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

Optimal. Leaf size=102 \[ \frac{1}{5} c^2 d x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} d x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac{b d \left (c^2 x^2+1\right )^{5/2}}{25 c^3}+\frac{b d \left (c^2 x^2+1\right )^{3/2}}{45 c^3}+\frac{2 b d \sqrt{c^2 x^2+1}}{15 c^3} \]

[Out]

(2*b*d*Sqrt[1 + c^2*x^2])/(15*c^3) + (b*d*(1 + c^2*x^2)^(3/2))/(45*c^3) - (b*d*(1 + c^2*x^2)^(5/2))/(25*c^3) +
(d*x^3*(a + b*ArcSinh[c*x]))/3 + (c^2*d*x^5*(a + b*ArcSinh[c*x]))/5

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Rubi [A]  time = 0.101588, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {14, 5730, 12, 446, 77} \[ \frac{1}{5} c^2 d x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} d x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac{b d \left (c^2 x^2+1\right )^{5/2}}{25 c^3}+\frac{b d \left (c^2 x^2+1\right )^{3/2}}{45 c^3}+\frac{2 b d \sqrt{c^2 x^2+1}}{15 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*d*Sqrt[1 + c^2*x^2])/(15*c^3) + (b*d*(1 + c^2*x^2)^(3/2))/(45*c^3) - (b*d*(1 + c^2*x^2)^(5/2))/(25*c^3) +
(d*x^3*(a + b*ArcSinh[c*x]))/3 + (c^2*d*x^5*(a + b*ArcSinh[c*x]))/5

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 5730

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1
+ c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int x^2 \left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{1}{3} d x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^2 d x^5 \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac{d x^3 \left (5+3 c^2 x^2\right )}{15 \sqrt{1+c^2 x^2}} \, dx\\ &=\frac{1}{3} d x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^2 d x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{15} (b c d) \int \frac{x^3 \left (5+3 c^2 x^2\right )}{\sqrt{1+c^2 x^2}} \, dx\\ &=\frac{1}{3} d x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^2 d x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{30} (b c d) \operatorname{Subst}\left (\int \frac{x \left (5+3 c^2 x\right )}{\sqrt{1+c^2 x}} \, dx,x,x^2\right )\\ &=\frac{1}{3} d x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^2 d x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{30} (b c d) \operatorname{Subst}\left (\int \left (-\frac{2}{c^2 \sqrt{1+c^2 x}}-\frac{\sqrt{1+c^2 x}}{c^2}+\frac{3 \left (1+c^2 x\right )^{3/2}}{c^2}\right ) \, dx,x,x^2\right )\\ &=\frac{2 b d \sqrt{1+c^2 x^2}}{15 c^3}+\frac{b d \left (1+c^2 x^2\right )^{3/2}}{45 c^3}-\frac{b d \left (1+c^2 x^2\right )^{5/2}}{25 c^3}+\frac{1}{3} d x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^2 d x^5 \left (a+b \sinh ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0825324, size = 78, normalized size = 0.76 \[ \frac{1}{225} d \left (15 a x^3 \left (3 c^2 x^2+5\right )+\frac{b \sqrt{c^2 x^2+1} \left (-9 c^4 x^4-13 c^2 x^2+26\right )}{c^3}+15 b x^3 \left (3 c^2 x^2+5\right ) \sinh ^{-1}(c x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(15*a*x^3*(5 + 3*c^2*x^2) + (b*Sqrt[1 + c^2*x^2]*(26 - 13*c^2*x^2 - 9*c^4*x^4))/c^3 + 15*b*x^3*(5 + 3*c^2*x
^2)*ArcSinh[c*x]))/225

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Maple [A]  time = 0.004, size = 105, normalized size = 1. \begin{align*}{\frac{1}{{c}^{3}} \left ( da \left ({\frac{{c}^{5}{x}^{5}}{5}}+{\frac{{c}^{3}{x}^{3}}{3}} \right ) +db \left ({\frac{{\it Arcsinh} \left ( cx \right ){c}^{5}{x}^{5}}{5}}+{\frac{{\it Arcsinh} \left ( cx \right ){c}^{3}{x}^{3}}{3}}-{\frac{{c}^{4}{x}^{4}}{25}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{13\,{c}^{2}{x}^{2}}{225}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{26}{225}\sqrt{{c}^{2}{x}^{2}+1}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*(d*a*(1/5*c^5*x^5+1/3*c^3*x^3)+d*b*(1/5*arcsinh(c*x)*c^5*x^5+1/3*arcsinh(c*x)*c^3*x^3-1/25*c^4*x^4*(c^2*
x^2+1)^(1/2)-13/225*c^2*x^2*(c^2*x^2+1)^(1/2)+26/225*(c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.15596, size = 196, normalized size = 1.92 \begin{align*} \frac{1}{5} \, a c^{2} d x^{5} + \frac{1}{75} \,{\left (15 \, x^{5} \operatorname{arsinh}\left (c x\right ) -{\left (\frac{3 \, \sqrt{c^{2} x^{2} + 1} x^{4}}{c^{2}} - \frac{4 \, \sqrt{c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac{8 \, \sqrt{c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} b c^{2} d + \frac{1}{3} \, a d x^{3} + \frac{1}{9} \,{\left (3 \, x^{3} \operatorname{arsinh}\left (c x\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac{2 \, \sqrt{c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*c^2*d*x^5 + 1/75*(15*x^5*arcsinh(c*x) - (3*sqrt(c^2*x^2 + 1)*x^4/c^2 - 4*sqrt(c^2*x^2 + 1)*x^2/c^4 + 8*s
qrt(c^2*x^2 + 1)/c^6)*c)*b*c^2*d + 1/3*a*d*x^3 + 1/9*(3*x^3*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sq
rt(c^2*x^2 + 1)/c^4))*b*d

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Fricas [A]  time = 2.70901, size = 234, normalized size = 2.29 \begin{align*} \frac{45 \, a c^{5} d x^{5} + 75 \, a c^{3} d x^{3} + 15 \,{\left (3 \, b c^{5} d x^{5} + 5 \, b c^{3} d x^{3}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (9 \, b c^{4} d x^{4} + 13 \, b c^{2} d x^{2} - 26 \, b d\right )} \sqrt{c^{2} x^{2} + 1}}{225 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/225*(45*a*c^5*d*x^5 + 75*a*c^3*d*x^3 + 15*(3*b*c^5*d*x^5 + 5*b*c^3*d*x^3)*log(c*x + sqrt(c^2*x^2 + 1)) - (9*
b*c^4*d*x^4 + 13*b*c^2*d*x^2 - 26*b*d)*sqrt(c^2*x^2 + 1))/c^3

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Sympy [A]  time = 2.75605, size = 126, normalized size = 1.24 \begin{align*} \begin{cases} \frac{a c^{2} d x^{5}}{5} + \frac{a d x^{3}}{3} + \frac{b c^{2} d x^{5} \operatorname{asinh}{\left (c x \right )}}{5} - \frac{b c d x^{4} \sqrt{c^{2} x^{2} + 1}}{25} + \frac{b d x^{3} \operatorname{asinh}{\left (c x \right )}}{3} - \frac{13 b d x^{2} \sqrt{c^{2} x^{2} + 1}}{225 c} + \frac{26 b d \sqrt{c^{2} x^{2} + 1}}{225 c^{3}} & \text{for}\: c \neq 0 \\\frac{a d x^{3}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**2*d*x**5/5 + a*d*x**3/3 + b*c**2*d*x**5*asinh(c*x)/5 - b*c*d*x**4*sqrt(c**2*x**2 + 1)/25 + b*d
*x**3*asinh(c*x)/3 - 13*b*d*x**2*sqrt(c**2*x**2 + 1)/(225*c) + 26*b*d*sqrt(c**2*x**2 + 1)/(225*c**3), Ne(c, 0)
), (a*d*x**3/3, True))

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Giac [A]  time = 1.44274, size = 200, normalized size = 1.96 \begin{align*} \frac{1}{5} \, a c^{2} d x^{5} + \frac{1}{75} \,{\left (15 \, x^{5} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{3 \,{\left (c^{2} x^{2} + 1\right )}^{\frac{5}{2}} - 10 \,{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} + 15 \, \sqrt{c^{2} x^{2} + 1}}{c^{5}}\right )} b c^{2} d + \frac{1}{3} \, a d x^{3} + \frac{1}{9} \,{\left (3 \, x^{3} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} - 3 \, \sqrt{c^{2} x^{2} + 1}}{c^{3}}\right )} b d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*c^2*d*x^5 + 1/75*(15*x^5*log(c*x + sqrt(c^2*x^2 + 1)) - (3*(c^2*x^2 + 1)^(5/2) - 10*(c^2*x^2 + 1)^(3/2)
+ 15*sqrt(c^2*x^2 + 1))/c^5)*b*c^2*d + 1/3*a*d*x^3 + 1/9*(3*x^3*log(c*x + sqrt(c^2*x^2 + 1)) - ((c^2*x^2 + 1)^
(3/2) - 3*sqrt(c^2*x^2 + 1))/c^3)*b*d