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

Optimal. Leaf size=230 \[ \frac {2 i b \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a-\sqrt {1-a^2}}\right )}{\sqrt {1-a^2}}-\frac {2 i b \text {Li}_2\left (\frac {e^{i \sin ^{-1}(a+b x)}}{i a+\sqrt {1-a^2}}\right )}{\sqrt {1-a^2}}-\frac {2 b \sin ^{-1}(a+b x) \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{-\sqrt {1-a^2}+i a}\right )}{\sqrt {1-a^2}}+\frac {2 b \sin ^{-1}(a+b x) \log \left (1-\frac {e^{i \sin ^{-1}(a+b x)}}{\sqrt {1-a^2}+i a}\right )}{\sqrt {1-a^2}}-\frac {\sin ^{-1}(a+b x)^2}{x} \]

[Out]

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

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Rubi [A]  time = 0.45, antiderivative size = 208, normalized size of antiderivative = 0.90, number of steps used = 11, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {4805, 4743, 4773, 3323, 2264, 2190, 2279, 2391} \[ \frac {2 b \text {PolyLog}\left (2,-\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )}{\sqrt {a^2-1}}-\frac {2 b \text {PolyLog}\left (2,-\frac {i e^{i \sin ^{-1}(a+b x)}}{\sqrt {a^2-1}+a}\right )}{\sqrt {a^2-1}}+\frac {2 i b \sin ^{-1}(a+b x) \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )}{\sqrt {a^2-1}}-\frac {2 i b \sin ^{-1}(a+b x) \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{\sqrt {a^2-1}+a}\right )}{\sqrt {a^2-1}}-\frac {\sin ^{-1}(a+b x)^2}{x} \]

Warning: Unable to verify antiderivative.

[In]

Int[ArcSin[a + b*x]^2/x^2,x]

[Out]

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

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 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 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^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sin ^{-1}(x)^2}{\left (-\frac {a}{b}+\frac {x}{b}\right )^2} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\sin ^{-1}(a+b x)^2}{x}+2 \operatorname {Subst}\left (\int \frac {\sin ^{-1}(x)}{\left (-\frac {a}{b}+\frac {x}{b}\right ) \sqrt {1-x^2}} \, dx,x,a+b x\right )\\ &=-\frac {\sin ^{-1}(a+b x)^2}{x}+2 \operatorname {Subst}\left (\int \frac {x}{-\frac {a}{b}+\frac {\sin (x)}{b}} \, dx,x,\sin ^{-1}(a+b x)\right )\\ &=-\frac {\sin ^{-1}(a+b x)^2}{x}+4 \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 )\\ &=-\frac {\sin ^{-1}(a+b x)^2}{x}-\frac {(4 i) \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 )}{\sqrt {-1+a^2}}+\frac {(4 i) \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 )}{\sqrt {-1+a^2}}\\ &=-\frac {\sin ^{-1}(a+b x)^2}{x}+\frac {2 i b \sin ^{-1}(a+b x) \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}-\frac {2 i b \sin ^{-1}(a+b x) \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}+\frac {(2 i b) \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 )}{\sqrt {-1+a^2}}-\frac {(2 i b) \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 )}{\sqrt {-1+a^2}}\\ &=-\frac {\sin ^{-1}(a+b x)^2}{x}+\frac {2 i b \sin ^{-1}(a+b x) \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}-\frac {2 i b \sin ^{-1}(a+b x) \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}+\frac {(2 b) \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 )}{\sqrt {-1+a^2}}-\frac {(2 b) \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 )}{\sqrt {-1+a^2}}\\ &=-\frac {\sin ^{-1}(a+b x)^2}{x}+\frac {2 i b \sin ^{-1}(a+b x) \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}-\frac {2 i b \sin ^{-1}(a+b x) \log \left (1+\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}+\frac {2 b \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}-\frac {2 b \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )}{\sqrt {-1+a^2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 208, normalized size = 0.90 \[ \frac {2 b x \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(a+b x)}}{\sqrt {a^2-1}-a}\right )-2 b x \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(a+b x)}}{a+\sqrt {a^2-1}}\right )-\sqrt {a^2-1} \sin ^{-1}(a+b x)^2+2 i b x \sin ^{-1}(a+b x) \left (\log \left (\frac {-\sqrt {a^2-1}+i e^{i \sin ^{-1}(a+b x)}+a}{a-\sqrt {a^2-1}}\right )-\log \left (\frac {\sqrt {a^2-1}+i e^{i \sin ^{-1}(a+b x)}+a}{\sqrt {a^2-1}+a}\right )\right )}{\sqrt {a^2-1} x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSin[a + b*x]^2/x^2,x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsin(b*x + a)^2/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsin(b*x + a)^2/x^2, x)

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maple [A]  time = 0.65, size = 333, normalized size = 1.45 \[ -\frac {\arcsin \left (b x +a \right )^{2}}{x}-\frac {2 b \sqrt {-a^{2}+1}\, \arcsin \left (b x +a \right ) \ln \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}+\frac {2 b \sqrt {-a^{2}+1}\, \arcsin \left (b x +a \right ) \ln \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right )}{a^{2}-1}+\frac {2 i b \sqrt {-a^{2}+1}\, \dilog \left (\frac {i a +\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a +\sqrt {-a^{2}+1}}\right )}{a^{2}-1}-\frac {2 i b \sqrt {-a^{2}+1}\, \dilog \left (\frac {i a -\sqrt {-a^{2}+1}-i \left (b x +a \right )-\sqrt {1-\left (b x +a \right )^{2}}}{i a -\sqrt {-a^{2}+1}}\right )}{a^{2}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^2,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 \[ \int \frac {{\mathrm {asin}\left (a+b\,x\right )}^2}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a + b*x)^2/x^2,x)

[Out]

int(asin(a + b*x)^2/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asin(a + b*x)**2/x**2, x)

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