3.15 \(\int \frac {a+b \cos ^{-1}(c x)}{x^3 (d-c^2 d x^2)^2} \, dx\)

Optimal. Leaf size=159 \[ \frac {c^2 \left (a+b \cos ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \cos ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}+\frac {4 c^2 \tanh ^{-1}\left (e^{2 i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{d^2}-\frac {i b c^2 \text {Li}_2\left (-e^{2 i \cos ^{-1}(c x)}\right )}{d^2}+\frac {i b c^2 \text {Li}_2\left (e^{2 i \cos ^{-1}(c x)}\right )}{d^2}+\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}} \]

[Out]

c^2*(a+b*arccos(c*x))/d^2/(-c^2*x^2+1)+1/2*(-a-b*arccos(c*x))/d^2/x^2/(-c^2*x^2+1)+4*c^2*(a+b*arccos(c*x))*arc
tanh((c*x+I*(-c^2*x^2+1)^(1/2))^2)/d^2-I*b*c^2*polylog(2,-(c*x+I*(-c^2*x^2+1)^(1/2))^2)/d^2+I*b*c^2*polylog(2,
(c*x+I*(-c^2*x^2+1)^(1/2))^2)/d^2+1/2*b*c/d^2/x/(-c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.26, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4702, 4706, 4680, 4419, 4183, 2279, 2391, 191, 271} \[ -\frac {i b c^2 \text {PolyLog}\left (2,-e^{2 i \cos ^{-1}(c x)}\right )}{d^2}+\frac {i b c^2 \text {PolyLog}\left (2,e^{2 i \cos ^{-1}(c x)}\right )}{d^2}+\frac {c^2 \left (a+b \cos ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \cos ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}+\frac {4 c^2 \tanh ^{-1}\left (e^{2 i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{d^2}+\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[c*x])/(x^3*(d - c^2*d*x^2)^2),x]

[Out]

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

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

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 4419

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dist[
2^n, Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[m]

Rule 4680

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

Rule 4706

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)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(p +
1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcCos[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^Frac
Part[p])/(2*f*(p + 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, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ
[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 1])

Rubi steps

\begin {align*} \int \frac {a+b \cos ^{-1}(c x)}{x^3 \left (d-c^2 d x^2\right )^2} \, dx &=-\frac {a+b \cos ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}+\left (2 c^2\right ) \int \frac {a+b \cos ^{-1}(c x)}{x \left (d-c^2 d x^2\right )^2} \, dx-\frac {(b c) \int \frac {1}{x^2 \left (1-c^2 x^2\right )^{3/2}} \, dx}{2 d^2}\\ &=\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}}+\frac {c^2 \left (a+b \cos ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \cos ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}+\frac {\left (2 c^2\right ) \int \frac {a+b \cos ^{-1}(c x)}{x \left (d-c^2 d x^2\right )} \, dx}{d}\\ &=\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}}+\frac {c^2 \left (a+b \cos ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \cos ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}-\frac {\left (2 c^2\right ) \operatorname {Subst}\left (\int (a+b x) \csc (x) \sec (x) \, dx,x,\cos ^{-1}(c x)\right )}{d^2}\\ &=\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}}+\frac {c^2 \left (a+b \cos ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \cos ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}-\frac {\left (4 c^2\right ) \operatorname {Subst}\left (\int (a+b x) \csc (2 x) \, dx,x,\cos ^{-1}(c x)\right )}{d^2}\\ &=\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}}+\frac {c^2 \left (a+b \cos ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \cos ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}+\frac {4 c^2 \left (a+b \cos ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \cos ^{-1}(c x)}\right )}{d^2}+\frac {\left (2 b c^2\right ) \operatorname {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{d^2}-\frac {\left (2 b c^2\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{d^2}\\ &=\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}}+\frac {c^2 \left (a+b \cos ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \cos ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}+\frac {4 c^2 \left (a+b \cos ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \cos ^{-1}(c x)}\right )}{d^2}-\frac {\left (i b c^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \cos ^{-1}(c x)}\right )}{d^2}+\frac {\left (i b c^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \cos ^{-1}(c x)}\right )}{d^2}\\ &=\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}}+\frac {c^2 \left (a+b \cos ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \cos ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}+\frac {4 c^2 \left (a+b \cos ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \cos ^{-1}(c x)}\right )}{d^2}-\frac {i b c^2 \text {Li}_2\left (-e^{2 i \cos ^{-1}(c x)}\right )}{d^2}+\frac {i b c^2 \text {Li}_2\left (e^{2 i \cos ^{-1}(c x)}\right )}{d^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.73, size = 217, normalized size = 1.36 \[ \frac {\frac {a c^2}{1-c^2 x^2}-2 a c^2 \log \left (1-c^2 x^2\right )+4 a c^2 \log (x)-\frac {a}{x^2}-2 i b c^2 \text {Li}_2\left (-e^{2 i \cos ^{-1}(c x)}\right )+2 i b c^2 \text {Li}_2\left (e^{2 i \cos ^{-1}(c x)}\right )+\frac {b c \sqrt {1-c^2 x^2}}{x}+\frac {b c^2 \cos ^{-1}(c x)}{1-c^2 x^2}-4 b c^2 \cos ^{-1}(c x) \log \left (1-e^{2 i \cos ^{-1}(c x)}\right )+4 b c^2 \cos ^{-1}(c x) \log \left (1+e^{2 i \cos ^{-1}(c x)}\right )+\frac {b c^3 x}{\sqrt {1-c^2 x^2}}-\frac {b \cos ^{-1}(c x)}{x^2}}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcCos[c*x])/(x^3*(d - c^2*d*x^2)^2),x]

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arccos \left (c x\right ) + a}{c^{4} d^{2} x^{7} - 2 \, c^{2} d^{2} x^{5} + d^{2} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccos(c*x))/x^3/(-c^2*d*x^2+d)^2,x, algorithm="fricas")

[Out]

integral((b*arccos(c*x) + a)/(c^4*d^2*x^7 - 2*c^2*d^2*x^5 + d^2*x^3), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccos(c*x))/x^3/(-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^3), x)

________________________________________________________________________________________

maple [A]  time = 0.52, size = 371, normalized size = 2.33 \[ \frac {c^{2} a}{4 d^{2} \left (c x +1\right )}-\frac {c^{2} a \ln \left (c x +1\right )}{d^{2}}-\frac {a}{2 d^{2} x^{2}}+\frac {2 c^{2} a \ln \left (c x \right )}{d^{2}}-\frac {c^{2} a}{4 d^{2} \left (c x -1\right )}-\frac {c^{2} a \ln \left (c x -1\right )}{d^{2}}-\frac {c^{2} b \arccos \left (c x \right )}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {c b \sqrt {-c^{2} x^{2}+1}}{2 d^{2} x \left (c^{2} x^{2}-1\right )}+\frac {b \arccos \left (c x \right )}{2 d^{2} x^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 c^{2} b \arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}+\frac {2 c^{2} b \arccos \left (c x \right ) \ln \left (1+\left (c x +i \sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d^{2}}-\frac {2 c^{2} b \arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}+\frac {2 i c^{2} b \polylog \left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}+\frac {2 i c^{2} b \polylog \left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}-\frac {i b \,c^{2} \polylog \left (2, -\left (c x +i \sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccos(c*x))/x^3/(-c^2*d*x^2+d)^2,x)

[Out]

1/4*c^2*a/d^2/(c*x+1)-c^2*a/d^2*ln(c*x+1)-1/2*a/d^2/x^2+2*c^2*a/d^2*ln(c*x)-1/4*c^2*a/d^2/(c*x-1)-c^2*a/d^2*ln
(c*x-1)-c^2*b/d^2/(c^2*x^2-1)*arccos(c*x)-1/2*c*b/d^2/x/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)+1/2*b/d^2/x^2/(c^2*x^2-
1)*arccos(c*x)-2*c^2*b/d^2*arccos(c*x)*ln(1+c*x+I*(-c^2*x^2+1)^(1/2))+2*c^2*b/d^2*arccos(c*x)*ln(1+(c*x+I*(-c^
2*x^2+1)^(1/2))^2)-2*c^2*b/d^2*arccos(c*x)*ln(1-c*x-I*(-c^2*x^2+1)^(1/2))+2*I*c^2*b/d^2*polylog(2,-c*x-I*(-c^2
*x^2+1)^(1/2))+2*I*c^2*b/d^2*polylog(2,c*x+I*(-c^2*x^2+1)^(1/2))-I*b*c^2*polylog(2,-(c*x+I*(-c^2*x^2+1)^(1/2))
^2)/d^2

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a {\left (\frac {2 \, c^{2} \log \left (c x + 1\right )}{d^{2}} + \frac {2 \, c^{2} \log \left (c x - 1\right )}{d^{2}} - \frac {4 \, c^{2} \log \relax (x)}{d^{2}} + \frac {2 \, c^{2} x^{2} - 1}{c^{2} d^{2} x^{4} - d^{2} x^{2}}\right )} + b \int \frac {\arctan \left (\sqrt {c x + 1} \sqrt {-c x + 1}, c x\right )}{c^{4} d^{2} x^{7} - 2 \, c^{2} d^{2} x^{5} + d^{2} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccos(c*x))/x^3/(-c^2*d*x^2+d)^2,x, algorithm="maxima")

[Out]

-1/2*a*(2*c^2*log(c*x + 1)/d^2 + 2*c^2*log(c*x - 1)/d^2 - 4*c^2*log(x)/d^2 + (2*c^2*x^2 - 1)/(c^2*d^2*x^4 - d^
2*x^2)) + b*integrate(arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x)/(c^4*d^2*x^7 - 2*c^2*d^2*x^5 + d^2*x^3), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {acos}\left (c\,x\right )}{x^3\,{\left (d-c^2\,d\,x^2\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(c*x))/(x^3*(d - c^2*d*x^2)^2),x)

[Out]

int((a + b*acos(c*x))/(x^3*(d - c^2*d*x^2)^2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c^{4} x^{7} - 2 c^{2} x^{5} + x^{3}}\, dx + \int \frac {b \operatorname {acos}{\left (c x \right )}}{c^{4} x^{7} - 2 c^{2} x^{5} + x^{3}}\, dx}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acos(c*x))/x**3/(-c**2*d*x**2+d)**2,x)

[Out]

(Integral(a/(c**4*x**7 - 2*c**2*x**5 + x**3), x) + Integral(b*acos(c*x)/(c**4*x**7 - 2*c**2*x**5 + x**3), x))/
d**2

________________________________________________________________________________________