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

Optimal. Leaf size=272 \[ -\frac {a b^2 \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )}{\left (a^2-1\right )^{3/2}}+\frac {a b^2 \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {a^2-1}}\right )}{\left (a^2-1\right )^{3/2}}+\frac {b^2 \log (x)}{1-a^2}-\frac {i a b^2 \sin ^{-1}(a+b x) \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )}{\left (a^2-1\right )^{3/2}}+\frac {i a b^2 \sin ^{-1}(a+b x) \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{\sqrt {a^2-1}+a}\right )}{\left (a^2-1\right )^{3/2}}-\frac {b \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{\left (1-a^2\right ) x}-\frac {\sin ^{-1}(a+b x)^2}{2 x^2} \]

[Out]

-1/2*arcsin(b*x+a)^2/x^2+b^2*ln(x)/(-a^2+1)-I*a*b^2*arcsin(b*x+a)*ln(1+I*(I*(b*x+a)+(1-(b*x+a)^2)^(1/2))/(a-(a
^2-1)^(1/2)))/(a^2-1)^(3/2)+I*a*b^2*arcsin(b*x+a)*ln(1+I*(I*(b*x+a)+(1-(b*x+a)^2)^(1/2))/(a+(a^2-1)^(1/2)))/(a
^2-1)^(3/2)-a*b^2*polylog(2,-I*(I*(b*x+a)+(1-(b*x+a)^2)^(1/2))/(a-(a^2-1)^(1/2)))/(a^2-1)^(3/2)+a*b^2*polylog(
2,-I*(I*(b*x+a)+(1-(b*x+a)^2)^(1/2))/(a+(a^2-1)^(1/2)))/(a^2-1)^(3/2)-b*arcsin(b*x+a)*(1-(b*x+a)^2)^(1/2)/(-a^
2+1)/x

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Rubi [A]  time = 0.59, antiderivative size = 272, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {4805, 4743, 4773, 3324, 3323, 2264, 2190, 2279, 2391, 2668, 31} \[ -\frac {a b^2 \text {PolyLog}\left (2,-\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )}{\left (a^2-1\right )^{3/2}}+\frac {a b^2 \text {PolyLog}\left (2,-\frac {i e^{i \sin ^{-1}(a+b x)}}{\sqrt {a^2-1}+a}\right )}{\left (a^2-1\right )^{3/2}}+\frac {b^2 \log (x)}{1-a^2}-\frac {i a b^2 \sin ^{-1}(a+b x) \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )}{\left (a^2-1\right )^{3/2}}+\frac {i a b^2 \sin ^{-1}(a+b x) \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{\sqrt {a^2-1}+a}\right )}{\left (a^2-1\right )^{3/2}}-\frac {b \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{\left (1-a^2\right ) x}-\frac {\sin ^{-1}(a+b x)^2}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a + b*x]^2/x^3,x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

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 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 3323

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c + d*x)^m*E
^(I*(e + f*x)))/(I*b + 2*a*E^(I*(e + f*x)) - I*b*E^(2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3324

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(c + d*x)^m*Cos[
e + f*x])/(f*(a^2 - b^2)*(a + b*Sin[e + f*x])), x] + (Dist[a/(a^2 - b^2), Int[(c + d*x)^m/(a + b*Sin[e + f*x])
, x], x] - Dist[(b*d*m)/(f*(a^2 - b^2)), Int[((c + d*x)^(m - 1)*Cos[e + f*x])/(a + b*Sin[e + f*x]), x], x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 4743

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

Rule 4773

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

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 314, normalized size = 1.15 \[ \frac {-2 a b^2 x^2 \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(a+b x)}}{\sqrt {a^2-1}-a}\right )+2 a b^2 x^2 \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {a^2-1}}\right )-2 \sqrt {a^2-1} b^2 x^2 \log (x)-2 i a b^2 x^2 \sin ^{-1}(a+b x) \log \left (\frac {-\sqrt {a^2-1}+i e^{i \sin ^{-1}(a+b x)}+a}{a-\sqrt {a^2-1}}\right )+2 i a b^2 x^2 \sin ^{-1}(a+b x) \log \left (\frac {\sqrt {a^2-1}+i e^{i \sin ^{-1}(a+b x)}+a}{\sqrt {a^2-1}+a}\right )+2 \sqrt {a^2-1} b x \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)-a^2 \sqrt {a^2-1} \sin ^{-1}(a+b x)^2+\sqrt {a^2-1} \sin ^{-1}(a+b x)^2}{2 \left (a^2-1\right )^{3/2} x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]^2/x^3,x]

[Out]

(2*Sqrt[-1 + a^2]*b*x*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x] + Sqrt[-1 + a^2]*ArcSin[a + b*x]^2 - a^2*Sqrt[-1 +
a^2]*ArcSin[a + b*x]^2 - (2*I)*a*b^2*x^2*ArcSin[a + b*x]*Log[(a - Sqrt[-1 + a^2] + I*E^(I*ArcSin[a + b*x]))/(
a - Sqrt[-1 + a^2])] + (2*I)*a*b^2*x^2*ArcSin[a + b*x]*Log[(a + Sqrt[-1 + a^2] + I*E^(I*ArcSin[a + b*x]))/(a +
Sqrt[-1 + a^2])] - 2*Sqrt[-1 + a^2]*b^2*x^2*Log[x] - 2*a*b^2*x^2*PolyLog[2, (I*E^(I*ArcSin[a + b*x]))/(-a + S
qrt[-1 + a^2])] + 2*a*b^2*x^2*PolyLog[2, ((-I)*E^(I*ArcSin[a + b*x]))/(a + Sqrt[-1 + a^2])])/(2*(-1 + a^2)^(3/
2)*x^2)

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arcsin \left (b x + a\right )^{2}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsin(b*x + a)^2/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsin(b*x + a)^2/x^3, x)

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maple [A]  time = 0.98, size = 526, normalized size = 1.93 \[ -\frac {i b^{2} \arcsin \left (b x +a \right )}{a^{2}-1}+\frac {b \arcsin \left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}}{\left (a^{2}-1\right ) x}-\frac {\arcsin \left (b x +a \right )^{2} a^{2}}{2 \left (a^{2}-1\right ) x^{2}}+\frac {\arcsin \left (b x +a \right )^{2}}{2 \left (a^{2}-1\right ) x^{2}}+\frac {2 b^{2} \ln \left (i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )}{a^{2}-1}-\frac {b^{2} \ln \left (2 i a \left (i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )-\left (i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )^{2}+1\right )}{a^{2}-1}+\frac {b^{2} \sqrt {-a^{2}+1}\, a \arcsin \left (b x +a \right ) \ln \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )}{\left (a^{2}-1\right )^{2}}-\frac {b^{2} \sqrt {-a^{2}+1}\, a \arcsin \left (b x +a \right ) \ln \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right )}{\left (a^{2}-1\right )^{2}}-\frac {i b^{2} \sqrt {-a^{2}+1}\, \dilog \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right ) a}{\left (a^{2}-1\right )^{2}}+\frac {i b^{2} \sqrt {-a^{2}+1}\, \dilog \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right ) a}{\left (a^{2}-1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a-1>0)', see `assume?` for mor
e details)Is a-1 positive, negative or zero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a + b*x)^2/x^3,x)

[Out]

int(asin(a + b*x)^2/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asin(a + b*x)**2/x**3, x)

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