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

Optimal. Leaf size=81 \[ d x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac{b \sqrt{c^2 x^2+1} \left (3 c^2 d-e\right )}{3 c^3}-\frac{b e \left (c^2 x^2+1\right )^{3/2}}{9 c^3} \]

[Out]

-(b*(3*c^2*d - e)*Sqrt[1 + c^2*x^2])/(3*c^3) - (b*e*(1 + c^2*x^2)^(3/2))/(9*c^3) + d*x*(a + b*ArcSinh[c*x]) +
(e*x^3*(a + b*ArcSinh[c*x]))/3

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Rubi [A]  time = 0.0697441, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {5704, 444, 43} \[ d x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac{b \sqrt{c^2 x^2+1} \left (3 c^2 d-e\right )}{3 c^3}-\frac{b e \left (c^2 x^2+1\right )^{3/2}}{9 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(3*c^2*d - e)*Sqrt[1 + c^2*x^2])/(3*c^3) - (b*e*(1 + c^2*x^2)^(3/2))/(9*c^3) + d*x*(a + b*ArcSinh[c*x]) +
(e*x^3*(a + b*ArcSinh[c*x]))/3

Rule 5704

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(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}, x] && NeQ[e, c^2*d] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \left (d+e x^2\right ) \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=d x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac{x \left (d+\frac{e x^2}{3}\right )}{\sqrt{1+c^2 x^2}} \, dx\\ &=d x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{2} (b c) \operatorname{Subst}\left (\int \frac{d+\frac{e x}{3}}{\sqrt{1+c^2 x}} \, dx,x,x^2\right )\\ &=d x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{2} (b c) \operatorname{Subst}\left (\int \left (\frac{3 c^2 d-e}{3 c^2 \sqrt{1+c^2 x}}+\frac{e \sqrt{1+c^2 x}}{3 c^2}\right ) \, dx,x,x^2\right )\\ &=-\frac{b \left (3 c^2 d-e\right ) \sqrt{1+c^2 x^2}}{3 c^3}-\frac{b e \left (1+c^2 x^2\right )^{3/2}}{9 c^3}+d x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \sinh ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0691694, size = 71, normalized size = 0.88 \[ \frac{1}{9} \left (3 a x \left (3 d+e x^2\right )-\frac{b \sqrt{c^2 x^2+1} \left (c^2 \left (9 d+e x^2\right )-2 e\right )}{c^3}+3 b x \sinh ^{-1}(c x) \left (3 d+e x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*x*(3*d + e*x^2) - (b*Sqrt[1 + c^2*x^2]*(-2*e + c^2*(9*d + e*x^2)))/c^3 + 3*b*x*(3*d + e*x^2)*ArcSinh[c*x]
)/9

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Maple [A]  time = 0.005, size = 109, normalized size = 1.4 \begin{align*}{\frac{1}{c} \left ({\frac{a}{{c}^{2}} \left ({\frac{{c}^{3}{x}^{3}e}{3}}+{c}^{3}dx \right ) }+{\frac{b}{{c}^{2}} \left ({\frac{{\it Arcsinh} \left ( cx \right ){c}^{3}{x}^{3}e}{3}}+{\it Arcsinh} \left ( cx \right ){c}^{3}dx-{\frac{e}{3} \left ({\frac{{c}^{2}{x}^{2}}{3}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{2}{3}\sqrt{{c}^{2}{x}^{2}+1}} \right ) }-{c}^{2}d\sqrt{{c}^{2}{x}^{2}+1} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.12198, size = 123, normalized size = 1.52 \begin{align*} \frac{1}{3} \, a e 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 e + a d x + \frac{{\left (c x \operatorname{arsinh}\left (c x\right ) - \sqrt{c^{2} x^{2} + 1}\right )} b d}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*e*x^3 + 1/9*(3*x^3*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sqrt(c^2*x^2 + 1)/c^4))*b*e + a*d*x +
(c*x*arcsinh(c*x) - sqrt(c^2*x^2 + 1))*b*d/c

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Fricas [A]  time = 2.47913, size = 208, normalized size = 2.57 \begin{align*} \frac{3 \, a c^{3} e x^{3} + 9 \, a c^{3} d x + 3 \,{\left (b c^{3} e x^{3} + 3 \, b c^{3} d x\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (b c^{2} e x^{2} + 9 \, b c^{2} d - 2 \, b e\right )} \sqrt{c^{2} x^{2} + 1}}{9 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(3*a*c^3*e*x^3 + 9*a*c^3*d*x + 3*(b*c^3*e*x^3 + 3*b*c^3*d*x)*log(c*x + sqrt(c^2*x^2 + 1)) - (b*c^2*e*x^2 +
9*b*c^2*d - 2*b*e)*sqrt(c^2*x^2 + 1))/c^3

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Sympy [A]  time = 0.668448, size = 109, normalized size = 1.35 \begin{align*} \begin{cases} a d x + \frac{a e x^{3}}{3} + b d x \operatorname{asinh}{\left (c x \right )} + \frac{b e x^{3} \operatorname{asinh}{\left (c x \right )}}{3} - \frac{b d \sqrt{c^{2} x^{2} + 1}}{c} - \frac{b e x^{2} \sqrt{c^{2} x^{2} + 1}}{9 c} + \frac{2 b e \sqrt{c^{2} x^{2} + 1}}{9 c^{3}} & \text{for}\: c \neq 0 \\a \left (d x + \frac{e x^{3}}{3}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d*x + a*e*x**3/3 + b*d*x*asinh(c*x) + b*e*x**3*asinh(c*x)/3 - b*d*sqrt(c**2*x**2 + 1)/c - b*e*x**
2*sqrt(c**2*x**2 + 1)/(9*c) + 2*b*e*sqrt(c**2*x**2 + 1)/(9*c**3), Ne(c, 0)), (a*(d*x + e*x**3/3), True))

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Giac [A]  time = 1.55539, size = 146, normalized size = 1.8 \begin{align*}{\left (x \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{\sqrt{c^{2} x^{2} + 1}}{c}\right )} b d + a d x + \frac{1}{9} \,{\left (3 \, a x^{3} +{\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\right )} e \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*log(c*x + sqrt(c^2*x^2 + 1)) - sqrt(c^2*x^2 + 1)/c)*b*d + a*d*x + 1/9*(3*a*x^3 + (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)*e