3.1.40 \(\int \frac {x (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^2} \, dx\) [40]

Optimal. Leaf size=55 \[ \frac {b x}{2 c d^2 \sqrt {1+c^2 x^2}}-\frac {a+b \sinh ^{-1}(c x)}{2 c^2 d^2 \left (1+c^2 x^2\right )} \]

[Out]

1/2*(-a-b*arcsinh(c*x))/c^2/d^2/(c^2*x^2+1)+1/2*b*x/c/d^2/(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {5798, 197} \begin {gather*} \frac {b x}{2 c d^2 \sqrt {c^2 x^2+1}}-\frac {a+b \sinh ^{-1}(c x)}{2 c^2 d^2 \left (c^2 x^2+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 5798

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

Rubi steps

\begin {align*} \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^2} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac {b \int \frac {1}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{2 c d^2}\\ &=\frac {b x}{2 c d^2 \sqrt {1+c^2 x^2}}-\frac {a+b \sinh ^{-1}(c x)}{2 c^2 d^2 \left (1+c^2 x^2\right )}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 74, normalized size = 1.35 \begin {gather*} -\frac {a}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac {b x}{2 c d^2 \sqrt {1+c^2 x^2}}-\frac {b \sinh ^{-1}(c x)}{2 c^2 d^2 \left (1+c^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.55, size = 61, normalized size = 1.11

method	result	size
derivativedivides	\(\frac {-\frac {a}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {b \left (-\frac {\arcsinh \left (c x \right )}{2 \left (c^{2} x^{2}+1\right )}+\frac {c x}{2 \sqrt {c^{2} x^{2}+1}}\right )}{d^{2}}}{c^{2}}\)	\(61\)
default	\(\frac {-\frac {a}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {b \left (-\frac {\arcsinh \left (c x \right )}{2 \left (c^{2} x^{2}+1\right )}+\frac {c x}{2 \sqrt {c^{2} x^{2}+1}}\right )}{d^{2}}}{c^{2}}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(-1/2*a/d^2/(c^2*x^2+1)+b/d^2*(-1/2/(c^2*x^2+1)*arcsinh(c*x)+1/2*c*x/(c^2*x^2+1)^(1/2)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*((2*log(c*x + sqrt(c^2*x^2 + 1)) + 1)/(c^4*d^2*x^2 + c^2*d^2) - 4*integrate(1/2/(c^6*d^2*x^5 + 2*c^4*d^
2*x^3 + c^2*d^2*x + (c^5*d^2*x^4 + 2*c^3*d^2*x^2 + c*d^2)*sqrt(c^2*x^2 + 1)), x)) - 1/2*a/(c^4*d^2*x^2 + c^2*d
^2)

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Fricas [A]
time = 0.35, size = 65, normalized size = 1.18 \begin {gather*} \frac {a c^{2} x^{2} + \sqrt {c^{2} x^{2} + 1} b c x - b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{2 \, {\left (c^{4} d^{2} x^{2} + c^{2} d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*c^2*x^2 + sqrt(c^2*x^2 + 1)*b*c*x - b*log(c*x + sqrt(c^2*x^2 + 1)))/(c^4*d^2*x^2 + c^2*d^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a x}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac {b x \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*x/(c**4*x**4 + 2*c**2*x**2 + 1), x) + Integral(b*x*asinh(c*x)/(c**4*x**4 + 2*c**2*x**2 + 1), x))/d
**2

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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