\(\int \arcsin (\frac {c}{a+b x}) \, dx\) [469]

Optimal result

Integrand size = 10, antiderivative size = 47 \[ \int \arcsin \left (\frac {c}{a+b x}\right ) \, dx=\frac {(a+b x) \csc ^{-1}\left (\frac {a}{c}+\frac {b x}{c}\right )}{b}+\frac {c \text {arctanh}\left (\sqrt {1-\frac {c^2}{(a+b x)^2}}\right )}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4916, 5359, 379, 272, 65, 212} \[ \int \arcsin \left (\frac {c}{a+b x}\right ) \, dx=\frac {c \text {arctanh}\left (\sqrt {1-\frac {c^2}{(a+b x)^2}}\right )}{b}+\frac {(a+b x) \csc ^{-1}\left (\frac {a}{c}+\frac {b x}{c}\right )}{b} \]

[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 65

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 212

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

Rule 272

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 379

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 4916

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 5359

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]

Rubi steps \begin{align*} \text {integral}& = \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 \text {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 \text {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 \text {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 \text {arctanh}\left (\sqrt {1-\frac {c^2}{(a+b x)^2}}\right )}{b} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(529\) vs. \(2(47)=94\).

Time = 2.78 (sec) , antiderivative size = 529, normalized size of antiderivative = 11.26 \[ \int \arcsin \left (\frac {c}{a+b x}\right ) \, dx=x \arcsin \left (\frac {c}{a+b x}\right )-\frac {(a+b x) \sqrt {\frac {a^2-c^2+2 a b x+b^2 x^2}{(a+b x)^2}} \left (\left (-c+\sqrt {-a^2+c^2}\right ) \sqrt {-a^2+2 c \left (c+\sqrt {-a^2+c^2}\right )} \arctan \left (\frac {b \sqrt {-a^2+2 c \left (c+\sqrt {-a^2+c^2}\right )} x}{a \left (\sqrt {a^2-c^2}-\sqrt {a^2-c^2+2 a b x+b^2 x^2}\right )}\right )+\left (c+\sqrt {-a^2+c^2}\right ) \sqrt {a^2+2 c \left (-c+\sqrt {-a^2+c^2}\right )} \text {arctanh}\left (\frac {b \sqrt {a^2-2 c^2+2 c \sqrt {-a^2+c^2}} x}{a \sqrt {a^2-c^2}-a \sqrt {a^2-c^2+2 a b x+b^2 x^2}}\right )+a \left (a \arctan \left (\frac {b^2 c \sqrt {a^2-c^2} x^2}{a^4+a^3 b x+b^2 c^2 x^2-a^2 \left (c^2+\sqrt {a^2-c^2} \sqrt {a^2-c^2+2 a b x+b^2 x^2}\right )}\right )+c \left (-\log \left (\sqrt {a^2-c^2}-b x-\sqrt {a^2-c^2+2 a b x+b^2 x^2}\right )+\log \left (b^2 \left (\sqrt {a^2-c^2}+b x-\sqrt {a^2-c^2+2 a b x+b^2 x^2}\right )\right )\right )\right )\right )}{a b \sqrt {a^2-c^2+2 a b x+b^2 x^2}} \]

[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]*((-c + Sqrt[-a^2 + c^2])*
Sqrt[-a^2 + 2*c*(c + Sqrt[-a^2 + c^2])]*ArcTan[(b*Sqrt[-a^2 + 2*c*(c + Sqrt[-a^2 + c^2])]*x)/(a*(Sqrt[a^2 - c^
2] - Sqrt[a^2 - c^2 + 2*a*b*x + b^2*x^2]))] + (c + Sqrt[-a^2 + c^2])*Sqrt[a^2 + 2*c*(-c + Sqrt[-a^2 + c^2])]*A
rcTanh[(b*Sqrt[a^2 - 2*c^2 + 2*c*Sqrt[-a^2 + c^2]]*x)/(a*Sqrt[a^2 - c^2] - a*Sqrt[a^2 - c^2 + 2*a*b*x + b^2*x^
2])] + a*(a*ArcTan[(b^2*c*Sqrt[a^2 - c^2]*x^2)/(a^4 + a^3*b*x + b^2*c^2*x^2 - a^2*(c^2 + Sqrt[a^2 - c^2]*Sqrt[
a^2 - c^2 + 2*a*b*x + b^2*x^2]))] + c*(-Log[Sqrt[a^2 - c^2] - b*x - Sqrt[a^2 - c^2 + 2*a*b*x + b^2*x^2]] + Log
[b^2*(Sqrt[a^2 - c^2] + b*x - Sqrt[a^2 - c^2 + 2*a*b*x + b^2*x^2])]))))/(a*b*Sqrt[a^2 - c^2 + 2*a*b*x + b^2*x^
2])

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(-\frac {c \left (-\frac {\left (b x +a \right ) \arcsin \left (\frac {c}{b x +a}\right )}{c}-\operatorname {arctanh}\left (\frac {1}{\sqrt {1-\frac {c^{2}}{\left (b x +a \right )^{2}}}}\right )\right )}{b}\)	\(47\)
default	\(-\frac {c \left (-\frac {\left (b x +a \right ) \arcsin \left (\frac {c}{b x +a}\right )}{c}-\operatorname {arctanh}\left (\frac {1}{\sqrt {1-\frac {c^{2}}{\left (b x +a \right )^{2}}}}\right )\right )}{b}\)	\(47\)
parts	\(x \arcsin \left (\frac {c}{b x +a}\right )+\frac {c \sqrt {b^{2} x^{2}+2 a b x +a^{2}-c^{2}}\, \left (a \ln \left (\frac {2 \left (\sqrt {-c^{2}}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}-c^{2}}-c^{2}\right ) b}{b x +a}\right ) \sqrt {b^{2}}+\ln \left (\frac {b^{2} x +\sqrt {b^{2} x^{2}+2 a b x +a^{2}-c^{2}}\, \sqrt {b^{2}}+a b}{\sqrt {b^{2}}}\right ) b \sqrt {-c^{2}}\right )}{b \sqrt {\frac {b^{2} x^{2}+2 a b x +a^{2}-c^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) \sqrt {b^{2}}\, \sqrt {-c^{2}}}\)	\(202\)

[In]

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

[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)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (45) = 90\).

Time = 0.26 (sec) , antiderivative size = 141, normalized size of antiderivative = 3.00 \[ \int \arcsin \left (\frac {c}{a+b x}\right ) \, dx=\frac {b x \arcsin \left (\frac {c}{b x + a}\right ) - 2 \, a \arctan \left (-\frac {b x - {\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + a}{c}\right ) - c \log \left (-b x + {\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - a\right )}{b} \]

[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

Sympy [F]

\[ \int \arcsin \left (\frac {c}{a+b x}\right ) \, dx=\int \operatorname {asin}{\left (\frac {c}{a + b x} \right )}\, dx \]

[In]

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

[Out]

Integral(asin(c/(a + b*x)), x)

Maxima [F]

\[ \int \arcsin \left (\frac {c}{a+b x}\right ) \, dx=\int { \arcsin \left (\frac {c}{b x + a}\right ) \,d x } \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (45) = 90\).

Time = 0.32 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.02 \[ \int \arcsin \left (\frac {c}{a+b x}\right ) \, dx=\frac {b {\left (\frac {c^{2} {\left (\log \left (\sqrt {-\frac {c^{2}}{{\left (b x + a\right )}^{2}} + 1} + 1\right ) - \log \left (-\sqrt {-\frac {c^{2}}{{\left (b x + a\right )}^{2}} + 1} + 1\right )\right )}}{b^{2}} + \frac {2 \, {\left (b x + a\right )} c \arcsin \left (-\frac {c}{{\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a}\right )}{b^{2}}\right )}}{2 \, c} \]

[In]

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

[Out]

1/2*b*(c^2*(log(sqrt(-c^2/(b*x + a)^2 + 1) + 1) - log(-sqrt(-c^2/(b*x + a)^2 + 1) + 1))/b^2 + 2*(b*x + a)*c*ar
csin(-c/((b*x + a)*(a/(b*x + a) - 1) - a))/b^2)/c

Mupad [B] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89 \[ \int \arcsin \left (\frac {c}{a+b x}\right ) \, dx=\frac {c\,\mathrm {atanh}\left (\frac {1}{\sqrt {1-\frac {c^2}{{\left (a+b\,x\right )}^2}}}\right )}{b}+\frac {\mathrm {asin}\left (\frac {c}{a+b\,x}\right )\,\left (a+b\,x\right )}{b} \]

[In]

int(asin(c/(a + b*x)),x)

[Out]

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