3.5.40 \(\int \frac {1}{\sqrt {1+c^2 x^2} (a+b \sinh ^{-1}(c x))^2} \, dx\) [440]

Optimal. Leaf size=18 \[ -\frac {1}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

[Out]

-1/b/c/(a+b*arcsinh(c*x))

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Rubi [A]
time = 0.03, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {5783} \begin {gather*} -\frac {1}{b c \left (a+b \sinh ^{-1}(c x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(b*c*(a + b*ArcSinh[c*x])))

Rule 5783

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

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 18, normalized size = 1.00 \begin {gather*} -\frac {1}{b c \left (a+b \sinh ^{-1}(c x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(b*c*(a + b*ArcSinh[c*x])))

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Maple [A]
time = 0.28, size = 19, normalized size = 1.06

method	result	size
derivativedivides	\(-\frac {1}{b c \left (a +b \arcsinh \left (c x \right )\right )}\)	\(19\)
default	\(-\frac {1}{b c \left (a +b \arcsinh \left (c x \right )\right )}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b/c/(a+b*arcsinh(c*x))

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Maxima [A]
time = 0.26, size = 18, normalized size = 1.00 \begin {gather*} -\frac {1}{{\left (b \operatorname {arsinh}\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*arcsinh(c*x))^2/(c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-1/((b*arcsinh(c*x) + a)*b*c)

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Fricas [A]
time = 0.38, size = 30, normalized size = 1.67 \begin {gather*} -\frac {1}{b^{2} c \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + a b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (14) = 28\).
time = 1.79, size = 36, normalized size = 2.00 \begin {gather*} \begin {cases} \frac {x}{a^{2}} & \text {for}\: b = 0 \wedge c = 0 \\\frac {\operatorname {asinh}{\left (c x \right )}}{a^{2} c} & \text {for}\: b = 0 \\\frac {x}{a^{2}} & \text {for}\: c = 0 \\- \frac {1}{a b c + b^{2} c \operatorname {asinh}{\left (c x \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/a**2, Eq(b, 0) & Eq(c, 0)), (asinh(c*x)/(a**2*c), Eq(b, 0)), (x/a**2, Eq(c, 0)), (-1/(a*b*c + b**
2*c*asinh(c*x)), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(c^2*x^2 + 1)*(b*arcsinh(c*x) + a)^2), x)

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Mupad [B]
time = 0.14, size = 18, normalized size = 1.00 \begin {gather*} -\frac {1}{c\,\mathrm {asinh}\left (c\,x\right )\,b^2+a\,c\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(b^2*c*asinh(c*x) + a*b*c)

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