3.4.54 \(\int \frac {1}{\sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))} \, dx\) [354]

Optimal. Leaf size=16 \[ \frac {\log (a+b \text {ArcSin}(c x))}{b c} \]

[Out]

ln(a+b*arcsin(c*x))/b/c

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Rubi [A]
time = 0.03, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {4735} \begin {gather*} \frac {\log (a+b \text {ArcSin}(c x))}{b c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[a + b*ArcSin[c*x]]/(b*c)

Rule 4735

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

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 16, normalized size = 1.00 \begin {gather*} \frac {\log (a+b \text {ArcSin}(c x))}{b c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[a + b*ArcSin[c*x]]/(b*c)

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Maple [A]
time = 0.09, size = 17, normalized size = 1.06

method	result	size
derivativedivides	\(\frac {\ln \left (a +b \arcsin \left (c x \right )\right )}{b c}\)	\(17\)
default	\(\frac {\ln \left (a +b \arcsin \left (c x \right )\right )}{b c}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(a+b*arcsin(c*x))/b/c

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Maxima [A]
time = 0.46, size = 16, normalized size = 1.00 \begin {gather*} \frac {\log \left (b \arcsin \left (c x\right ) + a\right )}{b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(b*arcsin(c*x) + a)/(b*c)

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Fricas [A]
time = 1.91, size = 19, normalized size = 1.19 \begin {gather*} \frac {\log \left (-b \arcsin \left (c x\right ) - a\right )}{b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-b*arcsin(c*x) - a)/(b*c)

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Sympy [C] Result contains complex when optimal does not.
time = 0.84, size = 42, normalized size = 2.62 \begin {gather*} \begin {cases} \frac {x}{a} & \text {for}\: c = 0 \wedge \left (b = 0 \vee c = 0\right ) \\\frac {\begin {cases} - \frac {i \operatorname {acosh}{\left (c x \right )}}{c} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {\operatorname {asin}{\left (c x \right )}}{c} & \text {otherwise} \end {cases}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (\frac {a}{b} + \operatorname {asin}{\left (c x \right )} \right )}}{b c} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/a, Eq(c, 0) & (Eq(b, 0) | Eq(c, 0))), (Piecewise((-I*acosh(c*x)/c, Abs(c**2*x**2) > 1), (asin(c*x
)/c, True))/a, Eq(b, 0)), (log(a/b + asin(c*x))/(b*c), True))

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Giac [A]
time = 0.44, size = 17, normalized size = 1.06 \begin {gather*} \frac {\log \left ({\left | b \arcsin \left (c x\right ) + a \right |}\right )}{b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(b*arcsin(c*x) + a))/(b*c)

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Mupad [B]
time = 0.18, size = 16, normalized size = 1.00 \begin {gather*} \frac {\ln \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{b\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a + b*asin(c*x))/(b*c)

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