3.1.49 \(\int \frac {x (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^3} \, dx\) [49]

Optimal. Leaf size=80 \[ \frac {b x}{12 c d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {b x}{6 c d^3 \sqrt {1+c^2 x^2}}-\frac {a+b \sinh ^{-1}(c x)}{4 c^2 d^3 \left (1+c^2 x^2\right )^2} \]

[Out]

1/12*b*x/c/d^3/(c^2*x^2+1)^(3/2)+1/4*(-a-b*arcsinh(c*x))/c^2/d^3/(c^2*x^2+1)^2+1/6*b*x/c/d^3/(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {5798, 198, 197} \begin {gather*} -\frac {a+b \sinh ^{-1}(c x)}{4 c^2 d^3 \left (c^2 x^2+1\right )^2}+\frac {b x}{6 c d^3 \sqrt {c^2 x^2+1}}+\frac {b x}{12 c d^3 \left (c^2 x^2+1\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^3,x]

[Out]

(b*x)/(12*c*d^3*(1 + c^2*x^2)^(3/2)) + (b*x)/(6*c*d^3*Sqrt[1 + c^2*x^2]) - (a + b*ArcSinh[c*x])/(4*c^2*d^3*(1
+ c^2*x^2)^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 198

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

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 )^3} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}+\frac {b \int \frac {1}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{4 c d^3}\\ &=\frac {b x}{12 c d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}+\frac {b \int \frac {1}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{6 c d^3}\\ &=\frac {b x}{12 c d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {b x}{6 c d^3 \sqrt {1+c^2 x^2}}-\frac {a+b \sinh ^{-1}(c x)}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 56, normalized size = 0.70 \begin {gather*} \frac {-3 a+b c x \sqrt {1+c^2 x^2} \left (3+2 c^2 x^2\right )-3 b \sinh ^{-1}(c x)}{12 d^3 \left (c+c^3 x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^3,x]

[Out]

(-3*a + b*c*x*Sqrt[1 + c^2*x^2]*(3 + 2*c^2*x^2) - 3*b*ArcSinh[c*x])/(12*d^3*(c + c^3*x^2)^2)

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Maple [A]
time = 0.53, size = 76, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {-\frac {a}{4 d^{3} \left (c^{2} x^{2}+1\right )^{2}}+\frac {b \left (-\frac {\arcsinh \left (c x \right )}{4 \left (c^{2} x^{2}+1\right )^{2}}+\frac {c x}{12 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {c x}{6 \sqrt {c^{2} x^{2}+1}}\right )}{d^{3}}}{c^{2}}\)	\(76\)
default	\(\frac {-\frac {a}{4 d^{3} \left (c^{2} x^{2}+1\right )^{2}}+\frac {b \left (-\frac {\arcsinh \left (c x \right )}{4 \left (c^{2} x^{2}+1\right )^{2}}+\frac {c x}{12 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {c x}{6 \sqrt {c^{2} x^{2}+1}}\right )}{d^{3}}}{c^{2}}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^3,x,method=_RETURNVERBOSE)

[Out]

1/c^2*(-1/4*a/d^3/(c^2*x^2+1)^2+b/d^3*(-1/4*arcsinh(c*x)/(c^2*x^2+1)^2+1/12/(c^2*x^2+1)^(3/2)*c*x+1/6*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)^3,x, algorithm="maxima")

[Out]

-1/16*b*((4*log(c*x + sqrt(c^2*x^2 + 1)) + 1)/(c^6*d^3*x^4 + 2*c^4*d^3*x^2 + c^2*d^3) - 16*integrate(1/4/(c^8*
d^3*x^7 + 3*c^6*d^3*x^5 + 3*c^4*d^3*x^3 + c^2*d^3*x + (c^7*d^3*x^6 + 3*c^5*d^3*x^4 + 3*c^3*d^3*x^2 + c*d^3)*sq
rt(c^2*x^2 + 1)), x)) - 1/4*a/(c^6*d^3*x^4 + 2*c^4*d^3*x^2 + c^2*d^3)

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Fricas [A]
time = 0.38, size = 98, normalized size = 1.22 \begin {gather*} \frac {3 \, a c^{4} x^{4} + 6 \, a c^{2} x^{2} - 3 \, b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (2 \, b c^{3} x^{3} + 3 \, b c x\right )} \sqrt {c^{2} x^{2} + 1}}{12 \, {\left (c^{6} d^{3} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\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)^3,x, algorithm="fricas")

[Out]

1/12*(3*a*c^4*x^4 + 6*a*c^2*x^2 - 3*b*log(c*x + sqrt(c^2*x^2 + 1)) + (2*b*c^3*x^3 + 3*b*c*x)*sqrt(c^2*x^2 + 1)
)/(c^6*d^3*x^4 + 2*c^4*d^3*x^2 + c^2*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a x}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx + \int \frac {b x \operatorname {asinh}{\left (c x \right )}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx}{d^{3}} \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)**3,x)

[Out]

(Integral(a*x/(c**6*x**6 + 3*c**4*x**4 + 3*c**2*x**2 + 1), x) + Integral(b*x*asinh(c*x)/(c**6*x**6 + 3*c**4*x*
*4 + 3*c**2*x**2 + 1), x))/d**3

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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)^3,x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)*x/(c^2*d*x^2 + d)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^3} \,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)^3,x)

[Out]

int((x*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^3, x)

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