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

Optimal. Leaf size=316 \[ \frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d \sqrt {d-c^2 d x^2}}-\frac {a+b \sin ^{-1}(c x)}{2 d x^2 \sqrt {d-c^2 d x^2}}-\frac {3 c^2 \sqrt {1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}+\frac {3 i b c^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{2 d \sqrt {d-c^2 d x^2}}-\frac {3 i b c^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{2 d \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {1-c^2 x^2}}{2 d x \sqrt {d-c^2 d x^2}}-\frac {b c^2 \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{d \sqrt {d-c^2 d x^2}} \]

[Out]

3/2*c^2*(a+b*arcsin(c*x))/d/(-c^2*d*x^2+d)^(1/2)+1/2*(-a-b*arcsin(c*x))/d/x^2/(-c^2*d*x^2+d)^(1/2)-1/2*b*c*(-c
^2*x^2+1)^(1/2)/d/x/(-c^2*d*x^2+d)^(1/2)-3*c^2*(a+b*arcsin(c*x))*arctanh(I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1
)^(1/2)/d/(-c^2*d*x^2+d)^(1/2)-b*c^2*arctanh(c*x)*(-c^2*x^2+1)^(1/2)/d/(-c^2*d*x^2+d)^(1/2)+3/2*I*b*c^2*polylo
g(2,-I*c*x-(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)/d/(-c^2*d*x^2+d)^(1/2)-3/2*I*b*c^2*polylog(2,I*c*x+(-c^2*x^2
+1)^(1/2))*(-c^2*x^2+1)^(1/2)/d/(-c^2*d*x^2+d)^(1/2)

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Rubi [A]  time = 0.44, antiderivative size = 316, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4701, 4705, 4713, 4709, 4183, 2279, 2391, 206, 325} \[ \frac {3 i b c^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{2 d \sqrt {d-c^2 d x^2}}-\frac {3 i b c^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{2 d \sqrt {d-c^2 d x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d \sqrt {d-c^2 d x^2}}-\frac {a+b \sin ^{-1}(c x)}{2 d x^2 \sqrt {d-c^2 d x^2}}-\frac {3 c^2 \sqrt {1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {1-c^2 x^2}}{2 d x \sqrt {d-c^2 d x^2}}-\frac {b c^2 \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 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])

Rule 4709

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

Rule 4713

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[
Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2], Int[((f*x)^m*(a + b*ArcSin[c*x])^n)/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a,
b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] &&  !GtQ[d, 0] && (IntegerQ[m] || 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 )^{3/2}} \, dx &=-\frac {a+b \sin ^{-1}(c x)}{2 d x^2 \sqrt {d-c^2 d x^2}}+\frac {1}{2} \left (3 c^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{x \left (d-c^2 d x^2\right )^{3/2}} \, dx+\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx}{2 d \sqrt {d-c^2 d x^2}}\\ &=-\frac {b c \sqrt {1-c^2 x^2}}{2 d x \sqrt {d-c^2 d x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d \sqrt {d-c^2 d x^2}}-\frac {a+b \sin ^{-1}(c x)}{2 d x^2 \sqrt {d-c^2 d x^2}}+\frac {\left (3 c^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{x \sqrt {d-c^2 d x^2}} \, dx}{2 d}+\frac {\left (b c^3 \sqrt {1-c^2 x^2}\right ) \int \frac {1}{1-c^2 x^2} \, dx}{2 d \sqrt {d-c^2 d x^2}}-\frac {\left (3 b c^3 \sqrt {1-c^2 x^2}\right ) \int \frac {1}{1-c^2 x^2} \, dx}{2 d \sqrt {d-c^2 d x^2}}\\ &=-\frac {b c \sqrt {1-c^2 x^2}}{2 d x \sqrt {d-c^2 d x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d \sqrt {d-c^2 d x^2}}-\frac {a+b \sin ^{-1}(c x)}{2 d x^2 \sqrt {d-c^2 d x^2}}-\frac {b c^2 \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{d \sqrt {d-c^2 d x^2}}+\frac {\left (3 c^2 \sqrt {1-c^2 x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{x \sqrt {1-c^2 x^2}} \, dx}{2 d \sqrt {d-c^2 d x^2}}\\ &=-\frac {b c \sqrt {1-c^2 x^2}}{2 d x \sqrt {d-c^2 d x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d \sqrt {d-c^2 d x^2}}-\frac {a+b \sin ^{-1}(c x)}{2 d x^2 \sqrt {d-c^2 d x^2}}-\frac {b c^2 \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{d \sqrt {d-c^2 d x^2}}+\frac {\left (3 c^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{2 d \sqrt {d-c^2 d x^2}}\\ &=-\frac {b c \sqrt {1-c^2 x^2}}{2 d x \sqrt {d-c^2 d x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d \sqrt {d-c^2 d x^2}}-\frac {a+b \sin ^{-1}(c x)}{2 d x^2 \sqrt {d-c^2 d x^2}}-\frac {3 c^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {b c^2 \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{d \sqrt {d-c^2 d x^2}}-\frac {\left (3 b c^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{2 d \sqrt {d-c^2 d x^2}}+\frac {\left (3 b c^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{2 d \sqrt {d-c^2 d x^2}}\\ &=-\frac {b c \sqrt {1-c^2 x^2}}{2 d x \sqrt {d-c^2 d x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d \sqrt {d-c^2 d x^2}}-\frac {a+b \sin ^{-1}(c x)}{2 d x^2 \sqrt {d-c^2 d x^2}}-\frac {3 c^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {b c^2 \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{d \sqrt {d-c^2 d x^2}}+\frac {\left (3 i b c^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{2 d \sqrt {d-c^2 d x^2}}-\frac {\left (3 i b c^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{2 d \sqrt {d-c^2 d x^2}}\\ &=-\frac {b c \sqrt {1-c^2 x^2}}{2 d x \sqrt {d-c^2 d x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )}{2 d \sqrt {d-c^2 d x^2}}-\frac {a+b \sin ^{-1}(c x)}{2 d x^2 \sqrt {d-c^2 d x^2}}-\frac {3 c^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {b c^2 \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{d \sqrt {d-c^2 d x^2}}+\frac {3 i b c^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{2 d \sqrt {d-c^2 d x^2}}-\frac {3 i b c^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{2 d \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 2.47, size = 404, normalized size = 1.28 \[ \frac {\frac {4 a \sqrt {d} \left (3 c^2 x^2-1\right )}{x^2 \sqrt {d-c^2 d x^2}}-12 a c^2 \log \left (\sqrt {d} \sqrt {d-c^2 d x^2}+d\right )+12 a c^2 \log (x)+\frac {b \sqrt {d} \left (\sqrt {1-c^2 x^2} \left (2 \left (\log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )-\log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )\right )+3 \sin ^{-1}(c x) \left (\log \left (1-e^{i \sin ^{-1}(c x)}\right )-\log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )\right )+6 i c x \sin \left (2 \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )-6 i c x \sin \left (2 \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )+2 \sin ^{-1}(c x)-2 \sin \left (2 \sin ^{-1}(c x)\right )-6 \sin ^{-1}(c x) \cos \left (2 \sin ^{-1}(c x)\right )-3 \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right ) \cos \left (3 \sin ^{-1}(c x)\right )+3 \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right ) \cos \left (3 \sin ^{-1}(c x)\right )-2 \cos \left (3 \sin ^{-1}(c x)\right ) \log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )+2 \cos \left (3 \sin ^{-1}(c x)\right ) \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )\right )}{x^2 \sqrt {d-c^2 d x^2}}}{8 d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a*Sqrt[d]*(-1 + 3*c^2*x^2))/(x^2*Sqrt[d - c^2*d*x^2]) + 12*a*c^2*Log[x] - 12*a*c^2*Log[d + Sqrt[d]*Sqrt[d
- c^2*d*x^2]] + (b*Sqrt[d]*(2*ArcSin[c*x] - 6*ArcSin[c*x]*Cos[2*ArcSin[c*x]] - 3*ArcSin[c*x]*Cos[3*ArcSin[c*x]
]*Log[1 - E^(I*ArcSin[c*x])] + 3*ArcSin[c*x]*Cos[3*ArcSin[c*x]]*Log[1 + E^(I*ArcSin[c*x])] - 2*Cos[3*ArcSin[c*
x]]*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] + Sqrt[1 - c^2*x^2]*(3*ArcSin[c*x]*(Log[1 - E^(I*ArcSin[c*x])
] - Log[1 + E^(I*ArcSin[c*x])]) + 2*(Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] - Log[Cos[ArcSin[c*x]/2] + S
in[ArcSin[c*x]/2]])) + 2*Cos[3*ArcSin[c*x]]*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]] - 2*Sin[2*ArcSin[c*x]
] + (6*I)*c*x*PolyLog[2, -E^(I*ArcSin[c*x])]*Sin[2*ArcSin[c*x]] - (6*I)*c*x*PolyLog[2, E^(I*ArcSin[c*x])]*Sin[
2*ArcSin[c*x]]))/(x^2*Sqrt[d - c^2*d*x^2]))/(8*d^(3/2))

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{4} d^{2} x^{7} - 2 \, c^{2} d^{2} x^{5} + d^{2} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.38, size = 474, normalized size = 1.50 \[ -\frac {a}{2 d \,x^{2} \sqrt {-c^{2} d \,x^{2}+d}}+\frac {3 a \,c^{2}}{2 d \sqrt {-c^{2} d \,x^{2}+d}}-\frac {3 a \,c^{2} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )}{2 d^{\frac {3}{2}}}-\frac {3 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) c^{2}}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c}{2 d^{2} \left (c^{2} x^{2}-1\right ) x}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{2 d^{2} \left (c^{2} x^{2}-1\right ) x^{2}}-\frac {2 i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} \arctan \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{\left (c^{2} x^{2}-1\right ) d^{2}}-\frac {3 i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} \dilog \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{2 \left (c^{2} x^{2}-1\right ) d^{2}}+\frac {3 b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{2 \left (c^{2} x^{2}-1\right ) d^{2}}-\frac {3 i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} \dilog \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{2 \left (c^{2} x^{2}-1\right ) d^{2}} \]

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)^(3/2),x)

[Out]

-1/2*a/d/x^2/(-c^2*d*x^2+d)^(1/2)+3/2*a*c^2/d/(-c^2*d*x^2+d)^(1/2)-3/2*a*c^2/d^(3/2)*ln((2*d+2*d^(1/2)*(-c^2*d
*x^2+d)^(1/2))/x)-3/2*b*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)*c^2+1/2*b*(-d*(c^2*x^2-1))^(1/2)/d^
2/(c^2*x^2-1)/x*(-c^2*x^2+1)^(1/2)*c+1/2*b*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)/x^2*arcsin(c*x)-2*I*b*(-c^2*
x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)*c^2/(c^2*x^2-1)/d^2*arctan(I*c*x+(-c^2*x^2+1)^(1/2))-3/2*I*b*(-c^2*x^2+1)^
(1/2)*(-d*(c^2*x^2-1))^(1/2)*c^2/(c^2*x^2-1)/d^2*dilog(1+I*c*x+(-c^2*x^2+1)^(1/2))+3/2*b*(-c^2*x^2+1)^(1/2)*(-
d*(c^2*x^2-1))^(1/2)*c^2/(c^2*x^2-1)/d^2*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-3/2*I*b*(-c^2*x^2+1)^(1/2)
*(-d*(c^2*x^2-1))^(1/2)*c^2/(c^2*x^2-1)/d^2*dilog(I*c*x+(-c^2*x^2+1)^(1/2))

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

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)^(3/2),x, algorithm="maxima")

[Out]

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

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

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)^(3/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asin}{\left (c x \right )}}{x^{3} \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

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)**(3/2),x)

[Out]

Integral((a + b*asin(c*x))/(x**3*(-d*(c*x - 1)*(c*x + 1))**(3/2)), x)

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