-c*x + 1) - 4*c)*sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^4*d^2*x^5 - 2*c^2*d^2*x^3 + d^2*x), x))*b/(c^2*d^2
*x^3 - d^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))/x^2/(-c^2*d*x^2+d)^2,x, algorithm="fricas")

[Out]

integral((b*arccos(c*x) + a)/(c^4*d^2*x^6 - 2*c^2*d^2*x^4 + d^2*x^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^{6} - 2 c^{2} x^{4} + x^{2}}\, dx + \int \frac {b \operatorname {acos}{\left (c x \right )}}{c^{4} x^{6} - 2 c^{2} x^{4} + x^{2}}\, dx}{d^{2}} \end {gather*}

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)**2,x)

[Out]

(Integral(a/(c**4*x**6 - 2*c**2*x**4 + x**2), x) + Integral(b*acos(c*x)/(c**4*x**6 - 2*c**2*x**4 + x**2), x))/
d**2

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)^2,x, algorithm="giac")

[Out]

Timed out

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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 )}{x^2\,{\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))/(x^2*(d - c^2*d*x^2)^2),x)

[Out]

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

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