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

Optimal. Leaf size=47 \[ \frac{c \tanh ^{-1}\left (\sqrt{1-\frac{c^2}{(a+b x)^2}}\right )}{b}+\frac{(a+b x) \csc ^{-1}\left (\frac{a}{c}+\frac{b x}{c}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.032551, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4832, 5251, 372, 266, 63, 206} \[ \frac{c \tanh ^{-1}\left (\sqrt{1-\frac{c^2}{(a+b x)^2}}\right )}{b}+\frac{(a+b x) \csc ^{-1}\left (\frac{a}{c}+\frac{b x}{c}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[c/(a + b*x)],x]

[Out]

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

Rule 4832

Int[ArcSin[(c_.)/((a_.) + (b_.)*(x_)^(n_.))]^(m_.)*(u_.), x_Symbol] :> Int[u*ArcCsc[a/c + (b*x^n)/c]^m, x] /;
FreeQ[{a, b, c, n, m}, x]

Rule 5251

Int[ArcCsc[(c_) + (d_.)*(x_)], x_Symbol] :> Simp[((c + d*x)*ArcCsc[c + d*x])/d, x] + Int[1/((c + d*x)*Sqrt[1 -
1/(c + d*x)^2]), x] /; FreeQ[{c, d}, x]

Rule 372

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

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

Rubi steps

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

Mathematica [B]  time = 0.137934, size = 140, normalized size = 2.98 \[ \frac{(a+b x) \sqrt{\frac{a^2+2 a b x+b^2 x^2-c^2}{(a+b x)^2}} \left (c \tanh ^{-1}\left (\frac{a+b x}{\sqrt{a^2+2 a b x+b^2 x^2-c^2}}\right )-a \tan ^{-1}\left (\frac{\sqrt{(a+b x)^2-c^2}}{c}\right )\right )}{b \sqrt{a^2+2 a b x+b^2 x^2-c^2}}+x \sin ^{-1}\left (\frac{c}{a+b x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[c/(a + b*x)],x]

[Out]

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

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Maple [A]  time = 0.017, size = 47, normalized size = 1. \begin{align*} -{\frac{c}{b} \left ( -{\frac{bx+a}{c}\arcsin \left ({\frac{c}{bx+a}} \right ) }-{\it Artanh} \left ({\frac{1}{\sqrt{1-{\frac{{c}^{2}}{ \left ( bx+a \right ) ^{2}}}}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(c/(b*x+a)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} x \arctan \left (c, \sqrt{b x + a + c} \sqrt{b x + a - c}\right ) + \int \frac{{\left (b^{2} c x^{2} + a b c x\right )} e^{\left (\frac{1}{2} \, \log \left (b x + a + c\right ) + \frac{1}{2} \, \log \left (b x + a - c\right )\right )}}{b^{2} c^{2} x^{2} + 2 \, a b c^{2} x + a^{2} c^{2} - c^{4} +{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - c^{2}\right )} e^{\left (\log \left (b x + a + c\right ) + \log \left (b x + a - c\right )\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctan2(c, sqrt(b*x + a + c)*sqrt(b*x + a - c)) + integrate((b^2*c*x^2 + a*b*c*x)*e^(1/2*log(b*x + a + c) +
1/2*log(b*x + a - c))/(b^2*c^2*x^2 + 2*a*b*c^2*x + a^2*c^2 - c^4 + (b^2*x^2 + 2*a*b*x + a^2 - c^2)*e^(log(b*x
+ a + c) + log(b*x + a - c))), x)

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Fricas [B]  time = 2.52404, size = 305, normalized size = 6.49 \begin{align*} \frac{b x \arcsin \left (\frac{c}{b x + a}\right ) - 2 \, a \arctan \left (-\frac{b x -{\left (b x + a\right )} \sqrt{\frac{b^{2} x^{2} + 2 \, a b x + a^{2} - c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + a}{c}\right ) - c \log \left (-b x +{\left (b x + a\right )} \sqrt{\frac{b^{2} x^{2} + 2 \, a b x + a^{2} - c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{asin}{\left (\frac{c}{a + b x} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asin(c/(a + b*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \arcsin \left (\frac{c}{b x + a}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsin(c/(b*x + a)), x)