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

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 = 0.04, 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 = {5675} \[ -\frac {1}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

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 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && 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 \[ -\frac {1}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

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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fricas [A]  time = 0.54, size = 30, normalized size = 1.67 \[ -\frac {1}{b^{2} c \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + a b c} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c^{2} x^{2} + 1} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}\,{d x} \]

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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maple [A]  time = 0.01, size = 19, normalized size = 1.06 \[ -\frac {1}{b c \left (a +b \arcsinh \left (c x \right )\right )} \]

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)

[Out]

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

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maxima [A]  time = 0.36, size = 18, normalized size = 1.00 \[ -\frac {1}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} b c} \]

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

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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sympy [A]  time = 3.06, size = 36, normalized size = 2.00 \[ \begin {cases} \frac {x}{a^{2}} & \text {for}\: b = 0 \wedge c = 0 \\\frac {\operatorname {asinh}{\left (c x \right )}}{a^{2} c} & \text {for}\: b = 0 \\\frac {x}{a^{2}} & \text {for}\: c = 0 \\- \frac {1}{a b c + b^{2} c \operatorname {asinh}{\left (c x \right )}} & \text {otherwise} \end {cases} \]

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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