3.3.37 \(\int \frac {x (a+b \sinh ^{-1}(c x))^2}{(d+c^2 d x^2)^2} \, dx\) [237]

Optimal. Leaf size=85 \[ \frac {b x \left (a+b \sinh ^{-1}(c x)\right )}{c d^2 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {b^2 \log \left (1+c^2 x^2\right )}{2 c^2 d^2} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5798, 5787, 266} \begin {gather*} \frac {b x \left (a+b \sinh ^{-1}(c x)\right )}{c d^2 \sqrt {c^2 x^2+1}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (c^2 x^2+1\right )}-\frac {b^2 \log \left (c^2 x^2+1\right )}{2 c^2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 5787

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[x*((a + b*ArcSinh
[c*x])^n/(d*Sqrt[d + e*x^2])), x] - Dist[b*c*(n/d)*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]], Int[x*((a + b*ArcS
inh[c*x])^(n - 1)/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 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 )^2}{\left (d+c^2 d x^2\right )^2} \, dx &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac {b \int \frac {a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{c d^2}\\ &=\frac {b x \left (a+b \sinh ^{-1}(c x)\right )}{c d^2 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {b^2 \int \frac {x}{1+c^2 x^2} \, dx}{d^2}\\ &=\frac {b x \left (a+b \sinh ^{-1}(c x)\right )}{c d^2 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {b^2 \log \left (1+c^2 x^2\right )}{2 c^2 d^2}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 145, normalized size = 1.71 \begin {gather*} -\frac {a^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac {a b x}{c d^2 \sqrt {1+c^2 x^2}}+\frac {b \left (-a+b c x \sqrt {1+c^2 x^2}\right ) \sinh ^{-1}(c x)}{c^2 d^2 \left (1+c^2 x^2\right )}-\frac {b^2 \sinh ^{-1}(c x)^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {b^2 \log \left (1+c^2 x^2\right )}{2 c^2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(205\) vs. \(2(79)=158\).
time = 3.42, size = 206, normalized size = 2.42

method	result	size
derivativedivides	\(\frac {-\frac {a^{2}}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {2 b^{2} \arcsinh \left (c x \right )}{d^{2}}+\frac {b^{2} \arcsinh \left (c x \right ) c x}{d^{2} \sqrt {c^{2} x^{2}+1}}-\frac {b^{2} \arcsinh \left (c x \right ) c^{2} x^{2}}{d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b^{2} \arcsinh \left (c x \right )^{2}}{2 d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b^{2} \arcsinh \left (c x \right )}{d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b^{2} \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d^{2}}+\frac {2 a b \left (-\frac {\arcsinh \left (c x \right )}{2 \left (c^{2} x^{2}+1\right )}+\frac {c x}{2 \sqrt {c^{2} x^{2}+1}}\right )}{d^{2}}}{c^{2}}\)	\(206\)
default	\(\frac {-\frac {a^{2}}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {2 b^{2} \arcsinh \left (c x \right )}{d^{2}}+\frac {b^{2} \arcsinh \left (c x \right ) c x}{d^{2} \sqrt {c^{2} x^{2}+1}}-\frac {b^{2} \arcsinh \left (c x \right ) c^{2} x^{2}}{d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b^{2} \arcsinh \left (c x \right )^{2}}{2 d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b^{2} \arcsinh \left (c x \right )}{d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b^{2} \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d^{2}}+\frac {2 a b \left (-\frac {\arcsinh \left (c x \right )}{2 \left (c^{2} x^{2}+1\right )}+\frac {c x}{2 \sqrt {c^{2} x^{2}+1}}\right )}{d^{2}}}{c^{2}}\)	\(206\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(-1/2*a^2/d^2/(c^2*x^2+1)+2*b^2/d^2*arcsinh(c*x)+b^2/d^2*arcsinh(c*x)/(c^2*x^2+1)^(1/2)*c*x-b^2/d^2*arcs
inh(c*x)/(c^2*x^2+1)*c^2*x^2-1/2*b^2/d^2*arcsinh(c*x)^2/(c^2*x^2+1)-b^2/d^2*arcsinh(c*x)/(c^2*x^2+1)-b^2/d^2*l
n(1+(c*x+(c^2*x^2+1)^(1/2))^2)+2*a*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))^2/(c^2*d*x^2+d)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (79) = 158\).
time = 0.38, size = 185, normalized size = 2.18 \begin {gather*} \frac {2 \, a b c^{2} x^{2} + 2 \, \sqrt {c^{2} x^{2} + 1} a b c x - b^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} - a^{2} + 2 \, a b - {\left (b^{2} c^{2} x^{2} + b^{2}\right )} \log \left (c^{2} x^{2} + 1\right ) + 2 \, {\left (a b c^{2} x^{2} + \sqrt {c^{2} x^{2} + 1} b^{2} c x\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, {\left (a b c^{2} x^{2} + a b\right )} \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))^2/(c^2*d*x^2+d)^2,x, algorithm="fricas")

[Out]

1/2*(2*a*b*c^2*x^2 + 2*sqrt(c^2*x^2 + 1)*a*b*c*x - b^2*log(c*x + sqrt(c^2*x^2 + 1))^2 - a^2 + 2*a*b - (b^2*c^2
*x^2 + b^2)*log(c^2*x^2 + 1) + 2*(a*b*c^2*x^2 + sqrt(c^2*x^2 + 1)*b^2*c*x)*log(c*x + sqrt(c^2*x^2 + 1)) + 2*(a
*b*c^2*x^2 + a*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^{2} x}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac {b^{2} x \operatorname {asinh}^{2}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac {2 a 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))**2/(c**2*d*x**2+d)**2,x)

[Out]

(Integral(a**2*x/(c**4*x**4 + 2*c**2*x**2 + 1), x) + Integral(b**2*x*asinh(c*x)**2/(c**4*x**4 + 2*c**2*x**2 +
1), x) + Integral(2*a*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))^2/(c^2*d*x^2+d)^2,x, algorithm="giac")

[Out]

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

[Out]

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

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