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

Optimal. Leaf size=50 \[ \frac {\log \left (1-(a+b x)^2\right )}{2 b}+\frac {(a+b x) \sin ^{-1}(a+b x)}{b \sqrt {1-(a+b x)^2}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {4807, 4651, 260} \[ \frac {\log \left (1-(a+b x)^2\right )}{2 b}+\frac {(a+b x) \sin ^{-1}(a+b x)}{b \sqrt {1-(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 260

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

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

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Mathematica [A]  time = 0.11, size = 66, normalized size = 1.32 \[ \frac {\log \left (-a^2-2 a b x-b^2 x^2+1\right )+\frac {2 (a+b x) \sin ^{-1}(a+b x)}{\sqrt {-a^2-2 a b x-b^2 x^2+1}}}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.45, size = 99, normalized size = 1.98 \[ -\frac {2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )} \arcsin \left (b x + a\right ) - {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + {\left (a^{2} - 1\right )} b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)*arcsin(b*x + a) - (b^2*x^2 + 2*a*b*x + a^2 - 1)*log(b^2*x
^2 + 2*a*b*x + a^2 - 1))/(b^3*x^2 + 2*a*b^2*x + (a^2 - 1)*b)

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giac [A]  time = 1.17, size = 83, normalized size = 1.66 \[ -\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (x + \frac {a}{b}\right )} \arcsin \left (b x + a\right )}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1} + \frac {\log \left ({\left | b x + a + 1 \right |}\right )}{2 \, b} + \frac {\log \left ({\left | b x + a - 1 \right |}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(x + a/b)*arcsin(b*x + a)/(b^2*x^2 + 2*a*b*x + a^2 - 1) + 1/2*log(abs(b*x
+ a + 1))/b + 1/2*log(abs(b*x + a - 1))/b

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maple [B]  time = 0.17, size = 155, normalized size = 3.10 \[ -\frac {-\ln \left (1-\left (b x +a \right )^{2}\right ) x^{2} b^{2}+2 \arcsin \left (b x +a \right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, x b -2 \ln \left (1-\left (b x +a \right )^{2}\right ) x a b +2 \arcsin \left (b x +a \right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a -\ln \left (1-\left (b x +a \right )^{2}\right ) a^{2}+\ln \left (1-\left (b x +a \right )^{2}\right )}{2 b \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.42, size = 160, normalized size = 3.20 \[ -\frac {1}{2} \, {\left (a {\left (\frac {\log \left (b x + a + 1\right )}{b^{2}} - \frac {\log \left (b x + a - 1\right )}{b^{2}}\right )} - \frac {{\left (a + 1\right )} \log \left (b x + a + 1\right )}{b^{2}} + \frac {{\left (a - 1\right )} \log \left (b x + a - 1\right )}{b^{2}}\right )} b + {\left (\frac {b^{2} x}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}} + \frac {a b}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}\right )} \arcsin \left (b x + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {asin}\left (a+b\,x\right )}{{\left (-a^2-2\,a\,b\,x-b^2\,x^2+1\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asin(a + b*x)/(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 \[ \int \frac {\operatorname {asin}{\left (a + b x \right )}}{\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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