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

Optimal. Leaf size=145 \[ \frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c d^3 \sqrt{c^2 x^2+1}}+\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{6 c d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (c^2 x^2+1\right )^2}+\frac{b^2}{12 c^2 d^3 \left (c^2 x^2+1\right )}-\frac{b^2 \log \left (c^2 x^2+1\right )}{6 c^2 d^3} \]

[Out]

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

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Rubi [A]  time = 0.145654, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {5717, 5690, 5687, 260, 261} \[ \frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c d^3 \sqrt{c^2 x^2+1}}+\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{6 c d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (c^2 x^2+1\right )^2}+\frac{b^2}{12 c^2 d^3 \left (c^2 x^2+1\right )}-\frac{b^2 \log \left (c^2 x^2+1\right )}{6 c^2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5717

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[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]

Rule 5690

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSinh[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a +
b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 + c^2*x^2)^FracPar
t[p]), Int[x*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ
[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 5687

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*Sqrt[1 + c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSinh[
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 260

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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && 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 )^3} \, dx &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}+\frac{b \int \frac{a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{2 c d^3}\\ &=\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{6 c d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}-\frac{b^2 \int \frac{x}{\left (1+c^2 x^2\right )^2} \, dx}{6 d^3}+\frac{b \int \frac{a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 c d^3}\\ &=\frac{b^2}{12 c^2 d^3 \left (1+c^2 x^2\right )}+\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{6 c d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c d^3 \sqrt{1+c^2 x^2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}-\frac{b^2 \int \frac{x}{1+c^2 x^2} \, dx}{3 d^3}\\ &=\frac{b^2}{12 c^2 d^3 \left (1+c^2 x^2\right )}+\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{6 c d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{3 c d^3 \sqrt{1+c^2 x^2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}-\frac{b^2 \log \left (1+c^2 x^2\right )}{6 c^2 d^3}\\ \end{align*}

Mathematica [A]  time = 0.24088, size = 152, normalized size = 1.05 \[ \frac{-3 a^2+4 a b c^3 x^3 \sqrt{c^2 x^2+1}+6 a b c x \sqrt{c^2 x^2+1}+2 b \sinh ^{-1}(c x) \left (b c x \sqrt{c^2 x^2+1} \left (2 c^2 x^2+3\right )-3 a\right )+b^2 c^2 x^2-3 b^2 \sinh ^{-1}(c x)^2+b^2-2 \left (b c^2 x^2+b\right )^2 \log \left (c^2 x^2+1\right )}{12 d^3 \left (c^3 x^2+c\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.095, size = 432, normalized size = 3. \begin{align*} -{\frac{{a}^{2}}{4\,{c}^{2}{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{2\,{b}^{2}{\it Arcsinh} \left ( cx \right ) }{3\,{c}^{2}{d}^{3}}}+{\frac{c{b}^{2}{\it Arcsinh} \left ( cx \right ){x}^{3}}{3\,{d}^{3} \left ({c}^{4}{x}^{4}+2\,{c}^{2}{x}^{2}+1 \right ) }\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{{c}^{2}{b}^{2}{\it Arcsinh} \left ( cx \right ){x}^{4}}{3\,{d}^{3} \left ({c}^{4}{x}^{4}+2\,{c}^{2}{x}^{2}+1 \right ) }}+{\frac{{b}^{2}{\it Arcsinh} \left ( cx \right ) x}{2\,c{d}^{3} \left ({c}^{4}{x}^{4}+2\,{c}^{2}{x}^{2}+1 \right ) }\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{2\,{b}^{2}{\it Arcsinh} \left ( cx \right ){x}^{2}}{3\,{d}^{3} \left ({c}^{4}{x}^{4}+2\,{c}^{2}{x}^{2}+1 \right ) }}-{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{4\,{c}^{2}{d}^{3} \left ({c}^{4}{x}^{4}+2\,{c}^{2}{x}^{2}+1 \right ) }}+{\frac{{b}^{2}{x}^{2}}{12\,{d}^{3} \left ({c}^{4}{x}^{4}+2\,{c}^{2}{x}^{2}+1 \right ) }}-{\frac{{b}^{2}{\it Arcsinh} \left ( cx \right ) }{3\,{c}^{2}{d}^{3} \left ({c}^{4}{x}^{4}+2\,{c}^{2}{x}^{2}+1 \right ) }}+{\frac{{b}^{2}}{12\,{c}^{2}{d}^{3} \left ({c}^{4}{x}^{4}+2\,{c}^{2}{x}^{2}+1 \right ) }}-{\frac{{b}^{2}}{3\,{c}^{2}{d}^{3}}\ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }-{\frac{ab{\it Arcsinh} \left ( cx \right ) }{2\,{c}^{2}{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{abx}{6\,c{d}^{3}} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{abx}{3\,c{d}^{3}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \end{align*}

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)^3,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{b^{2} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2}}{4 \,{\left (c^{6} d^{3} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}} - \frac{a^{2}}{4 \,{\left (c^{6} d^{3} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}} + \int \frac{{\left ({\left (4 \, a b c^{2} + b^{2} c^{2}\right )} x^{2} + \sqrt{c^{2} x^{2} + 1}{\left (4 \, a b c + b^{2} c\right )} x + b^{2}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{2 \,{\left (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 +{\left (c^{7} d^{3} x^{6} + 3 \, c^{5} d^{3} x^{4} + 3 \, c^{3} d^{3} x^{2} + c d^{3}\right )} \sqrt{c^{2} x^{2} + 1}\right )}}\,{d x} \end{align*}

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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.83139, size = 590, normalized size = 4.07 \begin{align*} \frac{4 \, a b c^{4} x^{4} +{\left (8 \, a b + b^{2}\right )} c^{2} x^{2} - 3 \, b^{2} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2} - 3 \, a^{2} + 4 \, a b + b^{2} - 2 \,{\left (b^{2} c^{4} x^{4} + 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} \log \left (c^{2} x^{2} + 1\right ) + 2 \,{\left (3 \, a b c^{4} x^{4} + 6 \, a b c^{2} x^{2} +{\left (2 \, b^{2} c^{3} x^{3} + 3 \, b^{2} c x\right )} \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + 6 \,{\left (a b c^{4} x^{4} + 2 \, a b c^{2} x^{2} + a b\right )} \log \left (-c x + \sqrt{c^{2} x^{2} + 1}\right ) + 2 \,{\left (2 \, a b c^{3} x^{3} + 3 \, a 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{align*}

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)^3,x, algorithm="fricas")

[Out]

1/12*(4*a*b*c^4*x^4 + (8*a*b + b^2)*c^2*x^2 - 3*b^2*log(c*x + sqrt(c^2*x^2 + 1))^2 - 3*a^2 + 4*a*b + b^2 - 2*(
b^2*c^4*x^4 + 2*b^2*c^2*x^2 + b^2)*log(c^2*x^2 + 1) + 2*(3*a*b*c^4*x^4 + 6*a*b*c^2*x^2 + (2*b^2*c^3*x^3 + 3*b^
2*c*x)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1)) + 6*(a*b*c^4*x^4 + 2*a*b*c^2*x^2 + a*b)*log(-c*x + sqrt
(c^2*x^2 + 1)) + 2*(2*a*b*c^3*x^3 + 3*a*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., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2} x}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx + \int \frac{b^{2} x \operatorname{asinh}^{2}{\left (c x \right )}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx + \int \frac{2 a 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{align*}

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)**3,x)

[Out]

(Integral(a**2*x/(c**6*x**6 + 3*c**4*x**4 + 3*c**2*x**2 + 1), x) + Integral(b**2*x*asinh(c*x)**2/(c**6*x**6 +
3*c**4*x**4 + 3*c**2*x**2 + 1), x) + Integral(2*a*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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{2} x}{{\left (c^{2} d x^{2} + d\right )}^{3}}\,{d x} \end{align*}

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)^3,x, algorithm="giac")

[Out]

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