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

Optimal. Leaf size=46 \[ \frac {\sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{2 b \sqrt {c-c (a+b x)^2}} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {247, 217, 203, 4643, 4641} \[ \frac {\sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{2 b \sqrt {c-c (a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 247

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rule 4641

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

Rule 4643

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

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 46, normalized size = 1.00 \[ \frac {\sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{2 b \sqrt {-c \left ((a+b x)^2-1\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-b^{2} c x^{2} - 2 \, a b c x - {\left (a^{2} - 1\right )} c} \arcsin \left (b x + a\right )}{b^{2} c x^{2} + 2 \, a b c x + {\left (a^{2} - 1\right )} c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 80, normalized size = 1.74 \[ -\frac {\sqrt {-c \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \arcsin \left (b x +a \right )^{2}}{2 b \left (b^{2} x^{2}+2 a b x +a^{2}-1\right ) c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.44, size = 206, normalized size = 4.48 \[ \frac {\sqrt {c} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )^{2}}{2 \, \sqrt {a^{2} b^{2} c^{2} - {\left (a^{2} c - c\right )} b^{2} c}} - \frac {\arcsin \left (b x + a\right ) \arcsin \left (-\frac {b^{2} c x + a b c}{\sqrt {a^{2} b^{2} c^{2} - {\left (a^{2} c - c\right )} b^{2} c}}\right )}{b \sqrt {c}} - \frac {\arcsin \left (-\frac {b^{2} c x + a b c}{\sqrt {a^{2} b^{2} c^{2} - {\left (a^{2} c - c\right )} b^{2} c}}\right ) \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(c)*arcsin(-(b^2*x + a*b)/sqrt(a^2*b^2 - (a^2 - 1)*b^2))^2/sqrt(a^2*b^2*c^2 - (a^2*c - c)*b^2*c) - arc
sin(b*x + a)*arcsin(-(b^2*c*x + a*b*c)/sqrt(a^2*b^2*c^2 - (a^2*c - c)*b^2*c))/(b*sqrt(c)) - arcsin(-(b^2*c*x +
a*b*c)/sqrt(a^2*b^2*c^2 - (a^2*c - c)*b^2*c))*arcsin(-(b^2*x + a*b)/sqrt(a^2*b^2 - (a^2 - 1)*b^2))/(b*sqrt(c)
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asin(a + b*x)/sqrt(-c*(a + b*x - 1)*(a + b*x + 1)), x)

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