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

Optimal. Leaf size=97 \[ \frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (c^2 x^2+1\right )^2}+\frac{b x^3}{12 c d^3 \left (c^2 x^2+1\right )^{3/2}}+\frac{b x}{4 c^3 d^3 \sqrt{c^2 x^2+1}}-\frac{b \sinh ^{-1}(c x)}{4 c^4 d^3} \]

[Out]

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

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Rubi [A]  time = 0.0861107, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {5723, 288, 215} \[ \frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (c^2 x^2+1\right )^2}+\frac{b x^3}{12 c d^3 \left (c^2 x^2+1\right )^{3/2}}+\frac{b x}{4 c^3 d^3 \sqrt{c^2 x^2+1}}-\frac{b \sinh ^{-1}(c x)}{4 c^4 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5723

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

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

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

Mathematica [A]  time = 0.14508, size = 79, normalized size = 0.81 \[ \frac{-3 a \left (2 c^2 x^2+1\right )+b c x \sqrt{c^2 x^2+1} \left (4 c^2 x^2+3\right )-3 \left (2 b c^2 x^2+b\right ) \sinh ^{-1}(c x)}{12 c^4 d^3 \left (c^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.016, size = 108, normalized size = 1.1 \begin{align*}{\frac{1}{{c}^{4}} \left ({\frac{a}{{d}^{3}} \left ({\frac{1}{4\, \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{1}{2\,{c}^{2}{x}^{2}+2}} \right ) }+{\frac{b}{{d}^{3}} \left ({\frac{{\it Arcsinh} \left ( cx \right ) }{4\, \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{{\it Arcsinh} \left ( cx \right ) }{2\,{c}^{2}{x}^{2}+2}}-{\frac{cx}{12} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{cx}{3}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{16} \, b{\left (\frac{4 \, c^{2} x^{2} + 4 \,{\left (2 \, c^{2} x^{2} + 1\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + 3}{c^{8} d^{3} x^{4} + 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}} - 16 \, \int \frac{2 \, c^{2} x^{2} + 1}{4 \,{\left (c^{10} d^{3} x^{7} + 3 \, c^{8} d^{3} x^{5} + 3 \, c^{6} d^{3} x^{3} + c^{4} d^{3} x +{\left (c^{9} d^{3} x^{6} + 3 \, c^{7} d^{3} x^{4} + 3 \, c^{5} d^{3} x^{2} + c^{3} d^{3}\right )} \sqrt{c^{2} x^{2} + 1}\right )}}\,{d x}\right )} - \frac{{\left (2 \, c^{2} x^{2} + 1\right )} a}{4 \,{\left (c^{8} d^{3} x^{4} + 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*b*((4*c^2*x^2 + 4*(2*c^2*x^2 + 1)*log(c*x + sqrt(c^2*x^2 + 1)) + 3)/(c^8*d^3*x^4 + 2*c^6*d^3*x^2 + c^4*d
^3) - 16*integrate(1/4*(2*c^2*x^2 + 1)/(c^10*d^3*x^7 + 3*c^8*d^3*x^5 + 3*c^6*d^3*x^3 + c^4*d^3*x + (c^9*d^3*x^
6 + 3*c^7*d^3*x^4 + 3*c^5*d^3*x^2 + c^3*d^3)*sqrt(c^2*x^2 + 1)), x)) - 1/4*(2*c^2*x^2 + 1)*a/(c^8*d^3*x^4 + 2*
c^6*d^3*x^2 + c^4*d^3)

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Fricas [A]  time = 2.33659, size = 209, normalized size = 2.15 \begin{align*} \frac{3 \, a c^{4} x^{4} - 3 \,{\left (2 \, b c^{2} x^{2} + b\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) +{\left (4 \, b c^{3} x^{3} + 3 \, b c x\right )} \sqrt{c^{2} x^{2} + 1}}{12 \,{\left (c^{8} d^{3} x^{4} + 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*c^4*x^4 - 3*(2*b*c^2*x^2 + b)*log(c*x + sqrt(c^2*x^2 + 1)) + (4*b*c^3*x^3 + 3*b*c*x)*sqrt(c^2*x^2 +
1))/(c^8*d^3*x^4 + 2*c^6*d^3*x^2 + c^4*d^3)

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

[Out]

(Integral(a*x**3/(c**6*x**6 + 3*c**4*x**4 + 3*c**2*x**2 + 1), x) + Integral(b*x**3*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 )} x^{3}}{{\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^3*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^3,x, algorithm="giac")

[Out]

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