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

Optimal. Leaf size=31 \[ a x+b c \tanh ^{-1}\left (\sqrt {1-\frac {c^2}{x^2}}\right )+b x \csc ^{-1}\left (\frac {x}{c}\right ) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4832, 5215, 266, 63, 208} \[ a x+b c \tanh ^{-1}\left (\sqrt {1-\frac {c^2}{x^2}}\right )+b x \csc ^{-1}\left (\frac {x}{c}\right ) \]

Antiderivative was successfully verified.

[In]

Int[a + b*ArcSin[c/x],x]

[Out]

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

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 208

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

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 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 5215

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

Rubi steps

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

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Mathematica [B]  time = 0.10, size = 89, normalized size = 2.87 \[ a x+\frac {b c \sqrt {x^2-c^2} \left (\log \left (\frac {x}{\sqrt {x^2-c^2}}+1\right )-\log \left (1-\frac {x}{\sqrt {x^2-c^2}}\right )\right )}{2 x \sqrt {1-\frac {c^2}{x^2}}}+b x \sin ^{-1}\left (\frac {c}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[a + b*ArcSin[c/x],x]

[Out]

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

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fricas [B]  time = 0.81, size = 75, normalized size = 2.42 \[ -b c \log \left (x \sqrt {-\frac {c^{2} - x^{2}}{x^{2}}} - x\right ) + a x + {\left (b x - b\right )} \arcsin \left (\frac {c}{x}\right ) - 2 \, b \arctan \left (\frac {x \sqrt {-\frac {c^{2} - x^{2}}{x^{2}}} - x}{c}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.71, size = 60, normalized size = 1.94 \[ a x + \frac {{\left (c^{2} {\left (\log \left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right ) - \log \left (-\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )\right )} + 2 \, c x \arcsin \left (\frac {c}{x}\right )\right )} b}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 37, normalized size = 1.19 \[ a x -b c \left (-\frac {\arcsin \left (\frac {c}{x}\right ) x}{c}-\arctanh \left (\frac {1}{\sqrt {1-\frac {c^{2}}{x^{2}}}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.93, size = 52, normalized size = 1.68 \[ \frac {1}{2} \, {\left (c {\left (\log \left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right ) - \log \left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} - 1\right )\right )} + 2 \, x \arcsin \left (\frac {c}{x}\right )\right )} b + a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.74, size = 32, normalized size = 1.03 \[ a\,x+b\,x\,\mathrm {asin}\left (\frac {c}{x}\right )+b\,c\,\mathrm {sign}\relax (x)\,\ln \left (x+\sqrt {x^2-c^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a + b*asin(c/x),x)

[Out]

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

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sympy [A]  time = 1.70, size = 32, normalized size = 1.03 \[ a x + b \left (c \left (\begin {cases} \operatorname {acosh}{\left (\frac {x}{c} \right )} & \text {for}\: \left |{\frac {x^{2}}{c^{2}}}\right | > 1 \\- i \operatorname {asin}{\left (\frac {x}{c} \right )} & \text {otherwise} \end {cases}\right ) + x \operatorname {asin}{\left (\frac {c}{x} \right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x + b*(c*Piecewise((acosh(x/c), Abs(x**2/c**2) > 1), (-I*asin(x/c), True)) + x*asin(c/x))

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