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

Optimal. Leaf size=248 \[ \frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \sin ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}-\frac {6 c^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^3}+\frac {3 i b c^2 \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}-\frac {3 i b c^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}-\frac {b c}{2 d^3 x \left (1-c^2 x^2\right )^{3/2}}-\frac {2 b c^3 x}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {5 b c^3 x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}} \]

[Out]

-1/2*b*c/d^3/x/(-c^2*x^2+1)^(3/2)+5/12*b*c^3*x/d^3/(-c^2*x^2+1)^(3/2)+3/4*c^2*(a+b*arcsin(c*x))/d^3/(-c^2*x^2+
1)^2+1/2*(-a-b*arcsin(c*x))/d^3/x^2/(-c^2*x^2+1)^2+3/2*c^2*(a+b*arcsin(c*x))/d^3/(-c^2*x^2+1)-6*c^2*(a+b*arcsi
n(c*x))*arctanh((I*c*x+(-c^2*x^2+1)^(1/2))^2)/d^3+3/2*I*b*c^2*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/d^3-3/2
*I*b*c^2*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))^2)/d^3-2/3*b*c^3*x/d^3/(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.35, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4701, 4705, 4679, 4419, 4183, 2279, 2391, 191, 192, 271} \[ \frac {3 i b c^2 \text {PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}-\frac {3 i b c^2 \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \sin ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}-\frac {6 c^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^3}-\frac {2 b c^3 x}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {5 b c^3 x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {b c}{2 d^3 x \left (1-c^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c)/(2*d^3*x*(1 - c^2*x^2)^(3/2)) + (5*b*c^3*x)/(12*d^3*(1 - c^2*x^2)^(3/2)) - (2*b*c^3*x)/(3*d^3*Sqrt[1 -
c^2*x^2]) + (3*c^2*(a + b*ArcSin[c*x]))/(4*d^3*(1 - c^2*x^2)^2) - (a + b*ArcSin[c*x])/(2*d^3*x^2*(1 - c^2*x^2)
^2) + (3*c^2*(a + b*ArcSin[c*x]))/(2*d^3*(1 - c^2*x^2)) - (6*c^2*(a + b*ArcSin[c*x])*ArcTanh[E^((2*I)*ArcSin[c
*x])])/d^3 + (((3*I)/2)*b*c^2*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/d^3 - (((3*I)/2)*b*c^2*PolyLog[2, E^((2*I)*A
rcSin[c*x])])/d^3

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 192

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

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 4679

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

Rule 4701

Int[((a_.) + ArcSin[(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*ArcSin[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*ArcSin[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*ArcSin[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 4705

Int[((a_.) + ArcSin[(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*ArcSin[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*ArcSin[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*ArcSin[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 \sin ^{-1}(c x)}{x^3 \left (d-c^2 d x^2\right )^3} \, dx &=-\frac {a+b \sin ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\left (3 c^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{x \left (d-c^2 d x^2\right )^3} \, dx+\frac {(b c) \int \frac {1}{x^2 \left (1-c^2 x^2\right )^{5/2}} \, dx}{2 d^3}\\ &=-\frac {b c}{2 d^3 x \left (1-c^2 x^2\right )^{3/2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \sin ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}-\frac {\left (3 b c^3\right ) \int \frac {1}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{4 d^3}+\frac {\left (2 b c^3\right ) \int \frac {1}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{d^3}+\frac {\left (3 c^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{x \left (d-c^2 d x^2\right )^2} \, dx}{d}\\ &=-\frac {b c}{2 d^3 x \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b c^3 x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \sin ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}-\frac {\left (b c^3\right ) \int \frac {1}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{2 d^3}+\frac {\left (4 b c^3\right ) \int \frac {1}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{3 d^3}-\frac {\left (3 b c^3\right ) \int \frac {1}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{2 d^3}+\frac {\left (3 c^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{x \left (d-c^2 d x^2\right )} \, dx}{d^2}\\ &=-\frac {b c}{2 d^3 x \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b c^3 x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {2 b c^3 x}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \sin ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}+\frac {\left (3 c^2\right ) \operatorname {Subst}\left (\int (a+b x) \csc (x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}\\ &=-\frac {b c}{2 d^3 x \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b c^3 x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {2 b c^3 x}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \sin ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}+\frac {\left (6 c^2\right ) \operatorname {Subst}\left (\int (a+b x) \csc (2 x) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}\\ &=-\frac {b c}{2 d^3 x \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b c^3 x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {2 b c^3 x}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \sin ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}-\frac {6 c^2 \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^3}-\frac {\left (3 b c^2\right ) \operatorname {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}+\frac {\left (3 b c^2\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}\\ &=-\frac {b c}{2 d^3 x \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b c^3 x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {2 b c^3 x}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \sin ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}-\frac {6 c^2 \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^3}+\frac {\left (3 i b c^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}-\frac {\left (3 i b c^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}\\ &=-\frac {b c}{2 d^3 x \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b c^3 x}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {2 b c^3 x}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \sin ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}-\frac {6 c^2 \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^3}+\frac {3 i b c^2 \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}-\frac {3 i b c^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}\\ \end {align*}

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Mathematica [A]  time = 1.69, size = 256, normalized size = 1.03 \[ -\frac {\frac {12 a c^2}{c^2 x^2-1}-\frac {3 a c^2}{\left (c^2 x^2-1\right )^2}+18 a c^2 \log \left (1-c^2 x^2\right )-36 a c^2 \log (x)+\frac {6 a}{x^2}+b c^2 \left (\frac {14 c x}{\sqrt {1-c^2 x^2}}+\frac {c x}{\left (1-c^2 x^2\right )^{3/2}}+\frac {6 \sqrt {1-c^2 x^2}}{c x}+\frac {12 \sin ^{-1}(c x)}{c^2 x^2-1}-\frac {3 \sin ^{-1}(c x)}{\left (c^2 x^2-1\right )^2}+\frac {6 \sin ^{-1}(c x)}{c^2 x^2}-18 i \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )+18 i \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )-36 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+36 \sin ^{-1}(c x) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )\right )}{12 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/12*((6*a)/x^2 - (3*a*c^2)/(-1 + c^2*x^2)^2 + (12*a*c^2)/(-1 + c^2*x^2) - 36*a*c^2*Log[x] + 18*a*c^2*Log[1 -
c^2*x^2] + b*c^2*((c*x)/(1 - c^2*x^2)^(3/2) + (14*c*x)/Sqrt[1 - c^2*x^2] + (6*Sqrt[1 - c^2*x^2])/(c*x) + (6*A
rcSin[c*x])/(c^2*x^2) - (3*ArcSin[c*x])/(-1 + c^2*x^2)^2 + (12*ArcSin[c*x])/(-1 + c^2*x^2) - 36*ArcSin[c*x]*Lo
g[1 - E^((2*I)*ArcSin[c*x])] + 36*ArcSin[c*x]*Log[1 + E^((2*I)*ArcSin[c*x])] - (18*I)*PolyLog[2, -E^((2*I)*Arc
Sin[c*x])] + (18*I)*PolyLog[2, E^((2*I)*ArcSin[c*x])]))/d^3

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \arcsin \left (c x\right ) + a}{c^{6} d^{3} x^{9} - 3 \, c^{4} d^{3} x^{7} + 3 \, c^{2} d^{3} x^{5} - d^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b*arcsin(c*x) + a)/(c^6*d^3*x^9 - 3*c^4*d^3*x^7 + 3*c^2*d^3*x^5 - d^3*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.41, size = 635, normalized size = 2.56 \[ \frac {c^{2} a}{16 d^{3} \left (c x +1\right )^{2}}+\frac {9 c^{2} a}{16 d^{3} \left (c x +1\right )}-\frac {3 c^{2} a \ln \left (c x +1\right )}{2 d^{3}}-\frac {a}{2 d^{3} x^{2}}+\frac {3 c^{2} a \ln \left (c x \right )}{d^{3}}+\frac {c^{2} a}{16 d^{3} \left (c x -1\right )^{2}}-\frac {9 c^{2} a}{16 d^{3} \left (c x -1\right )}-\frac {3 c^{2} a \ln \left (c x -1\right )}{2 d^{3}}+\frac {3 i b \,c^{2} \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{2 d^{3}}+\frac {2 c^{5} b \,x^{3} \sqrt {-c^{2} x^{2}+1}}{3 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {3 c^{4} b \arcsin \left (c x \right ) x^{2}}{2 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {2 i c^{6} b \,x^{4}}{3 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {c^{3} b x \sqrt {-c^{2} x^{2}+1}}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {9 c^{2} b \arcsin \left (c x \right )}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {2 i c^{2} b}{3 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {c b \sqrt {-c^{2} x^{2}+1}}{2 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right ) x}-\frac {b \arcsin \left (c x \right )}{2 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right ) x^{2}}+\frac {3 c^{2} b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d^{3}}-\frac {3 i c^{2} b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d^{3}}-\frac {3 c^{2} b \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d^{3}}+\frac {4 i c^{4} b \,x^{2}}{3 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {3 c^{2} b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{3}}-\frac {3 i c^{2} b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*c^2*a/d^3/(c*x+1)^2+9/16*c^2*a/d^3/(c*x+1)-3/2*c^2*a/d^3*ln(c*x+1)-1/2*a/d^3/x^2+3*c^2*a/d^3*ln(c*x)+1/16
*c^2*a/d^3/(c*x-1)^2-9/16*c^2*a/d^3/(c*x-1)-3/2*c^2*a/d^3*ln(c*x-1)-2/3*I*c^6*b/d^3/(c^4*x^4-2*c^2*x^2+1)*x^4+
2/3*c^5*b/d^3/(c^4*x^4-2*c^2*x^2+1)*x^3*(-c^2*x^2+1)^(1/2)-3/2*c^4*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*x^2
-3*I*c^2*b/d^3*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))-1/4*c^3*b/d^3/(c^4*x^4-2*c^2*x^2+1)*x*(-c^2*x^2+1)^(1/2)+9
/4*c^2*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)-2/3*I*c^2*b/d^3/(c^4*x^4-2*c^2*x^2+1)-1/2*c*b/d^3/(c^4*x^4-2*c^
2*x^2+1)/x*(-c^2*x^2+1)^(1/2)-1/2*b/d^3/(c^4*x^4-2*c^2*x^2+1)/x^2*arcsin(c*x)+3*c^2*b/d^3*arcsin(c*x)*ln(1-I*c
*x-(-c^2*x^2+1)^(1/2))+4/3*I*c^4*b/d^3/(c^4*x^4-2*c^2*x^2+1)*x^2-3*c^2*b/d^3*arcsin(c*x)*ln(1+(I*c*x+(-c^2*x^2
+1)^(1/2))^2)-3*I*c^2*b/d^3*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))+3*c^2*b/d^3*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1
)^(1/2))+3/2*I*b*c^2*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/d^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{4} \, a {\left (\frac {6 \, c^{4} x^{4} - 9 \, c^{2} x^{2} + 2}{c^{4} d^{3} x^{6} - 2 \, c^{2} d^{3} x^{4} + d^{3} x^{2}} + \frac {6 \, c^{2} \log \left (c x + 1\right )}{d^{3}} + \frac {6 \, c^{2} \log \left (c x - 1\right )}{d^{3}} - \frac {12 \, c^{2} \log \relax (x)}{d^{3}}\right )} - b \int \frac {\arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{c^{6} d^{3} x^{9} - 3 \, c^{4} d^{3} x^{7} + 3 \, c^{2} d^{3} x^{5} - d^{3} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a*((6*c^4*x^4 - 9*c^2*x^2 + 2)/(c^4*d^3*x^6 - 2*c^2*d^3*x^4 + d^3*x^2) + 6*c^2*log(c*x + 1)/d^3 + 6*c^2*l
og(c*x - 1)/d^3 - 12*c^2*log(x)/d^3) - b*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(c^6*d^3*x^9 - 3
*c^4*d^3*x^7 + 3*c^2*d^3*x^5 - d^3*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a}{c^{6} x^{9} - 3 c^{4} x^{7} + 3 c^{2} x^{5} - x^{3}}\, dx + \int \frac {b \operatorname {asin}{\left (c x \right )}}{c^{6} x^{9} - 3 c^{4} x^{7} + 3 c^{2} x^{5} - x^{3}}\, dx}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a/(c**6*x**9 - 3*c**4*x**7 + 3*c**2*x**5 - x**3), x) + Integral(b*asin(c*x)/(c**6*x**9 - 3*c**4*x**
7 + 3*c**2*x**5 - x**3), x))/d**3

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