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

Optimal. Leaf size=97 \[ -\frac {i \text {Li}_2\left (-e^{2 i \sin ^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \sin ^{-1}(a+b x)^2}{b \sqrt {1-(a+b x)^2}}-\frac {i \sin ^{-1}(a+b x)^2}{b}+\frac {2 \sin ^{-1}(a+b x) \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4807, 4651, 4675, 3719, 2190, 2279, 2391} \[ -\frac {i \text {PolyLog}\left (2,-e^{2 i \sin ^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \sin ^{-1}(a+b x)^2}{b \sqrt {1-(a+b x)^2}}-\frac {i \sin ^{-1}(a+b x)^2}{b}+\frac {2 \sin ^{-1}(a+b x) \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a + b*x]^2/(1 - a^2 - 2*a*b*x - b^2*x^2)^(3/2),x]

[Out]

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

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 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 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4651

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSin[c
*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n)/Sqrt[d], Int[(x*(a + b*ArcSin[c*x])^(n - 1))/(d + e*x^2), x], x
] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[d, 0]

Rule 4675

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

Rule 4807

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> Di
st[1/d, Subst[Int[(-(C/d^2) + (C*x^2)/d^2)^p*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A,
B, C, n, p}, x] && EqQ[B*(1 - c^2) + 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rubi steps

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

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Mathematica [A]  time = 0.39, size = 114, normalized size = 1.18 \[ \frac {\sin ^{-1}(a+b x) \left (\frac {\left (-i \sqrt {-a^2-2 a b x-b^2 x^2+1}+a+b x\right ) \sin ^{-1}(a+b x)}{\sqrt {-a^2-2 a b x-b^2 x^2+1}}+2 \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )\right )-i \text {Li}_2\left (-e^{2 i \sin ^{-1}(a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]^2/(1 - a^2 - 2*a*b*x - b^2*x^2)^(3/2),x]

[Out]

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

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} \arcsin \left (b x + a\right )^{2}}{b^{4} x^{4} + 4 \, a b^{3} x^{3} + 2 \, {\left (3 \, a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)^2/(-b^2*x^2-2*a*b*x-a^2+1)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*arcsin(b*x + a)^2/(b^4*x^4 + 4*a*b^3*x^3 + 2*(3*a^2 - 1)*b^2*x^2 +
a^4 + 4*(a^3 - a)*b*x - 2*a^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)^2/(-b^2*x^2-2*a*b*x-a^2+1)^(3/2),x, algorithm="giac")

[Out]

integrate(arcsin(b*x + a)^2/(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2), x)

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maple [A]  time = 0.31, size = 194, normalized size = 2.00 \[ \frac {\left (-\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, x b +i x^{2} b^{2}-\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a +2 i x a b +i a^{2}-i\right ) \arcsin \left (b x +a \right )^{2}}{b \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}+\frac {2 \arcsin \left (b x +a \right ) \ln \left (1+\left (i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )^{2}\right )}{b}-\frac {2 i \arcsin \left (b x +a \right )^{2}}{b}-\frac {i \polylog \left (2, -\left (i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )^{2}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(b*x+a)^2/(-b^2*x^2-2*a*b*x-a^2+1)^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)^2/(-b^2*x^2-2*a*b*x-a^2+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(arcsin(b*x + a)^2/(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a + b*x)^2/(1 - b^2*x^2 - 2*a*b*x - a^2)^(3/2),x)

[Out]

int(asin(a + b*x)^2/(1 - b^2*x^2 - 2*a*b*x - a^2)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(b*x+a)**2/(-b**2*x**2-2*a*b*x-a**2+1)**(3/2),x)

[Out]

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

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