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

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

[Out]

1/45*b*d*(c^2*x^2+1)^(3/2)/c^3-1/25*b*d*(c^2*x^2+1)^(5/2)/c^3+1/3*d*x^3*(a+b*arcsinh(c*x))+1/5*c^2*d*x^5*(a+b*
arcsinh(c*x))+2/15*b*d*(c^2*x^2+1)^(1/2)/c^3

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Rubi [A]
time = 0.07, antiderivative size = 102, normalized size of antiderivative = 1.00, 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, 5803, 12, 457, 78} \begin {gather*} \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} \end {gather*}

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 12

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

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 78

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]))))

Rule 457

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 5803

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]

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) \text {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) \text {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*}

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Mathematica [A]
time = 0.06, size = 78, normalized size = 0.76 \begin {gather*} \frac {1}{225} d \left (15 a x^3 \left (5+3 c^2 x^2\right )+\frac {b \sqrt {1+c^2 x^2} \left (26-13 c^2 x^2-9 c^4 x^4\right )}{c^3}+15 b x^3 \left (5+3 c^2 x^2\right ) \sinh ^{-1}(c x)\right ) \end {gather*}

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 = 0.96, size = 105, normalized size = 1.03

method	result	size
derivativedivides	\(\frac {d a \left (\frac {1}{5} c^{5} x^{5}+\frac {1}{3} c^{3} x^{3}\right )+b d \left (\frac {\arcsinh \left (c x \right ) c^{5} x^{5}}{5}+\frac {\arcsinh \left (c x \right ) c^{3} x^{3}}{3}-\frac {c^{4} x^{4} \sqrt {c^{2} x^{2}+1}}{25}-\frac {13 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{225}+\frac {26 \sqrt {c^{2} x^{2}+1}}{225}\right )}{c^{3}}\)	\(105\)
default	\(\frac {d a \left (\frac {1}{5} c^{5} x^{5}+\frac {1}{3} c^{3} x^{3}\right )+b d \left (\frac {\arcsinh \left (c x \right ) c^{5} x^{5}}{5}+\frac {\arcsinh \left (c x \right ) c^{3} x^{3}}{3}-\frac {c^{4} x^{4} \sqrt {c^{2} x^{2}+1}}{25}-\frac {13 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{225}+\frac {26 \sqrt {c^{2} x^{2}+1}}{225}\right )}{c^{3}}\)	\(105\)

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,method=_RETURNVERBOSE)

[Out]

1/c^3*(d*a*(1/5*c^5*x^5+1/3*c^3*x^3)+b*d*(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 = 0.26, size = 145, normalized size = 1.42 \begin {gather*} \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 {gather*}

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 = 0.38, size = 103, normalized size = 1.01 \begin {gather*} \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 {gather*}

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 = 0.34, size = 126, normalized size = 1.24 \begin {gather*} \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 {gather*}

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 [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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